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MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

16_Light_forces_Part_1.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: --equal forces of light. But before we get started with it, do you have any question about the last lecture, which was on open quantum system-- Quantum Monte Carlo methods-- and in particular, the conceptual things we discussed is what happens when you do a measurement, or when you don't detect a photon, how does it change the wave function? And this was actually at the heart of realizing similar quantum systems, and simulating them using quantum Monte Carlo methods. Any question? Any points for discussion? OK. Today is, actually conceptually, a simple lecture, because I'm not really introducing some subtleties of quantum physics. We know our Hamiltonian-- the dipole Hamiltonian-- this is the fully quantized Hamiltonian, how atoms interact with the electromagnetic field. And, of course, it's fully quantized because E perpendicular is the operator of the quantized electromagnetic field, and it's the sum of a plus a [? dega. ?] But when it comes to forces, it's actually fairly simple. We don't introduce anything else. This is our Hamiltonian. This is the energy. And the force is nothing else than the gradient of this energy. So I decided it today-- it's rather unusual-- but I will have everything pre-written, because I think the concentration of new ideas is not so high today, so I want to go a little bit faster. But I invite everybody if you slow me down if you think I'm going too fast and if you think I'm skipping something. Also another justification is at least the first 30%, 40% of this lecture, you've already done as a homework assignment, when you looked at the classical limit of mechanical forces. You actually had already, as a preparation for this class, a homework assignment on the spontaneous scattering force and the stimulated dipole force. PROFESSOR: Well, the only subtlety today is how do we deal with a fully quantized electromagnetic field? But what I want to show you in the next few minutes is in the end, we actually do approximations that the quantized electromagnetic field pretty much disappears from the equation. We have a classical field. And then, almost everything you have done in your homework directly applies. The only other thing we do today is the dipole moment. Well, we use, for the dipole moment of the atom, the solution of the Optical Bloch Equation. So I think now you know already everything we do in the first 30, 40 minutes. PROFESSOR: So we get rid of all of the quantum character of the electromagnetic field in two stages. One is the following, well, we want to describe the interactions of atoms with laser fields. And that means the electromagnetic field is in a coherent state. And you know from several weeks ago, the coherent state is a superposition of a [INAUDIBLE] states, coherent superposition and such, so they're really photons inside. It's a fully quantized description of the laser beam. But now-- and this is something if you're not familiar, you should really look up-- there is an exact canonical transformation where you transform your Hamiltonian-- you do, so to speak, a basis transformation-- to another Hamiltonian. And the result is, after this unitary transformation, the coherent state is transformed to the vacuum state. So in other words, the quantized electromagnetic field is no longer in the coherent state, it's in the vacuum state. But what appears instead is a purely classical electromagnetic field. A c number. So you can sort of say-- I'm waving my hands now-- the coherent state is the vacuum, with a displacement operator. All the quantumness of the electromagnetic field is in the vacuum state, and that's something we have to keep because this is responsible for spontaneous emission quantum fluctuations and such. But the displacement operator, which gives us an arbitrary coherent state, that can be absorbed by completely treating the electromagnetic field classically. So therefore, we take the fully quantized electromagnetic field and we replace it now by a classical field, and simply the vacuum fluctuation. So there are no photons around anymore. There is the vacuum ready to absorb the photons, and then there's a classical electromagnetic field, which drives the atoms. And, of course, the classical electromagnetic field is not changing because of absorption emission, because it's a c number in our Hamiltonian. So this is an important conceptual step, and it's the first step in how we get rid of the quantum nature of the electromagnetic field when it interacts with atoms mechanically. OK. So, by the way, everything I'm telling you today-- actually, in the first half of today-- can be found in Atom-Photon Interaction. Now, so this takes care actually, of most of the electromagnetic field. We come back to that in a moment. The next thing is we want to use the full quantum solution for the atomic dipole operator, and that means we want to remind ourselves of the solution of the Optical Bloch Equations. PROFESSOR: Now, talking about the Optical Bloch Equations, I just have to remind you of what we did in class, but I have to apologize it's a little bit exercise in notation. Because in Atom-Photon Interaction they use a slightly different notation then you find in other sources. On the other hand, I think I've nicely prepared for you, and I give you the translation table. In Atom-Photon Interaction, they regard the Bloch vector, and we have introduced the components, u, v, and w, as a fictitious spin, 1/2. Therefore, you find with this kind of old-fashioned letters, Sx, Sy, Sc, in Atom-Photon Interaction. The other thing, if you open the book, Atom-Photon Interaction is that their operators, density matrix-- the atomic density matrix-- sigma, and sigma head. One is in the rotating frame, and one is in the lab frame. So we have to go back and forth to do that, but the result is fairly trivial. So this is simply the definition. First of all, the density matrix is the full description of the atomic system, but instead of dealing with matrix elements of the density matrix, it's more convenient for two level system to use a fictitious spin or the Bloch vector. So the important equation-- the are the Optical Block Equations-- is the equation A20 in API. And it's simply the first order differential equation for the components of the Optical Bloch vector, u v, and w. Notation is as usual, the laser detuning, delta L. Omega 1 is the [INAUDIBLE] frequency. And this is the parametrization of the classical electromagnetic field which drives the system. Now, this equation is identical to the equation we discussed in the unit on solutions of Optical Bloch Equations. We just called the [INAUDIBLE] frequency, g, delta L was called delta. And then, vectors of 2r was confusing. The components, [? r,x,y,z, ?] are two times u, v, w. And the reason for that is some people prefer that it represents spin 1/2, so they want to normalize things to 1/2. Or if you want to have unit vectors, your throw in a vector of two. But that is all, so if you're confused about a vector of two, it's exactly this substitution which has been made. So these are the Optical Bloch Equations, and what we need now is the solution. What we need now is the solution for the dipole moment, because it is the dipole moment which is responsible for the mechanical forces. The dipole operator is, of course, the non-diagonal matrix elements, the coherencies of the density matrix. So all we want to take now from the previous unit, the expectation value of the dipole moment, which is nothing else than the trace of the statistical operator with the operator of the dipole moment. At this point, just to guide you through, we go from the lab frame-- from the density matrix in the lab frame-- to the density matrix in the rotating frame, and this is when the laser frequency appears, is into the i omega l t. And therefore, we want to describe things in the lab frame. If you use the u and v part of the Optical Bloch vector obtained in the rotating frame, in the lab frame everything rotates at the laser frequency. If you drive an atom, if you drive a harmonic oscillator with a laser frequency, the harmonic oscillator responds at the laser frequency, not at the atomic frequency. We discussed that a while ago. So therefore, the dipole moment oscillates at the laser frequency. And their two components, u and v-- since our laser field was parametrized by definition-- epsilon 0 times cosine omega t. We have now nicely separated, through the Optical Bloch vector the in phase part of the dipole moment, and the in quadrature phase of dipole moment. If you didn't pay attention what I said the last five minutes, that's OK, you can just start here, saying the expectation value of the dipole moment oscillates with a laser frequency. And as for any harmonic oscillator, or harmonic oscillator type system, there is one component which is in phase, and one component which is in quadrature. Fast forward. OK. And you know from any harmonic oscillator, when it comes to the absorbed power, that it's only the quadrature component which absorbs the power. And you can immediately see that when we say-- just use what is written in black now-- the energy is nothing else than the charge times the displacement. When we divide by delta t, you'll find that the absorbed power is the electric field times the derivative of the dipole moment. Well, if you ever reach over one cycle, we are only interested in d dot, which oscillates with cosine omega t. But that means since it's d dot, d oscillates with sine omega t. And this was the in quadrature component, v. So it is only the part, v, of the Optical Bloch vector which is responsible for exchanging energy with the electromagnetic field. And we can also, by dividing by the energy of a photon, find out what is the number of absorbed photons. But we did that already when we discussed Optical Bloch Equations. OK, so that was our look back on the Optical Bloch Equation, the solution for the oscillating dipole moment, and now we are ready to put everything together. So we want to know the force. The force, using Heisenberg's Equation of Motion for the momentum derivative is the expectation value of the operator, which is the gradient of the Hamiltonian. OK. So now we have to take the gradient of the Hamiltonian. This is the only expression, actually, today, I would've liked to just write it down, and build it up piece by piece. But let me step you through. We want to take the gradient. The dipole operator of the atom does not have an r dependence, it's just x on the atom, wherever it is. But the gradient with respect to the center of mass position from the atoms comes from the operator of the electromagnetic field. So in other words, what this involves is the gradient of the operator of the fully quantized electromagnetic field. But we have already written at the electromagnetic field as an external electromagnetic field-- classical electromagnetic field-- plus a vacuum fluctuations. But the vacuum fluctuations are symmetric. There is nothing which tells vacuum fluctuations what is left and what is right. So therefore, the derivative only acts on the classical electromagnetic field. So this is how we have now, you know, we first throw out the coherent state, and now we throw out the vacuum state. And therefore, it is only the classical part of the electromagnetic field which is responsible for the forces. Now we make one way assumption which greatly simplifies it. R is the center of mass position of the atom, and we have to take the electromagnetic field as the center of mass position. And if the atom is localized-- wave packet with a long [? dipole ?] wave-- we sort of have to involve the atomic wave function when we evaluate this operator of the electromagnetic field. However, if the atom is very well localized-- you would actually want the atom to be localized to within an optical wave length-- then, under those circumstances, you can replace the parameter, r, which is a position of the atom, by the center of the atomic wave packet. So therefore, the kind of wave nature of the atoms-- which means the atom is smeared out-- as long as this wave packet is small, compared to the wave lengths, you can just evaluate the electromagnetic field at the center of the wave packet. This assumption requires-- since the scale of the electromagnetic field is set by the wave lengths-- this requires that the atom is localized better than an optical wave length. And that would mean that the energy of the atom, or the temperature, has to be larger than the recoil in it. And if you're localized the, to within a wave length, your momentum spread is larger, by Heisenberg's Uncertainty Relation than h bar k, and that means your energy is larger than the recoil energy. So if you want to have a description or light forces in that limit, you may have to revise this point. But for most of laser cooling, when you start with a hot cloud and cool it down to micro-Kelvin temperatures, this is very [INAUDIBLE]. Colin? AUDIENCE: Vacuum field doesn't contribute in sort of in the sense that the expectation value is zero. But wouldn't the rms value be non-zero? So what's the--? PROFESSOR: Exactly. I mean, in the end, you will see that in a moment, when we talk about cooling limits, fluctuations, spontaneous emission, provides heating. It doesn't contribute if you want to find the expectation value for the force. The vacuum and the fluctuations-- I mean, that's also logical-- the vacuum has only fluctuations, it has no net force. But the fluctuation's heat. And for heating, it will be very important. But we'll come to that in a little bit later. AUDIENCE: So what is [INAUDIBLE]? PROFESSOR: If the light is squeezed, yes. That's a whole different thing. We really assumed here that the light is in a coherent state. Now-- AUDIENCE: The transformation that we had [INAUDIBLE]. PROFESSOR: Well, you know, you can squeeze the vacuum. And then it's squeezed. But there's so few photons, you can't really laser cool with that. So the only way how you could possibly laser cool is if you take the squeezed vacuum, and displace it. So now you have, instead of a little circle, which is a coherent state, you have an ellipse, which is displaced. I haven't done the math-- maybe that would be an excellent homework assignment for the next time I teach the course-- you could probably show that the displacement operator, which is now displacing the squeezed vacuum, again can be transformed into a c number, and the ellipse this squeezed vacuum fluctuation-- the ellipse has an even symmetry, has an 180 degree rotation symmetry. And I could imagine that it will not contribute to the force. So in other words, if you have a displaced, squeezed vacuum, my gut feeling is nothing would change here. But what would change, aren't the fluctuations, but which would possibly change is the heating. OK, so we within those approximations, Heisenberg's Equation of Motion for the force becomes an equation for the acceleration for the atomic wave packet. And it involves now, two terms. One is the atomic dipole moment as obtained from the Optical Bloch Equations, and the other one is nothing else than the gradient of the classical part of the electromagnetic field. And with that we are pretty much at the equations which you looked at in your homework assignment, which was a classical model for the light force. Because this part is completely classical, and the only quantum aspect which we still have is that we evaluate the dipole moment, not necessarily, it's just a driven harmonic oscillator. When we take the Optical Bloch Equations, we will also find saturation a two-level system can be saturated, whereas a harmonic oscillator can never be saturated. So let me just repeat, so this is completely classical except for the expectation value for the dipole moment. But for no excitation, of course-- and we've discussed it many, many times-- a two-level system is nothing else for weak excitation, than a weakly excited harmonic oscillator. So then we are completely classical. But we want to discuss also saturation and that's been the Optical Bloch Equations, of course, come in very handy. OK. Next approximation, we want to know-- approximate-- the dipole moment-- the expectation value of the dipole moment-- with the steady state solution. The steady state solution-- analytic expression-- we just plug it in, and you can have a wonderful discussion about light forces. The question is, are we allowed to do that? Now I hope you'll remember that's actually one reason why I did that when we discussed the master equation. And we discussed an [? atomic ?] [? cavity. ?] I introduced to you the adiabatic elimination of variables. We adiabatically eliminated [? of dipole ?] matrix elements. Whenever you have something which relaxes sufficiently fast, and you're not interested in this short time scale, you can always say, I replace this quantity by its steady state value. Steady state with regard to the slowly changing parameter. And we want to do exactly the same here. So let me just translate. It's the same idea, exactly the same idea, but let me translate. If an atom moves, we have two aspects, two times [? case. ?] One is the motion of the atom. The atom may change its velocity. The atom may move from an area of strong electric field to an area of weak electric field. But this is really all that is related to the motion, to the change of momentum of a heavy object. But now inside the atoms, you play with the, you know, Bloch vector, excited state. If the electric field is higher, you have a larger population in the excited state than in the ground state. But this usually adjusts with the damping time of gamma, with a spontaneous emission rate. And that's usually very fast. So you can make the assumption when an atom moves to a changing electric field, at every point in the electric field, it very, very quickly-- the dipole moment-- assumes the steady state solution for the local electric field. And this would now be called an adiabatic elimination of the dynamics of the atom. And we simply replace the expectation value for the dipole moment by the steady state solution of the Optical Bloch Equations. Well, this is, of course, an approximation. And I want to take this approximation and who you what simple conclusions we can draw from that. But I also want you to know, I mean this is what I focus off in this course, to sort of think about them and get a feel where those approximations break down. So this approximation requires that the internal motion, that this, you know, two-level physics, that the internal density matrix, it has a relaxation rate of gamma. Whereas the external motion, we will actually see that very soon when we talk about molasses and damping, has a characteristic damping time which is one over the recoil energy. The recoiled energy involves the mass. The heavier the object, the slower is the damping time. It has all the right scaling. And you will actually see in the data today, that we will naturally obtain this as the time scale over which atomic motions is changing. So, is that approximation, this hierarchy of time scales, is that usually fulfilled? It is fulfilled in almost all of the atoms you are working on. For instance, in sodium, the ratio of those two time scales is 400. But if you push it, you will find examples like, helium in the triplets statement-- metastable helium in the triplet state-- where the two time scales are comparable. Helium is a very light, so therefore, it's faster for a light atom to change its motion by exchanging momentum with electromagnetic field. And at the same time, the transition in helium is narrower, and therefore slower. So for metastable helium, the two time scales are comparable. | if you really want to do quantitative experiments with metastable helium, you may need a more complicated theory. But, of course, if something is sort of simple in one case, the richer situation provides another opportunity for research. And I know, for instance, Hal Metcalf, a good colleague and friend of mine, he focused for a while on studying laser cooling of metastable helium, because he was interested what happens if those simple approximations don't work anymore? Then the internal motion and the external motion becomes sort of more entangled. You have to treat them together, you cannot separate them by a hierarchy of time scales. Any questions about that? OK, now we have done, you know, everything which is complicated has been approximated away now. And now we have a very simple result. So I'm pretty much repeating what you have done your homework, but now with expressions which have still the quantum character of the Optical Bloch Equations. So let's assume that we have an external electric field. it's parametrized like that, and at this point, by providing it like this, I leave it open whether we have a standing wave-- if you have a traveling wave, of course, we have a phase factor, kr, here-- if you have a standing wave, we don't have this phase factor. So I have parametrized the electromagnetic field in such a way that we can derive a general expression. And then we will discuss what happens in a traveling wave, and what happens in a standing wave. I make one more assumption here, namely that the polarization is independent of position. I'm just lazy if I take derivatives of the electric field. I want to take a derivative with the amplitude of the electric field, and I want to take a derivative of the phase of the electromagnetic field. I do not want to take a derivative of the polarization. Of course, what I'm throwing out here, is all of the interesting cooling mechanisms of polarization gradient cooling, when the polarization of the light field spirals around in three dimensions. We'll talk a little bit about that later on, I just want to keep things simple here. OK, so now if you take the gradient of the electromagnetic field, finally, we have approximated everything which complicates our life away. And we have two terms. One is the gradient x on the amplitude of the electromagnetic field, or the gradient x on the phase. And since we have taken a derivative, when we take the derivative of the phase, because of the chain rule, the cosine changes into sine. So therefore, the gradient of the electromagnetic field has a cosine in phase and an in quadrature component. And now, for the force, we have to multiply this with the dipole moment. And, of course, we will average over one cycle of the electromagnetic wave. And, of course, that's what we discussed. There's is a u and v component of the Bloch vector, which gives rise to an in phase and in quadrature component of the dipole moment. And now, if you combine-- if you multiply-- the gradient of the electric field with the dipole moment, and average. The cosine part goes with the cosine part, the sine part goes with the sine part, and the cos terms average out. So therefore, we have the simple result that we have two contributions to the [? old ?] mechanical force of light. And the u component-- the in phase component-- of the dipole moment goes with the amplitude of the gradient. And the in quadrature component goes with the gradient of the phase. So in other words, when it comes to light forces, the cosine component and the sine component of the dipole moment talk, actually, to two different quantities associated with the electromagnetic field. It's just in phase and in quadrature. OK. Now I just have to tell you a few definitions and names. This component, which goes with the in phase component, is called the reactive force. Whereas the other one is called the dissipative force. The name, of course, comes that if you have a harmonic oscillator, the component which oscillates in quadrature absorbs directly energy, whereas the in phase component does not absorb energy. I'll leave it for a little bit later in our discussion how you can have a force and not exchange energy. That's a little bit a mystery when we talk about cooling with a reactive force. We will have to scrutinize what happens with energy, because at least in the most basic, the in phase component of the oscillating system does not extract energy from the electromagnetic field. So maybe just believe that for now, and we will scrutinize it later. Any questions? Am I going too fast? Am I going too slow? Does it mean about right? OK. So we have now obtained the reactive force, which is also called the dipole force or stimulated force, and the dissipated force, which is also called the spontaneous force. The next thing is purely nomenclature. We want to introduce appropriate vectors which point along the gradient of the phase, and which point along the gradient of the amplitude. And this is done here. So we want to say that the reactive and the dissipative force are written in a very sort of similar way, but the one off them points in the direction of a vector, alpha, the other one in the direction of a vector, beta. The beta vector is the gradient of the phase. If the phase is kr, the gradient of the phase is simply k, the k vector of the electromagnetic wave. So that will come in handy in just a few seconds. Whereas the reactive force points into the direction of the gradient of the amplitude, but we never want to talk about electromagnetic field amplitude. For us, we always use the electric field in terms of you parametrize the Rabi frequency. The Rabi frequency is really the atomic unit of the electric field. So therefore, this vector which involves now the gradient of the amplitude, becomes a normalized gradient of the Rabi frequency. So these are the two fundamental light forces, the reactive and the dissipative force. OK. Now we are going to have a quick look at those two forces. So therefore, I'm now discussing the two simple, but already very characteristic cases for those forces. One is we want to look at a plane travelling wave, and then we want to look at a pure standing wave. And the beauty of it is in a plane travelling wave, you've a plain wave, the amplitude of the electric field is constant everywhere, it just oscillates. And therefore, this alpha vector is zero. And in a few minutes, we look at this standing wave, and in the standing wave, the beta vector is zero. So the travelling wave and the standing wave allows us now to look at the two forces separately in two physical situations which are-- as you will see later-- highly relevant for experiments. Well, I don't think I have to tell anybody in this room-- maybe except some people who take the course for [? Brett's ?] requirement-- that standing waves or optical lattices are common in all labs. And similar, traveling wave beams are used, for instance, for decelerating atomic beams. So we illustrate the two forces now, but we already discussing two experimentally very important geometries. OK, plain travelling waves. I mean, I just really plug now the electromagnetic field for plain travelling wave, I plug it into the equations above. The important thing is-- and which simplifies things a lot-- is that the alpha vector is zero. The amplitude of the plain wave is-- that's what a plain wave is-- constant. And the beta vector is the gradient of the Rabi frequency is just the k vector of the light. So therefore, the dissipative or spontaneous light force was the Rabi frequency times the v component of the Optical Bloch vector times h bar k. And this has a very simple interpretation because the steady state solution of the Optical Bloch Equation is nothing else than gamma times the excited state population. And since the excited state scatters, or emits photons at a rate, gamma, this is nothing else than the number of absorbed emitted scattered photons times h bar k. We are in steady state here, so the number of photons which are emitted into all space, into the vacuum has to be the number of photons which has been absorbed into the laser beam. So therefore, the interpretation here is very simple, you have a laser beam. Every time the atom absorbs a photon, it receives a recoil transfer, h bar k. Now, afterwards, the photon is scattered, but the scattering is symmetric, and does not impart the force onto the atom. And this is what we discussed earlier. The quantum part of the electromagnetic field is symmetric. And here, we see sort of what it means visually. Spontaneous emission goes equally probable in opposite direction, and there is no net force, but there is heating as we discussed later. So we can plug in the solution-- the steady state solution-- from the Optical Bloch Equation. That's our Lorentzian. We have discussed power broadening already when we discussed the Optical Bloch Equation. And, of course, I just remind you if you saturate the system with a strong laser power, what you can obtain is that half of the population is in the excited state, and half of the population is in the ground state. Under those circumstances, do you get the maximum force. And the maximum spontaneous force is the momentum per photon times the maximum rate at which a two-level atom can scatter photons. So that's, in a nutshell, I think all you have to know about the spontaneous light force as far as the force is concerned. But we still have to discuss the heating associated with a spontaneous force, which are the force fluctuations. Questions? Well, then let's do a similar discussion with reactive forces. Which is a little bit richer because, well, you will see. So a standing wave is parametrized here. We only have an alpha vector, not a beta vector now, because everything is in phase, there is no phase phi of r. So therefore, there is no dissipative force, the only force is a reactive force. Since the reactive force depends only on u, there is no exchange of energy. So that raises the question, what is really going on it is only the reactive part with u, how can the atom really-- How can the motion change if it cannot exchange energy? Well, we'll come to that in different pieces. The one part I want to sort of discuss here is that you should-- I want to show you that at the level of our discussion we can already understand that the atom is actually doing a redistribution of energy and momentum. Well, if I only hold onto my standing wave, there is no energy exchange because the atom is always in phase with the standing wave-- the atomic dipole-- which is this one [INAUDIBLE] force. But I can now do what everybody does in the lab. We generate the standing wave as a superposition of two travelling waves. And if I have a superposition of two traveling waves, there is a point in space where the phases of the traveling waves are the same. Then one phase is advanced in phase, and the other side one of the waves is lagging in phase. And for pedagogical reasons, I pick the point where the two travelling waves are 90 degree phase shifted. OK. Now the situation is that we have our electric field and what is responsible for the reactive light force is the u part of the dipole moment of the Optical Bloch vector, which oscillates in phase. But nobody prevents me from analyzing it with respect to E1 and E2. And so if I ask now, I say, I have one laser beam. I have another laser beam. And if you want, after the laser beams have crossed, you can put in two photo-diodes, and you can not only ask, what happens to the standing wave, you can also ask, what happens to each traveling wave? Are photons absorbed? Or what happens? And now, what is obvious is if you have an harmonic oscillator, and you have an oscillating dipole moment, if the dipole moment is withing 0 and 180 degrees with a dry field, you absorb power. If it is in the two other quadrants, it emits, or delivers power. And now you see, that we are in a situation-- which I've peak here-- where one travelling wave loses energy, and the other traveling wave gains energy. And there is a classic experiment, which was done by Bill Phillips a while ago-- he had atoms in an optical lattice, and they were sloshing. And she could really measure that when the atoms were accelerated this way, one of the laser beams gained power, and the other one lost power. So while the atoms where sloshing in momentum space, the power between the two laser beams was distributed back and forth. So this is the character of the dissipative force, that it redistributes momentum and energy between the two traveling waves. Any questions? Boris? AUDIENCE: But if there's a net change in the patterns of motion, the energy still has to come from somewhere. Right? Can you make that [INAUDIBLE] PROFESSOR: This question has bothered in many different iterations. And at some point, I found a very, very easy answer to it. And this is one of your homework assignments for homework number 10. There's a very simple example where you will realize where the energy comes. What happens is, in steady state, if everything is stationary, of course, there can't be an exchange. But, if you're for instance, saying-- and I give you no part of the solution-- if you have atoms, and they are attracted by the dipole force, there's a strong laser beam and the atoms are accelerated in, and you're now asking, where does the energy come from? Because all of what the atoms do, at least according to what I'm telling you, they redistribute photons of energy, h bar omega, into photons of energy, h bar omega. So where does the energy come from? You have to go higher in the approximation here to understand where the energy comes from. Let me just tell you one thing, this is actually something which I've encountered several times. And it can be really confusing, I've really proven to mathematically the this is false. So in this level of treatment, with these very simple concepts, we correctly obtained the optical force. But if you're now asking, where does the energy go? You have to go deeper. And, for instance, what happens is if you take a cloud of atoms, and the cloud of atoms moves in an out of a laser beam-- just think of it as center of mass dipole oscillations of a Boson-Einstein condensate in an optical dipole trap-- what happens is when the atoms move in and out, they actually act as a phase modulator for the laser beam. The index of the refraction of your laser beam is changing. So therefore-- and we don't deal with that when he replace the electromagnetic field by a c number-- your laser beam, when the atoms move in will actually show a frequency modulation. And you will find-- and it's a wonderful homework assignment, I'm pretty proud that we created it, and that it works out so easily-- you will find that the kinetic energy of the atoms is exactly compensated by the energy shift the photons which have passed through the atoms. But I'm giving you a very advanced answer. It's often very difficult if you have a simple picture for the force to figure out what happens to the energy. Actually, let me give you the other example. When we have the previous example, where we have atoms and we have a laser beam which cools with a radiation pressure. This was our previous example. Where does the energy go? I have a wonderful correct-- within the assumptions-- an exact expression for the force. Where does the energy go? The atoms have a counter-propagating laser beam. They scatter photons with this rate. Every time they scatter a photon, they slow down by h bar k. They lose energy. We have a complete description what happens in the momentum picture and in forces. But where does the energy come? Or where does the energy go? The energy where the atoms have lost? AUDIENCE: They absorb a slightly high frequency with a Doppler shift. And then slow down and then re-release a photon with slightly lower energy. PROFESSOR: Exactly. So in that case, it actually depends which way. When the atom loses energy, the energy goes into blue-shifted photons. I'm just saying, in order to address the question, what happens to the energy, we have to bring in our physics, which we didn't even consider to address here. It is now the physics of the spontaneously emitted photons. We said for the force, spontaneous emission vacuum fluctuations can be eliminated. But if you try to understand where does the energy go, we have to go to those terms, and we even find the physics of Doppler shifts, which we didn't put in, which we didn't even assume here. But, of course, it has to be self consistent. So therefore, you sometimes have to go to very different physics which isn't even covered by your equation to see the flip side of the coin. One side is forces, which are very simple, but the other side of the coin can be more difficult. We'll actually come back to that when we talk about cooling [INAUDIBLE]. Cooling with a stimulated force. How can a stimulated force cool, because it's all u. Well, we'll get there. And it will again be-- the energy-- will be extracted through spontaneous emission from all those side [INAUDIBLE] and all that. So we have to trace down the energy, and we successfully will do that. But energy can be subtle. Forces are simple. Other questions? Actually, since I'm in chatty mood, when I explained to you where the energy goes in a dipole trap-- that it's a phase modulation for the laser beam-- I have been using optical dipole traps in my laboratory for many, many years, before I really figured out where the energy goes. And I'm almost 99% certain if you go to DAMOP, and ask some of the experts on cooling and trapping, and ask them to figure out this problem, most of the people will not be able to give you this answer. I've actually not found the answer in any standard textbook of laser cooling, until I eventually found it myself and posted and prepared it as a homework assignment. So many, many people may not be able to give you the correct answer, where does the energy go when atoms slosh around in a dipole trap. Try it out at DAMOP, it may be fun. OK. So back to the simple physics of forces. We don't need to understand the energy, because we know the forces, but we'll come back to that. So we have the reactive force. The reactive force is written here. What have we done? Well, you remember the alpha vector was the gradient of the Rabi frequency, and we have to multiply with the u component of the steady state solution of the Optical Bloch Equation. Here it is, just put together. It's a nice expression. There are two things you should know about it. The first one is that you can actually write this force exactly as the gradient of a potential. So there is a dipole potential, and is exactly this dipole potential which is used as a trapping potential when you have dipole traps. Probably 99% of you use dipole traps in the limit where the detuning is large. And then the logarithm of 1 plus x simply becomes x, and you have the simple expression you are using. But in a way, this is remarkable. The dipole force can be derived from a potential, even if the detuning is small, and you have a hell of a lot of spontaneous scattering, which is usually not regarded as being due to conservative potential. So this expression, that the reactive force is a gradient of a potential, even applies to a situation where gamma spontaneous scattering [INAUDIBLE] And you're not simply in-- and this is what this last term is-- in the perturbative limit of the AC Stark Effect. OK. That's number one you should know about this expression. The second thing is if you look at this expression, the question is, what is the maximum reactive force? We talked about the maximum spontaneous, or dissipative force, which is h bar k, the momentum of a photon times gamms over 2. What is the maximum force you can get out of the reactive force? Well, of course, if you want a lot of force, you want to use a lot of laser power. Therefore, you want to use a high Rabi frequency. And now, if you want to use your laser power wisely, the question is, how should you pick the detuning? And you realize if you pick the detuning small, there is a prefector which goes to zero, but if you pick your detuning extremely large, the denominator kills you. And well, if you analyze it and or stare at this expression for more than a second, you realize that the optimum detuning is when the detuning is on the order of the Rabi frequency. And if you plug that in, you find that under those optimum conditions, the reactive force can be written again as the momentum of a photon-- it must be, the momentum of the photon is the unit of force, the quantum of force, which has to appear-- but not the frequency is not gamma, but it is the Rabi frequency. So the picture you should sort of have is that under those situations, just assume the standing wave consists of two traveling waves, and the atom is Rabi flopping. But it is Rabi flopping in a way that it goes up by taking a photon from one laser beam, and then it goes down by emitting the photon into the other laser beam. So during each Rabi flux cycle, it exchanges a momentum of 2 h bar k with the electromagnetic field. So therefore, the reactive force-- the stimulated force-- never saturates if you increase the Rabi frequency-- depending, of course, on your detuning, but if you choose the detuning correctly-- you can get a force which is just going further and further. Questions? Yes, Jen? AUDIENCE: Is this the same concept as the Raman constants, except that the frequencies are just the same? Or is it just a-- PROFESSOR: It is a Raman process. Actually, everything is a Raman process here. Because in laser cooling, in everything which involves the mechanical-- the motion, the external degree freedom of the atoms-- we have an atom in one momentum state. We go-- light scattering always has to involve the excited state-- and then we go down to a different momentum state. So therefore, laser cooling is nothing else than Raman processes in the external degree of freedom of the atom. You may be used to [INAUDIBLE] Raman process more when you start in one hyperfine state, or for molecules in one vibration rotation state, and you go to another one, but, well, for vibration, and rotational states, and hyperfine states of it, Raman is used all the time. I usually used it also for the external motion, because I think it clearly brings out common features between all those different Raman processes. So you can actually say that the reactive force is a stimulated Raman process between two momentum states, where both legs are stimulated by laser beams. In a standing wave, it would be-- the two legs-- would be stimulated by the two travelling wave components of the standing wave. Whereas the dissipative force is spontaneous Raman process, where one leg is driven by the laser, and the other one comes with spontaneous emission. Nicky? AUDIENCE: [INAUDIBLE] when the atoms slow down, the [INAUDIBLE] PROFESSOR: So the spontaneously emitted photon? AUDIENCE: Yeah, I'm just thinking of actually the stimulated force. Because the atom must be stimulated. [INAUDIBLE] I'm confused how when the energy of the emitted photon must be slightly different due to the phase modulation? [INAUDIBLE] stimulated emission can be added to the frequency? [INAUDIBLE] PROFESSOR: I mean we had the discussion after [INAUDIBLE] question, if you go up and down, and you are stimulated, you go up in a stimulated way, you go down in a stimulated way, and both laser fields which stimulate the transition have the same frequency, omega, you would actually see that there is no energy exchange. So at our current level of description we have described where the force comes from, but we don't understand yet where the energy goes, or where the energy comes from. And this is really more sophisticated, and I don't want to sort of continue the discussion. We will encounter one situation, and this is in the [INAUDIBLE] atom picture, where we will discuss in a week or so, sisyphus cooling. And we will find out that there is symmetries in the mono-triplet. So again, we will find the missing energy in spontaneous emission. But if you don't have spontaneous emission, if you just have atoms moving in an optical lattice without spontaneous emission, it is really what I said, the phase modulation. And I would probably ask you at this point to do the homework assignment, and maybe have a discussion afterwards, because then you know exactly what I'm talking about. Does it at least-- You can take that as a preliminary answer. So these are the two limiting cases. One is spontaneous emission, and the other one is phase modulation at a certain frequency. One can be sustained in steady state, you can spontaneously emit in sort of an atom goes to a standing wave, and this is sustainable. Whereas the phase modulation thing is actually a transient. It's an oscillation which is added in. AUDIENCE: [INAUDIBLE] You're adding the force to the other one [INAUDIBLE] PROFESSOR: Yeah, actually-- AUDIENCE: [INAUDIBLE] PROFESSOR: If that helps you, but it's the following. I would sometimes say you know, I like sort of intuitive explanations of quantum physics. Let's assume I'm an atom. I'm in [INAUDIBLE] standing wave. And you don't know it yet, but I do sisyphus cooling. By just exchanging photons, a stimulated force, I'm actually cooling. And you would say, how is it possible? Because all the photon exchanges involve the same photon, the same energy. How can I get rid of energy? And I think what really happens is the following, the atoms can lose momentum without energy, but due to Heisenberg's Uncertainty Relationship this is possible. So the atom is first taking care of its momentum-- is losing momentum-- by stimulated force, and eventually, before it's too late, it has to do some spontaneous emission where it is paying back it debts in energy to Heisenberg. So and of course, after a certain time, everything is OK. The force was provided by exchanging photons of identical frequency. And the energy is provided by an occasional spontaneous emission event, using one of the [INAUDIBLE]. So everything is there. Everything is a perfect picture, but you have to sort of, in this description, allow a certain uncertainty that the force is the exchange of identical photons, and the energy balance is reconciled in spontaneous emission. And on any longer time scale there are enough events that the balance is perfectly matched. But if you would take the position, no, this is not possible. You know, the atom can only-- I mean, it's sort of in diagrams. It's not that one diagram which scatters a photon has to conserve energy. You have a little bit of time. You have Heisenberg's uncertainty time to make sure that another diagram jumps in, and reconciles energy conservation. That's at least my way of looking into it, but it's a very maybe, my personal interpretation of how all these diagrams and photon scattering events work together. OK. Let's do something simpler now. So I've explained to you the dissipative force, the reactive force, and in the next unit I want to show you simply applications of the spontaneous force. In every experiment on cold atoms, the spontaneous force is center stage. It is necessary to slow down atomic beams. It provides molasses, which is the colling of atoms to micro-Kelvin temperature, and the spontaneous force is also responsible for the Magneto-optical trap. So what I want to do here is, again, by showing you the relevant equations, how the spontaneous force, which we have just discussed, how these spontaneous force leads to those three applications. So this is one of the most experimental sections of this course, because this equation has it all in it. And I just want to show you how this equation can be applied to three different important experimental geometries. OK. So this is our equation. It has the momentum transfer per photon. It has the maximum scatter rate, gamma over two. And then, here, it has the Lorentzian line shape. And the important thing is when we talk about molasses and beam slowing is that the detuning is the laser detuning and the Doppler detuning. So the velocity dependence enters now the spontaneous force through the detuning in the Lorentzian denominator, and it enters through the Doppler effect. So you can pretty much say it like this, if you have a bunch of laser beam, and slow and cool your atoms, how can the lasers do the job? Well, they measure the velocity of the atoms through the Doppler shift. And if is the Doppler shift which tells the laser beams what to do, so to speak. And this is how laser cooling works. So just as an experimentalist, you should actually know what the scale is. If somebody asked you, how strong is the spontaneous scattering force? Well, a way to connect it with real time units-- with real life units-- is, what is the maximum deceleration? Well, this is 10 to the 5 G. You can ask, is 10 to the 5 G a lot? Or not? Well, for an astronaut, it would be a lot. It would. No living organism can sustain 10 to the 5 G. But I'm really surprised when I did this calculation. When you compare it to the electric force on an ionized atom, the same force, which is provided by the spontaneous light force, would be provided by an electric field of one millivolt per centimeter. So it's not easy if you have a stainless steel chamber to avoid petch effects, which create those electric fields. And if you've a battery with 9 volt, a centimeter apart-- just a 9 volt battery-- would accelerate an ion four orders of magnitude faster than your beloved strong spontaneous light force. So in that sense, 10 to the 5 G is a lot in the macroscopic world, but if you would look at microscopic forces-- which are often electric forces-- it's absolutely tiny. And sometimes I say, the fact that you can do 100 kilovolt per centimeter, that you can make electric forces which are seven, eight, or nine, orders of magnitude stronger, that's actually the reason why ion traps were invented before trapping of neutral particles. So some developments in the field of trapping particles-- and eventually laser cooling them-- it first started with ion traps, and then it proceeded to neutral atoms. And the reason is the ion trappers have forces at their disposal which are eight or nine orders of magnitude stronger. OK. Optical molasses. I know some of you will hate it, because I've just explained to you that in a standing wave, we don't have any spontaneous light force, all we have is the reactive force, the stimulated force, because it is the u, the in phase oscillation of the atomic dipole operator which is responsible for everything. But now I'm just, you know, wearing another hat, and I tell you that there is a limit where the total force can be regarded with some of the two forces. So therefore, I'm pretending now that if you have near resonant light in a standing wave, that the force in the standing wave-- which you know is a purely reactive force-- can now be written as the sum of the two propagating waves. And, of course, each propagating wave does a purely dissipative force. That is, what I'm telling you can be mathematically proven. That the stimulated force in a standing wave is equal to the sum of the two dissipative forces or each travelling wave. I just keep that in mind whenever you think you can rigorously distinguish between the dissipative force and the stimulated force. Keep in mind that a standing wave-- which has only a stimulated force, as I rigorously proved to you-- can alternatively be described as the sum of two spontaneous light forces, each of which provided by one of the travelling waves. Well, it sort of makes sense in a perturbative limit. You can just take one beam, you can take the other beam, and the combined effect of the two laser beams is higher order. So the fact that in some low intensity limit this has to be valid, is pretty clear. Anyway, but just a warning, if you really want to use a stimulated force here, you're in big trouble. It's much, much harder to get this result out of the stimulated force. Because to get a stimulated force with velocity dependence requires you to take solutions of the Optical Bloch Equations, which are not steady state, but non-adiabatic. So just a warning. You can do it, if you want with a stimulated force, but I would strongly advise you to first use the simpler formalism. You have to go to very different approximation schemes and much more technical complexity if you want to get it out of the stimulated force. OK. So with that assumption, you'll remember we had the dissipative force of each laser beam. Now we have two laser beams. We summed them up. Because the two laser beams come from different directions, we have a minus sign here. And for the same reason, we have plus and minus sign in the Doppler effect. So what we have is the following, each force of each laser beam has this Lorentzian envelope, shown in black, but the other force of the other laser beam has the opposite sign. And the two have opposite Doppler shifts. So if I add up the two black forces, I get the red force. And the red force is anti-symmetric with respect to velocity. So therefore, in the limit of small velocities I get a force which is minus alpha times v. And this is the force of friction. And in a whimsical way, people called this arrangement of two traveling waves Optical Molasses because the atom literally gets stuck in this configuration like a tiny ball bearing you throw into honey. So this is optical molasses. Actually, I think some of the Europeans had to learn the word molasses for it. In the US, everybody knows what molasses is. Well, in Europe, we know what honey is, but molasses is not a standard staple. But anyway, it's optical molasses. OK. Very simple result. And, again, it's not easy to get it out of the stimulated force. But it's possible. I will actually tell you how we get cooling out of the stimulated force later in some limit. AUDIENCE: Is there [INAUDIBLE] why this assumption of dissipative forces is easier [INAUDIBLE] and the stimulated force will be made difficult [INAUDIBLE]? PROFESSOR: I'm not sure if there is a simple answer why it's easier. Well, it can't be easier than that. AUDIENCE: I mean, there has to be something wrong with our assumptions of stimulated forces so that we cannot get [INAUDIBLE] PROFESSOR: Yeah. I think what happens is what I briefly said. To the best of my knowledge, but I'm not 100% certain. We have the simply expression because we have used the steady state solution of the Optical Bloch Equation. In other words, we have a laser beam, and it's just the power of the laser which tells us how much light scattering happens. And we use the steady state solution. And we sort of have folded that into the force for one laser beam. And then we have the same package for the other laser beam. And we have never considered that the two laser beam may have some cross talk. And this is indeed valid for all low laser power. However, if you want to get cooling out off the stimulated force, I will show some of you later, you can no longer use the steady state solution because it would miss the effect here. You wouldn't get any cooling out of it. In order to get something which is dissipative, out of a reactive force-- reactive force is by definition not dissipative-- you need a dissipation mechanism. And the dissipation mechanism is that you're not quite steady state, there is a relaxation time, a time lag. The atom is not instantaneously in its steady state solution. It needs a little bit of time to adjust. And it is these time lag of the atom which eventually gives rise to an alpha coefficient in the stimulated force. AUDIENCE: So then physically, you are, during cooling experiments, you can take care that you are actually not in the steady state limit? PROFESSOR: What I'm telling you is, for the spontaneous force, we get in leading order the effect by assuming the steady state limit, but if you want to get it out of the stimulated force, as far as I know, you don't get it with the steady state limit. AUDIENCE: OK. PROFESSOR: If you use different approaches-- I mean this is why you want to pick your approach. Sometimes you get something already in lower order. Sometimes you get it in higher order. And I think one favorite example is you solve this problem about [INAUDIBLE] scattering and Thompson scattering, pick your Hamiltonian, d dot e, or p minus a. In one case, you have to work harder than in the other case. And I think here it's similar. Anyway, let's just put in-- I think that's, yeah, we can do in the next few minutes. So we now want to put in the effect that the force fluctuates. And that means we have heating. Before I do that, let me just tell you what the solution is for force equals minus alpha times v. Well, it means we extract energy out of the system at a rate-- Well, energy per unit time is force times velocity. Force times velocity is now minus alpha times v square. But v square is the energy, the kinetic energy. In other words, this equation tells us that the atomic motion, the kinetic energy, is exponentially [? damned. ?] And if there were nothing else, you just need two laser beams-- optical molasses-- and you would go not just to micro-Kelvin, but to nano- and pico- Kelvin temperatures. It's an exponential decay to absolute zero. However, we don't each reach nano- and pico- Kelvin temperatures in laser cooling because there are other processes. And what is important here is spontaneous emission. OK. So the way you treat spontaneous emission is the following, every time an atom emits spontaneously, there is a random momentum kick of h bar k. If you have n photons scattered because the momentum kicks going random direction, they only add up in a random box. You get square root N. Or if you ask, what is the average of p square due to spontaneously emission? It is the momentum of the photon squared times the number of scattering events. So therefore, in the form of a differential equation, the heating rate, or the derivative of-- the TEMPO derivative-- of p square goes now with the number of photons per unit time, which is the scattering rate. OK. If you would stop here, and that's what many people do who explain heating in this situation, you would miss half of the heating, because what you have treated here is only the photon transfer and spontaneous emission. However, there is also fluctuation and absorption. I mean just look at two atoms in the [INAUDIBLE]. Kind of me and another atom. And we both scatter photons. On average, in the same laser beam, we absorb in photons and get in momentum kicks. But there is Poissonian statistics how many photons I absorb, and how many photons my twin brother absorbs. So therefore, due to the randomness in absorption, or the fluctuations in absorption. There is another square root n variance in the recoil kicks which comes from the absorption process. And it so happens, for exactly that reason, that the heating-- the derivative of kinetic energy, or the difference of momentum squared due to absorption is exactly the same as in emission. Well, if you now make different assumptions, spontaneous emission has a dipole pattern. And you can sort of factor on the order of unity, depending what the pattern of the spontaneous emission is, whether you make a 1D or 2D model of spontaneous emission that the photons can only go in one dimension, in two dimension, or three dimensions. So you have to get other prefectors, but the picture is that without going into numerical factors on the order of unity, you have fluctuations in the absorption, you have a randomness in spontaneous emission, and they both equally contribute. So that means now the following, that we have heating rate. The heating rate we just talked about. The increase in p square. Well, if you divide by 2 times the mass, it's increasing kinetic energy. So the increase in kinetic energy is given by this expression. And it is common if you have a heating process to introduce a momentum diffusion coefficient in that way. It's just the definition. It shows you how p square increases per unit time. And now we can get the cooling limit for spontaneous emission, namely by saying, in steady state when we have-- due to photon scattering-- the same amount of heating and the same amount of cooling, the temperature will have asymptotically reached a steady state value. The heating rate was parametrized by momentum diffusion coefficient, d over m. So this is sort of independent of the velocity of the atom. Whereas you'll remember the cooling rate was proportional to the energy because it was an exponential approach to zero energy. So therefore, when we have the heating equals the cooling, we have the energy of the atoms there, and therefore, we find that the energy of the atoms in steady state is given by, actually, the ratio of heating versus cooling. It's a simple expression. The energy or the limiting temperature for molasses is the ratio of heating versus cooling. Heating is described by momentum diffusion coefficient and alpha is described by the damping force. I want you, actually, to keep this expression in mind. A limit in temperature in laser cooling is always-- actually, in other processes in laser cooling-- is the ratio of heating over cooling. And momentum diffusion coefficient due to heating, over a friction force due to cooling. Because we will later find polarization gradient cooling, cooling in blue molasses, we will find other cooling schemes, where we will calculate-- with the appropriate model-- the heating and the cooling. And I don't have to repeat this part. The moment I calculate the heating, the momentum diffusion, and I calculate the friction, I know what the limiting temperatures is. Again, when I said, the energy is kt over 2, you know, of course, kinetic energy is kt over 2 times the number of degrees of freedom. I assumed 1D, here. So in everything I've said on this page, there are numerical factors which may change whether you assume one or two or three dimension. OK. I think that's the last thing I want to tell you. And then on Wednesday we do Zeeman's slowing and Magneto-optical trapping. We had an expression for alpha. Remember we had this Lorentz profile, for the other Lorentz profile, and alpha was just the slope. I mean, everything is known. I just didn't bother to calculate it. But it just involves derivative of Lorentzian. And now we can ask, what is the lowest temperature we can reach? Well, you can now analyze your expression for alpha for a two-level system solved by the Optical Bloch Equation. And you find that you will have the most favorable conditions for low temperature-- you get the minimum possible temperature-- in the limit of low laser power. And your detuning-- your optimum detuning-- is half a line width away. And then you'll find this famous result, that if you cool a two-level atom, the limiting temperature is simply given by the spontaneous emission rate, or the line width of your transition, gamma. And this is the famous result for the Doppler limit. For sodium atoms, it's 250 micro-Kelvin, for rubidium and cesium, because of the heavier mass, it's lower, is 10's of micro-Kelvin, 50 or 100 micro-Kelvin. So this is a famous limit. And the physics of it is the following, the narrower the line width is. Then you can say the better you can cool particle down to zero temperature-- and I think, Jenny, your idea of the Raman process comes now in very handy-- I can have an atom, and I can have a Raman process where I scatter photons and go to higher energy or I scatter photons and got to lower energy. And the difference between the two processes is actually, the one that whether I cool or whether I heat comes from the Doppler effect. So the more I can discriminate between through the Doppler effect, the better I can cool. But the Doppler effect-- the Doppler shift, kv-- how well I can resolve it, depends on the width of the atomic transition. The narrower the atomic transition, the more the Doppler effect can steer the laser cooling-- the Raman process-- towards lower velocity, not towards higher velocity. So that's the natural result that the-- or an expected result-- that the natural line width appears here. OK. I've talked a few extra minutes. Is there any question? Then I see you on Wednesday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

21_Ion_trapping_and_quantum_gates.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, so let's start to discuss ion traps. Now, in many situations, ions are just charged neutral atoms. And if the charge doesn't matter, you can do very similar physics with ions and neutral atoms. However, the charge makes major differences when it comes to confining and manipulating particles. So when I compare neutrals to ions, the forces which we exert on neutral atoms are magnetic forces, optical forces. And typically, the traps we can create have a depth of about 1 kelvin. Actually, you just need superconducting magnets. The magnetic traps we build here at MIT have millikelvin depths. So the trap depth is often on the order of microkelvin, millikelvin, and kelvin at the high scale. Well, this is very different in ions. In ions, you have the Coulomb force. If you take electric fields of 100 volts per millimeter, you can create confinement forces for 10,000 kelvin. So you have no problem in confining room temperature ions. When it comes to-- so this is trapping and confinement. If you talk about cooling, well, our workhorses are Doppler cooling and evaporative cooling. You can actually cool ions by evaporation. You can also cool ions by evaporative cooling. But what has been very common and special to ions is the ability to resolve sidebands, sidebands of the vibration and motion in the harmonic oscillator trap. And in problem set number 10, you have looked through the physics of sideband cooling. However, sideband cooling is now also coming back to neutral atoms. For instance, if you have neutral atoms in a very tightly confining dipole trap, you can resolve sidebands and use those for cooling. OK, what is the science? Well, a driving force for both ions and neutral atoms has been clocks. If atoms are cold, you can use them to have very accurate ways of keeping time. And I think currently, there is a race between the based neutral atom atomic clocks, neutral [INAUDIBLE] optical lattices, and ion traps in particular, the aluminum ion at Boulder. So we have clocks. In ions-- and this is what I want to talk a little bit more about today-- a major driving force was quantum information. You have two ions and get quantum gates between them, or you have eight ions and can have now eight qubits in a prototypical quantum computer. So quantum information science and quantum algorithms was quite important. But I would say some of the physics is at least discussed and partially pursued with atoms in optical lattices. We have talked a lot about, for neutral atoms, quantum simulations, studying super fluidity, studying phase transitions, magnetic properties of many body systems. But there are more efforts to do that with ions. Chris Monroe's group in Maryland is studying antiferromagnetic ordering of a spin chain. And the spin chain consists of a string of ions in an ion trap. So see actually on the science side that all three-- clocks, precision measurements, quantum information, qubits, quantum algorithms, and quantum simulation of many body systems-- are done with both neutral atoms and ions. So these are now two very different platforms. So I could spend a long time teaching about special aspects of ions. How do you do a trap? How do you cool ions? How do you do quantum state control? Because this is our last lecture, I decided to focus on two aspects. One is, I want to talk about a special way of trapping ions. Because there is one difference for ions. Ions, until recently, have not been trapped in stationary traps. They've only been trapped in dynamic traps. So you cannot trap ions by simply putting an electric potential around it. Because the divergence of the electric field is 0. And that means there's always one direction in space where the ions are dragged out. You cannot have an inward force from all directions. So therefore, when you want to use electric fields, you can only use dynamic traps. Well, for the experts I should say, this does not apply for the optical dipole trap. We know how we can create trapping potentials for neutral atoms using the optical dipole trap. And very recently, a group was able to trap ions not with electric and magnetic fields provided by electrodes or radio frequency, but by inner focused laser beams. So it's a very new development. But it uses exactly the mechanism which has been used for neutral atoms. So when it comes to dynamic traps, the standard two traps which have been used for ion trapping-- one is the Penning trap, which I do not want to describe in detail. The Penning trap uses a quadrupole configuration of two positive charges. And then you have a ring with negative charge. That means that an ion is confined here but would see anti-confinement in the outward direction. Now, what you do with the anti-confinement? Well, the solution is that you add a magnetic field. And what happens now is the particle is well-confined axially. But radially, when it wants to move out, it's turned into a cyclotron orbit, and therefore stays trapped. So this is the principal of Penning traps where the confinement in the axial direction is electrostatic, and the confinement in x, y is through cyclotron motion. OK, it's wonderful physics to describe the motion of ions in this trap. It's the combination of an axial motion, cyclotron motion, magnetron motion. It's very interesting physics. And for many years, Dave Pritchard here at MIT used such Penning traps to measure the masses, determine the masses, of ions with the highest precision ever possible. MIT still has the world record of the most precise mass measurements at the 10 to the minus 11 level. And this was done in Penning traps. But I want to talk more today about RF traps, or Paul traps, because the are used in the majority of experiments. And they're used in pretty much all experiments on quantum information processing with ions. So the neat thing-- and I try to teach you concepts in physics-- the interesting thing about Paul traps or RF traps is that you alternate at a radio frequency between confinement and anti-confinement. You can say you have a settle point potential, which is confining in x and anti-confining in y. But then, you switch x and y. So therefore, the situation is just imagine you're an ion, and you feel half of the time a confining harmonic oscillator and half of the time an anti-confining harmonic oscillator. The question I have for you, and this is a general physics question, well, is the sum of a confining and an anti-confining potential of the same strength give 0, or does it give something which in the end is confining? So if you were in a potential, and the potential switches permanently form parabolic upward to parabolic downward, at any given position, the average force is 0. Because in one phase, you have a force to the left. In the other case, you have the opposite. You have the same force but to the right with the opposite sign. But what's going on here? Yes. AUDIENCE: So I can't answer this mathematically or even physically. But I just remember this very cool example that someone showed me once where they took a saddle shape in space, put a ball in the middle, and then you spin the saddle. And as soon as the ball tries to fall to one side, the saddle will spin over and pick it up. So I know the answer has to be, it is confining. But I can't explain why. PROFESSOR: OK, Colin? AUDIENCE: Could it possibly be like a threshold? Because if you consider spinning it sort of slowly, then the particle can just sort of roll off. Maybe there's a threshold. PROFESSOR: Yeah, there is a threshold. I mean, if there's a confining potential for a long, long time, and then there's an anti-confining potential, and I just fall down the anti-confining potential, and I disappear to infinity, the confining part doesn't help me. So there will definitely be a threshold in time. We have to switch between confining and anti-confining sufficiently rapidly. Yes. AUDIENCE: Does it have something to do with fictitious force [INAUDIBLE]? PROFESSOR: No, but this would have been a possibility. Maybe there is a formalism. But the standard formalism does not use the rotating frame. Let me-- AUDIENCE: You could think of the classical analog where you have a bead on a wire, and you have a stable position at sort of a finite angle only at certain frequencies. PROFESSOR: Yes, I'm not sure. Anyway, you all-- I think everybody here works with optics or knows what optics is. So why don't I discuss the following. If you take a positive lens and a negative lens with equal strength, what happens now to your laser beam? You send a laser beam through, it gets an inward bend by the convex lens, an outward bend by the concave lens. Is the net effect nothing, or is it focusing? Well, it depends. What happens is the following. If the lenses are put in close proximity, the two bending actions on the rays just completely cancel. But now we can do the following. We can put the two lenses further apart. And now the following happens. We have a laser beam which is bent inward. And you probably all know the lens maker's equation, that the further out the rays are, the more they have to be bent inward to come to a focus. So what happens is those beams are now bent inward. But when they reach the other lens, they are now bent outward somewhat. I'm not very accurate in my drawing. But the angle by which they're bent inward is proportional to the separation here. The angle by which they're bent outward is proportional to the separation there. So the inward bend is stronger than the outward bend. So therefore what I'm telling you is convex and concave, with a certain distance between the two, results in focusing. Now maybe you can think about it. What happens if I reverse convex and concave? AUDIENCE: [INAUDIBLE] PROFESSOR: Is it now still focusing, or is it the opposite of the example above? Now the first lens bends the beam outward, yes? And the second lens bends it inward. But since we reach the second lens further out, the invert kick is stronger than the outward bend initially. So therefore, when you have positive and negative lenses of equal strength, no matter in what order you apply it, you always get an inward force. So this should actually explain-- AUDIENCE: As long as they're within the focal length. Could you imagine if I had positive and negative that was larger than the focal length away? PROFESSOR: Yeah, everything within reason here-- there will be a stability diagram. But I want to first make it very plausible what happens. And just to address the question which I raised before, if I'm in a potential, and I get an equal amount of inward and outward force, shouldn't that average to 0? The question is how you average. If I would keep the distance from the x as constant, I would say here as a function of-- let me call it x. The bending angle as a function of x is positive. The bending angle at the same x is negative. So if I would do the average at constant x, I would get 0. But what I do is I do the average following the trajectory of the beam. So in other words, if I'm sitting on this potential, which switches from attractive to repulsive, if I would be here at my position, the average force is 0. But when the potential is repulsive, in 1/2 cycle I'm pushed outwards. And further out, the attractive force is now stronger than it was before. And therefore, if you average the force at my position, you will always find an inward force. But if you average the force at a fixed location, it averages out to 0. So this is exactly how RF ion traps work. And well, the rest are just equations to describe that. So I think we have to make a choice. I could derive for you the equations, how you average over the rapid oscillations of the potential, and you get a harmonic oscillator potential, and this is how you trap ions. But it's pretty much showing to you how the concept I've just explained to you is reflected in equations. Or the second option we have that takes about 20 minutes is to explain to you how quantum computation is done with two ions, in particular how the famous Zoller-Cirac gate works. Do you have a clear preference? AUDIENCE: Second one. PROFESSOR: OK, fine, but there is one piece of mathematics you should understand. Really, and this is the only essence of the mathematical derivation. If I'm standing here, and I feel the alternating force, it's a cosine omega t, and it averages to 0. But if the force drives me, it shakes me. This is micromotion. Like the beam here, if you go through many lasers, the beam goes up and down, up and down. This is micromotion. The micromotion is at cosine omega t. And the net attractive force comes because the force which oscillates at cosine omega t, and I'm oscillating at cosine omega t, then you get cosine squared omega t. And this gives you the net force which provides the net attractive force. And it's the cosine square term averaged over the micromotion, which is one half, which is responsible for the trapping potential in ion traps. Anyway, you know the physics very well. You understand where the math comes form in that it's really just the product of the two. It's a rectification. It's an alternating force. But since the system itself, the particle itself, is driven at the micromotion, there's rectification going on, which leads to the cosine squared term. [? Timor, ?] yes? AUDIENCE: Two quick questions. So I think ion trappers always complain about the micromotion. But in fact, they need it for trapping to work-- is pretty much what's going on here, right? PROFESSOR: Yes, actually, the one interesting side of the mathematics is the micromotion is kinetic energy. The harmonic trapping potential is nothing else than the micromotion. In other words, if people show in an ion trap that you have a harmonic oscillator, the harmonic oscillator potential is high here. But mathematically, this harmonic oscillator potential is the kinetic energy of the micromotion. In other words, if you look at the micromotion, it has a kinetic energy. The kinetic energy appears as the potential energy in this slow secular motion. So in other words, the potential energy for the trap is the kinetic energy of the micromotion. So you can't avoid it. This is responsible for the restoring force. AUDIENCE: Is it a really conservative trap, or can I have an out of phase component to the micromotion response? Is there like a sine part to the response? PROFESSOR: Well, the secular motion is conservative. You're now asking-- AUDIENCE: I'm trying to think if there would be a reason why my response, my dipole response, would have a different [INAUDIBLE]. Is there a reason? PROFESSOR: Well, to the best of my knowledge, everything is wonderfully conservative if you have a separation of timescale. If you want to sort of get something strange and non-conservative out of the micromotion, you have to violate the adiabatic separation of timescales. There is a possibility that if particles move quickly through the potential, they're not following the micromotion adiabatically. And then there is a collision where micromotion is released as collisional energy. But on the similar particle level, it's a pretty good approximation. OK, we want to talk about quantum computation with ions. [INAUDIBLE] has a wonderful write-up on the wiki on the basic concept of quantum computation and how it applies to ions. But let me immediately jump to qubit operations. You've heard many, many times that a general quantum algorithm can be constructed out of single qubit rotations and two qubit gates. Single qubit rotations are just oscillations on the Bloch sphere. If you excite a particle, move it around on the Bloch sphere, this is how you can rotate a single qubit from ground state to excited state, any possible superposition state. So the physics of single atoms, Rabi oscillation, rotation on the Bloch sphere, this is all you have to know about single qubit operations. The interesting thing is how do we create two qubit operations? And two qubit operation would mean you want to do something to one ion here depending what the state of the other ion is. And this is much more subtle. So what I want to explain to you now is the famous Cirac-Zoller gate, which was really a pioneering paper which has thousands of citations now. Because it showed that quantum computation can be experimentally realized in a rather straightforward way. So it's a fundamental theorem of quantum information processing that if you have single qubit operations, and one to qubit gate, which is a CNOT gate, that this is a universal set of quantum gates, any arbitrary quantum algorithm can be constructed out of those gate operations. So let me just remind you what the CNOT gate is. We have qubits ion one, ion two. They can be in 0 or 1. These can be two different hyperfine states. Or in the case of the calcium ion, it can be the ground state or metastable excited state-- just two different states of ion one, two different states of ion two. The CNOT gate would now say the following. The first ion is the control bit. And whenever the control bit is 1, it flips the qubit, the second qubit. If the control bit is 0, it does nothing to the second ion. So therefore, the table which shows how the quantum states perform is 1, 1, 0, 0, 1, 1. It just shows that I'm writing down the matrix in these spaces. It's a truth table 0, 0, 0, 1, 1, 0, 1, 1. And it shows that when you have 1, 1 at the input, you get 1, 0 at the output. If you have 1, 0 at the input, you get 1, 1 at the output. Well, the two qubit gate I want to discuss, which can be realized in the simplest possible way with the Cirac-Zoller algorithm, is the controlled phase qubit, which has a truth table of 1, 1, 1, minus 1. And it means it simply changes the face of the two qubit states when both qubits are 1. Now, the CNOT gate can be constructed out of the controlled phase gate by having two single qubit rotations done with the Hadamard gate. The Hadamard gate is taking the bases 1 and 2 and creates 1 plus 2, 1 minus 2, the symmetric and anti-symmetric superposition state. So therefore, the CPHASE gate is S general as the CNOT gate. Because one can be used to construct the other. So how do we create the Cirac-Zoller gate? So how do we get now the controlled phase gate out of trapped ions? So the effect is the following. Remember, if we have two ions, we want to change the phase to minus 1 of the state only if the two ions are qubit up. So in other words, we want to do 180 degree rotation of the wave function, but only if both ions are up. And the question is, how do we do that? How does one ion feel what the other ion is doing? And let me explain that now. The critical element in the Cirac-Zoller case is a center of mass motion. We want at some point-- say when this ion is spin up, we swap the quantum information from the ion to the center of mass motion. In other words, when this ion is up, we use laser pulses. I will explain that to you. Then all the ions are moving together. And then, this ion, because they're all moving, knows that this ion was spin up before. So the center of mass motion, the common dipole oscillation of all ions together in the trap, this is the data bus. So this is the key idea. But in order to explain it to you, I have to go in two steps. So let's just take that as a genetic level structure. We have an S state. And it has two different hyperfine states, spin up and spin down. And these are our two qubits. There may be a P state, which we have used for laser cooling. But everything is in the ground state of motion using sideband cooling, which you've studied in your homework. And what I want to show you first is how we can create a phase gate. But I'm only focusing on ion one. So hold the thought. Ion two will come in a few moments, will come back. So what we can do is the following. Each qubit state has vibrational structure-- ground state, excited state without center of mass motion. But then, as a function of vibrational quantum number for the center of mass motion, we have g1, g2, e1, e2, and such. And what we can now do is, let's assume we have another state, a D state, which is an auxiliary state here. Fortunately, ions and atoms have many states. And we can drive a laser transition from S to D, which is only in resonance with the first vibrational excited state to the D state. So we use a [INAUDIBLE] sideband here, which only drives the transition from e1 to the non-vibrational state of the auxiliary state. And that means automatically that this laser does nothing to e0, does nothing to g0, does nothing to g1. So if you have this Hilbert space of g0, g1, e0, e1, the only transition which is effected here, or the only state which is effected here, is the e1 state. OK, I'm talking about a single ion. I'm just talking about talking to a single ion right now. It's really simple. So if we switch on this laser, what happens is we simply induce Rabi oscillations-- simple physics. The Rabi oscillation, however, after one Rabi period has changed the phase of the initial state by minus 1. If you carefully look what Rabi oscillations are, the wave function goes up, down, up, down. But we usually discuss what happens to the probability. And this is shown in green. So when we are back in the original state, we have changed the phase of the wave function by minus 1. So let's just assume that we switch on this laser to this auxiliary state, and we have a-- well, it's a 2 pi pulse in population. But it makes a phase shift of pi to the wave function, which is minus 1. So what we have accomplished now is a phase gate not between two ions. But we have a phase gate which, in this Hilbert space, between the qubit up, qubit down-- I call it ground and excited state. But the other qubit, so to speak, is the center of mass motion 0, 1, 0, 1. So this is the truth table of the phase gate. And so technically, we have now done a local phase gate with one ion where the two qubits are ground and excited state of the ion. This is our qubit we want to work with. But then, the second qubit to realize the phase gate was 0, 1 of the center of mass vibration. But once we have realized that, we can now immediately take this idea and get a phase gate between two ions. Because the center of mass vibration is everywhere. So therefore, what we can do is this was ion one. We can now-- and this is the missing step I want to show you-- go to ion two. And we can swap the qubit information-- ground, excited, spin up, spin down. We can swap that with the center of mass motion. So therefore, what we can do is, when the second ion is spin up, we do a swapping of the information from spin up, spin down to center of mass motion being 0 or 1. So the idea is the following. Ion two should control what ion one is doing. So we first focus on ion two. We swap the qubit of ion two into the center of mass motion. Then, we do what I just said, the controlled phase gate for ion one with the center of mass motion. And after this is done, we write the center of mass motion back into ion two. And then we have nothing to ion two. Ion two was just putting its information into the center of mass motion. It helped ion one to do the controlled phase gate. And then, we swapped the information back. So this is the sequence. After preparing the center of mass, the first of two lasers cool everything in the center of mass motion into the 0 state. Then we swap from ion two the information to this phonon state center of mass vibration. Then we do the controlled phase gate on ion one. And then, we swap the information back. So what I should show you-- you may have seen it. But what I should show you explicitly-- how do we swap the information ge from the internal state of ion two? How do we write this information into the data bus, which is the phonon, which is the center of mass vibration in the trap? Well, we again use [INAUDIBLE] side pulse. We do a pi pulse driving this transition. And just look how beautifully the math works out. We have an arbitrary qubit state, a superposition of ground and excited. And the center of mass motion was laser cooled to the ground state. So let me just multiply that out. So we have coefficient a times g0 plus coefficient b times e0. If we now do a pi pulse, we transfer the population from e0 into g1. And this is what I've done here. But now you see I can factor out the g. And what I have now is I have the same qubit information with coefficients e and b, but no longer in the internal state of the ion but in the phonon state in the qubit, which is the center of mass motion. So the idea here is you can have an arbitrary number of ions. And now you want to have a quantum gate. What you do with a quantum gate, you say, I want to now take this ion here and have a quantum gate with 10 ions further down. So you first go to your first ion, transfer the information of this qubit into the data bus, into the center of mass motion. And now with a local operation here, you can do something to this ion conditionally on the motion, the center of mass motion. But the center of mass motion reflects the ion over there, in which state it was. And after you have done the two qubit operation between those two ions, you take the center of mass motion, put it back into this ion. And at the end of the day what you have done is you have done a two qubit operation. You've changed the state of this ion. But all the other ions have remained the same. And with that, you can now, in a chain of ion, implement arbitrary two qubit operations. I think I said it clearer in words here. It is in a little bit more formal way the steps you will do to the wave function. You initialize. Your transfer the information to the data bus, you execute the controlled phase gate, and you put the information from the data bus back into the control ion. And now your data bus is free, is back again to the initial value and ready for the next operation. I should mention that this was pioneering paper. Right now, this quantum gate is not the most popular one. Because it has severe requirements. One is you really have to cool to the absolute ground state. The center of mass motion has to be cooled to the absolute ground state. There are other quantum gates which do not require such extreme cooling. And secondly, the center of mass motion, because all particles move together, is one of the slowest vibrational motions in the system. And if you want to scale up the system to more and more particles, the speed of the gate operation has to be slower than the vibrational period of the center of mass motion. So these are two disadvantages. But there are different gates, like geometric gates, like Sorensen gates. There are other possibilities to involve the same quantum logic with ions. But the one of Cirac-Zoller really stands out by it's pedagogy. It's really a wonderful way to see how we take the information, put it in a data bus, do the controlled phase operation, and then proceed to the next two qubit gate. OK, so that's the end of ions, the end of this class. Do you have any questions about the last topic? If not, let me say thanks. You were a fabulous class. I really enjoyed the classroom atmosphere. I enjoyed all the questions, whether it was after class or during class. And, well, I'm happy that most of you I will see around for a number of years. All right. [APPLAUSE]

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

3_Quantum_description_of_light_Part_1.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Good afternoon. Usually, my first question is do you have any questions about the last lecture? We did this fly over over 100 equations to go from first principles to the [INAUDIBLE] Hamiltonian. I've put up here for you the overview of this course. I won't go through it again. But I want to sort of make you aware that we started with the fundamental Hamiltonian for light atom interactions. But now we take a step back. In the next few lectures, we simply talk about light. We talk about photons. We talk about the quantum nature of photons. And then, after understanding photons in more depth, we come back and look at more depths at atom photon interactions. Let me just give you a pictorial summary of what I've just said. AMO physics is about atoms and photons. And in a cartoon like way, I can symbolize atoms by two level systems. Or sometimes, we have a two level system which is coupled by tunneling that's also equivalent to a two level system. And then we have photons, two state systems. And photons are different. Photons are not two state systems. They are actually harmonic oscillators. Usually, electromagnetic waves. But Hamiltonians for photons can be and have been realized by replacing the harmonic oscillator with the photon with vibrations, for example, of ions in an ion trap or micromechanical oscillators. So cartoon picture again is photons in a cavity. But this is just another harmonic oscillator. So we have the atoms. And most of what you should know about atoms has been discussed in 8.421. For those of you who will take 8.421 next year, we'll have a lot of fun talking about the structure of atoms. Photons, well, I would hope a lot of it, at least the harmonic oscillator aspects, you took in 8.05 or in 8.321. And actually, starting today in this unit, we want to talk about the quantum age of light. So you can say 8.421, we had a deep look at the atoms. We have now look at the photons. But eventually, what is very important is the interactions between the atoms and the photons. That's not all. This is sort of a very well defined system. A two level system and one mode of the harmonic oscillator. But it will become very important-- and that's what we'll do in a few weeks-- to discuss what happens when this system-, the system above the dotted line, becomes an open system and we interact with the reservoir. The reservoir will be relevant for relaxation. Atoms can decohere, can lose their decoherence. Photons will decay or decohere. Entanglement will get lost. And this phenomenon of relaxation requires the formalism of the master equation. And we will find the master equation in the form of the optical block equation. If you want me to add in cartoon picture of the reservoir, I would add many, many harmonic oscillators. So reservoir is characterized by having many modes and having a capacity to absorb energy without changing its state. So this is another kind of illustrated summary of what we will be doing in the next few weeks. And today, we want to talk about the quantum nature of light. We're actually talking mainly about already a very special form of light, namely monochromatic light in a single mode of the harmonic oscillator. And there is still a lot to be discussed about the statistics, about squeezed states, about coherent states. We want to understand how the quantum state of light is modified, is transformed, by propagation, or how it's modified when the light runs through a beam splitter. And all of this is already relevant for just a single mode. And then very soon, we'll also talk about two modes of light. Because we need two modes if we want to talk about entanglement of two modes. I hope you will realize in this discussion that the quantum nature of light is more than light is a wave in the form of electromagnetic wave, or light is a particle in the form of photons. Things go much, much deeper. So the unit on quantum light will actually evolve in six stages. Today, we will be talking about photons and statistics, fluctuation, coherent states. On next week, on Monday, we want to talk about non-classical light, states which have no classical analogy in squeezed states. Eventually, we want to focus on single photons. We will then talk about entangled photons. The fifth part is on interferometry, or more generally on metrology. We will talk about the short noise limit and the fundamental Heisenberg limit of performing quantum measurements. And the last one is now opening up the light to interactions with atoms. So we will talk about interaction of one photon and one atom in this specific situation of cavity QED. So this is an overview over the next, I don't know, three, four, five, six lectures. Today, we start with a very, very short review of the simple harmonic oscillator, just to set the stage. I then want to show you that in three minutes, I can discuss thermal light and the Planck Law by just writing down one or two equations in the formalism we have chosen. So we cover thermal states. When we do laser physics, when we have light atom in the action, we always need coherent states. And we talk about quasi-probability distributions. We then want to understand fluctuations, noise, and second order correlation functions. Probably this will spill over into the next class. We want to address some properties of the single photon. Which, as you will see, it seems a simple system, but it is the most non-classical system you can have, a single photon. But we'll come to that. Let's start with very, very short review of the simple harmonic oscillator. It just also connects me with the last lecture. So how do we describe light in a single mode? And single mode means it is perfectly monochromatic. Well, we know that-- and that's what we did last class-- that we have the Hamiltonian for the electromagnetic field, which is the quantum version of the electromagnetic energy, B square E square. And from that, we derived the quantized Hamiltonian. We introduced, just as a reminder, vector potentials, normal modes, and the normal mode operators became the a daggers. And in second quantization, this is the Hamiltonian for light, which is of course the Hamiltonian for harmonic oscillator. What I need today is I have to rewrite the harmonic oscillator. And we have to look at two variables-- the momentum and the precision. Actually, the momentum will later, I will show you that the momentum of this harmonic oscillator is actually the operator of the electric field. So we need that in a second. And of course, I assume you know from your knowledge of the harmonic oscillator that precision and momentum are linear combinations, symmetric and anti-symmetric, of a dagger a. And the unit is the prefactor square root of h bar omega over 2. Or here it is h bar over 2 omega. And of course, in precision and momentum operators, the Hamiltonian is simply p square, should remind you of kinetic energy. Omega square, Q square should remind you of potential energy. When we discuss light, we want to discuss the electric field. I had shown you in one of those 100 equations the operator for the electric field. The operator for the electric field was, of course, related to the vector potential. And the vector potential, which is a normal mode expansion. And then we put everything backward. And we found that the electric field operator can actually be written. Have to fill in a few things, but essentially, it's a minus a dagger. And a minus a dagger is nothing else than the momentum of the harmonic oscillator. So here, the momentum has a special role, because it is the operator of the electric field. If you have light in a single mode, it has a harmonic oscillator representation. And the operator for the electric field is-- I put in a few bells and whistles in a moment-- but is in essence the momentum operator of the mechanical and analogous harmonic oscillator. OK but the electric field has a polarization. We need that. We have the photon energy and the quantization volume. So what I'm writing down here is actually the electric field, this square root, is the electric field strings of a single photon in the quantized volume of our cavity. So v is the quantization volume. Then if you have a single mode of the electromagnetic field, well, in contrast to a harmonic oscillator, we have a prorogation kr plus omega t. And here we have e to the minus i. I hope I get the signs wrong here. Minus omega t. But, OK, we will mainly be looking at an atom at the origin. That's what we did in the dipole approximation. So when I said that the electric field operator corresponds to the momentum, I have to add now that it is at t equals 0 and at r equals 0. We'll come back to that in a few moments. OK, since I've given you the operator-- let me just put the operator symbol on it-- the operator of the electric field, I have to say now one word about Heisenberg operators versus Schrodinger operators. I assume you're all familiar with the two different representations of quantum mechanics. We often use the Schrodinger representation where the operators are time independent. But in the Heisenberg representation, the wave function, the state vector, is time independent. And the time dependence of the evolution of the quantum system leads to a time dependence of the operator. So what I've written down for you here is actually the Heisenberg operator, because it has a time revolution. But for a single mode, monochromatic electromagnetic field, the time evolution is just e to the i omega t. So if you want to obtain the Schrodinger operator, you have to eliminate the time dependence by setting t equals 0. At t equals 0, by definition, the Schrodinger operator and the Heisenberg operator are the same. But then, in the Schrodinger picture, the quantum state evolves and the operator stays put. In the Heisenberg picture, the quantum state stays constant and the operator evolves. So I'm usually writing down for you expressions for Heisenberg operators. And if you simply set t equals 0 in this expression, you have the Schrodinger operator. Questions? Yes. AUDIENCE: What happens when there's a factor of two that shows up in the electric field of a single photon, like the [INAUDIBLE]? I thought it was like-- [INAUDIBLE] it's just a factor of two [INAUDIBLE]. PROFESSOR: Yes, you have to be careful with factors of two. I think I got it right here. There is no extra factor of two. But the question is, OK, whenever you write down equations for the electric field, you want to sort of use a and a daggers photon number, or create photon numbers, so they are dimensionless. So you always need a prefactor, which has a dimension of the electric field. And for the quantisation of light, you have two choices. One is I can take the electric field associated with the single photon. That's one option. The other option is I take as a prefactor the electric field in the ground state, which you can say is the zero point fluctuations of the electric field and the two differ by a factor of two, because the energy of the vacuum field is 1/2 h bar omega, whereas 1 [INAUDIBLE] equals h bar omega. So that's one reason why you may sometimes find factors of two in some textbooks or not in others. But I think-- I checked. I think, as far as I know, it is consistent. AUDIENCE: So these operators you've written down for e, this is the Heisenberg operator with the time frames right now? PROFESSOR: This is the Heisenberg operator for the electric field. But it differs. I mean, at that point, don't-- there's nothing really to worry about or get concerned about. The question is, for monochromatic light, do we put the e to the i omega t into the state or the operator? It's really just an e to the i omega t factor has to be put either on this state or on the operator. These are the two choices. Actually, my advice to you is don't even develop an intuitive conceptual understanding. And whether you are in the Schrodinger, try to sort of see the structure of the operator. I sometimes like the omega t, because it shows you how the electric field evolves. And you see that explicitly. But in the end, it's only when you do real calculation that you have to think hard, OK, which picture am I using now? Which representation am I using now? I'm probably not 100% rigorous in this course when I write down quantum states and operators, whether I'm always in the Schrodinger picture or in the Heisenberg picture. But by simply looking, does it have a time dependence? Yes. It's a Heisenberg picture. If I write down a quantum state, which has a time dependence, then I've chosen the Schrodinger representation. It's as easy as that. There was other? Yeah. AUDIENCE: There's a sign wrong up there, because both a and a dagger goes into the i omega t as you have written down. So they should be kind of rotating right. PROFESSOR: Give me one second. Actually, I just wanted to say I was careful with my notes, but I didn't carefully look at my notes. This is what I have in my notes. Does that satisfy you? AUDIENCE: Yeah. PROFESSOR: OK, good. OK so we want to talk about quantum states of light. The simplest eigenstates are, of course, the harmonic oscillator eigenstates. But, as I've already mentioned, and we'll elaborate on it further, these are actually non-classical states. So quite often, we don't want those states. We want coherent states. But we need those number states said as one representation. So let me just introduce them. So we'll label by n the states with n photons. And the energy of those states is n plus 1/2 h bar omega. The states are normalized. Sometimes, out of laziness, because it doesn't matter, we can eliminate the 1/2. It just makes some equations more compact. And if I do that, I can write the Hamiltonian like this. You are familiar, of course, with commutators of a and a daggers. And the only non-trivial matrix element of a and a dagger that it raises and lowers the photon number by one. OK, this was a short review of the simple harmonic oscillator as it applies to a single mode of the electromagnetic field. Now, just as a short example, in subsection two, we want to discuss the thermal state. And the thermal state is nothing else than photons in a black body cavity. We have a photon field, which is in equilibrium at temperature t. So if you use the canonical ensemble, our statistical operator is given by this well known expression. And if I use-- I mean that's all. I mean, isn't it wonderful once you're trained in statistical mechanics? This is all you have to know about a thermal state of light? That's it. This is the Planck Law. All we have to do is take the Hamiltonian, which we have written down, upstairs. So this is e to the minus n photons. n times h bar omega is the energy over kt divided by the partition function. This is the density matrix, a statistical operator for single mode light at thermal equilibrium. The partition function is very simple, because this sum over Boltzmann factors this is a geometric series, which can be simply summed up in this way-- divided by kt. So therefore, this expression here can be written as-- the density matrix in thermal equilibrium can be written in the number bases. And it's diagonal. And the prefactor is the probability to have n photons populated in thermal equilibrium. And the probability for n photons is given here. We've evaluated the partition function. So then if you put it together, we simply get this-- you can call it Boltz-Einesten factor, which tells us what is the probability to find an harmonic oscillator with n [INAUDIBLE]. And this is the expression-- e to the minus h bar omega. So this result takes care of many situations-- actually, almost all situations-- you find in every day life unless you switch on a laser pointer. So this describes-- and resistive filament. It describes the light generated by gas discharges. It describes the light of the sun. And this kind of thermal light, because it has the-- I mean, light in thermal equilibrium maximizing the entropy under the given boundary condition, it's also called chaotic light. It's just another name for thermal light. Well, we should now lean back and take the attitude, if I know the statistical operator of an ensemble, I know everything about it. And every quantity I'm interested in can be calculated. So just as an example, we know what is a probability to find n photons. But now we want to ask what is the energy density, or what is the mean excitation number. What is the average of n? So for that, we have to take the number operator, a dagger a, and perform the trace with the product of the statistical operator. And this is, of course, nothing else when we look for the probability to find n photons. And this probability is weighted with n. And what we get is the very simple result. It's a Bose-Einstein distribution law with a chemical potential of 0, which is the familiar distribution. We can now use the quantity n to rewrite our previous expression for the probability to find n photons in thermal equilibrium. It is n bar to the power n divided by 1 over n bar times n plus 1. So all this should be very familiar to you. And I'm not doing the final step, but I could immediately get Planck's Law now. All I have to use it sum up now. This is a single mode of light, single monochromatic light. All I have to do is to get Planck's law is to sum overall modes using the density of modes. Any question? We will not be talking a lot in this course about chaotic and thermal light, because we are mainly interested in the purer form how atom interacts with light. And this will be when light is in a more specific state, not in a thermal ensemble, in a more specific state which are number states or coherent states. But nevertheless, I feel if I don't tell you a little bit about ordinary light, you will not be able to appreciate quantum light. So I want to use this example now of a thermal distribution of light and write down for you in the next 90 seconds expressions for the fluctuations. So you should know a little bit about what is the statistic of thermal light. If you find n photons in a black cavity and you look again, it's the standard deviation of the photon number square root. And is it Poissonian, or is it much larger? So you should know about that, because it will then make a difference when we talk about the purer states of light, coherent states, and number states, squeeze states, and all that. OK, so we want to look now at fluctuations. Let me just first mention that we have normalized correctly the probabilities. Just one second. What we are interested now is the fluctuations. So we want to know the variance. We want to know what is delta n squared. And delta n squared is of course n square average minus n average squared. And we want to find out what is that for thermal light, for chaotic light. So what we need now is we need an expression. We have already an expression up there for n average. And now we need an expression for n square average. We can get that as follows. If we take the probability, multiply it with n-- n minus 1. Sum over. So I have a brain cramp. This equation is in my notes, but I don't-- maybe step back-- I don't see where it comes from. But from this expression, of course, we have now an expression for n average squared. n squared-- the sum over it. Just a second. n average-- n thermal-- n average squared. Anyway, I don't want to waste class time. It made sense to me when I wrote the notes. I just can't reproduce the argument. So this should be obvious. It's not obvious to me right now. But let's just take it and run with it. The left hand side is n square minus n multiplied with the probability. So let me just take a note. I will show you where it comes from on Monday. But assuming that it's true, and I know it is true, that actually contains n square average minus n average is 2n average squared. So therefore, we have now our expression for the average of n square, which is 2n average squared plus n average. And therefore, for thermal light, for chaotic light, the variance, delta n square, is n average square plus n average. So the question is if this is our variance-- the variance of thermal light, of chaotic light, is that. What you should see immediately is that it is much more fluctuating then Poissonian statistics. Well, if you compare now with Poissonian statistics, Poissonian statistics would have fluctuations where, well, the RMS fluctuations are square root n. The variance is just n average. So thermal light is super Poissonian, has much, much stronger fluctuations than Poissonian distribution. OK, one question for you. If you look at the distribution of photons-- and we have written it down here. Let me just scroll up. Well, you probably don't have an auto-plot function in your brain. But if you plot this function, and I'm asking you what is the most probable photon number you will find for light in thermal equilibrium? AUDIENCE: Zero. PROFESSOR: Zero. It is a broad Gaussian centered at zero. So you have a broad Gaussian. The average photon number is related to the width of the Gaussian. But the nature of thermal light is that whenever you look into a black body cavity and ask, what is the most probable photon number for any mode, it is actually zero. No matter how hot the cavity is and how many photons you have in it, the photon number distribution is always a broad Gaussian with a maximum at zero. We'll come back to that later When we talk about coherent states. Then you will appreciate the difference. But I will be referring back to the result we have just derived for the photon number distribution. Let me just take a note. The most probably n is n equals 0. OK, I hope with that-- and this is III, the third subsection-- you can really appreciate the properties of coherent states. Coherent states were introduced by Roy Glauber in 1963. It's actually something you should wonder about. Coherent states seem so natural. I mean, we teach it in all courses. And when we think about harmonic oscillator, we often think in terms of coherent states. But they were not immediately used and invented when quantum mechanics was invented in the '30s. It really took somebody like Roy Glauber to invent coherent states. And that happened in the '60s. Maybe the reason why people didn't immediately jump at coherent states is because coherent states are not eigenstates of an Hamiltonian. The definition of a coherent state is actually an eigenstate, not of an Hamiltonian. It's an eigenstate of the annihilation operator. And of course, the annihilation operator is non-Hermitian. So it's a strange operator to look for eigenstates. But this is what defines the coherent state, namely that e acting on a coherent state alpha equals alpha times alpha. I'm using here the standard notation that alpha is both the label for the coherent state written down here, and is also the eigenvalue of the coherent state when acted upon with the annihilation operator. We normalize those states. And since we're talking about eigenvalues of a non-Hermitian operator, alpha can be any complex number. OK, so now we have introduced a second set of basis functions for light. Remember, we had the basis functions n, the eigenfunctions of the harmonic oscillator, and now we've defined a new set of eigenfunctions, the coherent state alphas. Of course, the next obvious question is how are the two representations connected? So what we want to do now is we want to figure out how do we write a coherent state alpha in the n basis in the basis of eigenfunctions of the harmonic oscillator? So we want to write-- this is just-- want to expand the coherent state into eigenfunctions n. So the coefficients we want to calculate are those matrix elements which are defined as cn. So what we have to use is in order to find this representation, we have to use the fact that alpha is an eigenstate of the annihilation operator. So we know that this is alpha times alpha. But we can also let the annihilation operator a, we can let the annihilation operator a act on this state and in this sum. And the annihilation operator, of course, does what it is supposed to do. It lowers the photon number by one. And we get the matrix element square root n. And then by just changing the summation index by one, we can write it as n times cn plus 1 times the square root of n plus 1. So this is when the annihilation operator acts on the sum. And now, I'll continue with the lower line, which is alpha cn times n. So what I've found now is I've found two ways to hide the same state, namely a acting on alpha. And therefore, those coefficients have to be the same. And what I just indicated in blue is a recursion relation, which expresses cn plus 1 by cn. So every time you go from cn to cn plus 1, you multiply with the square root of n plus 1. So if you stare at it for a moment, you find that cn is related to c0. Yes, cn plus 1 by alpha n over n factorial. So of course, we get only the relation between the coefficients. But if you throw in that we want to normalize the state, this determines c0. And then, we get as a final result the normalized representation of the coherent state in terms of number states. Excuse me, I think the square root is missing here, because it appears here times n. OK, so what we have found here is now how coherent states are written down in the number state basis. So now we know coherent states. And we can ask a number of questions. So what is this? So what did we define? And what have we now derived? Let's look at some of the properties. The first one is that if you look at this expression, this is the amplitude that will have n photons in a coherent state. The square of the amplitude is the probability. So the probability to find n photons is the square of the amplitude above. So this is that. And this is now a Poissonian distribution. So it's immediately clear that those coherent states have much, much smaller fluctuations in the photon number than the thermal state we discussed before. Because this is Poissonian, whereas the thermal state had a variance which was much, much larger than Poissonian. OK, that's number one. Number two is we would like to know what is the photon number. So we want to know what is the expectation value for n. That's actually very easy to find. You could, of course, get the average number of photons in by evaluating the expansion of the coherent state in terms of number states. But sometimes, you should also try to maybe directly evaluate the coherent state. And the photon number operator is a dagger a. And now you see here we are lucky, because a acting on alpha is just getting alpha. But a dagger acting on alpha on the left hand side gives us alpha star, the complex conjugate. So what we obtain here is simply the absolute value of alpha square. And this is our expression for the average photon number. I have already mentioned that the variance must be-- the variance must be n, because it's a Poissonian distribution. But since it is so nice to have creation annihilation operators acting on coherent states, let's just calculate-- let's verify that this distribution-- we know that distribution is Poissonian, but let's verify directly that the variance of the photon number in the coherent state is n, is Poissonian. If you need the variance, we need an average. And we need n square average. Let's just calculate n square average. So we need now the operator for n square, which is a dagger a, a dagger a. We know already if those two operators act on the [INAUDIBLE], that this just gives us alpha star alpha. And then we are left with a a dagger. Now, a dagger acts on alpha. We don't know what a dagger is acting on alpha. We know that alpha is an eigenstate of a, but not of a dagger. So therefore, we want to change the order of a and a dagger, use a commutator. And then of course, a dagger can act here. A can act here. And we just get another factor of alpha. So therefore, in a very straightforward way, we find now that n square is alpha square times alpha square plus 1. And using the result for the mean photon number, this is nothing else than n bar, 1 plus n bar. So if we now put it together, we want to know the variance. The variance is n square average, which we've just calculated, minus n average squared. And hooray, we find n bar. And we verify again that the fluctuations are Poissonian. What else can we ask? We've asked about the statistics of fluctuation. We've asked about the probability to find photons. Well, what else do you want to know about light? Electric field. Oh, that's something I forgot to ask. I will come back to that later. But just sort of to connect it, what is the average electric field for thermal light? Zero. And what else can it be? Because any value of the electric field, whether the electric field is positive or negative, requires that the-- I say it loosely-- that the photon has a phase. The phase will define whether it's positive or negative. But all different phases of the photon field, all different-- what I mean now is coherent states, but have different phases-- they are all degenerate. So in a thermal ensemble, the states are only populated due to their energy, they're populated by Boltzmann factors. And therefore, there is no distinction what the phase of the state are. So therefore, we haven't defined yet what the phase of a photon is, so therefore I'm a little bit guarded with my language. But in other words, whatever the state with the phase is in a thermal ensemble, you have all phases simultaneously. And therefore, you never have an expectation value of e. But of course, we had an average number of photons. And that means we have an expectation value of e square, because the expectation value of e square is the expectation value of the energy. The state has energy. But it doesn't have an expectation value for the electric field. OK, so maybe with that motivation, we want to figure out now does this coherent state, which we have constructed, does that have now an expectation value for the electric field? So there is one thing you of course know about this electric field. Whenever we have an electric field, it needs an amplitude and a phase. And what I just mentioned to you is that in a thermal ensemble, we were just looking at the partition function and such. We didn't even mention a phase. There was not any complex number in any of the equations we wrote down about thermal light. No complex number, no phase, means no electric field. But this is different now for the coherent state. Because I mentioned that alpha in general can be complex, the annihilation operator can have complex eigenvalues. So therefore, in the coherent state-- because it is characterized by a complex number-- we have now the possibility to include a phase. And what I want to show you is, in the next few minutes, is that yes indeed, the coherent state has an expectation value of the electric field. Instead of simply using the equation for the electric field operator-- remember electric field operator was a minus a dagger-- I want to introduce something else which will turn out to be very, very useful. And one of the application of this concept is to show you what the electric field is. This is sort of a visual representation, a visual representation of the states of the electromagnetic field. So I want to sort of give you, in a two dimensional diagram, a visual representation of thermal states, coherent states, number states. And this visual representation will make it immediately obvious what the electric field is. So the visual representation will represent what are called quasi-probabilities. So I have to define for you now those quasi-probabilities. So I denote the quasi-probability by Q. And I want to know what is the quasi-probability Q for any statistical operator which describes a light. It could be the operator for thermal light. It could be statistical operator also includes, of course, pure states. And the pure state could be a coherence state. So this is a way to define our quasi-probabilities for any statistical operator which is describing the light. And at the quasi-probabilities, well, the probability for what? Well, the probabilities for alpha. And alpha is related to the coherent state. So in other words, we say a statistical operator has a quasi-probability with a coherent state alpha by simply calculating this diagonal matrix element of the statistical operator with alpha. Well, that's an abstract definition, but it's exact. What we should now do is I want to show you some examples. And then you will actually see that it's actually a wonderful definition. So my first example is let's look at the quasi-probability for the vacuum state. So if the statistical operator is simply the ground state of the harmonic oscillator, no photon, no nothing, just empty vacuum. So the quasi-probability for the statistical operator is now a pure state. So I can say, what is a quasi-probability of the vacuum state to be in state alpha? Well, following the definition above, it's nothing else than the matrix element, the overlap, between the vacuum and the coherent state alpha. I've given you a path, the representation of the coherent state alpha in terms of number states. And all you have to do is look now upstairs, what is the amplitude that this representation includes a vacuum state. So it's just read of the amplitude c0 from the state above. And this was simply e to the minus alpha squared. So therefore, if I-- those quasi-probabilities can be plotted in a two dimensional plane, because it's a function of alpha. So if I use the real part of alpha, the imaginary part of alpha, now I can plot the quasi-probability Q. And since it's hard for me to do a 3-D plot on the tablet, I just sort of shade the region. Black means the amplitude is large. And now it's sort of here. It means the amplitude falls off. So what I'm plotting here is a Gaussian, a Gaussian with the widths of [INAUDIBLE] with the width of 1/2. I think in the standard deviation of the Gaussian, you have a factor of 2. Anyway, it's on the order of-- the Gaussian width on the order of [INAUDIBLE] here, I think it's 1 over square root 2, or whatever the definition of the standard deviation is. So here we have a Gaussian. Any question? So for pure state, we're simply asking what is the amplitude of the purse state to overlap to be in a coherent state. And what we are plotting is nothing else than the probability that this state is a coherent state with value alpha. You had a question? AUDIENCE: [INAUDIBLE]. PROFESSOR: The next example, and we have already prepared for that, is the thermal state. Example number two is the thermal state. So we want to know what is the quasi-probability for the thermal state as a function of alpha. The statistical operator for the thermal states, we had derived [INAUDIBLE] was the probability to find n photons. And in the number representation, the statistical operator is diagonal. And then the quasi-probabilities are nothing else, following the definition above, then those probabilities times the matrix element, the overlap between alpha and n. And if you look up what we had derived for the probability pn, you find a very simple result. What we obtain is again a Gaussian. It is also centered at the origin. So therefore, since I didn't label my axis, it looks exactly as a vacuum state. It's just that the Gaussian is much, much broader. And I will come to that, but an intuitive reading of this Gaussian is, well, this Gaussian has a peak at the origin. At the origin, this is where alpha equals 0. And this is directly related to the question I asked you before, that the photon number probability is peaked at n equals 0. But I hope this will become even clearer when we move on to the third example, which is now Q of alpha of a coherent state. Maybe just before I even discuss the quasi-probability of a coherent state, I have to address one thing up front. What would you expect the quasi-probability of a coherent state to be? AUDIENCE: The Gaussian [INAUDIBLE]. PROFESSOR: Pretty good. But I said before, the quasi-probability is-- let me just show you how it was defined. Here was my definition of the quasi-probability. And you can see the quasi-probability is the diagram matrix element of the statistical operator with a coherent state. But if our statistical operator is a coherent state, what would you expect this probability to be? Just naively. AUDIENCE: One. PROFESSOR: One or a delta function. I mean, it's sort of-- we ask, we ask-- if you expand this statistical operator in coherent states, what do we get? But if this operator is a coherent state, let's say it's a coherent state beta, you would expect, well, then we only get something non-vanishing if alpha equals beta. So you would naively expect that what you get is a delta function. The coherent state beta has a quasi-probability Q of alpha, which is peaked at beta. But it's peaked as a delta function. Now, I'm calculating it for you right now, and the result is this is not the case. The reason is-- and I will show that to you-- is that the coherent states are not your ordinary basis function. The coherent state forms a basis which is over complete. So we have sort of more coherent states than necessary. There is some redundancy. And therefore, the coherent state alpha and the coherent state beta, if there's only a small difference between alpha and beta, they have overlap. Coherent states which differ by only a little bit in their eigenvalue, they're not orthogonal. So this is a complication. It's one of the complications we have to deal with when we want to describe coherent states and when we want to describe laser lights. But this is one of the properties. So therefore, when I'm now discussing with you is what are the quasi-probabilities of a coherent state, it will not be a delta function. But we should simply follow the definition, and everything will fall into place. So our statistical operator for a pure coherent state is, of course-- this is a statistical operator with pure state beta. And the quasi-probability for the coherent state beta to be an alpha is this matrix element squared. And you can calculate that, for instance, by just putting in alpha, alpha and beta, the expansion of alpha and beta in number states. So just have to solve the integral and calculate it. And what you find is not a delta function, but a decaying exponential. So you can say, as long as alpha and beta do not differ by more than one, there is substantial overlap. So in other words, what we get for this quasi-probability is something which is centered at beta. But it is again a Gaussian. So the important message here is coherent states are not orthogonal to each other. They form an over complete basis. Maybe as a side remark, it may come as a surprise to you, because you have been trained too much in basis states of emission operators, of energy eigenstates. And those energy eigenstates form a complete basis. The coherent states are eigenfunctions of very strange operator, a non-emission annihilation operator. So therefore, the few things you took for granted do not apply here. So we have our nice diagram with-- [INAUDIBLE] diagram for quasi-probabilities. So now if you have-- that was the real part of alpha. That was the imaginary part of alpha. So if you take that as a complex plane, and I'm drawing now as a phase of the complex number beta, which is the eigenvalue of the coherent state beta, then the quasi-probability is centered near beta. And it's a decaying Gaussian. I'm not showing it in here explicitly, but it is simply the vacuum state we discussed before displaced by beta. My fourth example, just to show you. So far, I've shown you that everything is Gaussian. Thermal light is a broad Gaussian, coherent state is a narrow Gaussian, the vacuum state is a narrow Gaussian. Let me just show you something which is non-Gaussian, which is the number state. If you put into the definition of the quasi-probability the number state, and you calculate the quasi-probability as a function of alpha, then you find that what you get is actually not a Gaussian. What you get is a ring, a ring with a certain width. So this is the representation of a number state. And the radius of this ring is proportional to the square root of the number of photons in a number state. We'll come back to photon states in much more detail very soon. But I just wanted to show you that at this point. So quasi-probabilities are not always Gaussian. They really depict something, they show us something important about-- they show us important differences about the quantum states of light. Any questions? Yes. AUDIENCE: [INAUDIBLE] a delta function [INAUDIBLE] radius, [INAUDIBLE] little past [INAUDIBLE]? PROFESSOR: Can you hold the question for three minutes? OK. They quick answer is it's not. But in three minutes, I want to tell you that there are three possible definitions of quasi-probabilities-- the Q function, to W function, and the P function. In one of the functions, it's a delta function. In the Q function, it's not. But we'll come to that. But let's for now stick to the Q function. The next thing we want to put in is the time dependence. I want to show you that when we have quantum states of light-- and right now, we've just shown the quasi-probability distribution at a snapshot of t equals 0. What happens as a function of time? What I want to show you is-- and this makes those quasi-probabilities also nice and intuitive-- that as a function of time, the quasi-probability distribution simply rotates with an angular frequency of omega. So let me show that to you. We know that we have a coherent state. And what we want to understand is what happens when we act on the coherent state with the time propagation operator, which is the Hamiltonian in the exponent. Well, in order to evaluate the left hand side, we can simply take the expansion of coherent states into in a number basis, because we know that the time evolution of a number state is simply the energy of the number state with which it's given by the energy of the number state, which is n times h bar omega. So now if you look for a second at this expression, you realize that we can absorb the time evolution if you redefine alpha to become alpha times e to the i omega t. So in other words, the time evolution of the coherent state alpha preserves the character as a coherent state. It just means we get an incoherent state whose eigenvalue is now alpha times e to the i omega t. So that means the following. If you want to look at the time evolution in terms of quasi-probabilities, we have our diagram with the real part of alpha, with the imaginary part of alpha. And let's assume we had a coherent state, which happened-- this is sort of now my circle. It's, you can say, a high contrast representation of the Gaussian. So this is the quasi-probability of the initial state alpha. But as time propagates, it just gets multiplied with e to the i omega t. That means when the time evolution is just displacing the state over there, and as time goes by, the coherent state is just moving in a circle. And after one full period of omega, we are back where we started. Now let me give you now an intuitive picture what the electric field is. I will be a little bit more exact in two or three minutes. So we have a diagram of the real part of alpha and the imaginary part of alpha. For a harmonic oscillator, I can also label it as x and p. You know, in harmonic oscillator, when you can think about it classically, if you have some initial distribution of a mechanical oscillator in x and p, what the system actually does is it simply rotates on a circle in phase space. The classical phase space is x and p. And I haven't really told you that I was-- I didn't want to get lost in complex definitions. I defined for you the quasi-probability, but the quasi-probability is a generalization of the classical phase space function. Phase space function is a function of p and x. So take my word for a second, and allow me to identify the real axis with x and the imaginary axis with p. Well, then p is the electric field. So now, what you should visually take from those pictures is that if you want to know what the electric field is, you just sort of project this fuzzy ball on the imaginary axis. So in other words, when I ask you what is now the electric field as a function of time, at t equals 0, I project this onto the p axis. And what I get is something which is centered at 0. But it has also a certain fuzziness. The fuzziness is given by the size of the disk, or by the width of the Gaussian. And if I now use this picture as a function of time, as this fuzzy ball rotates around, the electric field goes up and down in one cycle. And the fuzziness moves with it. So the fuzziness here for the electric field is related-- and that's what we want to discuss next week-- is related to short noise. I also want to tell you that the coherent state is, to some extent, the best possible way to define an electric field. It's a minimum uncertainty state. I want to show you that the coherent state is sort of at the minimum of Heisenberg's Uncertainty Relation. So I think if you take this picture, you will immediately realize that you draw the quasi-probability, you project on the vertical axis. And you get the electric field that tells you immediately if you have a thermal state, which is a Gaussian centered here. You project it. You get an electric field 0. It has an enormous fuzziness, but it's 0. And this picture that everything rotates with angular frequency omega also tells you, as expected, that in a thermal state, it's already a circularly symmetric distribution. When it rotates, nothing changes. AUDIENCE: [INAUDIBLE], couldn't you find, eigenstates for that operator itself, and then get rid of this fuzziness? AUDIENCE: Could it just be, like, eigenstates of the harmonic oscillator, so not stationary at all? PROFESSOR: I'm a little bit confused now. Actually, two people mind if I teach a few minutes longer? I can just finish the-- I want to have one, I have one footnote. I really want to sort of give it to you, because I think it will help some of you. Pardon? AUDIENCE: We have a class starting at 2:30. PROFESSOR: There's a class starting at 2:30? AUDIENCE: Yes. PROFESSOR: Then I have to wrap up. Sorry. Then let me just tell you what I will do at the beginning of the next class. What I've told you here with the projection of the electric field, this is not 100% rigorous for the quasi-probabilities Q. Due to the non-commutativity of operators, when we generalize phase space function into quantum mechanics, we can order operators in a symmetric way, in a normal way, in an anti- normal way. And therefore, there are three different quasi-probabilities. The Q probability is the easiest to define. It comes in very handy. But what I showed you with the projection does not rigorously apply to the Q distribution. It applies to the p or w distribution. But the difference is, unless you really go into subtleties, are minor. So intuitively, what I did is correct. And to answer your question, an eigenstate of the electric field, I have a little bit problems with that. Because if you take this projection, what you need is you need something which is very, very narrow. But since this area is in Heisenberg uncertainty limited area, it means delta x delta p equals h bar over 2. The only way how you can create this electric field is you can completely squeeze it. So to the best of my knowledge, an electric field eigenstate would be a completely strongly squeezed state. We'll talk about that next week. But then you realize, if you've completed squeezed your state after a quarter period, this infinitely squeezed ellipse is now standing upside down. And after a quarter period your electric field is completely uncertain. So I think you can define an operator, you can find a state which has a sharp value of the electric field at one moment of time. But then it rotates around. And what was a highly certain electric field becomes highly uncertain later. OK, enough things to discuss. We'll go on next week.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

16_Light_forces_Part_2.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: I have a tiring day today. Just 10 minutes ago, I finished chairing the first part of a meeting of the COA advisory committee. And I gave a talk to them. And right after the lecture, I have to go back. So while they have a lunch break, I have to teach. But that's OK. That's life on the faculty. Yes. We're talking about applications of the spontaneous light force. We have used optical block recreations and such to learn about the spontaneous force, the stimulated force, in general. And this is our expression for the spontaneous light force. It's pretty much, the momentum of photons is transferred to the atom times to scattering rate. And what we started last class, and we want to finish today, is we want to discuss applications of this force to three situations which are of experimental importance. One can actually say that slow beams, more or less, in the magneto-optical trap are the key techniques which triggered a revolution in atomic physics. This is really the enabling technology into ultracold atom science. So I discussed with you, in the last class, molasses. And we discussed what happens when we have two counterpropagating beams. Maybe the key figure to remember is this one. When we have to counterpropagating beams, by just adding up the spontaneous light force of the two beams, we get the red curve, force versus velocity. And the expert eye immediately sees there is a linear slope. And that's a viscous force. This is the viscosity of light interacting with atoms. Of course, you'd better pick the detuning correctly. If you use the wrong detuning, everything is flipped upside down. And instead of a damping force, you have an anti-damping force. We then discussed the cooling limit. Any cooling limit is the balance between heating and cooling. And by just setting the heating rate equal to the cooling rate, this describes a steady state. We found the famous Doppler limit. OK. So that's where I want to continue. But do you have any questions about molasses, the Doppler limit, and what we discussed in the last class? Well, I know we had discussions about to take two beams which form a standing wave, and take it apart, and calculate the force for each beam. Well, sorry, this is what molasses is about. Molasses is about two beams. And if you want to treat two interfering laser beams in the full generality, there are a lot of other effects. Some of them, we will be uncovering in the next few lectures. But I'm happy now, when I discuss beam slowing, to give you when there is only one single laser beam decelerating atoms. That is the only situation I know about where you have an exact analytic solution. So I want to present it to you know. Actually, for a number of years, I presented the exact solution with one laser beam before I presented molasses, where I have to make approximations. But I always felt that beam slowing is more complicated, because you have to go to a decelerating frame and have to add fictitious forces. So some of the intuition is lost. So that's why I first presented you with molasses, with two beams, for which I need to make approximations. Because the interference pattern between two laser beams has richness and complexity. But now I discuss beam slowing with you. So this curve here could be the result of the very early beginnings of laser cooling in the late '70s, early '80s. And this is when people had an atomic beam with a Boltzmann distribution, and they had one laser been counterpropagating. But the laser beam had a fixed frequency. And what is this laser beam doing? Well, it resonantly interacts with the atoms which just have the correct Doppler shift to be in resonance. Those atoms are pushed-- scattered photons-- are pushed to lower frequencies-- to lower velocities. But they are pushed out of resonance. And eventually, the slowing becomes slower and slower. And eventually, things come, almost, to a standstill. So a laser beam which has a fixed frequency would modify the velocity distribution of an atomic beam in that way. People called it, already, cooling. And I think some pioneers in Russia who did those experiments were disappointed that they were not honored with a Nobel Prize in laser cooling. This was the first cooling. Because this distribution here is considerably narrower than here. And temperature is nothing else than a measure of the reach of the velocity distribution. But one can say that those experiments did not trigger a revolution in atomic physics. The revolution came when people eventually figured out how they can narrow down this peak from maybe a hundred or tens of kelvins down to a million microkelvin. So it's clear what is needed. If you want to do more slowing, you have to make sure that you are talking to the other velocities, and that you eventually sweep many velocities into one peak-- clear out the velocity distribution, collect all the atoms, and put them at a monochromatic velocity. So for that, you can say, well, if you use a comb of frequency, or a broad spectrum of frequencies, you could clean out a larger area. And, indeed, there are techniques like this-- white light slowing or diffuse light slowing. But since those techniques are not anymore of practical importance, I don't want to discuss them. Let me talk about the simplest technique-- at least, conceptionally-- which is chirped slowing. And this works like follows. You start with a laser beam which is initially in resonance with the velocity group. And as the atoms slow down, you change the detuning of the laser that the laser stays in resonance with the atoms. So it is a pulsed technique where you have a continuous atomic beam. You start with a laser and say, now I start slowing. And the laser is following a group of atoms as they are decelerated. So you can say it is like the atoms are riding the surf. The laser is always following the atoms as they change their velocity. And what is, of course, necessary for that to happen is that the chirp and the deceleration of the atoms are synchronized. You will see that in a moment, when I put this idea into equations. There is another technique which is used in many labs-- actually, in all my labs and Martin Zwierlein's lab. It's called Zeeman slowing. And this is discussed in your homework. It is a CW version of chirped slowing. And a lot of you know what I mean. The physics which goes on there is almost identical. But for conceptional reasons, it's easier not to use magnetic fields and such. Let's just keep it simple. Let's have one laser beam talking to a two-level atom. And the only other ingredient is the Doppler shift. Any questions about-- so I've surveyed, for you, beam slowing. And now I want to give you the exact description for one typical example of a slowing technique. And a lot of the results we find in chirped slowing apply to other slowing techniques. OK. I know I've taught it several times. And there is the conceptional point that you can only describe chirped slowing well when you go to the decelerating frame of the atoms. You want to describe it in the frame of the atoms which are continuously being decelerated. So the way I can introduce it is the following. I take the maximum-- I take the spontaneous light force. This is exactly the expression we have discussed several times. And now I'm saying we decide that we want to decelerate atoms with a deceleration, a, which is negative, because we slow them down. Of course, if you pick a too large that it would require more than the maximum spontaneous light force, you will not find a solution. But let's simply assume we want to decelerate atoms with a deceleration a. Well, what happens is this deceleration, a, requires a certain force. If the intensity of the laser would provide more force on resonance then it's not going to work. Because we want to decelerate the atoms at this deceleration. So therefore, the atoms, if they are decelerated, have to be at a detuning, delta prime, in such a way that the spontaneous light force is exactly providing the acceleration we decided we want to have for our atoms. In other words, this detuning-- in the frame of the atom, the atoms are on resonance when the light force is exactly providing the force to give you the acceleration a. But if you have more laser power than absolutely necessary, in the frame of the atoms, the atoms have to be detuned in such a way that the force, with the detuning, is exactly providing the acceleration we want. So that's the easiest way to explain it. We have to set the stage. We have to introduce this nominal detuning. OK. So you can say that this detuning is just a definition. And now we want to describe the slowing process. So we have said we want to start with atoms at an initial velocity-- which we'll actually cancel out in a moment. But let's do it in a systematic way-- and afterwards, we want the velocity to decrease, linearly, in time. I simply defined this parameter, delta prime, to be a solution of this equation. So this is a well defined quantity. And the laser detuning-- which, I've broken my laser to change its detuning as a function of time-- is now this nominal detuning minus the Doppler shift due to the velocity of the atoms. I'm really solving the problem backwards for you. I first decide which should be the velocity trajectory of the atoms-- how should the atoms slow down. And then, I define what is the detuning which will result in that. That's the easier way to understand it. Of course, if you pick this detuning, the atoms will exactly do that. OK. But now what we're doing is the following. This is sort of the acceleration and the velocity we want the atom to have. But there may be some atoms in the beam which do not have this velocity. They deviate from this velocity with v prime. So the goal of cooling would now be to reduce v prime to 0 to make all the atoms follow the trajectory we have designed for them. OK. Many words, many definitions. Now we can simply substitute what we have defined into the equation. And we get a wonderful result. So we take equation one, which was the equation for the light force. And we put all those definitions, which I went to great lengths to explain to you, into it. And we get this result. It's just that the detuning has been expressed by delta prime, and the velocity by this difference velocity v prime. It's pure mathematical substitution. But now comes the important point that we want to describe everything, now, in the decelerating frame. The atoms are meant to decelerate with a deceleration a. And you know, from classical mechanics, if you describe something in a decelerating frame, you have to add a fictitious force. This fictitious force, since the acceleration is constant, is a constant force. But I'm expressing it, now, by the other parameters. We've chosen our acceleration. We've chosen the detuning, delta prime, that-- everything is sort of connected. And this is just a way to use the correct-- the most easiest units, or the correct parametrization, for this constant force. So I have done-- and the important thing for you is, this equation is exact. It's exact for an arbitrary velocity v prime of the atoms. So let me just summarize. I've taken the spontaneous light force. I've given you a bunch of definitions, what delta prime is. That's how you program your laser. But then, in the decelerating frame, we need a fictitious force. And this is nothing else than a mathematically exact rewrite of the spontaneous light force in the decelerating frame. But now what happens is-- we had a decelerating-- we had a light force where, because we assume the laser is counterpropagating, there's a minus sign. The fictitious force has a plus sign. And that means that we can-- and if v prime is 0, the force is 0. So we can now make a Taylor expansion for small v prime. So we have, again, a viscous force, a friction force. And it turns out that this friction coefficient for beam deceleration is exactly 1/2 what we got before, for molasses. But it's clear we have one laser beam. Whereas in molasses, we had two laser beams. If you ask, what is the heating, well, if you have one laser beam and not two laser beams, the heating described by the momentum diffusion coefficient is also just 1/2. One beam heats exactly half of what we got for two beams. And therefore, the temperature, which characterizes the width of the velocity distribution in the frame of the decelerating atoms is exactly the Doppler temperature we had before. Because the two factors of 1/2 cancel. So now I have given you-- and I hope you take some pleasure in it-- an example where the laser cooling for a two-level system is exactly described. There's only a single beam. And the Doppler limit is the exact solution. I've not made any approximation other than assuming I have a two-level system for which the spontaneous light force has the expression we've derived. OK. I know, even with my explanation, you have to read through it once or twice. Because I made a few substitutions. And you really have to digest them at your own pace. So let me give you a summary of what we have done graphically. One laser beam scatters light. And it is this Lorentzian. And it's a negative Lorentzian because we have a counterpropagating beam. We push the atoms into the negative direction, they fly in the positive direction. But by going-- by chirping the laser, by going into the decelerating frame, the force doesn't go to 0. I had to add an offset. And this offset is the fictitious force due to the transformation of the decelerating frame. And now you realize that this force-- and it is the correct force in the decelerating frame-- has two 0 crossings. One is a stable point, and one is an unstable point. And the stable point says, if an atom happens to be too fast, it gets a correcting force. And this is like a lock point. It has a stable lock point. All the atoms which did not have the correct velocity we wanted them to have, they are sort of sucked into this lock point, and eventually, will pile up at a huge peak at this nominal velocity. And the width of this peak is simply described by the Doppler limit. Before I take your questions, let me show you what happens in the two frames. This is v prime. This is the velocity in the decelerating frame. And let's say our initial velocity is this. So what happens now is all the atoms within the atoms are decelerated towards v prime equals 0. The friction force puts them to v prime equals 0. But since we are in a decelerating frame, in the absence of the laser, the whole Maxwell-Boltzmann distribution is just shifted-- not because the velocity of the atoms changed-- because we are in a decelerating frame and we are accelerating away from the atoms. And what I just described for you is the following-- that, at v prime equals 0, we have our friction force. And the atomic distribution in this frame is now pushed through. And everything which goes through this resonant region is sort of collected. And you have a peak which is piling up at v prime equals 0. Probably, it's easier-- but mathematically, I needed the decelerating frame-- if you use the following picture. And just think of this beam slowing I explained to you. If you just switched on a laser here, you would just burn a hole into the velocity distribution and pile up the atoms. But now you sweep. And when you sweep, you take the peak, push it further. It gets a bigger peak here. Push it further. So eventually, what you're doing is, by chirping the laser in the correct way with the correct laser power and all that, you're just pushing the atoms all out. And eventually, they are piled up in a narrow distribution which is defined by the time when you just switch off your laser beam. So one chirp of the laser, from an initial detuning in the lab frame to a final detuning in the lab frame, will just sweep out the whole Maxwell-Boltzmann distribution. And if you suddenly switch off the laser, you have frozen in a velocity distribution which is exactly described by the Doppler limit. So what is easier to understand-- molasses with two beams, or beam slowing with a single beam? I think molasses is easier, because everything happens in the lab frame. But think about it. It's a nice example for which no approximations have to be made. Questions? Yes, Nancy. AUDIENCE: It's not really obvious to me how, in this mathematics, we are choosing a time-dependent detuning as opposed to a time-dependent deceleration. PROFESSOR: The way I presented it logically, we said we want a certain acceleration to happen. And then, we provide our laser that k/kv is the Doppler shift. ka is how the Doppler shift changes as a function of time. And we chirp our laser exactly in frequency with a chirp which is ka. So in other words, I started to define a constant a. And that would require that the laser is now chirped in a linear way. I could have started the other around and said, hey, let's assume we chirp the laser in a linear way. And then, we do corresponding substitutions. And the result would be the same. AUDIENCE: No, actually, like if we did not want to have a detuning which was time-dependent, would we have gotten-- we still have a force. And that force would have been just changing. PROFESSOR: Well then, it's more complicated, yeah. Then, you have to integrate a differential equation. But what happened is the following. What I've done is, by defining the acceleration and then the chirp, I have the situation that, in the reference frame of the decelerating atoms, the detuning is constant. So usually, you go to a decelerating frame if something else simplifies. And what simplifies is that, in the frame of the atom, the laser detuning is constant. So you can also say the frame in which the atoms decelerate is exactly the frame in which the chirp has disappeared. Because the decelerating frame compensates the chirp with a linearly varying, time varying Doppler shift. OK. But the physics, I think, is really-- the moment you have a force which has a 0 crossing, it's a lock point. You determine the slope. And alpha, together with the momentum diffusion, gives you the final temperature. So I hope you enjoy that it's actually the same physics-- exactly the same physics-- we discussed, with some approximation for molasses. OK. We had a discussion last time about, we have the force, we find exactly what happens to the atoms. But we may have to think about it in different ways to figure out where the kinetic energy goes from. And I think we mentioned it already, in the last class, that what happens is you have a red-detuned laser, which is, the photons are absorbed. But the atom is scattering light in all directions. And so if you go at 90 degrees, you don't have a linear Doppler shift. If you go forward-backward with the same probability-- a symmetric pattern-- the Doppler shift averages to 0. So on average, you emit photons which are on resonance. And therefore, every time you scatter a photon, you radiate away the energy which is equal to the Doppler shift. That means, initially, when the Doppler shift is huge, each photon transports away a lot of energy. And the slower the atoms are, the less a photon transports away. But that makes sense. If you have a huge velocity and you subtract h bar k, the momentum of the photon, the kinetic energy is quadratic in velocity. You know, h bar k, the momentum transfer of a photon, takes out more kinetic energy the faster the velocity of the atoms is. So everything makes sense and is consistent. OK. Let's go from the-- we started with two beams. We just discussed one beam, chirped slowing. And let's go now to what most of you use in the laboratory-- namely, six beams. You want to cool in x, y, z. So you have one-dimensional molasses with two beams times 3. So in the limit that the intensity is low, you just add up all the forces. You say each beam has a spontaneous light force. And you add the six spontaneous light forces up. At low intensity-- and you have to take my word for it-- you can ignore all the interference effect between the beams. They can actually lead to very interesting effects. I will give you at least a taste of what will happen-- actually, a pretty good taste-- what happens at higher power. We need the traced atom picture. And we'll discuss how the stimulated force can be used for cooling. But if you don't have a two-level system, you have sigma plus sigma minus transitions and the beams interfere, you get spatially varying polarizations of your interference patterns, and then you have much [INAUDIBLE] physics. But in the simplest case of a two-level system, there is a regime at low intensity where you can simply take the equations we got for molasses and apply them to XYZ. I put, on the website, a nice handout, a review paper by Bill Phillips, who is discussing 1D molasses versus 3D molasses, and if you enjoy it, you can read how numerical factors of 1 or 2 or 3 now appear in the friction coefficient, in the heating coefficient, and how they affect the final temperature. But you got the full [INAUDIBLE] concepts and the correct idea for one dimension molasses. Now, let me give you an outlook for what we will be doing next week. Next week, we want to understand the following-- what happens to the friction coefficient when you increase the power? Well, you remember when we have red-detuned light, then we had one Lorentzian, another Lorentzian. The slope was negative, and negative means positive friction, we cool. And if you put more power into it, the linear slope increases, and for low power, the friction coefficient is proportional to the laser intensity. But now look what happens. If you approach saturation, then the friction is not only saturating, it suddenly changes the sign, and this is qualitatively new. And you should a little bit wonder about it. Wasn't the reason why we have cooling that you have a laser which is red-detuned? And the photon which is spontaneously emitted is on resonance, and this is the energy difference. But now I'm telling you, at high intensity, it's not the red-detuned laser which cools, it's the blue-detuned laser. So you should wonder about it. There is really something new to be learned. So I want to motivate you to follow me today and next week through an alternative description of what light does to atoms using the dressed atom picture, and in the dressed atom picture, we will very naturally understand why now blue-detuned light provides cooling. But right now, for you, this should be a mystery, but it should be something where you say, hey, still something qualitative is missing in our understanding of light forces. OK, I will not be able to explain it [? physically ?] to you. I first have to introduce the dressed atom picture, and I will do that it in 10 minutes or so. But what I want to do here is if I'm telling you something goes terribly wrong and we completely miss even the sign of the force, blue-detuned light is now cooling, not red-detuned light, I at least want to tell you what went wrong. Because haven't we done an almost exact discussion of light forces? We used the gradient of d dot e. We used for d the steady state solution of the optical Bloch equations. Haven't we done everything right? Well, we did approximations. And what happens is, and this is the reason for those qualitatively different things, that we should, in order to get those effects, not use, for the dipole moment, for the off-diagonal matrix elements of the density matrix, the steady state solution for optical Bloch vector. We have to acknowledge that it always takes a finite time to reach steady state. For instance, if he takes, let's say, a spontaneous emission time, if the atom goes through a standing wave, it goes from high intensity to low intensity, it takes about one spontaneous emission time. When you are highly excited, but now you go through a node of the standing wave, the steady state solution is [INAUDIBLE]. At the node of the standing wave, there is no light. The steady state solution with no light is you better go back to the ground state, but that takes something on the order of the spontaneous emission time. So therefore, when I am excited because I'm in the middle of the [? pied ?] laser beam and I go fast into the dark region, I'm still excited because I didn't have time to get rid of my excitation. In other words, the internal degree of freedom, the population of grounded excited state, will lag behind the steady state solution by a lag time which is a spontaneous emission time. And that would mean if I multiply with the velocity, it's pretty much that there is a spacial displacement. I'm always a certain distance away from the steady state solution. And actually, this point, to introduce the lag time, we with do it in a wonderful, physical way when we have the dressed atom picture, but it can be done with optical Bloch equation. It's just a real pain because the optical Bloch equation is sort of like a matrix the equation. It's completely obscure, how it is done, but in order that you convince yourself, I posted the original paper from Gordon and Ashkin where this was done. You read the math. You understand the result, but you don't understand the derivation, the physical picture behind it. The dressed atom picture will give us a lucid explanation for exactly what a lag time does and why this results in colling. But anyway, this is the physics we have done wrong. At low power, it's OK. At high power, we really have to take care of effect. And if we would take care of effect-- And as I said, we don't need the dressed atom picture, we can fully do it within the framework of the optical Bloch equation, but it's messy. --we would actually get out, and you see that in the references, the result that the friction coefficient in a standing wave at weak power is exactly two times the friction coefficient for a travelling wave, and that justifies that we describe molasses by the spontaneous light force of two travelling waves, completely ignoring the standing wave. But the same treatment of the optical Bloch equation will then, eventually, give us the result that alpha changes sign at high intensity. But as I said, I want to describe that after we have introduced the dressed atom picture. Questions? OK, we have done molasses in one dimension. We have done beam slowing. We just did a short discussion about molasses in arbitrary dimensions. The one technique which is missing now is the magneto-optic trap. Let me just spend five minutes to discuss the background behind the magneto-optic trap because for many of you, I think you were just born when the magneto-optic trap was invented. Well, I joined the field in 1990, and that was three years after the magneto-optic trap has been invented. And many people say if there is one most singular development in laser cooling and such, it's molasses and the magneto-optical trap. So let me tell you why, the magneto-optical trap, you shouldn't take it for granted. It was actually in 1983 that Ashkin-- Was it Gordon or Ashkin? I think Ashkin, Art Ashkin. --that he presented an Optical Earnshaw theorem, and it is nicely described in Bill Phillips' Varenna notes which I have posted on the website. And it's pretty much a proof that you cannot build a trap using the spontaneous light force. So in other words, this is a proof which people took as a correct proof that something like the magneto-optic trap could not exist. So what this is proof like? Well, and that sort of teaches you what was the mental concept or the mental barrier people had to break through before they invented the magneto-optic trap. The proof is the following. If you think that you have beams-- I mean, a model for the spontaneous light force is you have photons and the photons are sand blasting the atom. Each photon is like a grain of sand, and the atom is in scattering descent in all directions. You can really make this completely classical picture of the spontaneous light force. Almost. If you do with that you would say the spontaneous light force is proportional to the stream of photons which is described by the Poynting vector. But now, you have in electromagnetism a continuity equation which says the divergence of the Poynting vector plus the change of energy density in a certain volume has to be 0. So if you are in steady state-- If you have an arrangement of laser beams and the energy density of the electromagnetic field is at steady state, that tells you that the divergence of the Poynting vector has to be 0. Well, if the divergence of the Poynting vector is 0, using this equation that says the divergence of the spontaneous light force has to be 0. But if the divergence of the force has to be 0, that means you can not have a force which is inward from all directions because such a force, a trap needs a force which is inward. A trap requires a negative divergence. Do you see the wrong assumption in this proof, or did I convince you that the magneto-optic trap is not possible because this proof would say you can not make a trap out of the spontaneous light force? AUDIENCE: Well, you through out the change in energy density, so that's-- PROFESSOR: But that's OK because if you switch on a laser beam, light travels at the speed of light, and in 10 to the minus 10 seconds, in picoseconds, you reach steady state. And you know when you operate the MOT, you're not switching on and off your lasers at picoseconds. It's not that. Yes, [INAUDIBLE]. AUDIENCE: --the spacial variation of c. PROFESSOR: Yes, exactly. The force is not just proportional to the Poynting vector. This would be true if I had grains of sand, but photons have to fill a resonance condition. You can maybe assume I'm an atom, but I have an spatially varying magnetic field around me. At the one magnetic field, I feel the force of the laser beam because I am resonant. At another magnetic field, the laser beam just goes through me because I'm not in resonance. So therefore, the proportionality between the Poynting vector of the laser beam and the light force ignores that the light force is resonant, that the light force depends on polarization and such. So the solution is the divergence of-- If the c parameter is spatially dependent, then we can have a divergence, and then, we can do trapping. And it was actually Dave Pritchard in '85 or in '86 who wrote a really influential paper. He said the Optical Earnshaw theorem can be circumvented. We can have a spatially dependent factor, and what can be spatially dependent is saturation. If you focus a laser beam, you have a spatially varying saturation parameter. There can be some optical pumping. There can be light-- So he discussed several possibilities how the Optical Earnshaw theorem was circumvented. At this point in 1986, he didn't realize that it can be done with a magnetic field gradient. When he talked about possible ways of circumventing the Optical Earnshaw theorem, Jean Dalibard, who many of who know, who's a famous researcher in our field, had the idea that applying a magnetic field gradient could work. But nobody thought, at this time, that it would work in three dimensions, so the fact that it works in three dimensions is almost a miracle. So when the idea was born, all those schemes, and also the original idea of applying a magnetic field gradient, were conceived as one dimension scheme. Most of the schemes can not be easily generalized to three dimension, but the MOT scheme works as well in three dimension as it works in one dimension. Anyway, so this is sort of the background. Since I was a postdoc when there were only two or three groups who were operating a magneto-optic trap, I put together, with my own hands, the first magneto-optic trap which was ever used at MIT. I mean, it's ancient history, so I wanted to share a little bit this history to you. But let me now simply describe to you how the magneto-optic trap works. Let me describe it for an atom which we are not using in the lab. We all usually use atoms which have hyperfine structure. But let me just for pedagogical simplicity assume that we have an atom which has an s to p transition, and in the p transition, we have the three different magnetic [? quantum ?] numbers. So it's a simplified model. And what we're doing right now is we are applying molasses, two beams which act as molasses, but there is one difference now. They have different polarizations, sigma plus and sigma minus, and in addition, we apply a linear magnetic field gradient. So what happens now is the following. The linear magnetic field gradient does Zeeman shifts in the opposite direction to the n plus 1 and n minus 1 level. Now, the transition to m minus 1 is done by sigma plus light. The transition to m plus 1 is done by sigma minus light. So what happens now is if you have an atom and it moves away from the origin, it will eventually come closer and closer in resonance with the sigma plus light, but the sigma plus light comes from the left side so the atom will be pushed back. If the atom moves out hear, it moves further and further away, in resonance, from the sigma plus light, but it moves now, in resonance, over the sigma minus light, and the sigma minus light is pushing the atoms back. So this geometry has an inward force from both directions and therefore, works as a trap. So let me now describe it mathematically which is very, very simple. What we have to do is we just have, like in molasses, two laser beams two times the expression for the spontaneous light force. And you'll remember we got the molasses equation by taking the two laser beams and putting in the Doppler shift which is a velocity dependent shift. Well, all we have to do is, for those two laser beams, we have to put now a spatially dependent Zeeman shift which varies linearly with position. So therefore, all we have to do is we sum up the total force as the spontaneous force for the right and for the left laser beam, and in addition to the molasses effect, we put now in the spatial dependence. So we don't have to do the math again. We did already a Taylor expansion in v, but let's now just do a Taylor expansion in the blue expression. And then, we find that there is a linearly restoring force both in velocity space and in configuration space which actually tells you something interesting. The more friction you have, the larger is the restoring force. So the trapping and the cooling are really connected, at least in the simple model of the spontaneous light force. OK, so if we have this kind of force, a linear force in velocity, a linear force in position, this is the equation of motion of a damped harmonic oscillator. You have a spatial force constant, k, this linear force which is linear in displacement. So this is a damped harmonic oscillator. Usually, you don't see atoms sloshing in your magneto-optic trap like in an harmonic oscillator because if you look through the [? typical ?] numbers and by just doing the Taylor expansion of the spontaneous light force, you can immediately get an alluded expressions for those numbers. In almost all cases, the harmonic oscillator is overdamped. So that's how the magneto-optic trap works. However, the magneto-optic trap is not so simple. I would actually say nobody fully understands the magneto-optic trap because we have multi-level structure. We don't have just one count level. A sodium and a rubidium atom has-- f equals one and f equals two-- has eight different hyperfine levels. So there is not one level and we can excite the atoms. There are eight different levels. And those laser beams to optical pumping among those eight levels, and you have to solve a rate equation for eight different levels. And the amazing thing is it still works. It still works pretty much, at least qualitatively, as well as I just described to you. So it's a combination of optical pumping and Zeeman shifts. It even works better than advertised because as I will tell you in the next few weeks, you can reach temperatures-- Because of the multi-level structure, you reach temperatures which are much colder than the Doppler limit. It's called polarization gradient cooling. It's just if you have laser beams and you have atoms with hyperfine structure, they are colder than the Doppler limit. First, nobody understood it, and then, Bill Phillips figured it out. And that he figured out why atoms get colder that was the main point emphasized by the Nobel Committee when he got the Nobel Prize for laser cooling. Of course, he had already invented Zeeman slowing before, so he had a few other things under his belt. But anyway, the fact that the MOT works even better than advertised is another miracle, and this is related to the hyperfine structure. And finally, the fact that you can simply extended it to three dimension, is-- I said the first, the second --maybe the third miracle. It also works. Because when you look at the field of two coils, a so-called Anti-Helmholtz field, there are some directions where the magnetic field is not radially outward. This is a much more complicated magnetic field. It goes outward here, outward here, but in between the quadrupolar field, it sort of curves. There are some directions where the magnetic field is not radial, and you would say, hey, the one dimensional scheme is [INAUDIBLE] [? set. ?] Actually, before the magneto-optic trap was demonstrated, Dave Pritchard's student, Eric [? Rob ?] did calculations what happens in these diagonal directions which have no resemblance to the one dimension scheme. The solution is in those directions, the trapping form is a little bit weaker, but the big picture is it doesn't matter. So there was an idea conceived in one dimension for a simple atom structure, and this idea worked even better for real atoms and in three dimensions. OK, in your homework assignment, you will put in some numbers which show what happens if you take a vapor, an alkali vapor at a very low density of 10 to the 8 per cubic centimeter, pressure of, typically, 10 to the minus 8 [? Tor, ?] and you arrange six laser beams around it. Well, those laser beams have a restoring force in velocity. That means they cool and increase the phase space density over your sample by 8 orders of magnitude. Those laser beams have also a restoring force in the spatial dimension. Therefore, you get a cloud which has 3 orders of magnitude higher density. So the temperature goes down by 8 orders of magnitude which means the phase space density is enhanced by 12 orders of magnitude. So if you're talking about the goal of reaching Bose-Einstein condensation which requires a phase space density of 1, just having a dilute vapor and these six laser beams with a little bit of magnetic field, increases your phase space density by 15 orders of magnitude. So this is why molasses and the MOT were a real milestone. Now, with 15 orders of magnitude, you are just 4 or 5 orders of magnitude from Bose-Einstein condensation, and we will discuss, in a week and a half, what techniques were needed to bridge the gap between those temperatures and densities and quantum degeneracy of gasses. So that's the story of the magneto-optic trap. I'm also telling you that because I'm proud that a lot of the pioneering work in this regard happened at MIT. It was Dave Pritchard who realized at the Optical Earnshaw theorem can be circumvented, and it was him, in a collaboration with Steve Chu, that they together realized the first magneto-optic trap, and this was in 1987. Questions? OK, I think simple applications of the spontaneous light force. So we have time to talk about the dressed atom. Now, I was waiting until this point to introduce the dressed atom picture for you because the dressed atom picture is the natural picture you want to think about, for instance, for the Mollow triplet, and we had already some discussion about it. But it is also the natural picture you want to think about it when you describe the stimulated light force. And I mentioned already, it is the dressed atom picture which will help me to give you the physical picture behind if you have strong laser beams, why blue-detuned light cools and not red-detuned light. So you should be motivated now to learn about the dressed atom picture. But first, I have to say it's not anything new, what I'm teaching you. It has exactly the same physics in it as the optical Bloch equations. It's just that the optical Bloch equations are often more complicated, and they don't provide the clear insight. And the reason is the following. Often when you describe a physical phenomenon, you can choose one basis set, or you can use another basis set. But if you're in the wrong basis, you get the correct results, but in the wrong basis, you just can't describe the physics easily. So in other words, with the dressed atom picture, I want to introduce dressed atom states for you which are no longer the ground state and the excited state. I want to treat the atoms dressed up by the laser beam, which is one mode of the electromagnetic field. So the atoms, together with this one laser beam, will be partially excited, partially in the ground state. And this partially excited atoms, this is the dressed atom basis. But if you think in terms of those states, you have a much easier way to formulate certain physical processes, exactly processes where you have high laser power. At low laser power, the dress states are identical to the naked states to the ground and excited states. But at high laser power, at infinite laser power, the two dressed states are the symmetric and anti-symmetric equal superposition of grounded and excited state. So that's what we want to describe now. So the difference to the Optical Bloch Equation is that we are now really looking for the combined state of the atom in the electromagnetic field. So what we want to do is-- this is sort of our picture that we have. This is our Hamiltonian. We have the atom interacting with the vacuum through spontaneous emission with a reservoir of empty modes. But now, it also interacts with the laser beam. And you remember when we did the master equation, we looked sort of at this. We derived a master equation for this. And then we added in the unitary time evolution due to laser beam. But what we want to do now is we want to first take care of the atom and the laser beam almost exactly in an exact diagonalization. And once we have re-diagonalized that, we have now eigenstates which we call the dressed states. Then we allow not the ground and excited state. We then allow the dressed states to emit into the vacuum. And this is the formulation we want to obtain now. Let me make one comment before I go through a few simple equation, and that's the following. If I want to treat this system almost exactly, I have two choices. I use a strong laser beam. So I could, as we have discussed several times, simply use the time dependent external electric field. The laser beam is E0 times cosine omega t. And I just introduced that in the Hamiltonian. And you can get everything I describe now for you out of this Hamiltonian. However, there is an alternative way. And this one is I simply assume that the laser beam is described by the fully quantized field. And I use the fog state. I just assume N, the number of photons, is very big. And this is, at least for that purpose, a valid description of the laser beam. So what happens is you really want to look through that. I've really learned a lot by comparing the purely classical treatment of the laser beam and the fully quantized description of the laser beam. The results are the same. They are Absolutely identical. And what you will see is, by a bunch of approximations, we reduce the total Hamiltonian to-- I've seen it many times-- a two by two matrix. And whether you start with the classical field or whether you start with the fully quantized field, if you make the right approximation, you get the same two by two matrix. But there's a difference. This Hamiltonian is time independent. a dagger a, time independent Hamiltonian. If you take a classical field, which is cosine omega lt, you don't have a time independent Hamiltonian. You have a driven system, and your wave function has an external time dependence. The laser beam is driving the wave function. So in other words, you have two choices when you want to describe how an atom interacts with a strong laser beam. You can use the fully quantized field, which is conceptionally a little bit more complicated, because we've quantized the electromagnetic field. But then everything is time independent, and everything I do here is really easy. I can go over it now in five minutes, and you understand everything. Well, in your homework, I've already asked you at an early homework to look at simply the two level system driven by classical field. And in your next homework, number 10, your last homework, I'm asking you now to interpret the solution with the classical field. And it's actually a little bit a challenge to take the time dependent wave function and read the physics out of this time dependent wave function, which corresponds to the physics we derive in the dressed atom picture. But you have two ways. One is the classical field. Then the wave function, the Hamiltonian, is time dependent. And the wave function is time dependent. Or here, I start with the quantum field and everything is time independent. But the solutions are identical. Well, one is a time dependent wave function. One is time independent. But if you interpret the solution in terms of what happens to the physical system, you will find exactly the same. So in other words, I decided I do the easy part here, which is a fully quantized part. And this is probably what will get stuck in your head. But since I think it's an important pedagogical exercise, you have to go through the homework and do the time dependent part. OK, so we want to know exactly-- well not-- we want to do the correct approximations. And after some approximations, exactly solve what the atom does with one laser beam. So the laser beam is described by the standard Hamiltonian for single mode light. The atom is described by, well, excitation energy omega 0. And b is the excited state. So if you don't have any light atom interaction, the eigenstates are just product states, an atom in the excited or the ground state. And N is the photon number. [? Collin? ?] AUDIENCE: You said that it's usually easier to understand physics in a dressed atom picture versus optical block equations. PROFESSOR: Yeah. AUDIENCE: Can you think of-- is that a general rule? Or is this just like it's only applicable in certain areas? [INAUDIBLE]? PROFESSOR: You're really asking an expert question. And I talked to Jean Dalibard and Claude Cohen-Tannoudji, who invented the dressed atom picture, when I wanted to understand what happens when we have a blue detuned lattice. And I used the dressed atom picture and I got confused. So what happened is the following. If you pick your right bases, the description is simple. But simple means if all the physics is your atom is in one dressed state or the other dressed state, and it does transitions between the dressed states and that is easy, and this happens in the limit of large detuning. But if you're in the regime where coherences between the dressed states matter. In other words, the density matrix in the dressed state basis has now off diagonal matrix element. The physics becomes complicated again. So in other words, what I'm telling you is you are the best description happens when we have a two level system, we always need a two by two matrix. But if you pick the description where the solution is diagonal, if you have to deal with coherence, using coherence is coupled to population, it's complicated. And there's some cases-- one is the blue detuned lattice, which I encounter where you have to consider the coherences to get the correct answer. And usually, also when the laser beams are weak and on resonance, the dressed atom picture is not providing any advantages. So you want to use the bare picture for weak excitations. And yes, in the resonance case, the dressed atom picture, if you're very close to resonance, the dressed atom picture gets more complicated. But to give you the answer in something we have discussed, you know already that when you drive an atom with a laser beam, you get the Mollow triplet. And the Mollow triplet shows you-- I hope we can finish that today-- shows you the energy splitting between the dressed energy levels, between the dressed energy levels. But if the Mollow triplet is not resolved because the generalized Rabi frequency is small, then you're not resolving your dressed atom levels. And then the dressed atom level is not providing a big advantage. For certain problems, the dressed atom picture will not help you. But I can reassure you, and this is what I learned from Jean Dalibard and Claude Cohen-Tannoudji, if you really look at the full matrix with the off diagonal matrix append and everything, you will always find exactly the same solution. Anyway, we were just writing down product states of the atoms and the eigenfunctions of the Hamiltonian for the light, which are fog states, number states with N, L N plus 1. But the important thing is now the following. In the limit of small detuning, the ground state with N plus 1 photon is almost degenerate with the excited state with N photons. And we call these two states a manifold. The manifold E of N. And N is the photon number in the excited state. So what happens is, if you look at all those states, they are grouped into manifolds. And the energy separation between two manifolds is, of course, I go from N to N plus 1, the energy of a photon. So now I introduce the atom laser coupling in the usual dipole approximation. And what happens is, this dipole approximation, this dipole light atom coupling will couple, of course, ground and excited state. And it will, of course, have a big effect when the energy splitting is small. Well, you will see that in the next line. But we have the co and counter rotating terms. This dipole Hamiltonian will also couple the ground state to the excited state in the other manifold. This is the off resonant term, which will be neglect. So here are the equations for that. Our electric field is, of course, a plus a dagger. The dipole operator is that. And if you take what is in front of the electric field and define this as the coupling constant, g, we have what I just told you, that within a manifold, the ground state with N plus 1 photons is coupled to the excited state with N photons. g is simply the coupling constant, which all parameters have been included in that. But then, because of the matrix element of a or a dagger, we have the square root of N plus 1. OK, so what we are now doing is we are going with this coupling within a manifold, which is strong because the two levels are energetically close. And we will neglect the non-resonant couplings to the other manifolds. And that's just another way of doing, again, the rotating wave approximation. Now, when we describe an atom in the laser field and the atom scatters light, or we have a laser beam, and I have to tell you, when we have a laser beam, we don't provide exactly N atom, N photons. The laser beam is in a coherent state. And there are Poissonian fluctuations. But now, we do the next approximation, namely we say when N is large, then delta N can be neglected with regard to N. Then at least for the range of important photon numbers for the problem, this coupling becomes independent of N. And we simply take the average number of photons in our laser beam and define now an electric field, which is independent of N. And the product with the dipole operator of the atom, that's what we call the Rabi frequency. So at that point, we introduce a classical electric field through this definition, and therefore, the dressed atom picture. Also, we start with the fully quantized Hamiltonian is now identical to the classical field with this electric field amplitude. But what we have gained is we could discuss everything in a time independent picture. And now, I promised you, we'll go back to a two by two matrix, which you have seen many times. It's a two level system. We have the detuning. We are in the rotating frame. So that means the ground and excited state with N and N minus 1 photons have the same energy. But if we have a detuning, there is a difference whether we excite the atom or whether we put a photon in the laser field. But you have seen that many, many times. And the coupling is now described by the Rabi frequency. So that's kind of our typical two level system where the two levels [INAUDIBLE] the rotating frame, everything is time independent. The two levels are split, and they are coupled. And well, how many times have you seen the solution of two by two matrix? The solution is that the coupling is now increasing the splitting. The splitting between the two levels is, again, the generalized Rabi frequency, which you have seen many, many times. But now we want to really get the wave function. We want to get something out of the dressed atom picture. And that's the following. We started out with the bare states-- ground state with N plus 1, excited state with N photons. And the two by two matrix after the diagonalization is now a linear combination of those. And it's convenient to use sine theta, cosine theta, because then the states are automatically normalized. If I introduce the solution is now that the angle theta is given by this equation, that's just the solution for the two by two matrix. And the physical picture, which we have obtained, is the following. If this here is the energy of the excited state with N photons, I've just sort of drawn this as a reference parallel. But now I want to draw it as a function of the laser frequency. Well, the ground state with N plus 1 photons has exactly the same energy on resonance. But if the photon is more energetic then this state, because it has N plus 1 photons versus n photons if the photon is more energetic than a resonant photon, this state has higher energy. And here, it has lower energy. In other words, we have now a level crossing between the two bare states. But a level crossing with interaction turns into an avoided crossing. And the solution which we have just derived are the blue curves. And if I go far away from resonance, this behavior here, which is perturbative, is nothing else than the AC Stark shift. So this is exactly what we have discussed at this point. Now, let's get some mileage out of it. And we will talk about light forces. That's really interesting. And I hope it's another highlight of this course to describe light forces in the dressed atom picture. But what I want to do here is already in preparation for that. I want to not yet introduce forces, simply figure out what happens in the dressed atom picture when the atoms scatter light. And you know already the solution. The solution is the Mollow triplet. And we have discussed with the Optical Bloch Equations that at low laser power, the Mollow triplet has three curves. It has three broadened curves. And they have different broadening. And I didn't prove it to you, but I mentioned to you, at low laser power there is-- perturbative laser power, we discussed at great length that there's a delta function. And at low laser power, the delta function remains. So this is sort of the spectrum we get out of it. And what I want to show you now is how this spectrum is described in the dressed atom picture. Atom photon interaction has a wonderful discussion of all the details. I will share with you a few highlights. But we will go much further in the description of the Mollow triplet then we've gone before. But I will not go into all excruciating details. So let's discuss first. That's something we hadn't done so far. What are the intensities in this Mollow triplet? So the way how we obtain it is the following. We have the manifold, the manifold with N minus 1 and the manifold with N photons. And this is the detuning. And if we allow for the atom laser coupling, the splitting here is a generalized Rabi frequency. And well, you immediately see that in this picture, if you allow spontaneous emission from the upper manifold to the lower manifold, you have four different combinations. One is the long photon is the blue sideband. This is the red sideband. This is the blue sideband. This is the red sideband. And those two are the carriers. But now, we can do the following. I've just written down for you the solution of the wave function with sine and cosine theta. If we want to emit the blue sideband, we go from the top to the bottom. But that means, since a photon in the rotating wave approximation can only be emitted when we go from b to a, we find that the matrix element for that involves cosine theta times sine theta. However, if you want to emit the carrier, you go from here to there. Well, we always have to find the excited state to begin with. And we go-- I must have made a mistake. Sorry. When I went to emit the blue side, when I go from all the way up to all the way down, and I always have to find an excited state which connects to the ground state. And this gives me a term, which is cosine square. The red detuned sideband gives me sine squared. And if I want to get the carrier, this one, I go from here to here. I get cosine theta sine theta. So in other words, we can immediately read off the matrix elements. And since the intensity, Fermi's Golden Rule, is the matrix element squared, we find immediately that the upper lower sideband and the carrier involve terms cosine to the 4 theta, sine to the 4 theta, cosine square sine square. So we got the matrix element. And the only thing we have to know now is that the width or the rate is given by the matrix element. The rate is given by the matrix element square. But now, what is the intensity of the line? Well, the upper sideband is emitted by the upper dressed state, because you have to go from the upper dressed state to the lower dressed state to get the blue detuned photon. So therefore, the intensity for this sideband is this rate times the population in the upper dressed state. So what is missing at this point, and we need that urgently for next week, what is the population in the upper and the lower sideband? Well, we know the transition rates between the upper and the lower sidebands-- well, between the upper and lower dressed state. We know we've just found here the rate. What is the rate? What is the matrix element squared when you go from one is the upper, two is the lower state of the manifold. So we have just gotten those rates. And rate equation says, the change of population in the upper dressed state is given by the departure from the upper state and what arrives from the lower state. So since we are interested in the steady state solution, we can just set the left side to zero. And then, the rate equations simply give us the population in terms of the rates. So in other words, if you ask, what is the steady state solution in the upper or lower dressed state, it's given by a ratio of those rate coefficients. But we just got those rate coefficients. And this gives us another bunch of sine and cosine functions. So in other words, we know now what is the population of the dressed states. I mean, everything is expressed by one non-trivial parameter theta. And all we get is sine square, cosine square, sine to the 4th, we just get those trigonometric functions. Yes. I think we can finish that. Good. So this is the steady state solution. But now, we can go back to the rate equation and say the time derivative of the rate equation has to relax to the steady state. This is nothing else than rewriting the equation above. But by rewriting the equation above in that way, we find now the prefactor in front of-- I mean, here, what you have is the time derivative is given by the difference from steady state. And this is the relaxation coefficient. And the relaxation coefficient is-- that's the solution. You just have to do the substitutions yourself. The relaxation coefficient, how populations relax to steady state, is now given by the sum of the two rates. One is going from the upper to the lower and from the lower to the upper state. And each of those rates has been expressed by sine square and cosine squares. I'm not doing it here, because it's a little bit more complicated. But you can also rewrite for the dressed atom picture what happens-- you can rewrite those equations in a way what happens not to the population, but what happens to the coherences, to the off diagram matrix element formulated in the dressed state basis. And what you find is, you find a correlation time, a relaxation time, for the coherences. Remember, for the naked state, population's relaxed with gamma. Coherence is relaxed with gamma over 2. This is for the Optical Bloch Equations. But now we have reformulated the physics in the dressed atom picture. And what the solution is the following. This is sort of a weak laser power. But in the most general case, what you will find is that the width of the central peak reflects the relaxation time for the population. And the width for the sidebands is the relaxation time for the coherences in the dressed atom picture. So we had before with the Optical Bloch Equations, I showed you different cases. We had a three quarter gamma. I need two more minutes and I'm finished with my unit. So we had weird factors of three quarters. And I showed you how they come from the eigenvalues of the matrix of the Optical Bloch Equation. But now we understand them physically, namely the central feature has a peak, which is the relaxation time for the populations in the dressed atom picture. And the sidebands have a width, which is the relaxation time for the coherences. They no longer differ by a factor of two, as in the naked bases in which we formulated the Optical Bloch Equations. By the way, I know a lot of you may be curious about the coherent peak. You can read about it in API. It's 10 pages about correlation function. But in the end, the rate of the delta function can be, again, described by values we have just calculated. We have just calculated steady state populations and transition rates. So this rate is nothing else than another combination of sine square and cosine squares of this angle theta. So everything is wonderfully and simply described the full physics of that by just this angle theta in the dressed atom picture. OK So with that, we are done. And I think I'll discuss the last page with you on Monday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

6_Entangled_states.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So good afternoon, everybody. You see the topic for today's class on the screen-- entangled photons. Just to remind you, when we talked about single photons and Mach-Zender interferometers, we realized that when we have a nonlinear interferometer where one mode, you can say one beam, affects the phase shift in the interferometer for the other photon. Then we get a nonlinear situation, and we can create photon states which no longer factorize. So we have a system a, a system b, and we can no longer write down the total wave function in a wave function for system a times system b. And this is something very interesting, and this is what we feature in this section. So we finished the last class by defining entanglement. And just to remind you, we said something is entangled if it is impossible to write it in a product of two wave functions. So therefore, if you have some correlation between the two systems, then we call the two systems and the state of the system entangled. Now this is a definition which needs explanation, so we went through some examples. And we showed that certain states which on first sight look entangled are not entangled, because if you try harder, you find the way to factorize the state. So I want to continue now in explaining different aspects of the definition, in particular what it means to have a system which has two subsystems, a and b. Before I do that, do you have any questions up to that point? So we want to talk about some standard entangled state. And the most basic state is a single state. So if you have a state which is 0 1 minus 1 0 normalized, this is an Einstein-Podolsky-Rosen state. This needs some explanation. We have often encountered, in physics, states which are simply a superposition of if you interpret 0 1 as spin up or spin down, you encounter that quite often. And I want to explain to you now what is not an entangled state for the reasons of using entanglement as a resource. So first of all, I want to point out this state is not a single photon, because we have entangled here two qubits. Let me just contrast it. If you have a single photon after beam splitter, the single photon after beam splitter is in a superposition of mode a, mode b, 0 1 minus 1 0 divided by square root 2. But now 0 is the vacuum. So we have not a system which can be decomposed into two partial systems a and b. You may even separate the system and then manipulate your single qubits individually by putting phase shifts on and doing other operations. If you have this state which is a superposition between having a photon, not having a vacuum, and having a vacuum state, you cannot-- the vacuum state itself is not a separate system. You cannot take the vacuum and perform operations on the vacuum. So I know it has been confusing for me when I heard about it for the first time. We have, so to speak, here an entangled mode, but not an entangled state. It's a singlet state. It has some aspects of entanglement, but it is not the entanglement we have defined is a resource. So let me just repeat-- this may be an entangled mode, but to be in an entangled state requires systems where you have two parts which you can separate. The second comment is we have situations where we have two different parts, but they can't be physically separated. So part of our definition of entanglement as a resource is that the state must be of two physically distinct systems. And we want to have parts which can be separately addressed, manipulated, and measured. In other words, if you read out one part of an EPR pair, the other part still exists. So that's sort of what we want. Let me illustrate that with examples. For instance, in the helium atom, you have two electrons. And the ground state-- if I use spin notation, is a singlet. So that's this. But does the ground state of helium fulfill our definition of an entangled state? Well, if you find a way to switch off the Coulomb potential, and then the electrons separate from each other, they still maintain their spin singlet character, then you can take them, measure them, take measurements, manipulate them. But since nobody has come up with a good idea how to switch off the Coulomb potential in an atom, you can never separate the states. And it is physically impossible to address this so the two electrons see them separately and such. So this state is not entangled, because it's not-- the two systems cannot be separated. What is usually a good choice for entanglement-- and this is why we discuss it here with photons-- is you have photon states. Photons always fly away. Photons are in a certain superposition state. You can always separate them, and take them individually. So let's assume we have two photons, two modes. We have two photons-- actually, the two photons can be in one spatial mode, but we are now playing with the polarization-- horizontal, vertical. So if we have a state with horizontal- vertical polarization, this is a nice entangled state of two photons. So what did means is that we have photon each. One is vertical. And one is horizontal. But you don't know which one. If you would just look at one photon, it would be 50% horizontal, 50% vertical. It would be completely unpolarized, which would be a random state, which would require and density matrix for its description. But if one photon is horizontal, the other one is vertical and vice versa. So it's a pure state, but all the pureness of the state comes from the entanglement and not from what one photon does by itself. So these are two photons in one mode, and they are polarization entangled. Or this brings us back to our dual [? Rail ?] single photon states. We have two photons, two qubits and each is in two modes. So our 0 state-- and just to make sure that you do not confuse it with no photon, our logic 0 state, so L means logic here-- is that the photon is in the second mode, and the 1 state of our logic state means that the single photon is in the other mode. So we can now have an entangled state, which is now this 0 1 1 0 state. But these are now logic states, which means that the 0 has one photon. It has a photon in one of the modes, and the one has a photon in the other mode. So each state here has two photons, but then the two photons have-- 0 1 and 1 0 are switched in the two parts of the wave function. And we actually saw in the last unit how Kerr medium and an interferometer can generate this state. So yes? AUDIENCE: [INAUDIBLE] PROFESSOR: There are two such states. I mean, I will later tell you what the four famous Bell states are. There is one which has a plus sign and one which has a minus sign. So when we talk about spins in singlet state, it's often more natural the minus sign. Here what naturally emerged was the plus sign, but they are both sort of Bell states and therefore, they have what Einstein-Podolsky-Rosen introduced into it. So tolerate both signs. They are two different states, but for the purpose of the current discussion, they have the same property-- they're maximally entangled. Other questions? OK, so let me point out that the properties of entangled states always involve two qualities. One is the non-local character, because we have correlation between two subsystems which may be together when they interact, but then they can be separated. So we have correlations which happen between the two systems which are a distance apart. And we later come back when we talk about Gauss inequalities and that we know that physics has non-local aspects. And secondly, if you can separate two parts and they interact with the environment, the environment may interact with them differently in those two different parts. And therefore, entangled states are always regarded as fragile against decoherence. And it's a technical challenge how do you find states? How do you implement entangled states which are robust? Just to give you one example, if you have an entangled state which is based on electron spin, you may be more than 1,000 times more sensitive to magnetic field fluctuations in your laboratory than if you have qubits which are entangled states which are based on nuclear spin. So that's a big research area to find states which are less sensitive, or even you immune against decoherence. OK, so I've mentioned to you that entangled states are states which you cannot factorize. But now we can sort of start playing with that definition. And we say, OK, if you have an entangled state which is up-down plus down-up, but now the contribution of down-up has only a tiny, little amplitude. So it's almost a pure state which can be factorized with a little bit of an extra configuration which prevents us from factorizing that. I mean, that doesn't look like good entanglement. It looks at the whole entanglement of the state depends on very small admixture to the wave function. And so what we want to address now is how can we quantify that? How can we look at this data and say, hey, this is sort of not a strongly entangled state. It has only a weak amount of entanglement. So let's not forget, entanglement is a resource. Entanglement allows you to do teleportation. Entanglement allows you to do more precise measurements. And what I want to convey you, if a state is only weakly entangled, it doesn't help you much to achieve precision beyond the standard limit, effective teleportation, and such things. Before I introduce several measures for entanglement, let me talk about entanglement purification. It's a very nice subject which tells you that if you have weakly entangled state, you can make them more entangled. And actually, the effort you have to spend to make a weakly entangled state more entangled can actually act as a measurement how entangled was your state in the first place. So the purification is introducing us with one measurement for entanglement, as I just said, but it also gives you an idea how quantum state can be manipulated. Purification is also the first example we encounter in this course for new insight into quantum physics. A lot of people thought quantum physics, at least non-relativistic quantum physics that was invented in the '20s in the last century, and by now we've understood it all. But there was an aspect of quantum physics, which I think nobody understood. And this is when you have a quantum state, which may decohere, a quantum state which may no longer be pure, that you have ways to error correction. You have ways to get the pure quantum state back. And ultimately, I mean, in the early days when I learned quantum physics, I thought if you have a quantum state, that's nice. But if the quantum-ness has decayed away, you can't get it back. But this is something we learn now from the new methods and new approach in quantum information science that you can do quantum error correction. You can have a stage which has decohered and you can get back to it. And there is something which-- it's not yet that which we're discussing today-- but in purification, you have states which have inferior entanglement, but you don't need to get stuck with that. You can take several of the purely entangled states and create the maximally entangled state out of it. So that's what I want to show you. So you should look at it as one example that wow, it's really cool how we can have quantum states with inferior quality. And by doing quantum operation on those states, we get something which is more entangled, and therefore, if that's our purpose, has a higher quality. So to introduce purification, I'm simply mentioning what are-- for two qubits-- the standard, maximally entangled states. In 10 minutes or so, we talk about how do we measure entanglement, and indeed, those states will come out as being maximally entangled. But you already get the idea. Maximally entangled means they're not factorizable and not just by a small margin. They may be two equal parts of-- you always need two equal parts of 0 1 and 1 0, and this state has maximally entangled. It's maximally non-factorizable. So the four states which are actually the Bell states-- the famous Bell basis-- are 0 1 plus minus 1 0 and 0 0 plus minus 1 1. Just to remind you if if you have two systems, each of them has two states. You can think of spin-- spin up, down for particle one, spin up down for particle two. That Hilbert space is four-dimensional. So you need a four-dimensional basis. And the trivial basis is up up, down down, up down, down up. But what is relevant for entangled states, we often use the basis of Bell states. And this is a new basis which spans the four-dimensional Hilbert space, but each of those bases function is maximally entangled. OK we should correctly normalize them. And that state is often called psi plus minus. And this here is phi plus minus. OK, so now let's get a purely entangled state. And let this state 0 0 plus 1 1. But we will have an example where b may be very, very small. So those states, this state is entangled for all choices of a and b. So the question is now how can we take such an arbitrary state where a and b-- one of them may be small, and create a standard Bell state? So what we have to assume for that is that we have a large supply of such states. And so let's assume we have large supply, identical copies. And now we want to take two such copies. And what I want to outline you is the following-- you take sort of two of your copies, and you do a measurement. And I will tell you what kind of qubit operation we need, what kind of measurement we perform. And then when the outcome of the measurement is such and such, you say-- OK let me be specific. So if you take two copies, we have a total of two states with two photons each. And now we perform a quantum operation on two of the photons, and the other two photons we leave untouched. So now it depends when the measurement of the two photons has a good outcome, we know those other two photons are maximally entangled. They're in a Bell state. If the outcome of the measurement is bad, it tells us the two photons are not entangled, and we throw them away. So therefore, we have a finite probability by performing measurements that a pair of our sample state with the coefficient a and b will result into a maximally entangled state-- in a Bell state. And I want to describe now what is the protocol, what is the procedure to implement that. And what we will find-- and this is what I think you should expect here-- if the initial state has very bad entanglement in the sense that b is very small, we will need many, many attempts doing many measurements on our pair of states before we produce a Bell state. It's probabilistic, but the probability to succeed in preparing a Bell state will depend on-- we will see-- the product of a and b. Any questions? Yes. AUDIENCE: Does the assumption of large supply of the state violate no-cloning in any sense? PROFESSOR: The question is, does it violate the no-cloning theorem? No, it doesn't, otherwise I would not say we have a large supply, because the no-cloning theorem is absolute. But it simply means we cannot have one state, and then clone and clone more and more copies. Let me be specific. If you have an experiment which produces a certain superposition state, you can just push the button on your experiment many times, and produce, in the identical create, as many states as you want. If you have spin-up state-- well, this is now for two photons, but if you have a spin-up state, you can make as many copies as you want of the state which has been rotated by a certain angle. So therefore, in state preparation, by going through the exact procedure, we can just create as many copies of a state we want. The no-cloning [INAUDIBLE] meant the following-- I give you one state which may be a spin which has been rotated at a certain angle. You know nothing about the state. And now you should try to make a copy out of it. And the answer is you can't, because any measurement you do is-- if the particle were spin up and you would measure spin up or speed down, you could say I got spin up. Now I produce 10 spin up particles. But you don't know along which axis the spin has been prepared. So unless you know which axis the spin has been oriented, and if you choose another axis, you have irreversibly lost information which cannot be retrieved. I don't know if it helps you, but if you have a certain state, and you're going to measure it without destroying it, you need a quantum non-demolition measurement. If you're in energy eigenstate, you can measure energy. If you're in a spin eigenstate which points along Z, you can measure the direction of the spin in the Z direction without destroying it. So if you can do a quantum non-demolition measurement on a state, you could clone it. But that violates the assumption that if I give you an arbitrary state, you do not know by definition what kind of measurement is a non-demolition measurement. You just take your chance, you try to take a Stern-Gerlach experiment, separate the spins in the Z component, and then it turns out I gave you a state which is polarized along x. So that's a subtle, but important difference. Other questions? So we take two copies and let's bring in Alice and Bob. So the first photon we associate with Alice. And the second photon is associated with Bob. So if you take two copies-- we have now is an Hilbert space, a four-dimensional Hilbert space, a direct product. And if you just take the state and calculate the direct product, you get four terms-- 0 0 1 1, 0 0 1 1, 1 1 0 0, and 1 1 1 1. And the coefficients are given here. Let me just underline that it if you think we have some state with some entanglement, and we separate the system, one goes to Alice and one goes to Bob. So the one which Alice has is the first part of it those states. And Bob has the other part. So the protocol is now that Alice and Bob first, they're not doing a measurement. They're not reading out the system. They perform a unitary operation. And what Alice and Bob have formed is the controlled NOT operation on the two qubits. Let me just write it down and then explain it. Perform what is called the controlled NOT or CNOT gate. And I've explained before in the last section how the controlled NOT can be performed using a non-linear Mach-Zehnder interferometer. So Bob and Alice both run their two photons with the non-linear Mach-Zehnder interferometer. So what is a controlled NOT? Let me remind you. If you have two qubits, the first one is the control. If the control is 1, you flip the second one. If the control is 0, you do nothing to the second one. So the controlled CNOT transforms. So since the first qubit is the control qubit, out of those four combinations, the controlled NOT only does something if Alice's controls and Bob's controls is 1, and then the second bit is flipped. If Alice's and Bob's controls are 0, they do nothing to the second bit. So therefore, the 1 1 0 is transformed into 1 1 1, and the 1 1 1 1 is transformed into 1 1 0 0. So that's the first operation. Let me just indicate it. So what has happened here, the 0 0 has been flipped into 1 1 and the 1 1 has been flipped into 0 0. That's the first step. The second step is that now Alice and Bob measure their target qubit. What does that mean? We have a controlled NOT operation. In the controlled NOT operation, we have the first one is the control qubit. The second one is the target of the operation. So in our, the way how we write it, Alice's target is the third in line, and Bob is the fourth in line. So now Alice and Bob measure the target qubits. Let me just be specific. So Alice has number one and number three. Bob has number two and four, and the target qubits are the third and fourth. So what is the probability that Alice and Bob measure both 1 1, that they both find the target qubit of 1? Well, it is this one and it is this one where the controlled NOT has flipped it. So one has the probability a squared b squared. The other one has probability a squared b squared. So with probability 2 a squared b squared-- well, let's allow a and b to be complex. They obtain 1 and 1. So in this case, what is left is here 0 0. What is here left is 1 1. You may just need an intermediate line to write down what the state is after the measurement. But if you read it off here, you find that when they obtain 1 1 1, that implies the post measurement state is then 0 0 plus 1 1 divided by square of 2, and this is one of our Bell states. So what we had is we had a system of four photons after a qubit operation, the controlled NOT. Alice and Bob do a measurement together on two of the photons. Collins. And if the outcome of this measurement is 1 1, the rest of the system is in the Bell state. So we assume that Alice and Bob have a large supply of those copies. So let's assume that they start with n copies of psi. And then, because the probability is m over n, they successfully obtain m copies of the Bell state. And the question is, what is the probability in the limit of a large ensemble. And we will see in a few moments that this is actually a measurement of how entangled the original space is. Any questions about purification? Well, then let's measure entanglement. The basic idea here is that if you have a state up/down plus down/up, it's a pure state. But this pure state has a correlation between the two subsystems. And the idea is now entanglement is that there is a correlation between the two subsystems. And you can say well, it would be a good way to characterize entanglement if I only look at one subsystem. In this case, you look at one subsystem, and you would just randomly see spin up, spin down. Spin up spin down is described by the unity density matrix, which is, therefore, the most random state on earth. So if you have a pure state and you only look at one subsystem, the more random the subsystem is, the more the pureness of the initial state comes from correlations, comes what is entanglement. So therefore, what I want to introduce now as a measure of entanglement is that we take the total system and then we perform the partial trace. We only look at one subsystem. And the purer the subsystem is, the less entangled it is, because in the ultimate limit that our system factorizes into two pure states, when we look at the subsystem, we still have a pure state. So the purity of the subsystem in terms of pure state is now a measure of entanglement. The purer the subsystem is, the less entangled it was. So the basic idea here is entanglement is related to correlations. And if you take half of a Bell state-- so if two particles which are an EPR pair and we take half of it-- then half of it is completely random. So let me illustrate that. So if you take one half of-- let's just take one of the Bell states, phi plus, which was the superposition of 0 0 plus 1 1. So let me be specific, because we need it for the definition. We describe this ensemble of this system in a pure EPR state by a density matrix. The density matrix is nothing else than you take your total system. So this is now the density matrix. So in our case, psi a b is just the state. And now we describe the subsystem by performing a partial trace on rho. The partial trace is over the system b. And that means we take all eigenfunctions k b of state b, sum over all k's, and this is our partial trace. So therefore, we would take our statistical operator from the line above, and perform the partial trace where those states are the state 0 and 1 of b. So these are the states of b. So when you do that, you just insert that. You find that what you get is 1/2. a has been traced out, so b has been traced out. And what we obtain for a is just from the two terms above. This gives that, and this gives that. And you immediately realize that this is 1/2 times the identity matrix. So therefore, we have shown that this is a completely random state. So now we can characterize the randomness of the partial trace of the density operator obtained by performing the partial trace with the standard von Neumann entropy. As a reminder, the von Neumann entropy for statistical operator rho is defined as the expectation value of rho log rho, where we take the logarithm with respect to the base 2. So this is the trace of rho log rho. Or if we use the eigenvalues of rho, we multiply eigenvalues with the logarithm of the eigenvalues. So for a pure state, the entropy is 0, because a pure state has one eigenvalue, which is 1, and the log of 1 is 0, so we get 0 for pure state. For completely mixed state, we're talking about a state which has two dimensions, so it can be up and down. A completely unique state has probabilities of 1/2 each. And then we say the entropy of this state is one, or we call it one bit. Yes. AUDIENCE: About the volume [INAUDIBLE] expectation of rho log rho is called a trace [INAUDIBLE] expectation of log rho, expectation of [INAUDIBLE] trace of the operator times density matrix. PROFESSOR: OK, so this is just a reminder of how we measure entropy of density matrix. And now we apply it to entanglement. We define now the entanglement for the entanglement e over state psi a b to be the entropy of the density matrix for system a after tracing out system b. And for pure state, this is-- it doesn't matter whether we trace out a or b if you start with a pure state. The entanglement, the entropy of the statistical operator rho a and rho b are the same. I tried for a moment to prove it. I saw it quoted somewhere. I didn't succeed in a split second, so either I overlooked something, or it's a little bit more involved to show that. So therefore, to use the inverts, our definition says that the entropy, so the entanglement, is nothing else than the entropy of the reduced density matrix. And we immediately see if we have any of the four Bell states, by performing the partial trace over one qubit, we obtain the identity matrix. So therefore, the entropy of all the Bell states is 1. Let me state without proof-- when we come back to the purification scheme the result is that the probability or the optimum probability-- if you do stupid measurements on your states, of course you get nothing. But the optimum strategy to create pure Bell states out of your reservoir of poorly entangled states-- so for an optimum strategy, the success probability m over n actually turns out to be not a different measure of entanglement. It is the entanglement which we have just defined through the entropy of the partial trace. So therefore-- and I think this is nicely illustrated with the purification scheme-- entanglement is a real resource. When you have better entanglement to start with, then you can get more copies. You can get more pure Bell states out of your supply of poorly entangled states. So therefore, you lose more if your states are not fully entangled. You lose more of them, and therefore, the success of the purification scheme makes it clear how entanglement is a resource. If you have entanglement, it's precious. You had to do something to get it. I didn't point out entanglement is not something which one number characterize it, it's all. We introduced to you already another measurement of entanglement through the Schmidt number, which was in homework number two. And usually when you have different measures for entanglement, they're not one to one related. It seems that, similarly when we measured non-classic light, we had a G2 function. We had [INAUDIBLE] bunching, antibunching. We have negative quasi probabilities. And it's often clear that one system which is truly non-classical, fulfills all the criteria, but how the quantitative measurements are related to each other is really subtle. In the case of entanglement, for a long time it has even been a big question in research-- if you have an arbitrary density matrix with a complicated many-body system, how can you even characterize the entanglement? We are focusing here on pure states where things are fairly simple. But in the general situation of a many-body system, it can be quite challenging just define and measure entanglement. Any questions? Yes. AUDIENCE: If you have a general [INAUDIBLE] that's not necessarily a pure state, can you show that it's always the reduced trace of the bigger [INAUDIBLE]? PROFESSOR: Say again, if I have a general state which is-- AUDIENCE: You have a subsystem-- let's say you have two spins, and then you have a density matrix for the first-- or actually just to say that you have one spin and you have a density matrix for that single spin, it's a matrix, not necessarily a pure state. Let's say it's a mixed state. Then you can introduce a fake second spin and show that this matrix is the trace of a matrix [INAUDIBLE] and maybe just put that one in a pure state and do some stuff, do some calculations more easily. PROFESSOR: Yes. If you have a density matrix, you can always regard it as a partial trace of a bigger system. That means you always represent your state as a pure state, but it is entangled with a bigger system. However, what the big system is, is by no means unique. I'm missing the technical word-- it's called unraveling the density matrix. You can always represent the density matrix written down in forms of pure state. But this unraveling of the density matrix-- no, actually, its related. When I said yes, I thought about the unraveling of the density matrix. You can always write down a density matrix as a mixture of pure states, but which are the pure states is not unique. So when you say my density matrix is half of the atoms are spin up and half of the atoms are spin down, somebody else would say, no, that's not true. Half of the atoms are spin [INAUDIBLE] and spin x and some are in spin minus x, and those representations are equivalent. So I was just thinking of that as to write down the density matrix in the pure state basis. But you are asking about-- AUDIENCE: [INAUDIBLE] what he's asking about [INAUDIBLE] PROFESSOR: Somehow, but I was thinking of that, but I think the answer to your question is yes. So you said, you confirmed that. AUDIENCE: Yes, but I am forgetting the name of the theorem here. AUDIENCE: Isn't it just purification again? AUDIENCE: Yeah, it's related to purification. You can prove that you're using purification. AUDIENCE: [INAUDIBLE] AUDIENCE: Using purification [INAUDIBLE] but I'm completely forgetting the names of [INAUDIBLE] PROFESSOR: But wait. We've talked here about-- just to be clear, we've talked here about purification of a pure state. We started with a pure state and we purified it to be a Bell state by doing certain measurements. So we've not talked here about density matrices, but it sounds very plausible that you can always construct a bigger system. But I should look it up and see if there is an exact proof. It's-- intuitively it sounds correct. Other questions? OK, so we have discussed the definition of entangled state. We've talked about purification of entangled states and how to measure it. I want to talk now about how we can create entangled states for atoms. Maybe let me say the following-- so by now we are convinced. Entangled states are great, and we want to create them. And for photons, I showed you that some simple element-- beam splitters, Kerr medium-- can create entanglement. It's much harder to do that with atoms. Now, we want to do it with atoms, because atoms-- in contrast to light-- they're pretty much staying still, whereas photons always move at the speed of light. And the only way to make photons stand still is you put them in a cavity and then they bounce back and forth. But even in super cavities with the highest reflectivity mirror, you get-- what are the longest ring-down times you get-- fraction of a second, milliseconds, depending kind of in which domain you work-- microwave domain, optical domain. Whereas atoms, you can hold onto your qubits for a long time. So therefore, if you want to use entanglement as a resource for certain protocols, you want to have entangled atoms where the entanglement would like for a long time. So the question is now for atoms, we do not have perfect beam splitters and perfect Kerr mediums. Also, we can control interactions between atoms for [INAUDIBLE] using VSEPR resonances, but that's another story. But now let me ask a question, how do we entangle atoms? And I want to first show you that if you had the right system, things can be fairly simple. This is a suggestion which was made almost 20 years ago, and it goes like follows-- if you have a diatomic molecule of two identical atoms, in this case mercury, and mercury doesn't have-- this molecule doesn't have any electron spin, but mercury has a nuclear spin. And if you now photo-dissociate mercury and two mercury atoms fly away, then you have separated a spin singlet into two parts. And now you have created the Bell state up/down minus down/up. So you could say this is sort of the example-- this realizes very closely the example I gave you with the helium atom, where I said you have an electron in spin up and spin down, but there was no way to separate the electrons. Therefore, you can say it is a state which has entanglement, but it's not entanglement as a resource. But right now, here it becomes a resource once you have found a method to separate the two parts of the wave function that you can give one to Alice, give one to Bob, and they can perform the operations on it. Well, if it looks so simple, why don't we have entangled atoms everywhere? This experiment has been suggested 20 years ago, but nobody has done it, or some several groups have worked on it. You need a molecule with suitable states. You don't want any electron spin which interferes with that. All you want to have is two nuclear spins. You want a singlet state here. So those requirements are not easily fulfilled with real atoms. Of course we liked-- and also who wants to work with mercury? Mercury has transitions in the ultraviolet. I think there's only one group in the world who has operated a magneto optic trap with mercury. So it's not your tabletop atom. So therefore, let's now talk about a method how we can entangle atoms by using light. So if you take atoms, maybe we can use the light which has been emitted from the atoms, performing measurement on the light, and then depending on the outcome of the measurement, we know the atoms are entangled. So Professor [? Swann ?] calls this the poor man's entangler. I don't know why poor man's. You still need quite a bit of equipment to do it. But at least it seems the poor man's solution if you can't make the above experiment work. So this addresses a question that, just for technical reasons matter is more difficult to entangle. Whereas photons are easy. Yeah so you can say the idea is related to the purification scheme. We can't take a system of two atoms, atom one and atom two which are unentangled, and we shine some laser light on them, excite them. Then they emit photons. So all we can do is-- we can only talk to the atoms with the photons. So the only thing we can do now is we can measure the two photons. And then the situation will be similar as in the purification scheme, where Alice and Bob did a measurement. If Alice and Bob said both of our target qubits are one, then what was left behind was in a pure, entangled Bell state. And similarly, what you want to do here is we have two atoms. There was nothing special about them but they scatter light. And if you now perform a measurement on the photons and the outcome of the measurement is positive, then we know for sure what has been left behind is entangled. That's the idea. So it shares with the purification scheme that it is a probabilistic entanglement. You run your experiment many times. You do a measurement. If the measurement is positive, you say now I have an entangled state. And maybe then you can move on to measure the entanglement. You can move on to do teleportation, other things you want to do with entangled states. But if your measurements says no, bad luck. The probability hasn't worked out this time. You just press a button again, scatter light again of your two atoms and/or your two ions and hope that the next outcome is positive. So the idea is we want to introduce now a probabilistic method. It's based on two atoms emitting light. And the result is with a certain probability that we get entangled atoms. Let me just scroll down and show you one thing I want to-- When I prepare the notes and everything is clear to me, but then I want to explain it to you and say hey, I have to motivate you. If I just go through a few lines, you wonder what it leads to. So let me maybe first give you the explanation I would have given you a little bit later. What is an entangled state is if the atoms are in maybe one of two ground states-- one atom is in one ground state, the other one is in the other one, or it is flipped. So we know one atom is in the ground state one. One atom is in the ground state two, but we don't know which one is in which. It is in the superposition state. So what you need now in this scheme is the following-- if the atoms scatter light, they can go to two different ground states. And we know to which ground state they have gone, because due to selection rules, they reach one ground state with polarization one. They reach one ground state with polarization two. So therefore, when we had two atoms, they emit light and we would measure the polarization of the light, we would know in which state they are. But if you know atom 1 is in state 1, and atom 2 is in state 2, this is not entangled. So what we have to do is when the two atoms have emitted photons, we have to mix the photon at the beam splitter. And after the beam splitter, when we measure that the photon is polarized, we know that one of the atoms is in one of the ground states, but we don't know which. So therefore, we can now use-- the photon carries through the polarization the information in which ground state the atom is. But now we have to perform operations to the photons that for fundamental reasons, fundamental quantum measurement, we know there is one photon in one state, but we have no way to ever figure out which atom has emitted the photon. So that's the idea, and the protocol I want to show you now is what do we have to do to the two photons to make sure that we never know, that we can't find out which atom has emitted the photon. And we have to do little bit more tricks also to make sure that when we do a measurement on the photon, we know the atoms are in the Bell state. Questions about that? So therefore, we need a beam splitter. So we can come back to what we have already introduced in the last section. So each atom will emit a photon. And I will actually show you at the end of the class that people have entangled with that scheme two ions which were in two different ion traps. So you have two distance atoms. They emit light, and after the measurement process, you know they are entangled. And that's pretty cool. So the situation how we do it is we have two photons which come. But the first thing we have to make sure is we have to scramble the photons. We have to make sure that we can't find out from which atom the photon has come, and this is done with the beam splitter. So there is one aspect of beam splitters and two photons which have to explain to you now. And this is this famous HUM, Hong-Ou-Mandel. This is the Hong-Ou-Mandel interference. It's a very famous effect, and it's actually very, very special. So let me explain what happens when we have two photons in the same state-- two identical photons, same frequency, same polarization-- coming to a beam splitter. So we have already all the tools. So we have a beam splitter characterized by this angle theta, which through cosine theta, sine theta determines what the beam splitter is doing. And all we want to do is we apply it now to the state 1 1. So what we will find is that there is a probability to get one photon each. Then there is a probability to get two photons in one output. And it will actually be the same probability with a minus sign in the amplitude to get 0 2. And the matrix which acts on each of the photons has cosines and sines. So what we get is products of trigonometric functions. And here for getting two photons in one arm, it's square root 2, cosine theta, sine theta. And the spectacular thing here is what happens when we have a balanced beam splitter. If you have the 50-50 beam splitter-- and this is called the Hong-Ou-Mandel interference, you have the situation where you have a beam splitter, you have one photon, you have two photons. And it is absolutely impossible that afterwards, you have one photon in each arm. So you send two photons on a beam splitter, and after the beam splitter is, you have either two photons coming out here, or two photons coming out there. You can say it's Poissonic stimulation photons and bosons. That all plays a role. If one photon goes one path, the other photons follow suit. This is now very powerful, because it already happens, of course, when the two photons are identical. We had to use in this formalism the photon is in the same mode on each part. But that means now the following-- if you set up photo detectors here, and each photon detector makes click, you know you had one photon each. And that tells you that the two photons were not identical-- for instance, because they have different polarizations. So if you want to now, with the atoms, get a Bell state, where atoms decay-- one atom decays to state one, one atom decays to state two-- the signature of that would be that we have one atom each. And if you do it right, one atom each is an ingredient for 1 2 plus 2 1 for Bell state. We can detect that we have one photon each because at such a beam splitter, it's only in this situation that we can get one photon after each beam splitter if we start with two non-identical photons. Of course, by the way, experimentally there are quite some challenges, even if you have identical photons of the same polarization, if they arrive as a nanosecond pulse, and they don't arrive exactly at the same time at the beam splitter, then you first split one photon, and then the next photon, and the two photons cannot influence each other. So there are a lot of experimental requirements to realize this ideal experiment. So as I was preparing for today's class, I actually saw a paper which just came out in "Science." So in this month's "Science" they discussed the Hong-Ou-Mandel interference experiment with fermions. So let me just explain what happens. OK, now it fits. So if you have two identical photons, if you have two particles which impinge on the beam splitter-- they are the two input beams for the beam splitter-- you would say you have four different outcomes. One is both come out here, both come out here. Or they are both reflected, or they are both transmitted through. And what happens is bosons-- as I just pointed out-- bosons characterized by photons can only do that. Identical photons want two bunch up. They appear in pairs-- 50% left output, 50% right output. Well, for fermions, they just do the opposite. For fermions, you will always get one particle each. You may immediately, of course, explain it with the Pauli Exclusion Principle, which does not allow two particles to be in the same state after the beam splitter. And you should contrast it with classical particles. When you have two classical particles, you will actually find that all of those four possibilities, each of them has a 25% probability. So it was a major experiment which was featured in "Science" when people realized that with electrons, they created electrons in a semiconductor structure, and showed through some statistical measurement that this was the physics which happened. The measurement they did is, I forgot details-- if you have bosons and you get either two here or two there, you have more fluctuations in your system than for fermions. And they conclusively showed that they had realized the Hong-Ou-Mandel interference for fermions. I think we have to stop. Let me just add one more thing and then we are done. So this Hong-Ou-Mandel interference is at the heart of how perform the measurement which is ultimately entangling the atoms. But we need one more element. We need sort of to scramble the photons in one more way, and this is by adding circular polarizers at the input. So we assume for now that we start out with linear polarization in those states. And here is our beam splitter. Here is mode a and mode be, which we detect. And before we measure this Hong interference, we put in quarter-wave plates which provide circular polarization. If you start with linearly polarized light-- we put in a polarizer-- after the circular polarizer, we have this linear superposition of horizontal and vertical at mode one, and we have linear polarization-- superposition of linear polarization in mode two. So this is a situation at the input of the beam splitter. And if we expand it, we have probabilities where the polarization is different, and where the polarization is the same. So we have two detectors here. And if both detectors click, then we know that the input to the interferometer was not H H nor V V, because in that case, the Hong-Ou-Mandel interference would have directed both photons to one output, and we would not have obtained clicks from both. So therefore, when both detectors click, we know that the quantum state before the interferometer-- before the beam splitter-- was H V or V H-- or, of course, in another basis, one of those. And this is sort of the ingredient, which I will show you on Wednesday, which can lead to probabilistic entanglement of atoms. Any questions? OK, one announcement-- we post this week's homework assignment later today. It will be due in a week. See you on Wednesday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

9_Diagrams_for_lightatom_interactions.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Before I continue with the material, I want to show you at least the title of a recent paper in Nature, because it's related to material we have covered in this course. It's about the Kerr effect, the effect that one photon can create a phase shift for another photon. And one goal, of course, for quantum computation where things are about single photons is to have a single-photon Kerr effect that one photon can change the phase of the other photon in a strong, noticeable way. So maybe one photon should create a phase shift on the order of pi. And this was reported here in this paper. Of course, the non-linearity created by nonlinear crystals is much too weak for that. But what those authors did is they used microwave photos, had microwave photons into cavities. And they were coupled through a sapphire substrate with a Josephson junction. So the non-linearity here is the non-linearity of a Josephson junction, which is actually realized with a superconducting qubit. I can't explain you many more details, but I just thought it sort of cool to see how the Kerr effect, which we discussed and which we discussed for single-photon, is realized, at least in the microwave domain. And also, just sort of to illustrate that I hope this course enables you to read recent research papers, what those people measured is a [? key ?] representation. This is a coherent state. And then they showed, and this is the subject of the paper, that the coherent state which has a well-defined phase, lost its phase through the Kerr medium. And you clearly see there is a big phase uncertainty. But then after certain time-- this is experiment and this is simulation-- there is a re-phasing, and the phase is back. There is a revival of the coherent state. All right. Now I want to address one question which Cody asked about the g2 function and fluctuations of single-mode light. Let me just summarize. I told you that if you do the thermodynamics of a single mode, we find Bose-Einstein distribution of photons. And we have a thermal distribution. And a thermal distribution means sometimes we have more photons, sometimes we have less photons, depending on the thermal distribution. And when we calculated what the intensity fluctuations were, we found they're characterized by a g2 function of 2. OK. Then a little bit later in this course-- actually, just this week-- I told you that single-mode light always has a g2 function of one. And what I meant here is the following, rather trivial. If you have a single mode, that means that you [? align ?] [? it ?] is simply e to the i omega t, the intensity is constant. There are no intensity fluctuations. And also, because everything is sort of predictable, it's just one wave. The Gn function factorizes into-- can be re-expressed by the g1 function. So therefore, this is the most trivial case. But the question now is, how do we reconcile those two statements, that a single mode, e to the i omega t, does not have intensity fluctuations? Therefore, is a g2 function of one. And our earlier treatment about single-mode black-body radiation. And of course, the answer is, what is the single mode in one context is different from the single mode in the other context. Maybe let me explain that. Let's just create an ensemble of cavities. We put them in thermal contact with a reservoir. And then we break the thermal contact with a reservoir. Each cavity has now a perfect single mode, e to the i omega t. But each cavity is filled with a different photon number according to the thermal statistics. So therefore, if you just look at one cavity, we find no intensity fluctuations. The g2 function is one. But if you extend the ensemble average over all the different cavities, we find that there are fluctuations in intensity. Well, we can now keep that in mind. But now we can say, well, let's just take one cavity which is weakly coupled to a thermal reservoir. And instead of looking at the ensemble average of many cavities, we look at the long-time average of this one cavity. And what will happen is thermal photons will be created, will disappear, and such. So now this one cavity fluctuates. But technically, what that means now is it means that the sharp mode of the cavity is interacting with the environment, and it becomes broadened. It has a broadening delta omega. And this can be regarded as that we mix in modes of the environment. So in that case, strictly speaking, it's no longer a single-mode cavity. So you have to consider those things. And depending what point of view you want to take, you get the different result. Other questions? Then let me ask you a question. Last class, I explained to you-- well, at least, tried to explain to you that g2 function for bosons and fermions with the counting statistics, with permutations and such. I wasn't sure, at least from one question I got, whether this was completely clear. Do you have any question about that? [? Teroy. ?] AUDIENCE: This seems very obvious. But during class, I was trying to something with thermal state. What is the definition of our thermal state in terms of any basis, just generally speaking? We write it as-- I thought it'd be something like e to the minus beta light Hamiltonian. PROFESSOR: Yeah. So our definition of the thermal state-- when we had thermal light, we say that the statistical operator is given by that. And H is the Hamiltonian for a single-mode light, which is n plus 1/2 nn. Well, with suitable parentheses and summations. Other questions? OK. Then let's get to the main subject we want to discuss now. And these are actually Feynman diagrams. I wanted to give you an exact definition and a deep understanding, what does it mean when we talk about processes of absorption and emission, but also about absorption, emission processes which violate energy. And some people refer to them as virtual photons. The reason is that virtual photons cannot really exist for a long time. When you emit a virtual photon, another photon has to be absorbed immediately to reconcile energy conservation, as we want to see in a moment. So the goal of this presentation is I want you that you realize that each of those doodles has an exact mathematical meaning. Each of those diagrams represents one term, or a class of terms, in an exact solution for the time evolution of the system. So in other words, if you would ask me, we have a ground and excited state. Is it possible that the ground state emits a photon, goes to a virtual state, emits another photon of another frequency, and then somehow absorbs the photon, goes to here, and eventually takes another photon, and is back to the ground state. Is that a possibility? Can that happen? And I think what the message here is yes, everything happens. The system is trying out all of its possibility. And the two time evolution is the sum of all those possibilities, of all the amplitudes related to those diagrams. But what I want to show you is how sort of the weirder the diagrams get, the more you go in energy below the ground state, the more you go away from real atomic states, the bigger is your energy denominator. And that means those diagrams have a smaller and smaller weight. And in all practical calculations, we neglect those. But I want you sort of to be able to see that and realize, I exactly know what it means. It means this and this term in a summation over all the amplitudes which the quantum system is exploring. And I think with that, we really learn something about physics. We learn about what is actually inside the Schrodinger equation. A lot of people actually, before they take this class, think that this is just nonsense, that this has no physical reality. But I hope after this class, you see that pretty much everything you draw has physical reality. It's just-- it may not contribute a lot. So what we have done-- and let me just start here and invite your questions. We have figured out how an initial state evolves with a time evolution operator to another basis state, toward the final state. And formally, this is the formal solution and exact solution of the Schrodinger equation. We have to sum over all orders, orders in the interaction. Well, I will immediately tell you what is the first and second order. We are not going much higher. But if you want, here, you can. And then what you have to do is in the time evolution, you have to sum over intermediate times. You have to allow the system to propagate, to change its state, propagate again, change its state again. And the times where the change of state happens, that can happen at any time between your initial and your final time. And we integrate over all possible times. And I showed you-- and I think this was the very last thing we did on Wednesday. I showed you how this diagram can be translated into mathematical equation. And I think I picked the second order diagram. But I think from the way how I presented should be clear now how any such diagram can be translated into an equation. And eventually, you have to perform an integral over all intermediate times. And this is part of the time evolution of the system. Questions about that? AUDIENCE: We allowed these to be [INAUDIBLE] right? Because we haven't done anything with that? PROFESSOR: Good question. I assume that it's time-independent. Actually, right now, I assume, just to assume something, that it's a dipole interaction. My understanding is if it has an explicit time dependence, it would just appear there, and it would not change the-- would it? Wait, would it? Actually, when be derived the differential equation for the time evolution operator, did we assume that v was time-independent or not? I don't think we did. We integrate over time. I think the formal solution. Remember, I wrote down the differential equation for the time evolution operator and then say, this is a formal solution. My gut feeling is nothing changes when v is time-independent, but this one step should be confirmed. Other questions? OK. Then let me just spend a few minutes on connecting what we have done with standard first and second [? auto-perturbation ?] theory. I want to sort of throw a few definitions at you. S-matrix, T-matrix. But I'm not really going into any details. I just want to sort of wrap up the perturbative treatment by connecting it with the standard first and second [? auto-perturbation ?] theory. But after that in a few minutes, I want to have a discussion about the nature of intermediate and so-called virtual state, and then talk about the interaction v, whether it's d, dot, e; or the p minus e interaction. OK. So far, I've presented the formalism that we started in initial time and ended at final time. But usually, these are microscopic times. And in the experiment, we observe a system for macroscopic time. So for that purpose, we usually go to the limit that initial and final times are infinitely apart. And that actually means we have energy conservation. The initial and final energy has to be the same. And that can be, for instance, even if you restrict ourselves to second order, remember we had all those propagators, e to the i energy over H bar times t. And when we integrated over long times, it will just average out to zero unless the initial and final energy are the same. And technically, you have seen that in undergraduate derivation second order perturbation theory. You integrate the exponential function. And eventually, for sufficiently long time, capital T is the difference between initial and final times. It approaches the delta function. So that's how energy conservation comes in. So the fact that we have energy conservation is then used to define s and T-matrix. The transition amplitude from the initial to the final state, what we have just calculated and discussed, is called the S-matrix. It's just how it is called. In first order, of course, the time evolution is the unitary matrix. So therefore, we get [? conical ?] delta. And then we discussed that in the limit of large times, we have a delta function. So therefore, if we take the S-matrix, which is the transition amplitudes we have calculated, and sort of take out of the S-matrix, the unity matrix, and factor out the delta function, then what is left is the so-called T-matrix. When we talk about transition amplitude, transition probabilities, we are asking, what is the probability that the system has gone from an initial state, maybe the ground-- the excited state to the ground state through spontaneous emission? Well, probability is an amplitude squared. So we take the matrix element of the S-matrix and square it. And from the line above, this is now involving the matrix element of the T-matrix squared. There's a delta function which becomes a delta function squared. But if you integrate over all final states, that's-- I mean, a delta function is always requiring that you do some integration later. Otherwise, the delta function doesn't make sense. The delta function squared. You can actually explicitly see from the sine function above. The delta function squared just turns into the time t. So therefore, if you divide the probability by the time, we have our transition probability. We have our transition rate. And what we obtain is the second order expression for the transition rate, which is essentially Fermi's Golden Rule. So anyway, this is just finishing the formal derivation. But now I want to discuss the nature of those intermediate states. And maybe what you should have in mind-- the intermediate state, which comes about when we have the system in the ground state, it emits a photon and goes down to this intermediate state, or this weird state which seems to be lower in energy than the ground state. Well, what happens is those intermediate states, when they appear after the vertex, they propagate with the energy. But if their energy is less than the initial energy, here, the energy-- the intermediate state k has a phase factor in its propagation which is determined by the difference of its energy with the initial energy. So when we violate energy conservation in intermediate state, delta Ek is non-zero, and it is the larger, the more we violate energy conservation. And then in the solution of the time evolution, we have to integrate over all intermediate times. So what we have here is we have an oscillating phase vector. And when we integrate it, when we integrate something oscillating over longer than an oscillation period, it averages out to zero. So therefore, those intermediate states which are off the energy shell, which seem to violate energy, can only noticeably contribute over a duration which is 1 over the energy defect, delta Ek. So it is correct to say that the system in its time evolution for short times can, so to speak, violate energy protected by Heisenberg's uncertainty relation. Or you can say, the system can do whatever it wants. It can spontaneously create 10 photons. But this is pretty much like taking money, taking a deposit out of Heisenberg's bank. And after very, very short time, you have to pay it back by having another process which brings you back to the correct energy. I should put it under quotation marks when I said we violate energy, because energy is not sort of defined or measurable in this intermediate times. We start a process with a quantum system, we look what happens afterwards. And whenever we assess energy, it is assessed when the final time is much, much larger than the initial time. And I just showed you that eventually, the system has to be back to a final energy, Ef, which is identical to the initial time, Ei. So in other words, I just want you to keep it in mind. I've proven to you energy conservation in the limit that t final minus t initial is large. And when I'm now talking about non-conservation of energy, I do that in quotation marks, because we know energy is conserved in the end. It's just that for very, very short times in the time evolution of the system, there appear virtual states which seem to violate energy. But you can think about it in that way. Questions about that? Finally, let me now address the question, is everything I'm explaining to you really happening? Is it really happening in a physical system? Well, the first answer is, I wouldn't tell you about it if it had no reality. So yes. You can imagine that this is what your atom is doing. You can imagine that the hydrogen atom that's [INAUDIBLE] [? shift ?] is permanently emitting and absorbing photons and all sorts of weird photons from the ground state, lower, up, and back again, and such. Yes. This is the way how we derive some of the most precise predictions in physics, namely QED. However, we can represent systems in different gauges, in different representations. And we had discussed earlier that we often take the dipole representation for the light atom interaction. But there's also the p minus a representation. And if you look at the two, both the e, dot, d and the a, dot, p interaction are the product of something which creates and annihilates single photons-- a plus a [? dega. ?] And then the p operator or the d operator connect the ground with the excited state. So those matrix element would tell you, you can only emit and absorb a photon when you go from the ground to the excited state. However, in the p minus a representation, we also have the a squared term. And the a squared term allows you-- because there is no atomic operator in front of it-- allows you to scatter two photons without changing the quantum state of the atom. But this does not contradict anything. You can take either approach. And in your homework, you will actually do that when you calculate Rayleigh scattering and Thomson scattering. And you will find out that the two approaches give identical results. So if you know ask the philosophical question, can an atom scatter a photon without changing its quantum state, well, the answer to this question is actually gauge-dependent. But maybe to lift away some of the confusion, one can, of course, also say, the a squared term is really important. And it's a simple description of Thomson scattering. Thomson scattering is about photons which have much, much more energy than the energy difference between the ground and excited state. So therefore, if you want to completely describe the system-- and you can, it's just more complicated-- with a dipole approximation, we have to scatter a photon. But because the photon has so much more energy, it's so far away from resonance, this intermediate state has a huge energy defect. And that means what I just explained to you-- the system wants to immediately pay back the money to Heisenberg's bank. So it will remain in this state only for very, very short time. And now the other gauge tells you, this very, very short time can also be zero. So see, it's not as dramatically different as you might assume. But this is just something which is common that, if you have different representations, the physics tastes different, but you always have to remind you that when you calculate an observable result, the two different gauges must exactly agree. Questions? All right. So this is all I wanted to say about the diagrammatic solution for the time evolution of a quantum system. But I want to use it now. I think this is interesting stuff. It's a powerful method. And I want to illustrate to you how this method can be used in the next two sections. Our next section is van der Waal's interactions. So the chapter on van der Waals interaction is quite interesting for the following reason. It really tells us something about the vacuum. I think modern physics has come to the conclusion that the vacuum is one of the most interesting subjects to study, because the vacuum is alive. It's filled by virtual photons, virtual particles. And as we know now, by a condensate of Higgs boson. So there is a lot of stuff, a lot of structure, a lot of phenomena associated in the vacuum. And the subject of van der Waals interactions is it's a nice way to talk about the vacuum, but it's nice also for the following reason. There is a completely, I would say, semi-classical way. Just use the Schrodinger equation and calculate what is the Van der Waals interaction between two atoms. So you can just calculate in perturbation theory. And in your whole calculation, you never use the quantized electromagnetic field. Photons never appear. You just use perturbation theory. And you do that in your homework. It's rather straightforward. And you could have Van der Waals interaction. On the other hand, I started to derive for you the theory of the quantized electromagnetic field by saying, we really look at everything. We have charges, and we have the electric and magnetic field. And we described everything. And the way how we separate it is we had this Hamiltonian H naught, which gives us the atom. And the longitudinal Coulomb field became part of it. So this is the degree of freedom of the atom. And we have H naught for the other atom. And then we had radiation field and the coupling to the atom described by our operator V, which can be the dipole operator or can be written down in other gauge. But what I'm telling you is the way how we have fundamentally divided the world into atoms, atom's neutral objects, and the rest is interactions with the radiation field. That should tell you that the Van der Waals interaction between two atoms must have a description in QED where the Van der Waals indirection between atoms comes from the exchange or photons. So there are two very simple pictures at the same physics. One is you don't even know that they're photons, you just do perturbation theory. But in a more comprehensive description where we include the photons, you should be able to understand the Van der Waals action between two atoms due to the exchange of virtual photons. In other words, one atom is in the ground state, emits a photon that's going down from the ground state. But until now, if you have only one atom, it has to reabsorb the photon again and be back in the ground state. Otherwise, energy would be violated. But if you have two atoms, one atom can emit a photon, and the other atom can absorb it. And then the other atom can send a photon back. And if we consider that process, we will actually find the Van der Waals interaction. So I hope this is showing interesting physics from two angles-- that something which looked maybe trivial a long time ago now looks much richer, because those forces are really mediated by virtual photon pairs. So that's sort of the discussion I want to go through. There is another aspect to it. And this is when we go from the Van der Waals force to the Casimir force. The Casimir force has one exact derivation, which I want to share with you, which relates the Casimir force to the vacuum fluctuations of the electromagnetic field. So eventually, for those forces between two metal plates-- a neutral atom in the plate, two neutral atoms-- we will have three different pictures. One is we use the semi-classical dipole field as a perturbation operator. You don't think about it. It's trivial, and you check it off in your homework. The second one is you look at the exchange of virtual photons. And the third one is you only look at the zero-point fluctuations of the electromagnetic field. What is real here? What causes it? We'll see.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

1_Introduction_to_Atomic_Physics.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So this first lecture is a spirited but lighthearted introduction into the course. I usually never talk about technicalities first. I want to talk about the excitement of the field first. So I will talk about technicalities a little bit later. So when you take a course in AMO physics, the obvious question is, what is AMO physics? What defines our field? Actually, by far the best definition of what AMO physics is-- it is what AMO physicists do, which is defined by a community of AMO researchers. And this is really characteristic for our field. I will tell you in the next 10 minutes or so about these enormous dynamics of the field, how AMO physics has changed in a fraction of my lifetime. And what happened is, I felt, whatever I was doing-- and it turned out to be very different of what I did 10 and 15 and 20 years ago-- has stayed in the center of AMO physics. So myself and the whole community, we're moving, and took the field along. So therefore AMO physics is what gets AMO researchers excited. It's not a joke. It's the way it sort of happened. Well, a little bit more mechanically, we can define AMO physics. AMO physics is what is made out of the building blocks we have in AMO physics. So defined by the building blocks. And in AMO physics, we build systems out of atoms, or molecules, and then photons, or light, and in general, electromagnetic fields-- electric, magnetic fields, microwaves and all this. So in other words, everything which is interesting and we can put together with those building blocks, this is AMO physics now. And this may redefine AMO physics in the future. Now, historically this meant that those building blocks, atoms and molecules, were available in the gas phase. So you had gases of atoms, gases of molecules, and you studied them. And almost all of AMO physics was actually about individual particles, individual atoms, individual molecules, because in the gas phase at high temperature, pretty much, you learned that in statistical mechanics each particle is by itself. And the partition function of the whole system is just factorized into partition functions of the individual particles. Well, maybe with some exception-- occasionally, particles collide. And the field of collisions-- collisions between atoms and atoms, atoms and ions, atoms and molecules, molecules and photons-- this was widely started in AMO physics. Well, now the field has really moved away from just individual particles and collisions between two particles. What is in the center of attention is few body physics. And that of course takes us to entanglement. For instance, people study entanglement of eight ions in ion traps. And of course this is deeply related to quantum information science. Or if you want to go from few body to many body, this is now starting to overlap with condensed matter physics. And this is now widely studied in the field of [INAUDIBLE] atoms and quantum gases. Now, when we talk about many body physics, we get, of course, into overlap with condensed matter physics. In condensed, I would actually say now a larger fraction of AMO physics and condensed matter physics overlap strongly, that we speak the same language, we study the same Hamiltonian, a lot of theorists apply the same methods to topics coming from either field. However, often in technology, how the systems are studied, there's a different culture, different tradition, and still two different communities. And the overlap comes about because there are systems in nature which you can say-- they are natural two or few level systems. Of course, we've helped nature a little bit by engineering. Those systems are, for instance, quantum dots or NV centers in diamond. Or another overlap with condensed matter physics is that AMO physics is now-- one frontier of AMO physics is the optical control of mechanical oscillators. Micro cantilevers, membranes, tiny mirrors in cavities, they have mechanical motion. And the mechanical motion is strongly coupled to the photon field. Of course, for fundamental AMO physicists, a mechanical oscillator is nothing else than an harmonic oscillator. So one can say-- and this is sort of my third attempt in defining for you what AMO physics is-- AMO physics is almost defining itself by low energy quantum physics. So all of the quantum mechanics which doesn't take place at giga and tera electron volt, which takes place at low energy, this is AMO physics. Of course, maybe not all of it. Usually when it is in [INAUDIBLE] solution, it's biophysics, and it's distinctly different from AMO physics. Or if the solid state is involved, then it's more condensed matter physics. But I would say there is one part of solid state physics which is already becoming interdisciplinary with AMO physics. And this is when the control of the system is not done by wires and carbon probes but it is done by lasers. So if you have a solid state system and you use all of the methods you apply to atoms, you use electromagnetically-induced transparency, coherence, all those concepts, then AMO physicists feel at home and they don't care if the two level system or the harmonic oscillator is an ion oscillating in an ion trap or whether it's a small cantilever, the methods and concepts are the same. So therefore, one could say AMO physics is sort of the playground where we can work on extensions of simple systems which we understand and cherish. And of course, our exactly solvable problems are the hydrogen atom-- my colleague, Dan Kepner, would have said, maybe the hydrogen atom, if you understand the hydrogen atom, you understand all of atomic physics. I'm not so sure. I would actually say, in addition to the hydrogen atom, you have to know the two-level system. And of course, you have to understand the harmonic oscillator. So these are three paradigmatic Hamiltonians, and a lot of understanding of much more complicated systems really comes from taking the best features of those three systems and combining them. If you have questions or comments, this is an interactive class. Feel free to speak out or interrupt me. OK so, now we know, or we don't know, what AMO physics is. Let me now address-- how has AMO physics developed? And I mentioned to you that AMO physics has done breathtaking evolution in my lifetime, or even in the shorter part of my life, which is my research career. Well, traditionally, almost all fields in science started with observing nature. The pursuit of science was born out of human curiosity to understand the world around us. And atomic physicists, well, they started to observe atoms and molecules, usually in the gas phase, and what they were doing. And already there was some evolution, because original observations at low resolution were taken to a completely new level when high-resolution methods were developed, when lasers came along, when people had light sources which had fantastic resolution and eventually finer and finer details of the structure of atoms and interactions between atoms were resolved. But AMO physics is a field which has taken the pursuit of science much further. So there is not just observation of nature-- and I want to write that with capital letters-- there is CONTROL OF NATURE. And you maybe take it for granted, but you should really appreciate it, that controlling nature, having control over what you study, modify it, advance it, take it to the next level, is really something wonderful. It is completely absent in certain fields, like astrophysics. In astrophysics, all you can do is, you can observe. In atomic physics, we create the objects we can observe. So the control of nature, the control of our atomic physics system, developed in stages. The first kind of control was exerted about internal states. If you have an atom at thermal energies, it would only come in hyperfine states which are thermally populated, or molecules come in rotation states, and well your limited control was simply to raise and lower the temperature. But with the advent of optical pumping-- this actually happened already with classical light sources before the invention of the laser-- so with optical pumping, you can pump the internal population of molecules into, let's say, a single rotational state. So this is control over the internal Hilbert space. And this was actually rewarded with a Nobel Prize to Alfred Kastler in 1966. Of course, the next step after controlling the internal degrees of freedom is have control over the external degrees of freedom, and this means control motion. This was of course pursued by understanding the mechanical aspect of light. How do photons mechanically interact with atoms? This eventually led to laser cooling and Bose-Einstein condensation. And those developments were recognized with major prizes. Well there is more to it than controlling internal and external degrees of freedom. You can then also say, well, how much can we control the number of building blocks? And eventually AMO physics advanced to exert control onto single quantum systems-- single photons, single atoms. A single atom in a cavity exchanging a photon with a cavity thousands of times. So this control of single quantum systems was actually just recognized with a Nobel Prize a few months ago. Well, at this point I sometimes make the joke, we have gone from big ensembles in many, many quantum states to single photons, single atoms, in a single quantum state-- a single quantum state for many particles is Bose-Einstein condensation. So we have really gone down to single atoms, single photons, single quantum states. Well, what comes next? To have no atoms and no light in vacuum. Well, the vacuum has some very interesting properties. And if you talk to Frank Wilczek, the nature of the vacuum and dark energy is one of the big mysteries in physics and in science in general. But the study of that is definitely outside the scope of AMO physics. So what happens is, when we have gone down and have now control over the building blocks, now we can sort of go up again in a controlled way, create complexity by assembling a few photons, a few atoms into new entangled states, so we can now take our system into very different regions of Hilbert space. So what is defining now the control is, we want to use this pristine control over the building blocks to now put in something which hasn't existed naturally before, or when it existed it was completely obscured, completely hidden by thermal motion or by you can say homogenous broadening our lack of control. And the best buzzwords are here now entanglement and many-body physics. It's hard to capture that as in a diagram, but let me try that. I don't know actually what I'm drawing, but I think you get the message that this is sort of Hilbert space. And I have 2 x's. One is sort of entropy, high temperature, and the other one is complexity. And for quite awhile, people studied hot gases. So these are gases, there's a lot of entropy in it. The complexity is actually not particularly high. And everything is described by a statistical operator-- the density matrix. The pursuit of cooling, and actually, control-- gaining information about a system is also a way of cooling. If you know in which state the atom is, the entropy of the system is zero even if you haven't changed the state of the atom. So control and cooling, control measurement and cooling, has now taken us to the point where we have systems which have no entropy anymore. They are very, very well defined. And our goal now is to take these systems to much more complexity where wave functions become entangled and we have strong correlations in many-body systems. But this is now, maybe here, this is now described by wave functions. But here is the wave function of a single particle, and here we have highly correlated, highly entangled many-body wave functions. So at least for me-- but all predictions are notoriously incorrect when you look at them in a few years from now-- but for me, this is where the future of our field is moving, to get into interesting regions of Hilbert space where no person has been before. As an experimentalist, and with a lot of experimental graduate students in this room, I want to emphasize that a lot of those rapid developments of the field are driven by technology. So it's driven by technology advances. In the '50s and early '60s, people thought AMO physics is pretty much dead. Only a few people with gray hair continue what they have done and the field will eventually die. But then technological developments made it possible to do major conceptual advances. I've mentioned the conceptual advances-- let me now say a few words about the technology which has driven it. There was one phase of developing lasers which I experienced when I was a student. But those lasers were fantastic. They were very narrow already, very stable, but they were very expensive and very complicated. So if you had one laser, this defined your laboratory and then you studied a lot of things with the single laser you could afford. Well, you're probably not used to one big dye laser pumped with a big argon ion laser. It was a $250,000 investment for the lab. And of course, you couldn't afford a second laser. It required 50 kilowatts pf electrical power. And all this power had to be cooled away by gushing water through thick pipes. So it was an expensive undertaking but you could really do wonderful science with it. So what I've seen in the last 20 years is the proliferation of solid state lasers, starting with diode lasers and continuing until the present year. So now, in a lot of our laboratories here at MIT, we have 10 lasers. And we've stopped counting them, because adding a laser to the system is almost like adding an amplifier to a circuit or adding another circuit to a data acquisition system. But it's not just the simplicity of the lasers which we have now, the robustness-- to have 10 lasers in the lab, it's fine. Previously, if you had three lasers in the lab, you spent 90% of your time just keeping the lasers running. Those lasers, not so much continuous-wave lasers, but pulse lasers have also very, very different properties. Laser pulses got very, very short-- femtosecond or even attosecond. Shorter pulses means that the energy is now focused to a much shorter temporal window. Therefore, laser pulses of very, very high intensity-- if you focus a short pulse laser on an atomic system, you can easily reach an electric field of the laser which is stronger than the electric field of the atom. So in other words, if you have protons and electrons-- well, maybe the outer electrons, not get down to the single protons. But maybe an ionic core, and you've electrons. Now, you should first look at the motion of free electrons in the strong field of the laser and add the atomic structure as a perturbation. It really takes the hierarchy of effects upside down. So the appearance of high intensity lasers has given rise to a whole new field of atomic physics. Lasers got more precise. The invention of the frequency combs, recognized by the Nobel Prize in 2005, meant now that we can control laser frequencies at a level of 10 to the minus 17. And this has completely redefined precision metrology and has advanced the control over atoms and molecules I've mentioned before. Finally, another technical development which plays a major role in research being pursued here at MIT and elsewhere are the development of high finesse cavities. High finesse cavities in the microwave range-- then they're superconducting-- or high finesse optical cavities by having super mirrors. It is actually those super cavities which have enabled the study of single photon physics. Because after all, photons move away with the speed of light. And if you want to observe a photon in your laboratory, it has to bounce around zillions of times in order to have enough time for the photon to do something interesting. So sometimes a field at the frontier of science is defined by paradigms. If you want to explain to somebody why your field of interest is cool and exciting, you usually do it by picking a few really exciting examples. And I want to show you how, over the years, it has advanced. Definitely in the '50s and '60s, you would have mentioned that we understand now atomic structure of multi-electron atoms. Optical pumping just started. So these were flagship developments of AMO science. The cool thing to do in the '60s and '80s was, use the new tool, the laser, applied to atoms and do laser spectroscopy. Sub-doppler spectroscopy, sub-natural spectroscopy, resolving hyperfine structure-- wow. I mean, this was really exciting in those days. And well, the older people I have met, my teachers, my thesis adviser, these were people who started their research career before the laser was invented but then, as a young researcher, embraced this new tool and helped to redefine the field. Definitely in the '80s and '90s, the cool pictures were those of magneto-optical traps, atoms standing still and hovering around. So the new aspect where mechanical forces of light which led to laser cooling and trapped atoms. In the late '90s, of course, the excitement was about Bose-Einstein condensation. And it was really Bose-Einstein condensation which drove AMO physics from single atoms, maybe two atoms colliding, to many-body physics. It's always easier to analyze those things by looking backward, so if I'm now getting closer to the past, I have to be a little bit vague. But in the 2000s, I think hot topics were ultracold fermions and the study of entanglement and correlations. And what is the paradigm now or in the near future? Well, I think you have to help to define it. If you make an interesting discovery, this is what people will be pointing to and would say, this is what now defines AMO physics. Some candidates are, of course, if there is a major breakthrough in quantum computation-- let me put question marks here. In the field called atom science, we may actually do some progress towards topological states which have different symmetries and different properties. And another emerging frontier is micro-mechanical oscillators. The last couple of years, we just had the breakthrough for the first time. Mechanical objects were cooled to the absolute ground state. So at least for that community it was, but for many of us, Bose-Einstein condensation was 15 years ago. Any questions? Well, eventually we have to talk about this course. So I've told you about, at least, a snapshot of where AMO physics is, how it has developed, and on what trajectory it is. In this course I want to present you the concepts behind many of the major advances in the field. So over the years, quite often a topic was added to the course, because I felt, hey, that's getting really exciting, that's what people want to do in research, that's what graduate students want to do here. And then the subject was added and other subjects were dropped. I know in the '90s, I was teaching aspects of laser cooling, sub recoil laser cooling, which was the latest excitement. This year, I may mention it for 30 seconds. So the course has evolved. It wants to stay connected to what is exciting, what is hot, and what prepares you for research at the frontier. 8.422, the second part of the two-course cycle in the graduate course in AMO physics, is somewhat different, not radically different, but is somewhat different from part one, from 8.421. First of all, 8.421, 8.422 can be taken out of sequence. We alternate between AMO 1 and AMO 2. And whenever you enter MIT, you're probably in your second semester, take whatever we offer. So I expect-- let me ask you, who of you has taken AMO 1? Yes, statistically, it should be about half of the class. It can be taken out of sequence because that's the way how we've structured it. But to give you one example is, in 8.421 you really have to learn about hyperfine structure. You have to learn about atoms, you have to learn about Lamb shift and all that. So you have to learn what all these atomic levels are. And here in that course, in 8.422, I will say, here's a two-level system and then I run with that. And we do all sort of entanglement, manipulation of two-level systems. So it helps you if you know where those two levels come from, but you don't really need the detailed knowledge of atomic structure, for instance, to understand that. So this is why the different parts of the course are connected, but in terms of learning the material, somewhat decoupled. I've spoken to many students who said there was no problem in starting with 8.422. The only sort of critical comment I've heard is that taking 8.422 first and then 8.421 is anti-climactic. You see all this excitement in the modern physics, and then, eventually, you have to work on the foundation. Prerequisites for this course-- the course announcement said 8.05. It is actually 8.05 and 8.06. The main part of 8.06 which we really need here is perturbation theory-- time independent, time dependent perturbation theory, and this is usually covered in 8.06. However, I've had students who took the course without 8.06. If you're really determined and want to acquire certain things by self-study, you can follow this course. So the topics we will cover include QED. I really want to talk about light-atom interactions from first principles. Sure, 95% of what we're doing is just done by saying we have a matrix element, which maybe the dipole matrix element. But you really have to know what are the approximations, what are the conditions, which lead to the dipole approximation. And I want to do that from first principles, and we do that starting on Monday. So a discussion of light-atom interaction has two parts. One is the simple part-- excitation and stimulated emission. Because this can be simply described by a unitary time evolution, and you can do a lot, if not everything, by using Schroedinger's equation. Things get much more complicated and richer if you include spontaneous emission or, more generally, if you include dissipation. Then we talk about open systems. And for fundamental reasons, we need a formulation using the density matrix, a statistical operator, and a master equation. One major part of the course is discussion and [INAUDIBLE] derivation of the mechanical forces of light. This will include a discussion of important experimental techniques using those mechanical forces. So various simple and sophisticated methods of trapping and cooling. We will spend some time in not talking about atoms at all. We're just going to talk about photons, about single photons. We want to understand where the photon nature of light makes light very, very different from a classical electromagnetic wave. Also, it's not the focus, we will come across basic building blocks of quantum information science. Pretty much when atoms and photons interact, this is a fundamental quantum gate. And we'll talk about the many-body physics off quantum gases. So maybe it becomes clearer what we are covering by saying what we are not covering. And this tells you that there is at least some selection of topics. It's not that we talk about everything. We will not talk about the physics above 10 electron volts. We will not talk about collisions-- or maybe I should say, high-energy collisions. We will, of course, talk about nanokelvin collisions, which is the physics of the scattering links and some s-wave collisions which are really relevant to understand quantum gases. We're not talking about any advanced topic in atomic structure. All we do about atomic structure is done in 8.421. And if you want to graduate in atomic physics at MIT, yes, you have to understand atomic structure at the level of the hydrogen atom. And maybe know a little bit about a new phenomena-- when another electron enters, when electrons interact and that's a helium atom-- but we're not going beyond it. Let me just mention here that there is, of course, more interesting things in atomic structure. For instance, if you go to highly charged ions, you have QED effects. You can discuss very interesting correlations between two electrons in an atom. And you can have very relativistic effects if you have highly ionized atoms. If you have bare uranium, then the electron in the lowest orbits becomes relativistic. You can even see, if you scale the fine structure constant-- which has an e squared in it-- with the charge of uranium 92, well, 1 over 137 times 92 gives about 1, so you really get into a new regime of coupling for atomic physics. But we won't have time to talk about it. And this may be an omission, because there is really interesting work going on. We're not talking about high intensity lasers and short laser pulses. This choice maybe mainly determined by that the experimental program in the physics department is not overlapping with that. But of course, you know we have world class researchers and short pulse lasers in the electrical engineering department. Any question about the syllabus? Questions about what to expect in the next 12, 13 weeks? Well, then let's talk about some technicalities. I'll keep it short because all this information is available on the website. We run the website of the CUA server. So it is cua.mit.edu/8.422. When I say all the information is available, I have to say-- not yet. I realized yesterday night that [INAUDIBLE] has re-modified the server. I didn't have access to the server. I just got it an hour ago. What you will find under this URL is the website from when the class was taught two years ago. And actually, more than 90 percent of the information is the same. I always try to improve the course, but I know it will be more incremental changes and not dramatic changes. So if you look at the website, you'll already get a taste what the course is, but within the next day or two, we'll have updated information. I know, at least for most of you, I have contact information for MIT registration. But if you are a Harvard student or a listener just sitting in and you haven't registered for the class, I would ask you to send an email to our secretary, Joanna, j_k@mit.edu, because I would like to have an email list for the class for corrections to the homework assignment or last minute announcements. And then we'll add you to the mailing list. The schedule of the class is, well, we meet at this time on Monday, Wednesday. And let me know disclose, I plan to occasionally teach on Fridays at the same time. Over many years, it has been my experience, if you have a class on Monday, Wednesday at one o'clock, you don't have another class Friday at one o'clock. Is that assumption correct? So within the next few days, I will tell you on the website which Friday I would like to teach. The reason is rather simple. Like every faculty member who's active in research, I have to go to funding agency workshops and conferences. I try to keep it to a minimum, but on average I will miss two or three classes. And instead of asking a [INAUDIBLE] or a graduate student to teach the class, I would like to teach the class myself. So I will have makeup classes on Fridays. OK, that's the schedule. Finally, talking about requirements, one requirement is homework. We have 10 wonderful problem sets with a lot of problems which I actually designed from current research. So you will actually recognize that a lot of problems are created based on some research papers which came out of my own group or other groups at MIT in the last few years. The good news is, there is no mid-term, there is no final, but there will be a term paper. And the term paper is due on the last day of classes. The teaching team are myself, Joanna Keseberg, our secretary-- that's also where you will be dropping off your homework-- and then we have assembled a wonderful team of five TAs. They are all advanced graduate students. And actually, I've picked TAs from all of the active groups here at MIT. From Martin Zwierlein's group, I. Chuang's group, Vladan Vuletic's group, and my group. Are any of the TA's around? Nick, Alexei, Jee Woo, Lawrence, and Molu. Each TA will be responsible for two weeks in the course, and the TA will indicate availability and office hours on the weekly homework assignment. Other details about the term paper, how long the term paper should be, the, kind of, what I regard as an honesty code, that you're not trying to get access to solutions from previous courses, all that is summarized on the website. And once I've updated the website, I'm sure you will read it. Any questions? OK, well, with that we will actually be ready to jump into the middle of the course and start with some heavy duty equations of QED. But no-- just joking. I think this would maybe spoil the introduction. What I actually like to do is, to make the transition from the introduction to the discussion of atom-photon interactions and quantum electrodynamics, I always like to start the course by giving you an appetizer by talking to you in a lighthearted but hopefully profound way about the system, which helps to showcase what are we doing in this course. And until a few years ago, I would have started with the simplest example of laser cooling, simply beam slowing or optical molasses, in the simplest possible picture, just to give you a taste of what we will be doing together during the semester. But as I said, AMO physics is moving along. And what I now want to use as an example which clearly synthesizes many aspects of this course are atoms in an optical lattice. So let me, in the last, next 20 minutes or so, make some connections to different topics of this course by using as a starting point a concrete, very simple system, but very, very rich and profound, and these are atoms in optical lattice. So the situation I want to use here is that you have two laser beams which interfere. And those laser beams form an optical standing wave. So next week we will learn how this electromagnetic wave interacts with atoms. So we have to put a few atoms into the system. And we will derive, from first principles, the QED Hamiltonian, which, after a lot of manipulation and eliminating complicated terms, will be the dipole interaction. But of course, each symbol here is an operator, and there's a long story behind that. In about two months, we will describe light-atom interaction with a formalism which uses a Bloch vector and the optical Bloch equations. There is a vector with three components which describes what is the state the atom is in. One component will tell us if the atom is in the upper or the lower state, whereas the other components tell us whether the dipole moment of the atom is in phase or out of phase with a driving electromagnetic field. Well, if there is part of the dipole moment which is in phase with a driving electric field, then the suitable expectation value defines a mechanical potential. And if the electric field is a standing wave, then we generate, through light-atom interaction, a periodic potential for the atoms. We will learn everything which we have to know about this potential. In a simple case, it is simply the Rabi frequency divided by the detuning of the electromagnetic wave. But we will find it very interesting to look at it from very different point of view. And maybe let me use this example to point out that I'm really a big friend of explaining the same physics from very, very different perspectives. And when we talk about optical standing waves, we will use a picture of a classical potential, like a mechanical potential, and the atom is just moving around. We will use a photon picture that every time the atom feels a force, photons are involved. You don't see them in the classical potential, but photons are behind it, and ultimately, the forces of the classic potential come from stimulated absorption emission of photons. I may go down to the microscopic level. And I may ask you, but in the end, it's an atom, but the atom consists of electrons. And the electrons are simply oscillating because it's driven by the electromagnetic field. It's actually something which most people are not aware of, but you can ask the question now-- is the force in the optical lattice on the atom, it's ultimately a force on the electron? Well, if you have a charged particle, you can have two kinds of forces-- an electric force or the Lorenz force. And I don't know if you would know the answer, but an atom is in an optical lattice, is the force just the [INAUDIBLE] the potential which the atom experiences, is that fundamentally due to the Coulomb force exerted by the electric field of the light, or is it due to the Lorenz force exerted by the magnetic part of the light? Who knows the answer? Who doesn't know the answer? OK, great. I was actually surprised when I derived it a few times for the first time. And it's not in the standard textbooks. So anyway, I hope, even for many of you who know already a lot about the subject, I hope I can add other perspectives for you. OK, so what I've discussed so far is that the standing wave of light creates now a periodic potential for the atom. And we will understand it from many, many different perspectives, from the photon type, quantum optics perspective, to the most classical description. So what immediately comes to my mind is that we now want to look at this periodic potential in two different situations. We can ask, well, whenever you have light, spontaneous emission is a possibility. And we may ask, what happens when spontaneous emission is not negligible? And we discuss that approximately in week nine of the course. Then what happens is, an atom has an excited and a ground state. And those in the excited and the ground state, do the atoms feel the periodic potential? Well, eventually, we have to generalize the notion of ground and excited state into [INAUDIBLE] states, and I will tell you all about it in a few weeks. But the situation can now be, when an atom is in the ground state, it has to move up the standing wave potential. Then it's getting excited with a laser. It has to move up again the standing wave potential. And then there may be spontaneous emission. So what I'm just describing here is a situation where the atom is mechanically moving up the potential. It's excited at the top of the potential. And it emits when it's on top of the potential. So the atom is doing mechanical work, and this is called Sisyphus cooling. This is the way, this is one method, one mechanism, of laser cooling which leads to the lowest temperatures in the laboratory before evaporative cooling is used. So that's some cool aspect we will come along by discussing motion in a standing wave, but taking into account that photons are emitted and really understanding when and where are the photons emitted. Well, the second situation is, of course, if spontaneous emission is completely negligible. Then we have a situation where all what matters is the potential, and we can completely forget that it's photons, it's quantum optics which has created this potential. We can simply use a classical potential in our description in our Hamiltonian. Now, again, we have two limiting cases. One case is when this potential, or optical lattice, is really deep. Deep means that atoms are sitting deep in the potential and they can oscillate around, but they cannot jump over to the next potential. So in this situation, you would say, well, that's boring. Nothing happens, the atom just stays put. But what is boring for some of you is exciting for some others, because these atoms are exquisitely controlled. They cannot collide, they cannot interact with other atoms. So these are really the most ideal situation you can imagine for atoms. Of course, if you have less than one atom per site. And this is the way how you want to prepare atoms for the most accurate interrogations. And you can build atomic clocks, optical atomic clocks, based on atoms insulated in such a lattice, which approach now 10 to the minus 17 accuracy. Well, if you drive a clock transition, if you take the atom from the ground to the excited state, you may face a situation I mentioned earlier that the periodic potential for the ground and excited state are different. And that would actually interfere with your clock, because the clock frequency depends now on what the lattice is doing to your atoms. However, and we discuss that in week four, there are what people call magic wavelengths, where you pick a certain wavelength for your optical lattice where the periodic potential is absolutely the same for ground and excited state. So that means you have, then, the perfect decoupling between the ticking of the clock, the internal structure of the atom and the mechanical motion of the atoms in the potential. And in that situation, you create a situation which has been studied since [INAUDIBLE], namely with trapped ions. If you have a single ion in an ion trap-- and ion trapping is pursued here at MIT in Ike Chuang's group-- you have just a single object completely isolated. And in the form of the aluminum ion, this has just been demonstrated to be the most accurate atomic clock in the world with 10 to the minus 17 accuracy. So anyway, with optical lattices, avoiding spontaneous emission using magic wavelengths, we can only engineer with neutral atoms what has been available with trapped ions but we can simultaneously have 10,000 copies and look at 10,000 atoms which are all identical copies, identical systems with each other. So you see already from that example that there will be intellectual overlap and synergy between talking about neutral atoms in optical lattices and talking about trapped ions and how they are cooled and how they are manipulated. And we'll talk about trapped ions at the end of the course, sideband cooling of trapped ions, which is week 12. OK, so this is the simple but pristine situation that we have a deep lattice. The other limit is now that we have a weak, or shallow lattice. And the new physics which now comes into play is that atoms can move around-- they can tunnel from side to side. So towards the end of the course, I think the last months, approximately week 10, I will give you a short summary of what some of you may have already learned in the solid state course-- namely, band structure of atoms in optical lattices, Bloch states, effective mass, et cetera. This has now become language of atomic physics, because there is a very clean and straightforward realization of this physics using cold atoms. What we are mainly interested in, in our research, is when we have atoms which tunnel in this optical lattice. And for bosons, if the interactions get strong enough, the Bose-Einstein condensate is destroyed and what forms is a Mott insulator. This is a phase transition. And for fermions, we have a crossover from a metal to a fermionic Mott insulator. And with that, we are already overlapping conceptually with condensed matter physics, because the Mott insulator is a paradigm-- one of these paradigmatic examples where you understand some deep physics. It's a paradigm of condensed matter physics where you have only a partially-filled band. Common sense, or undergraduate textbooks, would say, partially-filled band, this means you've a metal, you've a conductor. But because of the interactions of the atoms, the system is an insulator. So this is where, really, the many-body physics profoundly changes the character of your system. Any questions? Well then, let me add one last aspect to my introductory example of optical lattices. In a sort of flyover, I described to you what is the physics we encounter when we have an optical lattice switched on and the atoms move around or they don't move around, and both cases are interesting. But if you think you've explored everything, well, then you think harder and say, hey, there's another angle we can get out of it. And this is, we can take the optical lettuce and simply pulse it on, switch it on and off. Well, what happens is, and then it becomes a time-dependent problem. It becomes something where we can shape and control the wave function of atoms in time-dependent way. Let me give you one example. If we start with very cold atoms-- can be a Bose-Einstein condensate-- and then, for a short time, we switch on the lattice, afterwards, we observe that we have still some atoms at zero momentum. But now we have atoms which have a momentum transfer of plus minus 2 h bar k. You can understand that-- and this again exemplifies that we want to look at the physics from different angles-- this can be described that you have an two-photon transition from the ground state with zero momentum to the ground state with two-photon recoil. So you can understand it in the photon picture. But you can also understand it by saying you have some matter waves which are now exposed to a periodic potential. And you simply ask what happens to waves in a periodic potential? Well, that's the same what happens to optical waves when they encounter a grading, and that's the physics of diffraction. So in a nutshell, this happens when we use a pulse on an optical potential. And let me now finish by telling you that, again, in this situation, we discover the ambiguity or the two-sidedness of a lot of things we do in atomic physics. I've mentioned to you that the same experiment, the same experimental setup, cold atoms in optical lattice-- you turn up the lattice and you have the world's best atomic clock, atoms just isolated from each other. You turn down the lattice and you have an interesting condensed matter physics system. Let me just show you that the same two-sidedness of atomic physics-- precision and pristine control and interesting many-body physics-- happens when you pulse on optical lattices. So one application of those optical lettuces is atom optics. You have atoms and when they encounter an optical lattice, some atoms will just continue, will not be diffracted. They have zero momentums. Others are diffracted with a transverse momentum. You can then expose the atoms to a second zone where, eventually, momentum is transferred, and then the atoms come back together. So what we have here is an atom interferometer where atomic matter waves first go through a beam splitter, are reflected back towards each other, all by photon transfer, and then they're recombined. So this pulsed optical lattice acts as a beam splitter. It is the way how, today, the most precise atom interferometers are built. And this is used for precision measurement, for measurement of inertial forces, gravity, rotation, acceleration, and it is used for navigation and accurate observation of the changes of the gravitational field of the Earth. Since we just had a talk by Holger Mueller from Berkeley day before yesterday over at Harvard, he talked about the combination of atom interferometer and frequency combs, another development-- let me just mention that. If you derive those laser frequencies from an optical frequency comb, and with an optical frequency comb you can pretty much count the frequencies. So this laser is mode, I don't know-- 100,000 or whatever you are in the comb. By doing that, by combining it, you can actually build an atomic clock where-- let me just say, plus combs-- can give an atomic clock which is called the Compton clock because the frequency is now given by the rest mass of the atom, but divided by a big integer number. So using completely unrelated developments in AMO science which I've mentioned, the optical frequency comb, combining it with the physics of a pulse standing wave, you now have an atomic clock which is directly related to the energy of the rest mass. But finally, if you pulse on an optical standing wave and your object are not individual atoms, non-interacting atoms, your object is the Bose-Einstein condensate or atoms which strongly interact, then you're not transferring recoil to individual atoms, you're transferring momentum to a complicated many-body system. And this means what we are measuring is the dynamic structure factor. If you have atoms, you want to do spectroscopy, you want to know what energy levels are there, and then you know your atom. If you have a strongly-interacting system, you also want to know what energy levels are there. But each energy level in a homogeneous system or in a periodic potential is associated with momentum and quasi-momentum. So in other words, if you have a more complicated system, you want to figure out what are the possible states in terms of momentum and energy. And the optical standing wave, the pulsed optical standing wave is the way how we impart momentum and energy to a system. What I actually just told you is a story in my own research career. I was a post-doc with Dave Pritchard. He had trained at MIT in the '90s by a pioneer in laser cooling. And when we had Bose-Einstein condensates in the late '90s, Professor Pritchard and myself, we teamed up. I was the expert on Bose-Einstein condensation and he was the expert on atom interferometry. So just by sort of exploring things, we took Bose-Einstein condensates and we pulsed on a standing wave. What was on our mind was, hey, let's build an interferometer. But I'm more of a many-body physics person. I suddenly said, yes, but if we now change the momentum and the frequency of this standing wave, what happens? And I suddenly realized that this is a way to measure properties of a Bose-Einstein condensate in a way which hadn't been done before. So I realized-- and this is maybe the last thing I want to tell you today-- I realized in my own research and my collaboration with Dave Pritchard, that we'd built an experiment, and we just turned one knob at our experiment, and the following day we were no longer doing atomic interferometry, we were doing many-body physics. So this is, I think, what makes our field exciting. We are using the tools, the precision, and the control of atomic physics, which leads to the most accurate atomic clocks in the world-- atomic clocks, which are the most accurate in the world. And we are using those tools to do entanglement and many-body physics. And I think it's just a compelling combination. Anyway, that's an appetizer, that's an outlook over the semester. Do you have any questions about the last examples I gave you or the course in itself? OK. No homework assignment today. And we'll meet Monday at the same time here in this lecture hall.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

11_Casimir_force.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Good afternoon. Welcome to April fool's day. But we'll be serious about atomic physics. I hope you enjoyed your spring break. At least I did. It's just a little bit change of pace. So let me remind you what we have been discussing. We have been talking about interactions between neutral atoms. And that can be trivial and very profound. Trivial because you've heard about van der Waal's forces. Polarization forces between neutral atoms. And if you describe them phenomenologically, you just have 1 over r to the 6 or whatever power of an interaction. On the other hand, if you take the position that everything which happens between neutral objects has to be mediated, has to be created by the photon field, then it becomes a question of the vacuum, the vacuum which surrounds the two neutral atoms. And I've tried in this chapter to show you both sides, to similar or the same physics from two different perspectives. We talked mainly about the van der Waal's interaction between two neutral atoms. I don't want one to-- we pretty much finished it, so I just give you a bird's view. I don't really repeat a lot of things. Because today we want to talk about the forces which happen between neutral atoms in metal plates and then two metal plates, which is the Casimir force. But just to remind you that we showed that the forces between neutral objects are actually created by quantum fluctuations. Classical objects would not create fluctuating dipole moments, would therefore not create-- would not give rise to any force between neutral atoms. And I showed you that there are two aspects where quantum physics comes in, namely through the fluctuations or the zero point motion of the atoms as oscillators. And we even had a circuit model for that. But then I also said that, well, assume that you have two neutral atoms and you have the vacuum fluctuations of the field. This vacuum fluctuation is driving coordinated synchronized dipole moments between the two atoms, and then the two atoms can attract each other. So we have seen the dual role which quantum physics plays here in terms of quantum fluctuations of the atomic oscillators and quantum fluctuation of the electromagnetic field, which is another harmonic oscillator. And at least when I use very simple pictures, it seemed that the 1 over r to the six potential, the standard van der Waal's potential at short range, comes from the vacuum-- from the fluctuations of the oscillator. I didn't do the derivation, but the simplest derivation for the long range Casimir Polder retarded potential, 1 over r to the 7, focused on the fluctuations of the electromagnetic field. So these were some preliminaries. Then I thought I want to show you the power of diagrams by showing to you that there are two different kinds of diagrams dominant at short range and at long range. And with that we realize diagrammatically that, because of the nature of virtual photons which are exchanged by the two neutral atoms, there is a change in power law. The power-- the inverse power of the distance between the two neutral atoms is one higher at long range. And long range meant-- that also became clear diagrammatically. Long range means the distance is longer than the wavelengths of the resonant radiation for the atom. Any questions about that? So the goal for the first hour today is to go from two atoms to many, many atoms, which then form a metal plate, and make the transition from forces between neutral atoms to forces between two metal plates, which is the famous Casimir force. And I want you to sort of hold the thought about what is really the quantum aspect here? Is it the vacuum? Is it the zero point fluctuations of the electromagnetic field? Or are the atoms the zero point fluctuations of the atomic oscillators responsible for those forces? So we'll come back to it after we have discussed the Casimir interaction between two neutral plates. So maybe one highlight for you today is that maybe in an hour you have should at least have an expert opinion whether zero point fluctuations of the electromagnetic field are real or just a convenient way to describe some physics. But let's get there. So the way how I want to make the transition from two neutral atoms to two metal plates is by reminding you that the potential was proportional at long range to 1 over R to the 6, at short range at 1 over R to the 6. And if you had the model that the long range interaction comes, because the vacuum fluctuations drive dipole moments in the two atoms, then we know that the potential is proportional to the polarizability of atom one times the polarizability of atom two. So let's now extend that to the interaction between an atom and a wall. Well since I don't know-- at least not yet-- how to describe the wall. I take the position that, well, we have an atom come close to a wall, but now I extend the wall to a sphere of radius z. And it's not obvious that if an atom interacts with the wall, if the wall were flat, it is of course this part of the wall which is most important. So we should actually get a quantitative or semi quantitative result out of that. But now I can use a nice result from electrostatics, which I also discuss in 8421, in the [? MO1 ?] course, namely when we have a conducting sphere of radius c. Do you know what the polarizability of a sphere is? You must have also had this problem in classical electrodynamics. You apply an electric field. You calculate with the boundary condition of the sphere some spherical harmonic function. You calculate what the dipole moment is, and you find the proportionality constant. AUDIENCE: It's like z squared or [INAUDIBLE]. PROFESSOR: Yes. The one thing I want is that it's proportional to z cube. It's the volume of the sphere. And the reason why I discuss that in atomic physics is when we talk about the dipole moment at the polarizability of the hydrogen atom, we find that the polarizability of the hydrogen atom is the pole radius cube. So it seems that when it comes to electric fields and polarizability, an atom, a hydrogen atom behaves exactly like a conducting sphere, and the size of the sphere is now the size of the atom. But that means now that the long range potential will now be proportional the polarizability of the atom. And we lose three powers in the power law, because we have three powers of z in the polarizability of the sphere. OK, if that worked so well once, we can now use that at least as a bridge to discuss what happens when we have two metal walls. Well, we can now say we have two spheres, and the polarizability of each sphere is proportional to z cubed. Well when we have two walls, we want to use the potential per area. So what we had is we had 1 over R to the 7. We lose three powers of R or z, because of one sphere. Another three powers of the other sphere, so it's 1 over z. But when we normalize by area, it becomes 1 over z squared. And now we find that the potential is 1 over c cubed. And at least with it I have-- I wouldn't say derived, but motivated for you the famous Casimir potential, which is 1 over c cubed. And I want to give you an exact derivation of the Casimir potential in the next few minutes. But before I do that-- yes? AUDIENCE: Why did you choose the spheres to be the same size as the distance between them. That seems kind of-- PROFESSOR: It's all just a trick. [LAUGHTER] You know what would happen is-- let me just go back here. If I would choose this sphere smaller, I would not have given the atom the whole exposure. I have to make it at least larger than c to make sure that the curvature of the sphere doesn't dominate and what the atom sees is it sees there is a wall. But then you can see when I make this sphere larger than c, then I delocalize the dipole moment. And the bigger sphere has a big polarizability, but the atom which is so close to the sphere doesn't care about it. So you would say the right choice is just when you think either picture gets corrupted. But that's often what it is. So if you think about it, it's the only choice. Both other choices would lead to an unphysical result, and you would immediately realize why. In one case it's the curvature, and the other case you're not really sampling the dipole of the sphere because you want to do a dipole field, and if you have a dipole of size z, the field with the dipole starts to be an approximation if you go distance z away. Other questions? Yes? AUDIENCE: Can you explain where the first term [INAUDIBLE] z to the [INAUDIBLE] come from? PROFESSOR: We had it. Last class we had the result that the Casimir-- that the van der Waal's-- the retarded van der Waal's potential is 1 over R to the 7. But with the physical picture in mind that the van der Waal's potential at long range, which is 1 over R to the 7 comes because a vacuum fluctuation drive to dipoles, and the two dipoles attract each other. When each dipole is alpha 1 times a vacuum field, the other one is alpha 2 times a vacuum field. But now for this sphere, alpha is z cubed. So therefore three powers in the denominator cancel, and I get z to the 4. So you can say the alpha 2, I absorbed-- I used the kind of result that alpha 2 is z cubed, and then the alpha 2 cancelled with three powers of the distance in the denominator. Other questions? I have a question for you now. If you have two metal plates, what about short range and long range? We expect something under the rock. If we assume a metal plate, what is the vessel-- we had the distinction between long range and short range, which came from the resonant radiation from the resonant excitation of the atom. And when we are closer than the wave lengths, we're in short range, if you're further than the wavelengths of the resonant transition, we're in long range. Between an atom and a plate, well, we still have the resonance of the atom. But what happens when we have two plates? There's clearly no resonance, so therefore there is only one power law, 1 over z cubed. But I want you to sort of think about it for a moment whether this is now-- whether everything is long range or everything is short range. In other words, you say effective resonant radiation, you say effective wavelengths of the plate lambda equals 0 or lambda equals infinity. Any opinion? AUDIENCE: [INAUDIBLE] ability, so it must be infinite [INAUDIBLE]. Take the two metal plates, the vacuum energy between those plates, and compare it to vacuum energy at the same volume free space and subtract the [INAUDIBLE] energy difference. And high frequency components cancel out, so you have the-- PROFESSOR: OK, we'll actually do exactly what you suggest, to compare vacuum energies, in the next few minutes. It's not quite obvious, but I want to tell you now we have made-- by assuming the metallic boundary condition, by assuming that we have idealized metal plates which fulfill the boundary condition that the electric field is 0 at the surface. The parallel component of the [INAUDIBLE] at the surface. We've made an assumption here, and the assumption is actually that the resonant wavelengths goes to 0, and resonance frequency goes to infinity. And the power law we obtain is actually the long distance power law so to speak, and for two metal plates, there is no short distance. I'm not sure if I can give you a simple explanation because the boundary condition is so idealized that you don't recognize, you don't smell any atomic resonance anymore. But maybe the one thing I would say is that you say the boundary condition, metallic boundary condition for infinitely high frequency. And sometimes that would also mean that the plasma frequency of the metal is pushed to infinity. We come back to that in the next chapter. Any questions? AUDIENCE: [INAUDIBLE]? AUDIENCE: Excellent question. I will go for you now through-- its mainly classical e and m. This is why I use pre-written slides. We sum up over all modes with a boundary condition, with a metallic boundary condition. If you have in general a dielectric or semiconductor which is not an ideal metal, has different boundary conditions, the Casimir force will change. And this is actually a pretty outstanding difficult problem where people have really chewed on hard. I think Lifschitz has some famous solution on it. What is the Casimir force for an arbitrary dielectric constant of the surface material? But absolutely yes. If you don't have a metallic boundary condition, your Casimir force changes. I think this is sort of an interesting point, and it may trigger immediately discussions what does it really mean. But let me go now through school the derivation where we have metallic boundary condition, and this actually leads to a quantitative derivation of the Casimir force. Quantitative in the way that we get-- we're not just getting the power law. We get the prefector. We everything 100% of this derivation. Collin? AUDIENCE: So what is the mechanism for the Casimir force changing? Because if think of it in the picture of two metal plates, the lower density and modes inside and then do outside. Even with the arbitrary boundary condition, I [INAUDIBLE]. But what is the physical mechanism for changing the form? PROFESSOR: To the best of my knowledge, it is the density of modes. But it's very subtle. I suggest, since not everybody is probably familiar with the way how I will count all the modes in the metal plate, let me just do that, and then we should-- and I hope then it's obvious, or we should then discuss the boundary conditions being modified. And you have a question? AUDIENCE: Yes, but maybe we can discuss it after this. PROFESSOR: So the idea is the following, if the Casimir force is only due to zero point fluctuations of electromagnetic field-- let me say up front this is a valid assumption. I get a valid result. But I tell you in half an hour that you can take another perspective and say this is just a holistic description. There are other derivations of the Casimir force which do not use the concept of zero point energy at all. So in other words, I'm presenting you one presentation of the problem, one solution of the problem, but there's a whole different formulation. But here I assume-- mean I can easily modulate. OK, well here we have two metal plates. We want to understand if there is potential energy and interaction between the two metal plates. And let's just calculate what is the total energy of the system. And if I have two metal plates, which are two mathematical boundary conditions, all I have to calculate is the ground state of the electromagnetic field in between. And that means I have to find all of the modes, and each mode contribute h bar omega over 2. So this is our agenda now. We say the metal plates are not real objects. They're mathematical boundary conditions. Looks like a good assumption. And now based on those boundary conditions we calculate the zero point energy of the world, at least of the electromagnetic world. And then we ask if the plates are a little bit closer, tool has the total zero point energy changed? And when it has changed that means there is a force between the atoms because of potential changes as a function of distance. So this is the agenda I went to execute now. Any questions about what we want to do? As you know when you write out modes in e and m and you want to do it exactly the many, many indices. But this is a result of which could actually come straight out of Jackson. For your convenience, this is also discussed-- a wonderful discussion by Serge Haroche. In some these summer school notes from the '70s or '80s. I've posted those on the web. So if there is a mathematical detail you want to look up, this is posted on the web. So anyway, we have two metal plates with a distance l. And we have metallic boundary conditions. That means that when we calculate the modes, we get TE modes and TM modes. TE modes means there is no longitudinal electric field. TM modes mean there is no longitudinal magnetic field. And the frequency of each mode is discrete in the z direction. In the z direction we have standing waves, which are labelled with an intentional value of m, whereas parallel to the plate the modes can have an arbitrary transverse wave vector k. And when we look at the dispersion relation, the frequency of those modes is in quadrature. You can see the component k squared c squared of the transverse propagation plus the component of the standing wave. So for each m, for each integer value, we have a TE and TM mode, with the exception of m equals 0. For m equals 0 we don't have a TE mode. We only have a TM mode. So that's just fresh out of Jackson. But it's very important now-- and I want to spend a little bit more time with it so it's just a purely classic result. What is the density of states? Well, let's assume we have a given m. So we have a standing wave perpendicular to the capacitor plates, and therefore now the density of modes only comes-- the density of modes only comes from the transverse. Let me just get my toolbar back. So it is from this k squared. And you know that the number of modes is proportional to the volume in k space. And the important part is that I can write this as dk squared. But k squared is d omega squared. And this gives me omega d omega. So for each n value, for each discrete mode perpendicular to the plates, we have now a density of modes which goes as omega d omega. Before I show you the mathematical expression, let me just sort of plot that, what that means. The density of modes was-- the number of modes was omega d omega. That means the density of modes was proportional with omega. That means for m equals 0 we started omega equals 0, and we get a straight line. For m equals 1-- m equals 1 shown in green, and m equals 2 shown in purple. The frequency omega-- we have omega squared is sick the quadrature sum k squared and m squared. So therefore if you want to ask what is the density for modes for m equals 1 or me equals 2, you have to start at the lowest frequency for m equals 1, at the lowest frequency for m equals 2. But then as we just derived, the density of modes is proportional to omega. So what you we to do in the end, we want to sum those modes all up. And what we will observe is that here we have a line with slope 1. Here we have to sum two lines with slope 1. Here we have to sum three lines with slope 1. So progressively, the slope gets higher and higher. But the slope, which increases linearly with the coordinate axis. If it were continuously, would just be a parabola. So in other words, if I would [INAUDIBLE] the density of states which comes from the exact mode spectrum, if I would [INAUDIBLE], it would be parabola. And parabolic density of state, the number of modes is omega squared d omega. This is three dimensional space. So in other words, what we observe is what exactly the effect of the boundary condition is. Instead of getting this smooth parabola omega squared, we have sort of broken up the parabola into piecewise linear pieces. And the fact that we have a boundary condition, that we plates in free space, and not just the space, is actually the difference between the two. So it is the difference between the two, which leads to the fact that a boundary condition modifies the vacuum energy. Questions? OK with that the rest is mathematics, but I want to show you the steps because it is ingenious mathematics. It makes a few approximations. It creates a few infinities and gets rid of it. It's actually rather wild, but amazing that, in the end, we have a simple exact result. OK so the most important thing is we have a density of states. And the density of states for one m is proportional to omega, but now we have to multiply-- when we want to know what is the density of state in a given omega, we have to multiply with the integer number of discrete modes which are available at this omega. So this is because there is only one m equals 0 node. We have a one here. And then for m equals 1 and higher, we have two [INAUDIBLE] between TE and TM. So this is just counting the modes. And this is now a mathematical trick that, instead of counting the modes-- it's an integer-- we can sum up heavy side step functions and sum over all m. This is just an [? exact v, ?] right? But this is something which helps us in the next step to do an integral. So now so far we haven't made any approximation. We've just reminded ourselves what is the exact result of classical e and m. OK, but now we want to calculate the zero point. So now it's getting interesting, because we take the classical result, and take it into quantum physics. The first step is very innocent. We say each mode contributes h bar omega over 2. And then we integrate with the density of states. And the next line is also an exact result, just using the density of modes from the previous page. But of course, now we are asking for trouble. We are integrating with omega squared d omega, and there is uv catastrophe. Now I can say I can put on my mathematical head and say let's add a convergent term, something which exponentially decreases with omega, which is just cutting off the high frequency part. And I hope that in the end-- and this will come out too-- that I get a result which is independent of the cutoff. So by choosing this cut of parameter lambda to zero, I have-- if this is zero, I have no cut off. So maybe I can introduce a cut off, and then at the end let lambda go to got to zero. And if this gives me a physical result, I've sort of mathematically tricked around. But I also like the physical justification for that. I mean you know that a metal plate tennis has not metallic boundary conditions at infinitely high frequency. If you go deep to the x rays and gamma rays, it becomes transparent. So therefore, we would argue that, once we go to very, very high frequencies, the modes do not feel that there is a boundary condition, and therefore we may be allowed to do some physical cut off. OK so we do this cut off by-- so now we have to find the zero point energy. The zero point energy has a sum over m. And I call each integral I sub n. And this integral is the omega square from the density of states, but now multiplied with a convergence term. Whether it's mathematical or physical. And by introducing this cut off, I can now exactly solve the integral. The omega squared can be taken out of the integral by taking the second derivative. And this is just an exponential function which can be integrated. And therefore we have a mathematically exact result. So we are now able to do the integral because we have avoided the divergence. OK. Now we come back and say we are really interested if the cutoff is pushed to higher and higher frequency. And this means lambda equals 0. So therefore, we want to take this expression and expand it around lambda equals 0. Well, you could say that's physically very well motivated. But if I take this function, which is lambda e to the lambda over x times e to the x, and expand it, I get divergence is 1 over x squared. So you would say how can you expand it. Let's see what happens. Let's just do it. Let's say we are interested around lambda or x equals 0. We have some divergencies now, but we have to find a way to deal with it. So we get rid of some of the divergencies in the following way. By saying, well, if you simply calculate the zero point energy between those two capacitor plates, and then when, let's say, if there is-- if the plates come closer and there is less zero point energy, we would see the plates attract each other. But what happens is the plates are not moving, and there is no world behind them. The world is sort of as it is. And that means we have to sort of now look more carefully at the zero point energy of our capacitor plates where l is small. And the rest of the world. So we are now representing the world as a big capacitor of size l 0. And then our plate is just moving within that, because we are not assuming that then the two capacitor plates attract each other and move towards each other that the size of the universe is changing. So we want to move a boundary condition within the universe. So therefore, the correct expression for the zero point energy is not the zero point energy of capacitor l. It's the zero point energy of a capacitor size l plus of a capacitor l0 naught minus l. And so we take the result and simply add the expressions for l and l0 minus l. And what is now a nice is that the first highly divergent term with 1 over lambda to the 4-- reminder that we want to let lambda go to zero. It becomes independent of l. So we would say, well, if there is an infinity, it doesn't change with l. It doesn't provide a potential. It's just one of those infinities which is constant, which is not affecting the physics. Also here, this is independent of l. And then the next term gives us the famous Casimir potential, 1 over l cubed, and it is independent of lambda. So we have the result which was independent of the cut off we introduced earlier. So therefore, we have derived, with a little bit of hand waving to navigate about infinities, that the potential between two metal plates is 1 over l cubed. a squared was the area, the size of the capacitor plates. And this is the exact prefector. Or if you want to figure out what is the pressure off the vacuum. Pressure is, of course, force per unit area. So you take the derivative of the potential that's the force, that's the force per unit area, and this is now the vacuum pressure. Sort of funny now you can put in the unit. It's 10 to the minus 5 millibar at a distance of one micrometer. So anyway this is, I think, the classic deviation of the Casimir potential by, I would say, exactly summing up the zero point energy within a certain boundary condition. Good. Maybe to address Collin's question. What would happen if the metal plates were maybe semiconductors or had a different dielectrical constant. The mode spectrum would change. AUDIENCE: Like if you had mixed boundary conditions. You don't have these nicely spaced-- PROFESSOR: I don't know how to calculate. That's a hard problem, and a lot of people have really worked very, very hard on it. It seems for idealized boundary condition you can do it easily, but it has been clearly a challenge to mathematical physics. And I'm not even sure if it has been solved in all generality or only in special examples to extend that to every type of surfaces. But we can, for instance, assume if you have a conductor, which is not-- conductor which has not infinite conductance, 0 resistivity, then the electric field can penetrate a little bit into the conductor. And if the electric field penetrates into it, it changes the frequency of the mode. So it's clear that the boundary conditions have an effect here. But let's now have some discussion. We had focused before on the force between two atoms. And if we use a microscopic model for the metal plates, it should be, it must be possible to obtain the Casimir potential by just summing up all the forces between atoms. Because ultimately if you have no metal plates, you don't have any force. Where s-- in the last few minutes, we focused on the derivation which was simply summing up the zero point energies. And there are two papers which I've posted on the wiki which argue that both results are equivalent. So I would really recommend to you-- he expresses it much better than I can. Read the introduction and the conclusion from Bob Jaffe's paper. He clearly says in the introduction that there are a lot of books and references by famous people who have said the Casimir force is the manifestation of the zero point energy. The fact that people have observed the Casimir force means that h bar omega over 2 is real. Those authors take the opposite approach. They say this is just heuristic. It's convenient. You can get the correct result. But there is an equivalent derivation which gives exactly the same result, which focuses only on pair wise interactions between atoms, and it does not include any-- it doesn't make any reference to the zero point energy of the electromagnetic field. So therefore, one, I would agree with those authors that the conclusion is there is absolutely no observational evidence for the zero point energy, that the zero point energy is sort of not just convenient counting, but that it is real energy. And the Casimir force, of course, also it has often been quoted as a counter example that this is a direct observation, is not-- should not be regarded as such. Now people at MIT probably know Bob Jaffe. He's a high energy physicist. And why is he interested in atomic physics? Why is he interested in the Casimir force? Well, the question of zero point energies is very, very important in these days for dark energy. I mean, you know that 80% of the energy of the universe is dark energy. Then there's dark matter, and then there is baryonic matter. But this dark energy is really mysterious. And one candidate for dark energy is the dark energy simply comes from the zero point energy. So therefore, people are now really interested is the dark energy real. And if it is real, does it have a gravitational effect? Does it change the metric of space? And can it accelerate the expansion of the universe? Of course, you have a problem if you calculate the zero point energy, you don't introduce a cut off. You have a spectacular uv divergence. You can maybe now concatenate the divergence by saying, OK, eventually if the wavelengths become shorter and shorter and shorter, there is a cut off. And if you have no idea at all, you would introduce a cut off at the Planck lengths, because beyond the Planck lengths all physical descriptions break down. But if you cut it off at the Planck lengths, you find that the zero point energy of all modes is larger than the observed dark energy in the universe for acceleration of the cosmic expansion by 124 orders of magnitude. And if you think you want to introduce a cut off which is maybe the classical electron radius. The classical electron radius is the radius of a charged sphere for which the electrostatic energy would just be the rest energy of the electron. Well, then you're off by 43 orders of magnitude. So some people have called those discrepancies, which are completely unresolved as of now, the biggest failure ever of theoretical physics. So anyway, that's why there is renewed interest in the Casimir force. And at least in electromagnetics, we have a very, very simple model. And we have experiments which measure the Casimir force, but, as I've said, it cannot be regarded as a direct observation of the zero point energy. OK, let me now come to a question we started discussing earlier, namely if we-- here we have atoms. Here we have a metalized boundary condition. So maybe let's just talk for one minute about it. What kind of assumptions do we make about atoms when we replace them by a metallic boundary condition? From the whole derivation of the Casimir force it's not obvious that there are no atomic properties which have entered the calculation of the Casimir potential. What Bob Jaffe argues in this paper-- and again, it's wonderful reading. I really recommend Just if you want to enjoy some physics, open up the document and read the first and the last page. The Casimir force from an idealized boundary condition is completely independent of atomic properties. So therefore, it must be a limit where-- well, we talked about the resonant wavelengths, but we can also talk about the fine structure constant, because the fine structure constant in a hydrogenic model provides what that resident transitions in hydrogen are. So you can argue if something is completely independent of atomic properties, somehow by sweeping things under the rug we must have set alpha to either zero or infinite. Because if it is set to any value, it would correspond to some version of the hydrogen atom which has a resonant radiation. So the question is, is alpha set to zero or is alpha set to infinity when you derive the Casimir force? And I think this is now easy to discuss, because the fine structure constant alpha is e squared over h bar c. If you set alpha to zero, you do it by setting the charge of the atoms to zero, and then you have no interaction at all. So therefore, if you want to have the Casimir force in a limit which is independent of atomic properties, the alpha equals zero limit gives no force. So therefore, the equivalence this picture, summing up over the electrons and obtaining the Casimir force for the metallic boundary condition, this equivalence happens when you do this derivation and set the fine structure constant to infinity. How big is the fine structure constant? 1 over 137. In some expansions of alpha and alpha squared, it's actually a smaller parameter. And now I'm telling you when you have two metal plates and measure the Casimir force, this corresponds to the infinite alpha limit. So this is now getting complicated and I can refer here only to Bob Jaffe's paper. But he shows that when you go to distances of 0.5 micron that people have measured the Casimir force. You are already the alpha equals infinity limit. If you are-- I wish I could read my handwriting. I think that's minus 5. If alpha is larger than 10 to the minus 5. So therefor, alpha equals 1 over 137 is for that purpose already very, very well in the large alpha limit. But I just hate this result here. This is rather complicated. Bob Jaffe says that to get the Casimir potential by summing up over atoms is actually a quite challenging mathematical exercise. Questions? Yes? AUDIENCE: If these calculations of summing over atoms has been [INAUDIBLE], then what is it-- we can just talk about the plates being dielectric from that atom [INAUDIBLE]? PROFESSOR: There are solutions for dielectric plates. And sometimes when people do experiments, the best surfaces you can get is a silicon surface. Because they want you to do qualitative measurements, they're discussing corrections due to the dielectric constant. I know the problem has been worked out for dielectric surfaces. I assume not by doing the transition from summing up atom by atom, but by rather looking at the boundary condition for electromagnetic fields in the presence of a dielectric. But I don't know more details about that. Yes? AUDIENCE: This question may not make sense because I haven't thought [INAUDIBLE]. But again, when we consider two plates, if you have room for two atoms, we get the two 1 over R to the 6, 1 over R to the 7 cases. Once of which is the retarded Casimir potential, but for two plates-- again, we conclude that it was more like the short range-- PROFESSOR: Well, long range. We are always retarded. AUDIENCE: OK, but then I'm a bit confused. Because then if you say alpha goes to infinity, don't you have an h bar or c to zero? Oh wait. That makes sense. PROFESSOR: Alpha infinity is-- we know that this is the vastness of the electron. The [? Whitbeck ?] constant is alpha squared times smaller. And the [? Whitbeck ?] constant-- well, a quarter of the [? Whitbeck ?] constant is alignment alpha radiation. So anyway, the [? Whitbeck ?] constant is a scale factor in the transition frequencies of the hydrogen atom. And so therefore, it is consistent that, if you make alpha large, we actually get-- we push the frequency of the atomic transition to very, very high frequencies. And that would mean the resonant wavelengths goes to zero. And therefore any finite distance between the capacitor plate is, in the long range, is longer than the wavelengths of the transition. AUDIENCE: I think equivalently you could say if the speed of light is really slow, you always will have [INAUDIBLE]. PROFESSOR: Yeah, sure. You can do the transition in many ways by tuning the speed of light, or tuning the charge, but it all has the same implication. Other questions? Lindsay? AUDIENCE: Is there any other observable reason why we should constantly be thinking about Casimir forces and dark energy? Like experimenting [INAUDIBLE]. PROFESSOR: As far as I know, this question about dark energy and whether it is related or not to zero point energy of the field is completely open. I think if you had an idea how to do a decisive-- if any of you had an idea how to make a decisive experiment measuring whether zero point energies are real or not, that would have a big impact. As far as I know, it's completely open. Nobody has really good idea. So some people feel, mathematically, the zero point energy has the right, correct physically characteristics. Zero point energy could explain the extra term in Einstein's field equation, which is often called the cosmological constant term. But the quantity, it's so many orders of magnitude off that people don't even know if that is correct or not. One could also say, well, maybe zero point energies do not give rise at all to any gravitation potential. And therefore maybe if energy doesn't give rise to gravitational potential, maybe the energy is not real. And what we find, what is responsible for the expansion of the universe is gravity-- so Einstein's-- what's the constant called again? I just mentioned it. AUDIENCE: Cosmological. PROFESSOR: Cosmological. So that is just the cosmological constant, but the cosmological constant reflects something else in the universe and is not related at all to zero point energy. To the best of my knowledge, these are the possibilities. OK, so let's go back to atoms. Our next big chapter 12 is on resonant scattering. Just that you see the structure of the course. I wanted to introduce sort of some mysteries and subtleties of electromagnetic interactions to you by using diagrams, by using an exact formulation of QED. And then I gave you one example where we could use those diagrams, and these were van der Waal's forces. So we got a little bit side tracked with the vacuum, with van der Waal's and Casimir forces, but now we want to go back to the diagrams. And there is one aspect of the diagrams which is sort of fascinating, and this is where we really need a treatment which goes beyond perturbation theory. And this is when we have resonant radiation. When resonant light interacts with atoms. Because, as I'm going to show you, we have then in any perturbative expression zero in the energy denominator, and we would have divergence. So I want to show you now, partially also for the beauty of the physical theory, how can we deal-- how can we fix those divergencies? How can we fix those infinities we encounter in perturbation theory? The answer will be we get rid of these infinities by summing up an infinite number of diagrams. So I'm going to show you now-- it actually sounds so much more harder than it is. How, with rather modest mathematical effort, we can perform an infinite sum over diagrams. And I want to show you what those diagram are. And the result of that is that the divergencies disappear, and we have a valid description of what happens when atoms encounter resonant radiation. So I'm following here, almost religiously, the atom photon interaction book. So this in production is a summary of those pages. So when we discussed diagrams, we wanted to figure out what is the amplitude for a system to go from an initial to a final state. And those were the matrix elements of the t matrix, the transition matrix. And if we do it in second order-- well, remember the structure is in second order. We sort of emit a photon, virtual or real. We have another photon. This gives us second order. And in between the system propagates with its eigen energy. So this was the structure of the diagrams we had. And this term here, let's expand it in one wave functions of the unperturbed Hamiltonian. And then we obtain this following structure. So this is just rewriting in a very suggestive way what we had discussed before. And the case I focus on now is the interaction of atoms with resonant light, because this is when we have divergencies. If we have state a, state b. We have a laser and we scatter light. If the frequency is approximately resonant, then those perturbative results have a divergency. So for resonant light scattering, the typical matrix element we are interested in, we start and end of course in the ground state. We scatter photon. We go up to an excited state and go back to the ground state. But we started initially with a photon of wave vector k and polarization epsilon. And this may be different for the scattered light. The interaction happens by the interaction operator, which, for the purpose of this discussion, can either be the A dot p term or the d dot E term. We've discussed about the special role of the a squared term. We discussed it earlier. I don't want to include it in the discussion right now. And then you would say-- just want to make sure they do not confuse. In second order perturbation theory, you always have an energy denominator. But as I also explained to you, the energy denominator comes from the free propagator. You had a dot propagate and a dot-- and what is the time evolution of this free propagation when you integrate it with time? This gave rise to the energy denominator. Just a side remark. So you should be very familiar that we have an energy denominator. But now this energy denominator-- the initial energy is the initial atomic state plus the photon. And then if you want to sum over all states, I can just put in here the operator H0. But you may immediately think about it, the only state which really matters is the state b. And what matters is only when we go through intermediate state b, then H0 is Eb, and then we have a divergence because the denominator is 0. So let's see how we can fix it. But let me first give you the perturbative result, which you have seen many, many times. It is the product of two matrix elements. We have an intermediate state. The excited state without photon. We started with a photon, and we end up with a photon. And here we have our dipole or A dot p term. And our energy denominator diverges when the frequency of the photon approaches the resonance frequency. Well, how can we fix it? You know that in many cases you're just adding an imaginary part to it, which reflects that, due to the coupling with the vacuum of the electromagnetic field, the excited state is broadened. And it gives its energy an imaginary part. This of course is very phenomenological, but at least technically it avoids the divergence. Now what I want to show you, in the next 15 minutes and we continue on Wednesday, is how I can get this result of a very rigorous approach. I want to use the diagrammatic approach. I want to sum an infinite number of diagrams. And will tell you what are the approximations to get this. For instance, as you know, when you have an imaginary part, when you have just an imaginary part of the energy, the physics of it is exponential decay. Well I hope, not today but on Wednesday, you realize what assumptions lead to exponential decay, and that, in principle, non exponential decay is rather the standard than the exception. But the effects are small. But I want to show you. So in some sense I want to give-- we have to go through a half an hour of work to derive this result. So you'd say, what's about it? You just get the phenomenological result. Well, we have put it on a rigorous basis. And number two is we do understand what assumptions really go into this phenomenological result. So the way how we will actually obtain this result is that in-- the second order diagram is that you start in state a, you go in state b, and you're back in state a, and you exchange photons. So the way how we-- the physical picture we want to use now is after the photon is absorbed right here, the system is not propagating with the Hamiltonian H0. It will propagate with the Hamiltonian H. If we do that, then we avoid the divergence. We get a result which can be applied toward-- the formalism can be applied to many situations. And in one limiting case, we obtain that. So the physics behind it is the following. The divergence which we encounter in resonant physics is only a divergence because we assume the atom in state b is naked. It's an excited state of an isolated atom. But in reality, the state b interacts by virtual photon emission, interacts with all the modes of the vacuum. And once we throw that in by saying that between two-- between absorbing and emitting a photon, the state b propagates not with a free Hamiltonian, but with the total Hamiltonian, we avoid the divergence. But this is no longer a perturbative result. And let me illustrate that to you. If you take the energy denominator, omega minus and omega 0, and we add the imaginary part to it, I can Taylor expand it, assuming the imaginary part is small. Well, that's sort of a trivial expansion, but now you should know that gamma, the rate of spontaneous emission, the simplest expression is obtained in [INAUDIBLE] is called null in the second order in the interaction Hamiltonian. Spontaneous emission is proportional to the atomic dipole matrix element squared. So it's second order in the Hamiltonian. But that means that this harmless addition of an imaginary part means that if you look at the Taylor expansion, that we are using now infinite orders. Infinite orders in this expansion. So the phenomenological addition of [INAUDIBLE] gamma to the excited level b is actually highly nontrivial, because it's corresponds technically or physically to a non perturbative result which involves infinite order in the interaction matrix element. Question about it? So this is more setting up the agenda. I wanted to remind you of the perturbative result. I want to sort of show you how we can fix it. I want to show you what it means. But with that motivation I want to now back to our diagrammatic approach, our formal solution, our exact formal solution of the time evolution of a [INAUDIBLE] system, and show you one how I can implement this agenda in particular by not allowing the atom to propagate with a naked Hamiltonian H0, but sort of staying dressed with photons and propagating with the Hamiltonian H. Questions about that? OK, so the derivation is now focusing on two chapters in API. I need one result which I hope you remember from about a week and a half or two weeks ago. We formulated the time evolution of the system using the time propagation operator u. And we found an exact expression. I'm dropping all of the arguments here, t and t prime. You can look them up either in the book or in the previous lecture notes. We found that the propagator in the interaction representation is the unperturbed operator u0. Now I need, actually, the terms. It was a time integral over some time t1, which involved u0, v, and u. And I introduced it to you as you could plug that into whatever is the equivalent of Schrodinger's equation for the time evolution operator, and you really find this is a solution for u. But well, it's not really a solution, because it's a solution of u expressed in terms of u. But then I showed you that it allows an iterative approach, and this is why we did it. If you have a first order result, the zero order result is u0. When you plug the zero order result in here, you get the first order correction. When you put the first order result here, you get the second order correction. So you get a recursive formula which solves the problem. And this is now our starting point. This is an exact formulation of QED. And this is now our starting point to address the problem of divergencies. We want to describe what many of you do in the lab, namely have resonant laser light interact with atoms. And I want to show you how can you describe it with QED with the formalism we've introduced. OK, now for this chapter we want to have one big simplification. Things aren't only getting complicated. They're getting simpler. If you have a time integral which is a convolution, and you can often simplify it by Fourier transforming it, because the Fourier transform of a convolution is simply the product of the two Fourier transforms. There is one glitch. So one subtlety I want to feed in a moment. But let's just assume for a moment, we go from time before the Fourier transform to frequency or energy. So G of E is the Fourier transform of-- after filling a few corrections of u. That would mean, if that were true, the equation above would now be an equation where the integral, V sort of constant, instead of a convolution of U0 and U, we get simply the product of the two Fourier transforms. That's much, much simpler. It's an algebraic equation instead of an integral equation. And remember, when we get into nth order, we get n integrals with some time ordering. We were able to do it, but it's mathematically more complicated. The problem is that this is not a convolution. It would be a convolution if the time limits were infinite. So what we can do instead is we can define G of E. Not the Fourier transform of U. If we say it is the Fourier transform of k, which uses a step function, then if you insert now k in the formula above, you see that because of this step function we can now take the integral limits from minus infinity to plus infinity. So therefore, when I say G, G is called the resolvent. It's really the mathematical quantity we have to work on now. And I wanted to tell you very simply hey, it's just a Fourier transform of the time evolution operator. Not quite. It's the Fourier transform of a time evolution operator with a theta function, just to make sure that formally we can extend the upper and lower integration limits to infinity. Or this is also a name-- what I'm really kind of discussing are Laplace transformations. But I want to convey to you the physics. It would take many weeks to do the mathematical accurately, but you can get the idea and really profound feeling for how QED works and what we are doing with not so much mathematical rigor. So with that grain of salt we have Fourier transformed equation. We have the Fourier transform of the time evolution operator. But if you look at it mathematically more exact, it's a Laplace transform, and we are little bit navigating in the complex place. The Fourier transform does not exist for real values. We always have to add a small imaginary part to make things converge. But you've seen that in many other places, I assume. OK, so the quantity which allows us to describe resonant interaction in a very straightforward way is called the resolvent. I can simply define the resolvent. All I gave you was some motivation. But if I had wanted, I could just say the resolvent is defined by this operator equation. z is the complex number. H is the Hamiltonian including interaction between atoms in electromagnetic field. And actually if you would take that-- if you take this and plug it into the equation above-- so if you plug this into the equation, you find this expression solves the equation. So I just want to say you can forget about all the subtleties with Laplace transform and such. You can just say let me define some interesting object. But it is connected to what I said earlier, because this solves this equation. And also I like to sort of show you the mathematical structure. If you write the time evolution operator as e to the minus the Hamiltonian, and now you Fourier transform it. If you Fourier transform it, you multiply it with e to the I omega t, and you integrate-- let me integrate from 0 to infinity. There are some issues with 0 or infinities. But if you simply solve this integral with time, you have sort of an exponent e to the i omega minus h. So if you Fourier transform it, what you obtain is nothing else than the energy minus h. And this looks very different. This looks very, very similar to the definition of the resolvent. The only mathematical thing I'm not fully discussing here is z real, or has it a small imaginary part? The small imaginary part-- actually, you have to add a small imaginary part here to make sure the Fourier transform exists. But you should clearly see how things are connected. So OK, I'm focusing now on the resolvent. It is equivalent to the time evolution operator. It is a sort of Fourier transform of it. And we want to formally write down a diagrammatic solution for the resolvent. And with that I think I should stop now. We start on Wednesday? Any questions to that point? OK, then see you on Wednesday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

7_Metrology_shot_noise_and_Heisenberg_limit_Part_1.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WOLFGANG KETTERLE: So, good afternoon. Let me start by presenting something I learned today during lunch. I had lunch with three colleagues and we discussed entanglement. And, well, I'm telling you the story a little bit differently, but the question which came up is you have two harmonic oscillators. One harmonic oscillator is in the ground state, one is in the excited state. And then you have the harmonic oscillators where the excitation is here and there. Is that an entangled state or not? It's a real question. I want to think about it. AUDIENCE: What's the physical nature of these oscillators? AUDIENCE: Can you distinguish these? WOLFGANG KETTERLE: OK. Who's working with ion types? There's a few people I know, a few ion types here at MIT. If this is ion-- what is a good ion? [INAUDIBLE]? AUDIENCE: [INAUDIBLE]. WOLFGANG KETTERLE: Strontium plus, OK. So if it's strontium plus, which can be in the quantum state v equals 1 or v equals 0, and you have another strontium plus ion in a second ion [INAUDIBLE]-- and it can be in the [INAUDIBLE] first excited state-- isn't it wonderfully entangled? Two systems far away. You can manipulate them, you have two particles, and then two quantum states. It fulfills with flying colors all the qualities you want to see of an entangled state. But now I can say, if I call this harmonic oscillator, I have two optical fibers. And you know, a single-mode fiber defines a harmonic oscillator, namely the electromagnetic field in the single mode. And if I put one excitation into this fiber, I have a photon here and here I have zero photon. Or here, I have one photon and I have zero photons. So if you look at the fiber, you would say the fiber has two states. It can have an excitation or not. And if you use now that approach, you would say, well, I have an entangled state between two fibers. However, I can realize exactly that state by having a beam splitter, by having one photon coming on the beam splitter, and then the photon is coupled into one of the fibers. So in other words, there is an ambiguity which I want to point out between whether you have a state of one particle-- one photon into two fibers-- or whether you have two fibers which can be in two different states. So there is often an ambiguity, what do you call the state and what do you call the particle? So at our lunch conversation it came up in a somewhat different context, but people would actually say that splitting a photon-- sending one photon across a beam splitter fulfills, if you interpret it in that way, fulfills the definition of entanglement. I know our wiki page and Professor Schwan would probably not agree, but welcome to the frontier of research where different people have different opinions and certain definitions are still being worked out. But there is one thing which is real. We have here two different harmonic oscillators, ions oscillating in a harmonic oscillator potential or photons in a single mode. I remember from 15 years ago, there were papers that show if you put the ion in a cavity or such, you can really transfer the quantum state from the single photon from this harmonic oscillator-- the harmonic oscillator of the electromagnetic field-- to the harmonic oscillator of the ion. So if we want to hold onto the definition, which I'm not sure if you should that this is a single photon, a single particle cannot be entangled, well I know and people reminded me at lunch the there is a protocol to [INAUDIBLE] map the single photon in the two modes after the beam splitter onto an entangled state of ions. So if you focus here on the photon and say a single particle cannot be entangled, you're facing the problem that there is a protocol which would transfer something which you call not entangled to something which is entangled. I hope you enjoy that. It's not clear what is the particle and what is the excitation. There's really ambiguity. Any questions? OK, so that's what part of my lunch was about. I thought since I wanted to remind you of the schedule. In a few hours I will actually be heading to Europe and say there for next week, so therefore no classes next week. We've already made up for one class with the Friday class, and the week later we also have three classes and we've made up for the week. I've just put down here for you the PSET schedule. I kept the Monday due date so there is one PSET every week. We have wonderful material for those PSETs. The due date is now for this week not the day of the class, but you will find a way to get your homework delivered or email it to the TA. Any questions about schedule? I think that has been clearly announced. There was a question in the last class about if you have a density matrix, if any arbitrary density matrix can be regarded as a partial trace of a pure state. I was not immediately clear-- I thought it was, but I wasn't sure. But the proof is so simple that it fits on three lines. You can simply take a density matrix, you can double up the Hilbert space, and define our pure state in a Hilbert space of twice the dimension, and you immediately see that this density matrix is the partial trace of a pure state. So therefore, if you want, you can always say the density matrix is sort of entangled-- you are entitled to the opinion that the density matrix always originated into a pure state, but the entanglement [INAUDIBLE] to the other subsystem have been broken. And I showed you last class that if you are one of the Bell states, the most entangled states in a fully-entangled state between two particles, but you just look at one particle, one particle isn't just a random states with a density matrix which is a unity matrix. Now let's get back to where we left it on Monday. I introduced to you-- and I just want to remind you of that, because [INAUDIBLE] I introduced to you this famous phenomenon of Hong-Ou-Mandel interference which is the following situation. If you have 50/50 beam split and you have two identical photons, you will never get single photons out. The photons are bosons. They want to be together. And so you will always get two photons, but you don't throw on which side of the beam splitter. Now, this is important because we can now use the Hong-Ou-Mandel interferences effect to entangle atoms. So the element we will take from this beam splitter is the following. We will have atoms emit photons. But then we create entanglement probablistically. When these photons have passed through beam splitters-- and I tell you everything about it-- and both detectors make a click, if both detectors make a click you know for sure that the two photons at the input of the beam splitter were not identical. Because if they were identical, only one detector receives light. So we will actually used this home interference to project out photon states where the two photons are not identical because they have difference polarization. Any questions? Yes? AUDIENCE: Does [INAUDIBLE] include the phase. Like if you were in phase lab with those photons, would it be still? WOLFGANG KETTERLE: Frequency very important. Polarization important, timing is important. A single photon has no phase. It's a global phase. And if you have many single photons, your many photon state is a product state of those single photon states. And all of the individual phases-- if you want to hold onto this concept for a second-- just become one big multiplicity phase. Yes, Nancy? AUDIENCE: [INAUDIBLE]? WOLFGANG KETTERLE: Oh, everything [INAUDIBLE]. easy experiment. But I think the beam splitter is probably not-- I mean, beam splitter is a piece of wonderfully polished optics. And in Germany and elsewhere, people have really learned how to do exquisite optics. I don't think you're limited by optics right now. What I would think is difficult, if you have two photons on a beam splitter, they have to be in the same spatial mode. So if you have to fibers and the fibers are not fully aligned, or you have another mode which is [INAUDIBLE], or if you have a lens which is distorting your Gaussian mode, you mix other modes in so all this creates non-distinguishability. But I think equality of optics is probably the least of your concerns. So actually for today, because I'm out of town next week, I want you to have something to think about it, I already pre-wrote some of the slides. So I can go a little bit faster. Give me feedback. If you think I'm going too fast, I will not do it again. But I felt I can probably have a reasonable speed by making annotations to what I've already prepared. So the situation is the following, we want to entangle two atoms. We have two atoms which are both in the excited state in the scatter light. And they can go to two different ground states. Think about two different type of ion states. I call them U and D, Up and Down. And when because of selection rules, one state can be reached with one polarization. The other state can be reached with the other polarization. And I call this polarization H and V, horizontal and vertical. In reality, this will be circular polarization, so you may want to transform from linear to circular after the fact. But for just the discussion right now, I assume that horizontal polarization leads to one state, vertical polarization leads to the other state. So therefore, if you have two atoms-- and Chris Monroe in Michigan did the experiment where he really had two different vacuum chambers with two different ions, they both emitted photons, and then the photons came together on a beam splitter. So that's what you should sort of have in mind, two ions here and there. They both emit a photon. And after the have both emitted a photon, each ion is in a superposition before you make any measurements. It can be in up with a horizontal photon or it can be in down with a vertical photon. So that is the state of the ion. And so we have two of those. And then you want to detect photons. And the goal now is by using beam splitters and all those tricks, that the detection of photons and the outcome of a measurement will project this product state of two atoms with their photon into a bell state for the atoms. So measurement on photons can take two atoms which were completely remote, had nothing to do with each other, and suddenly they're in a Bell state. And this is done by the probabalistic measurement. And so let's-- so lets develop it. Our initial state is UH plus DV. This is system one. And we have the direct product with our second ion trap apparatus which has done the same. The ion has emitted a photon and we don't know anything about it at this point. And now it is important, if you want to have entanglement you cannot go and measure the polarization, because this would project an individual ion into a state and would not have entangled it. So what we are doing is we are allowing the two photons, one photon here, another photon here, they come together at a beam splitter. Actually, before I do it let me just perform the product here. So what we get is-- well, it's just multiplying it out. UU for the atoms, HH for the photons. UDHV plus DUVH, plus DDVV. So what you want to do is we want to send this beam-- the photons of course, the atoms stay put-- onto a beam splitter. Now, I can do it more rigorously by keeping all the terms until the end of the calculation. But I hope it's rather obvious that we want to have it-- if we want to send it on a beam splitter and then we want to detect that both detectors make click. That requires the photons to be distinguishable. So therefore, those two terms will not contribute because the photons have the same polarization. So we are now using a protective measurement. We want to find out what is the wave function of the atoms. Measurement of one photon in the modes A and B, which are the output modes of our beam splitter. So it needs one more line. The output of the beam splitter is-- so I'm only focusing now on the two terms which can give rise to two clicks at the two different detectors. So I will factor out UD. And now we have the two photons hitting the beam splitter and each photon goes into a 50/50 superposition. So I take know this horizontal photon. And since we don't mess around with the polarization, it's still a horizontal photon, but the photon can be now in mode B or in mode A in one of the output modes of the beam splitter. And I label that as 0,1 plus 1,0. Now for the second photon, it had a vertical polarization, so it's a vertical photon. The second photon is coming onto the beam splitter in the other input port, because we have taken the two photons from two different experiments, from two different experimental setups, and now we combine them so that we'll also be now in a superposition of 0,1 and 1,0. But you know that one mode transforms with a plus sign and the other mode transforms at the beam splitter with a minus sign. So that's how you should look at those terms. And of course, from of this here, from this term we get something very analogous, 0,1 plus 1,0. And for the other spatial mode, the minus sign from the beam splitter, 1,0. OK, now we are done. Now we are detecting photons. So we want to look at this wave function and ask, what happens if in port A-- the output port-- we find one horizontal photon, and in B it must be vertical because the photons have to be distinguishable. So where do we have one photon in mode A? In horizontal it is here. And here it is this product with the minus sign where we have-- no, sorry, the plus sign. Here we have one photon in the vertical, so this is with a plus sign. But we have a second possibility, one photon in mode A horizontal happens here with a minus sign. And one photon in mode B vertical happens here. So in other words, form the term which gives rise to this measurement process, we have a plus UD here, and because of the minus sign, a minus DU. So therefore, if you would detect the photon with the polarization at this point, which we don't do, this polarization gives rise to UD minus DU. We have a second possibility to have a click at each detector, and this is when we reverse the two polarizations. But in this case, you can just add the wave function. You get the atomic state, the sign doesn't matter. So you have to measurements, two possibilities in polarization, but they both have the same outcome. They generate this state, so the result of all that is I've shown you that the two atoms are left now in-- and now I put back in the correct normalization-- in one of the Bell states. So whenever you detect the two single photons in coincidence on both counters, at that moment you know what you have is an entangled state, a Bell state of atoms. Now, the big issue here is efficiency. Your photon detectors are not highly efficient. And I was sort of only looking at combinations where we have this double detection. Most components of the wave function will give the Hong-Ou-Mandel effect. The two photons go in one way. So therefore, you have to prepare your system, have photons emitted, and in most of the cases you will not get the state you want, so you have to keep on trying. And usually those experiments are heavily limited by the very, very dismal success probability of creating the state you want. And when you want to extend it, not just to two atoms entangled, to three atoms and four atoms entangled, then you get a smaller number to the power n for your efficiency. So this probabalistic preparation doesn't scale up, but it is a very simple scheme, it's a very powerful scheme, and it has been used to create teleportation of atomic states and do some tests of Bells inequality and the EPR Paradox. And here is the abstract of a paper. So-- well, this was just 2008, so just a few years ago. And what you see here is that here are one meter apart two [INAUDIBLE] ions. They emit light. The light goes through an optical fiber. And now, after the beam splitter, when you detect one photon you don't know which ion has emitted the photon. So this is exactly the setup which I described. Of course, in an optical fiber you have the two good polarizations are horizontal and vertical in the polarization maintaining fiber. So therefore, the light which originally was emitted in a circular base is-- angular momentum selection rules-- transformed into linearly polarized light using the [INAUDIBLE]. Any questions? Yes, Nikki? AUDIENCE: [INAUDIBLE] create the initial state [INAUDIBLE] atom [INAUDIBLE] one state [INAUDIBLE] make sure that the ions state is a pure state, not a mixed state, [INAUDIBLE]? WOLFGANG KETTERLE: Good question. So first of all, if you think about it you will discover more and more experimental challenges. What people must have used there is sort of a storm laser pulse that you have more than overkill to make sure that with a very, very short time window both ions are excited. The ions are in a pure state, and then they emit a photon. And if you have one system prepared which can emit a photon, but it has a branching ratio of 50/50, if it's an isolated system, it will have a superposition state of photon in one polarization-- let's say a sigma plus photon going to a magnetic quantum number state m equals plus 1 and a sigma minus photon going to m equals minus 1. And this is a pure state. The mixture only comes if you're not careful. If you have a magnetic field and you don't shield your magnetic field well, or you have some magnetized materials and you were not aware of it, then that means that you get different phase shifts which you can't calculate for, so now you have a random phase. You don't know it, and if you have ignorance you have to trace out or you have to average over the phase. And then your pure state becomes a density matrix. But the quantum mechanical process itself of a particle having different branching ratios is a pure system. It undergoes unitary time evolution, and it stays in a pure state. And the state which is populated by two ions, both emitting a photon simultaneously, is exactly the state I've written down for you. But you're absolutely right that decoherence is an issue. You have to be very careful with magnetic fields. You probably want to work with atoms which are non-magnetic. So you want to where the spin is only a nuclear spin which is much less sensitive with a magnetic field, and so on. Other questions? AUDIENCE: So as soon as both detectors click, then [INAUDIBLE] so that you can no longer make use of that entanglement? WOLFGANG KETTERLE: No, no wait. Look here. We have a state which has atoms in a state and two photons. So this state has four particles. We detect two particles which are the photons. And the atoms are untouched. The atoms are then afterwards in an entangled state. So we have a bigger system, we do a measurement of part of the system, and we have arranged things in the skillful way that the moment we know the outcome of the measurement is such and such, we know in which quantum state the rest of the system is. And this protocol means that the atoms are left after the outcome of the measurement is such and such, the atoms are left in a pure Bell state. OK I've used the word Bell, Bell states so often, I think it's time to talk about what Mr. Bell is famous for, namely the Bell's Inequality. So I said already in the introduction for the quantumness of light and entanglement that it is the EPR Paradox in the Bell's inequalities which a lot of people, including myself, think is the deeper essence of quantum physics. It really shows that quantum physics is not just classical physics with wave character, it goes way beyond it. So I want to demonstrate that with Bell's inequality. And the formulation which is very simple, and I want to present it here, is an inequality, which is-- I mean the name. Many Bell's inequalities now, you have different states, different detectors, and you can derive Bell's inequalities which are then violated by quantum physics. And what I want to present here is the CHSH inequality. That's Clauser, Horn, Shimony, and-- AUDIENCE: Holt. WOLFGANG KETTERLE: Yes, thank you. So the situation is the following. We have something which decays, something which emits two photons. Maybe an atom in an excited state, and it does a click clack, a two photon cascade. One photon goes to Bob, one goes to Alice. Or I always try to stress similarities between light and atoms. We discussed the experiment where you take a mercury molecule, you dissociate it, and then Bob and Alice each get an atom. And you can say these photons has a polarization or the atom has a spin. But Bob and Alice can now use different Stern Gerlach figures in x and y, at obscure angles, circular bases, I mean you name it. But what happens is because it's a spin up, spin down, horizontal, vertical polarization, it's a two-level system which is immediate. And after a Stern Gerlach filter, you have only two combinations. We call it plus and minus. So in its most general form, what we assume is that Bob does measurements in a basis, if you think. Think about a spatial orientation of a Stern Gerlach filter called S, where the outcome is plus minus 1, t where the outcome is plus minus 1, and Alice has her own choices. And we want to assume now that this is everything is classical probability. That if you just write down this expression-- don't ask me where it comes from. This is probably something which people wrote up after they found something interesting and try to prove it or simplify it. You just write down QS, RS, RT minus QT. You can then rewrite it by factoring out S and T. And now the next step is because Q and R are-- they're are in each measurement 1 or minus 1, either this is 0 or that is 0. No sorry, one is 0, one is 2. And therefore, this kind of funny combination of letters for every measurement is either plus or minus 2. Now we do many measurements, and the probability for a certain outcome that the variable Q-- there's a certain probability that the outcome is Q, so you sort of use this pretty much probabilistic thing that the system has a certain probability to B. And this is of course the assumption, the probability is that the particular comes with a certain probability in the state q, r, s, t. Of course you should scream, q, r, s, t commute, but this is classical now. They don't commute with each other quantum mechanically, but [INAUDIBLE] in a moment. So then you simply put-- by multiplying each event with its portability, you put now brackets around it. These are expectation values, and what you have is an inequality that this expression, which is the correlation between certain measurements between the quantity QS, RS, and so on is smaller or equal than 2. I mean it's very, very basic. And that's why I wrote it down. I don't want to spend a lot of time on it. It's just classical probabilistic reasoning. So now what are we doing quantum mechanically? We want to prove that quantum mechanics cannot be reduced to this classical reasoning. So we want to show that this inequality is violated. We have a source or entangled photons. We've talked about how we can entangle photons. And now we measure the quantities Q, R, S, and T. And just to give you an example about the many possible choices, Q can be a linear polarizer and R can be a circular polarizer. S can be a linear polarizer at 45 degree, and T can be a [INAUDIBLE] plate followed by a linear polarizer at 45 degrees. Well, that sounds like many trig functions, but we're not going into it. You have to choose your polarization. There's a certain scheme that linear polarization is not orthogonal to circular polarization, so there's a certain theme behind it. And if you would work it out by just looking at entangled photons, the entangled photons are in a state, let's say HV plus VH. They are correlated in polarization or spin up, spin down with the down spin [INAUDIBLE] state. And you can just work out what is the polarization when you detect it. And what happens is that a simple but tedious calculation says that all those quantities QS, RS, and and RT are equal. Here we have a minus sign and they are all 1 over square root 2. And that means that instead of the classical inequality, that this quantity is smaller than 2. We find that this is 2 times square root 2. And the fact that this is larger than 2 has been experimentally confirmed with larger and larger precision. Actually, the person who gave this [? CUA ?] seminar a week ago, [INAUDIBLE], was part of the team with [INAUDIBLE] who did one of the very, very first measurements violations of Bell's inequality in the '80s some 30 years ago. So I mean, this happened rather recently. So the math is trivial, the result seems trivial. It just shows that the world is quantum mechanically and not classical. And a lot of papers have been written and discussions have been had about what does it really mean? What does it mean about the world? What does it tell us about the world? So so the implication from the established violation of Bell's inequality and the CHSH inequality is that when we assumed that the state has definite values of Q, R, S, T before observation, we have to give up that. Or we have to give up that a measurement performed by Alice does not influence Bob's experiment, Bob's measurement. So in other words, what I formulated here is the locality principle that what happens in one location cannot influence what happens in the other location. Well you can say maybe there's some secret channel sending [INAUDIBLE] the speed of light. But people [INAUDIBLE] to create links to put the detector so far away that even a [INAUDIBLE] at the speed of light could not have influenced the other measurement. So there has been after the first Bell's inequality experiment, there have been a series of experiments to avoid all of those loopholes that there is some unknown secret communication between the two detectors. The first one is the principal of-- sometimes I think a more philosophical word than physical word, "reality". That those quantities are real and they should exist before we measure them. So the outcome of Bell's inequality is-- the violation of Bell's inequality is, at least one of them, at least one of these principles-- reality-- at least one of these principles does not hold in nature. So either reality or locality are violated. Any questions? And maybe just to connect it to what I've said before, one reason you saw that here in the title, Bell's inequality with two remote atomic [INAUDIBLE]. Those experiments got a lot of attention because Bell's inequality is something all physicists talk about. And when you do it with photons, there is always the detection loophole. You don't have perfect photon to detectors. And some people said, well, maybe the violation of Bell's inequality comes only from the photons we detect. The undetected photons would make up for the violation of Bell's inequality, and Bell's inequality is not violated. I mean, that almost sounds like that nature wants to fool us. The photons team up and say the photons which are detected behave very differently from the photons which are not detected. But these are at least logical loopholes. And graduate students spend half of their Ph.D., or maybe several graduate students spend their whole Ph.D. In building such an experiment which is now-- you get a famous paper out of it. And your research has attention, so you know. I didn't sleep so much, so maybe I'm not too serious now. But this is great research. I mean, this is really pushing the limit of our understanding of quantum physics. And Chris Monroe is a wonderful physicist. And we are in the same [INAUDIBLE]. So if you have a Bell state of atoms, and I told you how you get it also with a lousy efficiency but you get a Bell state. If you now do a measurement of those Bell state and you measure violation of Bell's inequality, atoms, they don't run away. You can detect the atoms with 100% probability. You can shine a laser light on them, hundreds, thousands, millions of photons until they have scattered enough light that you know 100%, I've detected the atom in it's quantum state. So Bell states with atoms, one real world application of them is to test violations of Bell's inequality with atoms which do not fall into the detection loophole. Our next unit is partially motivated to show you what all this entanglement can do for us. So I mentioned that entanglement is a resource. It's something useful like energy is a resource. And so maybe ultimately there should be a stock market, and you can buy so and so many bits of entanglement. And you have to pay for it because there's something you can do. And let me introduce what can be done with it. If you have one photon at frequency omega, and you have an observation time-- you do a measurement over time T, then the precision at which you measure the frequency is fundamentally limited by, you can say time energy uncertainty by the Fourier theorem, that the uncertainty and the frequency of a single measurement is 1 over the time T you had to detect or to measure the frequency of the photon. But now we have n photons. And that means that the uncertainty in the measurement, which is now small delta omega, is delta omega for a single photon. But you know, when you do n measurements, and average n measurements, you gain by the square root of n. And this is regarded as the fundamental shot noise limit of measurements. But now assume we can do something fancy. We can make a super photon. We take our n photons of frequency omega and make one photon of frequency n omega. Now we have only one photon. The frequency uncertainty of the measurement is delta omega, but this is now the uncertainty off the n times more energetic photon. So therefore, if you're interested in the quantity omega, which is the frequency of the single photon, we have now made an improvement over the standard shot noise limit by square root n. So everybody follows? So that's just a [INAUDIBLE] experiment. If you can take n photons-- I've told you how we've talked about the [INAUDIBLE] oscillator, how we can pump a crystal. And in your homework assignment you do a nice calculation with a Hamiltonian. One big photon in, it breaks into two photons. The reverse process is frequency dot doubling. So if you would not measure n photons individually, but first ate up their frequency by making a photon of n times the frequency and then look a single photon, you have now, you measure all the photons together. And therefore, your accuracy improves by a factor of n and not just by a factor of square root n. So that tells us something. If we do something with the photons, if we entangle them, there is a possibility to vastly improve the standard limit of measurements. And so people who are really interested in it are people who push the limits of precision, people who build atomic clocks and want to get the last little bit of accuracy which is possible. So they have already exhausted all technical possibilities, and the next thing is now, well maybe entanglement and sort of subtleties of quantum physics can come to their help. So this shows you what is possible. But now I want to tell you how it can be done. So what I first want to do is, before we can improve over the fundamental shot noise limit, I have to show you how the fundamental shot noise limit naturally emerges. So I want to introduce to you beam splitter interferometers. Interferometers are there to measure phase shifts. You split a beam, you combine it, and if there's a phase shift you notice if. You get interference fringes. And I want to show you that very naturally to standard quantum limit, the shot noise emerges. But then we are ready to look at our description of the interferometer and say, where can we now change the rules to put in more quantum-ness or entanglement and eventually get higher precision. Since the standard quantum limit is well know, I want to rather quickly go over that. So the goal is that we want to measure a phase shift. And my first simple deviation of the phase shift is that in this picture of the quasi probabilities, the coherent state is a circle like that. The width of the circle in natural units is 1, but the radius of that is alpha of the coherent state, which is square root n. So now if you ask how well with this uncertainty can I observe the phase of the photon which circulates in this quasi probability plane, you find that the phase is 1 over square root n. Or based on your homework assignment number one, you can say we have some Heisenberg uncertainty between photon number and the phase. And in a coherent state, the standard deviation in the photon number is square root n, and you again get the standard quantum limit. Let's now obtain the standard quantum limit from a real measurement device, because this is what we want to generalize for entangles, [INAUDIBLE] state, and all the special things. So an interferometer is the following. It consists of two beam splitters. After the second beam splitter we have two detectors. And the quality we will measure will be the difference of the two photo currents. And if you don't do any phase shift, we know the two beams splitters are just the identity transformation. But now we put in a phase shift, and the question is, what is the smallest phase shift we can measure? What is the accuracy measuring this phase? We have discussed at length all the elements. So we have a beam splitter, a phase shifter, and a second beam splitter. You know that the phase shifter is a rotation in z, the beam splitter is a rotation in-- was it X or Y? Looks like I think Y because of the i's. Just a warning here, in this section I use a beam splitter which uses a different phase convention. But they are all equally apart from some phases. So by multiplying the three matrices, we get the transformatrix for the [INAUDIBLE] interferometer. So the output CD is this matrix here which is a simple rotation times the input state. And our measurement is the difference in photon numbers in the two output modes. So it's D dega D minus C dega C. So what we can do now is we can obtain C and D from the input modes A and B. So therefore we can now express our signal in terms of the input modes. When we know what we put into the interferometer, because we know all its elements we know what we can get out. And this is done here. And the phase which appeared in the rotation matrix, the phase shift in the interferometer appears now as a cosine phi and sine phi contribution. And we have sort off- the cosine and sine phi have two operators as a pre-factor. In one case, it's A dega A minus B dega B. In the other case it's a cross-term, A dega B plus B dega A, Now we will be more specific what is signal and noise. I just want to tell you in the standard way of operating the interferometer you have only one beam in the mode A. You take a beam, you split it, you recombine it. B is nothing or the vacuum. It may introduce some noise. But if you have a lot of intensity in the beam A, it is this term which dominates, and you find fringes as a function of the phase shift cosine phi. Whereas what comes here is sort of more the vacuum mode. It gives rise to noise terms. So if you ask, what is the expectation value for x, this x operator, it varies co-sinusoidally. And there is a special point for a phase shift of 90 degrees when we have the steepest slope and the highest sensitivity. So these are sort of all just setting the stage. Which we are interested in is, if we now have an input state of light-- and as you know, there's all of this noise. There's the fundamental noise of the vacuum, or coherent state, of Heisenberg's Uncertainty. And this means when we do repeated measurements we will have a variance in the phase. And we want to know, what is the fundamental limit on the standard deviation in the measurement of the phase. Well, the standard deviation in the phase is nothing else than the standard deviation of what we measure in divided by how sensitive it is to the phase. So by taking this expression for M, M is an operator X plus cosine phi plus something times sine phi. We can now evaluate this expression. This will actually appear several times today, and it's also a key question to one of your homework problems. So this is sort of what defines the accuracy of the interferometer. And what we have here are expectation values of operators. And now we can-- and that's what we will do for the rest of this lecture. We will look at this expression for different inputs states. Coherent state, single photons, [INAUDIBLE] light, entangled state. So that's what you're going to do. So if you take the derivative of dn d phi, the cosine becomes a sine. A minus sign, the sine becomes a cosine. And if you specialize to the situation, which is where the slope is very steep, a 90 degree phase shift around this point, our phase sensitivity is given by the expectation value of the variance of the operator y divided by the operator x. So now we are ready to plug things in. The first thing is of course the coherent state. For the coherent state, our input to the beam splitter, one is a coherent state, the other one is a vacuum in mode A and B. We had expressions-- let me just scroll back-- for x and y, expressed by the operators a, b, a dega b dega. So therefore we can calculate that now. And x is nothing else than the number of photons in the coherent state. And this is our signal. And y is 0. This was the noise term. And the variance in y-- we're just doing commutators-- is given by n. So therefore, if we calculate the square root of y square over x for the coherent state input, we take the square root of y squared, which is square root n, we divide by n, we obtain the standard shot noise limit. Sure, what else should we expect? So now we know how to use the formalism. We can now apply it too-- we can go from the most classical state of light, the coherent state, to what I regard at least for a single mode the most quantum state, namely a single photon. Remember when we talked about the G2 function, G2 function can only be smaller than 1 for non-classical states. And for the single photon it's 0. That's the biggest violation of the classical equation that G2 has to be larger or equal than 1. So now we're really dealing with a quantum system. And the question is, what will we get for the single photon? So for the single photon we want to use the dual-rail representation. We want to use the powerful formalism we have used. After the beam splitter, the photon can be in one mode or the other mode. We call this in the dual-rail representation the logical 0 or the logical 1. So in this representation, when we start with a photon in one input mode, this is the logical 0, the beam splitter is directing the photons into one mode or the other mode, but this is a single qubit rotation. The phase shifter is another single qubit operation. The second beam splitter is another operation. So by just using the rotation, we'll find out what is the output state. So we start with the logical 0 with the photon in one mode. The beam splitter creates a superposition state. Remember, we have the interferometer, and in the lower arm we put in a phase. So therefore, the lower arm gets multiplied with a phase shift. And then it goes again through a beam splitter, which is just another rotation, and with that we obtain the output state. And so the output state is now a superposition of the logical 0 and the logical 1, and we've picked up a phase shift. So no we ask-- we have the beam splitter, phase shift, recombiner, and now we are asking, what is the probability to detect the photon in one of the two output modes. We just put a counter there. And this probability is nothing else than the probability to have a logical 1, because a logical 1 in the dual-rail representation means the photon is in one mode. The logical 0 would be the other mode. I felt if I would spend three or four times as much time on it, it wouldn't increase your understanding. You may have to sit down and look through it, but it's just really putting matrices together, [? single ?] photons, mapping it. Every step is trivial and is what we have done in a different context before. So the result is that the probability of finding the [? single ?] photon in our detector has a cosine phi factor. So that's how we measure phase. Our counter, the probability that photons arrive. If phi is 0, the probability is one. If phi is 180 degrees, all the photons go to the other mode. I mean, that's what you'd expect an interferometer to do. But now comes the interesting question. How precise can be measured the phase? Remember, in the coherent state, the coherent state had fluctuations in the photon number of square root n, and this gave rise to the standard quantum limit. A single photon is a single photon. There is no noise. We always measure a single photon if you would simply detect the number of photons in the input state. But we are not doing that. We have sent the photon through an interferometer in order to measure the phase. And now we have-- and this is a result of this, I would say, trivial calculation that this is now the probability to observe a photon. Well, there is not a whole lot we can learn from one photon, so we run the experiment n times. And what we get is a binomial distribution, just a coin toss with probability P. What is the variance in P? It's not a Poissonian distribution. It is a binomial distribution. Of course, if the probability is 1 you detect all your photons. There is no uncertainty. If the probability is 0 you detect 0 with 0 uncertainty. So therefore, the expression for the variance of the binomial distribution is p times 1 minus p. If not, I admit I had to refresh my memory about the binomial distribution yesterday evening. So therefore, when we repeat the experiment many times and we have sent into our interferometer m photons and we measure n clicks. And n over m is our measurement for the probability. This measurement of the probability has noise. And the noise is a variance of the standard deviation of the binomial distribution. This expression for P put into the variance of the binomial distribution, and a little bit of triganomic manipulation gives us the result on the right-hand side. And we want to know what is the uncertainty in the phase. I mean, this is what we want to measure. Well, the uncertainty in the phase is the uncertainty in our measurement, which is delta P. And then we have to divide by how sensitive the probability is with respect to the phase. So we take this expression of the probability as a function of the phase, take a derivative, and then by doing the ratio of this over that we find 1 over square root n. So it doesn't make any difference if you put the n photons into a coherent beam and run our interferometer or if you go through the great pain of preparing each photon in a non-classical flux state and just operating the interferometer with single photons. In both cases do we obtain the standard quantum limit 1 over square root n. Now we are really ready to see how can we improve on the shot noise? It seems that unless we do something special, we will always get square root n. I've already shown you at the beginning that's the shot noise limit is not fundamental. Just take this thought experiment that you take the n photons through a highly nonlinear process, you create one photon of frequency n omega, and then you have more precision of this measurement. So there must be a way to have a precision which scales 1 over n and not 1 over square root n. The mathematical augment related to our treatment of the interferometer is the following, our signal was A dega A minus B dega B. Well, if one of the input mode dominates, A dega A is the [? photon ?] [? number. ?] So this sort of looks like the [? photon ?] [? number ?]. But if we now find a scheme-- and I will show you that this is possible-- that this cos term A dega B plus B dega A, which is some form of quantum noise, as we will see, is 0, well what do we get now? So let's hope that maybe by some squeezing-- I will show you several versions which show you how quantumness can give us more than short noise. And one example will be that by squeezing light. We've learned already that squeezed light can suppress the noise. If you squeeze light, something which comes from the modes where we apply squeezed vacuum has been reduced. I mean, it's what we discussed already. So the best we can do is that the noise term is 0. So what do we have now? If we have a signal which is finite and the noise is 0, what is our precision of measurement? Well, it first looks like the signal to noise ratio is infinite. But it's not quite that, because if our signal is x, it's a photon number. And the signal x was sensitive to the phase by cosine phi. The sensitivity of our measurement of the quantity m with phase is n times sine phi. And this is smaller or equal than n, where the equal sign is obtained for 90 degrees. So therefore, we have a [? photon ?] number n. We may absolutely know what the initial [? photon ?] number is. But now we want to measure whether the phase is different from pi over 2. And the smallest change in our signal which we can resolve is that we get one photon less. So that's the smallest resolvable change due to the [? photon ?] nature of our detection. And that implies that the smallest phase shift we can resolve is 1 over n. So this is more a thought experiment, now looking at the math and seeing what is the best signal? A dega A can never be larger than [? photon ?] numbers, so that's a resource. We put in energy in the terms of photons, and then the best we can do is that we don't have any noise. And then we are simply limited by the fact that when we deviate from the maximum output signal because there is a phase shift in the interferometer, our sensitivity comes in grains, is grainy by the [? photon ?] number. So I'm not telling you how we achieve to get y equals 0. The question is only that at least the math seems to make it possible. So now we ask how to achieve sub-shot noise precision. So here we know this is at least mathematically possible. So we want to improve this interferometer, and the question is we have to change something. This interferometer, [INAUDIBLE] we've seen [INAUDIBLE] state is not giving us any better result, so we have three options now. We can do something fancy in the input state, use entangled state. We can use some very fancy beam splitters, or we can do something special to the way how we read out the interferometer. And all those three are possibilities. We can put in some extra quantumness into each of the three steps. OK AUDIENCE: [INAUDIBLE]? WOLFGANG KETTERLE: OK, the question is can we do something with the phase shift? AUDIENCE: [INAUDIBLE]. WOLFGANG KETTERLE: I think if we would do something with the phase shift, we would do-- you know, we would change apples with oranges, because we want to compare different interferometers by measuring the same thing. And maybe as an experimentalist say, what I want to be able to measure I take a very, very thing glass plate which just makes a very small phase shift, and I'm now comparing different interferometers. And when I put the glass plate in and pull it out, I want the person who reads out the interferometer to see a change. So that's how I want to compare the interferometers. Actually, the question you are raising was in another way also discussed today at lunch with three of my wonderful colleagues. Namely what we discussed was some recent papers, one of them published in Nature, which claims precision better than Heisenberg. I mean, what I'm sort of indicating to you is Heisenberg should be the best which can be achieved. How can we do even better than Heisenberg? Well, and this was actually related, Nancy, to your question. If you want to measure magnetization and you use some nonlinear physics where the photons, where the magnetization involved affects the photon field by some higher power of what you measure, you really change what you measure. Maybe I'm not expressing it clearly. If you measure magnetization and you have a certain quantum limit in measuring the magnetization, that's one thing. But if you now bring another material close to your magnetization and this material goes through a phase shift, you sort of amplify by a physical process by a nonlinear Hamiltonian what you want to measure. And then of course you can-- depending on what kind of nonlinear process you're using-- you can get a signal which scales tremendously with the number of photons you put into your system. So in other words, there are loopholes like this, and some of them led to very fleshy papers. And our lunch discussion was that some of it is really completely trivial. And some papers who claim that they have seen a scaling of the precision with photon number which is better than Heisenberg, not 1 over n, maybe 1 over n square, that some of those papers were purely classical and the only quantum character of this paper was the name of Heisenberg in the title. So it's not related to any uncertainty relation, but that's not what we want to discuss. Let me just spend quickly-- let me see, yup. We should the week with something interesting and not just shot noise. So what I want to use now is I want to replace the beam splitter by something which involves Bell states. So as the beam splitter we do a massive creation of bell states. It's our entangler. And our second beam splitter is a Bell analyzer. It is a disentangler. So what I mean is the following, we will actually just put one photon into this input beam, and we will only read out one channel. So all these here are only auxiliary modes. This is how I make a special quantum beam splitter. And what we need for this description is essentially two gates. We need a single qubit. The single qubit is the [INAUDIBLE] which we have already discussed. And in the dual-rail representation where its photon can be in one of the two modes, the [INAUDIBLE] is simply connecting the two modes in this way. And we discussed that the [INAUDIBLE] can be described by a beam splitter with a phase shift. So in other words, we need one element which is a beam splitter with a phase shift. But this is only acting on one qubits. And now we want to connect qubits with a controlled NOT gate. To remind you, controlled NOT gate is something where the photons stay where they are. But if the control bit is 1, it flips the target bit. If the control bit is 0, nothing happens to the target bit. And we discussed already that we can realize one qubit with one interferometer. These are the dual-rails. We always have one photon in two states, this mode or that mode. And know the other qubit, one of the rails can go through the nonlinear Kerr medium. If the bit is in c through the phase shift in the nonlinear Kerr medium, it flips the [INAUDIBLE] down there, and this the controlled knot. So now we want to use two qubits. So I'm talking about, in our fancy entangler I'm just talking about the first two rails here. So we have those two rails. The [INAUDIBLE] puts us into a superposition of the logical CO and 1. And now after the c-not gate, so we are in a superposition of-- just give me a second-- we start here with 1. Sorry, we start here with 1. AUDIENCE: Excuse me. WOLFGANG KETTERLE: Yeah? I'm wrapping up. This is my last. AUDIENCE: One [INAUDIBLE] minus 1 [INAUDIBLE]. WOLFGANG KETTERLE: The 0 [INAUDIBLE] in 0 plus 1. And let me just finish that. This is the target, year. Sorry, it's correct. We start with 0, we start with 0. So what we have here is the product state of 0 plus 1 upstairs and 0 downstairs. But if we have a 1, we flip the 0 to 1. So therefore, what we have now is the state 00 plus 1 1 over square root 2. And then we apply our phase shifter. And what we get out of this state is 00 plus e to the 2i phi times 1 1 over square root 2. So now by using sort of-- and now I should stop. I know people are waiting. But by using two of those, I suddenly have multiplied the phase by 2, so something is now sensitive to 2 phi. And if I use n such, more and more entanglement, I will show you-- no class next week, but in the following week-- that we suddenly have a term in our quantum state which is e to the n phi. And this gives us effect of an in precision. OK, have a good rest of the week and we meet again in a week and a half.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

2_QED_Hamiltonian.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: My first question is, do you have any questions about the last class, about the introduction, about the course overview, the syllabus, any requirements of the course, formalities? It should be pretty clear. So today, we start from first principles. I want to give you a rigorous derivation of what this course is based, namely, the Hamiltonian to describe the interaction between atoms and light. I mentioned to you last week that atomic physics can be defined about everything that is interesting about the building blocks of nature, atoms interacting with light electromagnetic fields, Coulomb fields, and today, we'll talk about the first principle, Hamiltonian. The reference for that, and I'm closely following that derivation, is the book Atom-Photon Interaction, the green book by Cohen-Tannoudji, Grynberg, and others, and it's taken from the appendix. And if you flip through the appendix, you will find out that there are about 100 equations. Today, we'll discuss every single of them. However, the good news is the result in the end is simple and intuitive. In the end, we have what we want, that the electromagnetic field couples different energy levels, and then we can play around with quantum case, with laser cooling. All we need is a coupling matrix element. And for most of the course, I will not even elaborate what this matrix element is about. But here, we derive it from first principles. The result we will use most often is actually the electric dipole approximation, which I'm sure you have already seen many, many times. So in that sense, I sometimes have second thoughts. Should I dedicate a whole lecture to derive something you know already? On the other hand, this is sort of the meat of atomic physics, and I want to go as deep as possible into it, that whenever you wonder what form of other interactions exist, you have a reference to look it up. And so to some extent, I want to also encourage you to read more about the fundamental nature of how light and atoms interact. I will give you another reference a little bit later on. But as a motivation, I want to tell you what I learned when I prepared this lecture, and I hope that there is something that you will learn from this treatment. One is a rigorous separation of local fields from radiation fields. Let me just write it down, rigorous separation of local Coulomb fields from radiation fields. So to clearly separate the terms in the Hamiltonian, let me explain that. When we introduce electromagnetic fields into our world, we would say we introduce them by saying particles have charge. But some parts of the charged particle physics is something we want to include in the description of atoms, and that's the Coulomb field of protons. The structure of atoms is electromagnetic, but when we now introduce electromagnetic field as a new degree of freedom-- photons can be admitted, absorbed, and such-- we only want to talk about the electric field which belongs to the photons and not the electric field which belongs to the atoms. And you may wonder, can you distinguish between two kinds of electric fields? The answer is yes, and I want to show you how mathematically you distinguish when you introduce electromagnetic fields between the fields which stay with the atom, which don't have an extra degree of freedom, and if you do canonical quantization, they don't enter as an extra variable, and the fields which are the photons, and these are the extra objects we have to consider. What is also related is when you do second quantization, if you wanted, I could quantize electromagnetic field for you in five minutes. Just say it's harmonic oscillator. You're familiar with harmonic oscillator, and let's just do it. Yes, this is correct, and I will remind you of that a little bit later, but there is one non-trivial aspect if you want to do quantization of a field theory, you have to make sure that before you quantize it, you know which degrees of freedom are really independent. So the question is, how many independent fields does the electromagnetic field have? The naive answer is three electric field components, three magnetic field components, that makes six. But of course they're not independent, and we will discuss that. So we'll spend some time in the identification of truly independent field components, or degrees of freedom of the electromagnetic field. So all that is an excursion into classical physics because it is just reformulating classical electrodynamics to be ready for quantization. Actually, all the work is done to discuss and derive the appropriate classical description. That means to eliminate all redundant variables, and then at the end, have variables which, yes, will look like a harmonic oscillator. And once we have reduced it to degrees of freedom which look like a harmonic oscillator, it is very straightforward to do the field quantization following the recipe of the harmonic oscillator. There's one more highlight, if I want to say so, and this is a truly rigorous derivation of the electric dipole approximation. In most quantum mechanics textbooks, you do a dipole approximation and you wind up with the electric dipole Hamiltonian, which I have already put up there, but you do actually unnecessary approximations in terms of neglecting an a squared term, and there are often confusing discussions what happens with the a squared term. I want to show you a more rigorous deviation of the electric dipole approximation, which is including the quadratic term. That actually means-- that's the take home message-- the electric dipole approximation is actually better than many of you have thought so far. Since I don't want to write out 100 equations, I copied them and summarized them out of the book. You'll find them all, as I mentioned before, in the appendix of Atom-Photon Interaction. But I sort of want to walk you through. What I'm aiming at is that you have sort of a complete overview how is everything coming together, and if it's just mathematical, well, mathematics is always exact. I can go faster, but I really want to highlight here every single physical concept and approximation we are doing. When we start out with Maxwell's equations, we have actually six components of the electromagnetic field, and we will see in a moment that this is redundant. What will turn out to be very important to separate the local fields from the radiation field is a Fourier transformation, so we introduce an expansion into plain waves. And what then happens, of course, is the derivative operator becomes the k vector. So therefore, when you have Maxwell's equations with the curl, del cross B, it turns into k cross B. This is now important because we have now separate equations for the component of the electric and magnetic field, which is parallel to k, projected onto k, and of course, the cross product takes out the component which is perpendicular to k. So that's something which is sort of nice, that the Fourier transform it allows us to separate the fields into longitudinal fields, where the vector of the Fourier component is parallel to the k vector, and into transverse components. AUDIENCE: [INAUDIBLE]? PROFESSOR: The difference between curly and non-curly vector is one is a Fourier transform, the other one is other spatial components. I have to actually say the book API tries to be super accurate in choosing [INAUDIBLE]. Claude Cohen-Tannoudji is really a master of elegance and perfection, so they don't want to use the same [INAUDIBLE] for a spatial component and for the Fourier component. On the other hand, I have to tell you one thing. For the next hour, use a little bit your intuition. If you see e, it means electric field. Whether it is curly, whether it's italic, whether it's bold face, don't let those differences clutter your view. So I would say don't ask me too often, what exactly does it mean now because you have gone from curly to non-curly? It is explained in the appendix. I will tell you everything which is necessary, but these are subtleties which are more to present the mathematics in a more rigorous way. We have the spatial Fourier components, and as I mentioned, what you can do only because of the Fourier transform, you have now the distinction, a rigorous separation between the parallel and the longitudinal fields. So of course, you can also now get back what is the longitudinal field, not in Fourier space but in position space, by just taking the longitudinal or transverse components of the Fourier transform and transforming them back. Why I'm elaborating on that is we will see in the next few minutes that the transverse field is the field which propagates. The longitudinal field is the Coulomb field. It belongs to the atoms and will not become a new degree of freedom. And we see that immediately in the next equations. When we look now at Maxwell's equations, we find that the transverse and longitudinal fields decouple. We have two kinds of equations which are completely decoupled. The longitudinal electric field is the Coulomb field associated with a charge density, whereas the transverse fields are creating themselves, and this is radiation, how the electric and magnetic fields create themselves as they propagate. Now, the fact that the parallel electric field can be expressed by the charge density, or it's expressed by the momentary position of charges, and this means immediately that the longitudinal field is not an independent variable or an extra degree of freedom. Canonical, of course, with how we describe particles by their position, and here you see explicitly that the longitudinal field which we defined the way we did it, that this part of the field only depends on the momentarily electric field, the Coulomb field associated with the charges. The next thing, of course, is to define the vector potential and the scalar potential. By using the standard substitution, we can now express electric and magnetic fields by a scalar potential and by a vector potential. We continue to have the separation between longitudinal and transverse fields, namely the transverse electromagnetic fields depend only on the transverse component of the vector potential. And the transverse component is gauge invariant. In other words, you know there are different gauges we can consider when we introduce electromagnetic potentials, but the choice of gauge does not affect the transverse vector potential. Now we want to pick our gauge and repeat the gauge which is most convenient. This is typically, for low energy quantum physics, for describing atoms and radiation, the Coulomb gauge. So we have the freedom of gauge, and the Coulomb gauge is written by the divergence of the vector potential is 0. If you now think about Fourier transform, this means k dot the Fourier component is 0, and that means the vector potential does not have any longitudinal component. So with that, we have now reduced equations for six variables, three electric fields, three magnetic field components. By introducing vector and scalar potential, you go to four. We've eliminated one more with the Coulomb gauge, so we have three. The scalar potential is the Coulomb potential. It's not an extra degree of freedom. And what is left are the two field components of the transverse vector potential. And these are now, as we will see, the independent variables of the radiation field. So in other words, when we talk about the electromagnetic field in the remainder of the course, the part of the electromagnetic field we are interested in are the fields generated by the transverse vector potential. Now, what is next is to go into normal modes. The reason is that we want to identify each normal mode of the electromagnetic field as an independent harmonic oscillator. Now, I'm showing you those equations, but you sort of have to sit back because now there will be alphas, a's, lower a's, upper A's. They all mean pretty much the same. In one case, you are normalized, maybe something has to be factored out or not. The derivation in the appendix of API is exactly [INAUDIBLE] distinction, but let me just give you a flyover here. And that is we are now taking the transverse components from the electromagnetic field, and we define new parameters, which are the normal modes. What we actually want to see in a moment is that those normal mode variables are harmonic oscillators. Those normal modes can be defined by the original field, which are transverse, but since we have expressed the transverse field by the transverse vector potential, they also simply can be expressed by the transverse vector potential. So to remind you, we've done a Fourier transform in the spatial coordinate, but not yet in the temporal coordinate. What we obtain now is an equation of motion for the normal mode variables. And here, I have reminded you that the transverse vector potential, how it can be expressed by the normal mode variables. In other words, yes, it looks complicated and a little bit messy, but it's classical physics, and all we have done is introduce normal modes in a just more complicated notation, as you may have seen it in 803 or some other course where for the first time you talked about normal modes of a pendulum or a chain of springs. So we have identified normal modes by the equation above, and then it's a mathematical identity that our radiation field, our transverse vector potential, can simply be expended into normal modes. So the equation of motion for the transverse fields involves of course, the transverse vector potential. I should have scrolled back. Here we have our equations for the transverse fields. This is what we want to describe, and if we express everything in terms of the vector potential, then we have an equation which involves the transverse vector potential, but in addition, because it's a differential equation, the first and second derivative. That means that if you have a second order differential equation, which is the full description of the electromagnetic field, then you know that the solution of the second order equation needs as an initial condition the field itself and the first derivative. Therefore, our classical fields are determined by a perpendicular and its derivative at the initial time. So we need a and we need a dot, and they are coupled, but what we are doing right now is-- and this is the idea behind the normal modes-- the normal modes coordinates combine a and a dot. It's the same when you have a harmonic oscillator and you want to introduce coordinates for the normal mode, they are a combination of position and momentum, and then they are decoupled. But position and momentum always couple or oscillate back and forth, and the same happens here between a and a dot, and the normal mode coordinates are the linear combination. In other words, the equation of motion coupled source and normal mode means that we have introduced he decoupled normal modes. And whenever, in classical physics, we have normal modes, normal modes means that time dependence is e to the i omega t, then we have distilled the problem of coupled components to decoupled harmonic oscillators. So at this point, each mode acts as an independent harmonic oscillator. So what I wanted to show you here is clearly, as a flyover, and as an appetizer to read more in the book, that we can start with the electromagnetic field components. We haven't just assumed we have now empty space and only [INAUDIBLE] radiation. We have rigorously separated the electromagnetic field in what belongs to the atoms and what belongs to radiation, and the trick was a spatial Fourier component. At that moment, we had a description in terms of the transverse vector potential, and by using purely classical physics, by combining the vector potential and its derivative, we found normal modes, and they are now completely independent harmonic oscillators. This approach, and I wanted to show that to you, involved a lot of notation-- a's, little alphas, normal modes, and such. Let me just simply show you another much, much shorter pathway how you see that everything looks and smells like a harmonic oscillator, and then we'll do the last step, which is rather straightforward. We quantize the electromagnetic field. I could quantize it right now, but we should take a short break from the appendix in Atom-Photon Interaction and do a more intuitive shortcut to the same physics. Any questions at this point? If you feel it's confusing, remind yourself this is just classical physics. We are just rewriting Maxwell's equations in new variables. So what we do now is, as I said, let's repeat some of this deviation by focusing on energy. If you write down energy for our systems of atoms, photons, and Coulomb fields, we have particles alpha with mass m alpha and velocity v alpha. This is our kinetic energy. And then we use the well known expression for the electromagnetic field, which is the spatial integral over E squared plus B squared. This integral over the electromagnetic field energy density can actually be nicely separated into integral over the longitudinal field which we have introduced and a second integral, which involves the transverse field. Of course, the magnetic field is only transverse because the divergence of the magnetic field is 0, and that means the magnetic field does not have any longitudinal component. So this part here, as we actually have shown, is given by simply the Coulomb energy where we have charge density at position r and r prime interacting with the Coulomb interaction. Eventually, if you want to treat that further, this Coulomb energy should be split into some divergent self energy, which is the energy of the electron interacting with itself, and people know how to deal with that, and the interaction energy, for example, between the proton and the electron, which is responsible for the atomic structure. So in other words, we know how to deal with that. This actually becomes atomic structure. This is in contrast to this part here, which is the classical energy of the radiation field, which is sort of the transverse part of the energy. To describe now radiation, we have to focus on this transverse part, and we are now expressing the transverse part by the vector potential and its derivatives. So we introduce the vector potential as before, and I don't want to go through the re-derivation. I just want to show you how the total energy now appears in terms of the vector potential. And what we need for that is, of course, the vector potential and its derivative. The vector potential depends on polarization, depends on the Fourier component, and we call this derivative of the vector potential the conjugate momentum. So now focusing on energy, we, of course, reduce everything as before to the vector potential and its derivative. So we sum over polarization. We integrate over all Fourier components. and the integral involves now the vector potential, the square of it, or in complex notation, the complex conjugate, depends on polarization. We have c squared k square. This part, of course, comes from the magnetic field, which depends on the vector potential. It's a spatial derivative, the curl, and this gives the k squared, whereas the part of the electromagnetic energy, which is related to the electric field, the electric field is a temporal derivative of the vector potential, and therefore, it involves the temporal derivatives which are now the canonical momenta. So this equation should really remind you now-- it's an energy equation-- of the energy of a harmonic oscillator because the energy is now a sum over all oscillators, but this is sort of x squared, the amplitude of the oscillator squared, this is the potential energy. And here we have the derivative of x, the velocity or momentum, and this should remind you of the kinetic energy of a harmonic oscillator. In other words, this should tell you that by focusing on the transverse component of the vector potential, it's about lots of summation-- k summation, polarization summation. We in the end find that each such mode is a harmonic oscillator. Any questions? Let me just show you now, and it may help you to go through the previous derivation, let me know introduce what I did before, the normal modes, and then show you how the energy looks like defined in normal modes. So we have these normal mode variables, which are defined as a superposition of A and A dot. And I'm not telling you here with all the indices whether it's the polarization component, the k component. I just want to tell you the structure. The normal mode is a superposition of A dot and A. And with that, the energy of the electromagnetic field can be written-- well, we always have to sum over Fourier components. We always have to sum over the polarization. But then we have something which is very, very simple and intuitive. It's just the square of alpha with the correct polarization, the pre-factor is h bar omega over 2. It looks like quantum mechanics, h bar omega times A dagger A plus 1/2, but this is purely classical. There is no quantization, there is no operator. We've simply defined something new, namely alpha in terms of the vector potential. In that sense, if you're going to go back to the derivation, the alphas are nothing else than some elaborate combination of the transverse electric and magnetic field. Colin? AUDIENCE: Where does the h bar come in? Is that the coefficient of alpha? PROFESSOR: h bar only enters through the constant in the definition of the normal mode parameter. So actually, I have introduced h bar by choosing this parameter wisely in such a way that it connects with quantum mechanics. But it's a completely arbitrary introduction here. I could have said h bar to 1. I hope you enjoyed, or at least did not dislike this excursion into classical physics. We have now two equations for describing the energy of a harmonic oscillator. So both of these equations look like a harmonic oscillator, but if something looks like a harmonic oscillator, it is a harmonic oscillator. And I think I should always remind you that so far, everything has been purely classical. And also, let me write down that h bar enters solely through the constant in the definition of the normal mode parameter, alpha. If you want to know more about it, there is a second reference. I will actually show you the cover page in a few moments. But there is a second book by Claude Cohen-Tannoudji and collaborators, not Atom-Photon Interaction, but called Photons and Atoms. It's a whole book on rigorously defining QED. So a whole book has been written about the subject of this lecture, and on page 27, you can read more about that. So enough of classical physics. All we have done so far-- I'm sorry for repeating it, but I think I can't repeat it too often with so many complicated equations on the screen-- all we have done so far is we have rewritten Maxwell's equations. We have rewritten Maxwell's equations in Fourier space, with the vector potential, eventually with normal modes. That's all we have done. But now what we are doing is we do the step in quantum mechanics which you have already seen a few times, and this is you have written equations in such a way that they look like harmonic oscillators, and then you postulate that the classical quantities become operators. The transverse vector potential and the conjugate canonical momentum fulfill now commutators, and we use those commutators to define them as operators. Now, it looks particularly easy when we use normal mode operators because the normal mode operators after quantization become our a's and a daggers. That was one of the things I wanted to show you, that you can go through everything introducing classical normal modes, and then the normal modes turns into operators, and now we have all your creation and annihilation operators. In other words, quantization cannot be rigorously proven. I mean, you cannot prove quantum theory from first principles. You can have a mathematical framework and check it against nature. What we have done here is to formulate the quantum theory. At this point, we've made a postulate that we have operators which fulfill a commutation relation. And it is now your choice if you want to formulate postulate the commutator for the transverse vector potential and its conjugate momentum, or if you immediately want to jump at the normal modes, and that would mean you have the commutator for a and a dagger. We are almost done with that. We started with the electromagnetic fields. We went through transverse fields, vector potential, normal modes, now we quantize. But now all the equations, all the substitutions we have made, we can go backward. Therefore, we can now express the transverse vector potential, the electric and the magnetic field, by the normal modes, or that would mean in quantum mechanics, after quantization, we can express them by the a's and a daggers, and this is how we define the operator of the electric, the magnetic field, and the vector potential. So now, our fully quantized theory has operators, a of a dagger, or if you want now, the first defined operators of vector potential, electric and magnetic field. And just to remind you, since atomic physics is between fields and particles, we have exactly the standard definition of the particle operators. Each particle is described by its momentum and its position. So with that, we have our Hamiltonian, and a lot of what we discuss in this course are understanding this Hamiltonian, understanding its solution, understanding what is the physics described by that. So this Hamiltonian, this is the kinetic energy. The canonical momentum minus the vector potential is the mechanical velocity. This is kinetic energy. And we have separated the energy of the electromagnetic field into the Coulomb energy, which is written again here, and the radiation field. There is now one more term which we need later on, which is this one here. And at this point, I would say-- I'm just heuristically adding it by hand-- classical we don't have spin, but our particles have spin, and now we need [INAUDIBLE] coupling, how does the spin couple to the rest of the world? Well, the spin, if you multiply it with the g factor, the Bohr magneton has a magnetic moment, and what we simply add here is mu dot p, be the interaction of the magnetic moment with the magnetic field. So in that sense, we have been classical all the way. Now at the end we said, we need the spin. Let's put it on by simply taking an interaction, which is mu dot p. If you don't like that because I try to be very fundamental today and go from first principles, for electrons, you can get that by taking the non-relativistic unit of the Dirac equation. So if you have electrons, you can start with the Dirac equation and do the Pauli approximation, which is the non-relativistic limit of the Dirac equation. Any questions? So these are all the terms in the Hamiltonian. I mentioned, but I don't want to dwell on, that the Coulomb energy has to be separated into a Coulomb self energy and an interaction energy between the charged particles, but all of that becomes just one term in our Hamiltonian. This is the atomic structure. When we assume that atoms have energy levels, all the energy is included in that, so we will not discuss any further the Coulomb energy. We will simply assume we have an atom which has certain energy levels, and that includes all the Coulomb terms. But we will talk a lot about the Hamiltonian for the radiation field. The Hamiltonian for the radiation field can be conveniently written in a and a dagger, but I have a few equations up there I rigorously defined for you operators E and B in terms of a and a daggers, and those two equations are identical. So this looks very quantum, this looks very classical, but if you interpret the electric and magnetic field as operators, we have identical equations for the Hamiltonian of the radiation field. What we want to study are interactions between light and atoms, and of course, this comes from the vector potential. And in particular, when we square it out, the canonical momentum of the atoms times the vector potential has this cross term, p dot a. So let me know just take this Hamiltonian and write it in the way how we will need it for this course. We want to take the Hamiltonian, and I will often refer to that. We want to take the Hamiltonian and split it into three parts, the atoms, the radiation field, and the interaction between the two. The Hamiltonian for the particles, H subscript P, has the momentum squared, and it has the part of the electromagnetic fields which are longitudinal, the energy of which can be described by a Coulomb integral. We have already discussed the radiation field. The radiation field was nothing else than h bar omega, a dagger a, but the new part which we need now is the interaction term. I want to show you now, or remind you by just summarizing the term, that the interaction part has actually three different terms. The first one is the cross term between p and A. The second one, when we had p minus A and squared it, is the A squared term. And the third one is the interaction of the spin with the magnetic component of the radiation field. This is the mu dot B interaction. I don't think it's an exaggeration when I tell you that with those three equations, you can understand all of atomic physics. This term here that's important is second order in A, which of course also means it will be very important for very strong laser palaces, but there's more to be said about it. If you're interested in this subject, and I couldn't do full justice to it, you may want to look at this book from Claude Cohen-Tannoudji, Jacques Dupont-Roc, and Gilbert Grynberg about photons and atoms. The whole book is dedicated how to describe the interaction of atoms with electromagnetic fields. In this book, they go through different ways of formulating the electromagnetic field classically. They use Lagrangian formalism for the electromagnetic field and show the Euler-Lagrange equation or Maxwell's equation. There are lots of different approaches, and also, one can say what we did here, which is the simplest way of describing light-atom interaction, will work in the Coulomb gauge, which is not the gauge you would choose when you want to describe relativistic physics. There is the other gauge, the Lawrence gauge. It makes the field quantization much more difficult, but when you ever wondered, why is the quantization done in the way I did, and you think something I did was arbitrary-- why did I pick the Coulomb gauge? Read this book. There are hundreds of pages which explain it to you. And I'm not joking here, it's wonderful reading. I got the book just looked up a few things and I almost got hooked on it and read more and more. It's a fascinating story how deeply people have thought about how to describe this aspect of the course and how profound the thoughts are. What this book will emphasize-- and this is what I want to tell you in closing about the quantization of the electromagnetic field-- is the following. We were really working hard with classical field equations to describe the electromagnetic field classically by completely eliminating redundant variables. We started with six components of the electric and magnetic field and reduced it to two components, the two components of the transverse vector potentials. And as those masters say, as a result, the field can be quantized with a great economy in the formalism. If you want to have more symmetric formulations, where you don't eliminate the variables using the Coulomb gauge, the problem is you have variables, which are not independent. And now I think if you quantize, you have to formulate auxiliary conditions between the not independent operators. Aesthetically, it may be pleasing because the approach is more symmetric, but mathematically, it's much more involved. Anyway, if you're interested, this book has a wonderful discussion on the different ways how you can approach interactions of atoms with electromagnetic fields. I hope nobody has a question about that, because I'm not able to explain to you much more than that. Anyway, questions? We have derived from first principles the Hamiltonian, and the result is something many of you are already familiar with. Let me now spend the next 10 or 15 minutes on what is called the dipole approximation and the dipole Hamiltonian. We are talking know about one further expansion of our expressions, and this is the multipole expansion. Let me first give you the simple derivation, which you may have seen, but then you can appreciate the rigorous derivation. Either directly or through the vector potential, we have formulated our theory in terms of the electric and magnetic field as a function of position. But what we now want to exploit is that atoms are tiny, and usually, the electric or magnetic field is not changing over the extent of an atom or molecule. In other words, what I'm telling you is the relevant frequencies, the relevant components of the electric and magnetic fields will, of course, be the components which are in resonance with the atoms and molecules. What I'm telling you is that in many, many situations, the wavelengths of the relevant modes of the electromagnetic field are much, much longer than the size of an atom. So therefore, if we pinpoint our atom at the origin, we may be able to neglect the spatial dependence of the electromagnetic fields and replace it by electric and magnetic fields at the origin. Or, if you want to go higher in a Taylor expansion, we use gradients of it. We've talked more about that, actually, in 8.421, when we talked about multipole transitions, but here I want to only focus on the lowest order, which is called the electric dipole approximation, because this is the most important case which is used in atomic physics. So for the electric dipole approximation, we make actually several assumptions. One is we neglect the quadratic term, the a squared term. And now if you look at the A dot p term, and we are looking for matrix elements between two atomic levels, we expand the vector potential into plain waves. Well, it's called the electric dipole approximation, so maybe we want to express things by the electric field, which is the derivative of the vector potential. So with that, we have now described things by the electric field. And what is needed now is if we assume that kr equals 0, or r over lambda is very, very small, the exponential, the plain wave vector, is approximated by 1, and we are simply left in the electric dipole approximation with the a element of the momentum operator. We can simplify it further by writing the momentum operator as the commutator between the position operator and the Hamiltonian. If you take the commutator between r or x and p squared, you'd simply get p and p factors, so that's what we are using. And now the matrix element becomes-- so what we have here now is rH minus Hr, but H acting on 1, because 1 is an eigenstate, gives the energy, E1. And for the flipped part, we have Hr, and now H acting on state two gives us the energy, E2. So therefore, by taking care of the Hamiltonian, which gives us, depending on which side H appears, E1 or E2, we have reduced it to now a position matrix element. And what we have here is, of course, the transition frequency between the levels one and two. I know you have seen it, but if I can derive it in two minutes, it's maybe worth the exercise. What I've shown you is that the momentum matrix element can be replaced by the position matrix element multiplied by the resonance frequency. But now we want to put things together, and what I want to point out for you is that we have an omega here, which is the frequency of the electromagnetic field, and we have an omega 12 here, which is the transition energy. So therefore, if I put it together, I started with a p dot a Hamiltonian, but the p matrix element became an r matrix element. The a was expressed by the electric field. So therefore, we have now the electric field times the position matrix element, or the dipole matrix element, if you multiply the position of the electron with the charge two. But in addition, we have this prefactor. So this becomes the dipole Hamiltonian E dot d when we replace that by 1, assuming that we are interested anyway only in interactions with atoms where the radiation field is near resonance. So it seems that we have actually made three assumptions to derive the electric dipole approximation. One was the long wavelengths approximation. The second one was that we are near resonance, and the third one was that we neglected the quadratic term in the light-atom interaction. There are often people who are wondering about it. When I said we are only interested when the atoms interact with electromagnetic radiation near resonance, that's OK, but atomic physics is the area of physics with the highest precision. We can make measurements with 10 to the minus 16 and 10 to the minus 17 precision. So therefore, if you have an electromagnetic field which resonates with atoms-- the typical frequency of visible light is 10 to the 14 Hertz-- and you're just one megahertz away from the resonance, it's 10 to the minus 8. So the question is, would we observe a correction to the dipole approximation if omega 12 is not exactly omega? Well, the answer is no because, as I want to show you now in the next few minutes, and this is also the end of our flyover over the appendix of atom-photon interaction, the last two assumptions are not really necessary. In other words, we go back to atom-photon interaction, to this long appendix on the derivation of the QED Hamiltonian. We go back to our fundamental Hamiltonian, and we want to do one approximation and this is the only approximation which is needed. It is the dipole approximation, and that means that the vector potential-- of course, we only need the transverse part of it-- is replaced by its value at the origin. That's the only approximation. So here is our Hamiltonian in its full beauty, and the only thing we have done is we've put in the origin here. For later convenience, we introduce the dipole operator, which is the position operator of particle alpha times the charge of particle alpha. And then we do simply a transformation. So this is a unitary transformation, and we transform from the original Hamiltonian, which includes the p dot a term and the a squared term to a transformed Hamiltonian, H prime. And this transformed Hamiltonian has now the electric dipole approximation. It's sort of written in polarization Fourier space, but this is the electric field, and a, a dagger gives position. I'm sorry. These are normal modes. It's a dipole operator times the electric field, and the A squared term has disappeared. So after the transformation, we have no quadratic term, and all we are left is this is nothing else than d dot e, the dipole interaction. And there is no prefactor omega 12 over omega. This is a rigorous unitary transformation to another basis. In full disclosure, it involves a dipolar self energy, which is sort of a constant energy. And if you look at the transformed velocities, the transformed velocities v prime are now identical to the original canonical momentum. So after that, what was the canonical momentum in the original Hamiltonian, H, becomes now velocity times mass. So p becomes now the mechanical momentum. And therefore, p squared is the kinetic energy. Anyway, I think this is important that we have this much more rigorous deviation of the dipole approximation. I don't want to go through that. I just want to make sure you have seen it. If you want to do it really rigorously, then you have to distinguish between the electric field d and E. So you have an electric field here, and here you have the polarization. It's pretty much the same as in classical e and m. At the end of the day, we have now, for the prime operator, for the Hamiltonian after the canonical transformation, after the unitary transformation, the only interaction term which remains is the electric dipole interaction between the dipole and this field d, but this field d prime corresponds to the original a transverse electric field. I'm not going to explain the difference between E, E prime, d, and d prime. It can be done rigorously. But the take home message is-- and with that, I want to conclude-- that for most of this course, the interaction of the electromagnetic field with our atoms is described by this term. We take the transverse electric field. This is our radiation field, and it couples to the operator of the atomic dipole moment. I hope you got the take home message that whenever you have any doubts about any part of this deviation, or if you have any doubts whether it's rigorous or not, you know now where to look it up and learn everything about it. Any questions? OK, then that's it. We'll probably post the first homework assignment tomorrow-- I will meet with the TAs-- and it will be due in about a week, but you will see all of that on the website and hear from me on Wednesday. Have a good afternoon.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

5_Single_photons_Part_1.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. Now, before we start with the prepared questions, I have another question for you. It's sort of taking a straw poll before and after discussing something. And the question for you, I want to ask you is, what is the nature of shot noise? The whole set of questions will build up to teach you something about what is shot noise. In other words, when you measure in photons, you start with coherent beams, you always have sort of square root n. And the question I want you to ask is, when we measure square root n fluctuations, when we detect n photons, is this caused by the nature of the measurement process-- sort of how we detect photons-- or is it an intrinsic quantity of the quantum fields? So I just want to get your intuition, your gut feeling, with the shot noise-- it's maybe not black and white, but if you think it's more related to the measurement-- Well, let me write it down. Shot noise. OK, so the question is, shot noise, is it caused by the quantum measurement process, projection of whatever you know about it? Second part, is it a property of the quantum fields? For instance, if you don't make a measurement, if you run the quantum fields through a beam splitter, we can split it. We can split the shot noise even before we measure it. That would mean quantum fields. And maybe then, third is both. And the fourth is none of the above. So let's just take an opinion. And I'm not discussing the answer at this point. I will come back to that at the end of those 10 or 15 minutes and ask you the question again. So I just want to see what your intuitive understanding is when you measure the number of photons in a laser beam in an optical field and you see fluctuations. Quantum measurement versus properties of a quantum field. Quantum measurement means mainly it caused the moment you detected. And quantum fields means it's there already before, and it can be, whatever, modified. Quantum fields or both. All right. Good. Fine. Let's now go through the question. We first have to make sure that we use the same definition and measure things in the same unit. We want to talk about coherent states. So the first is just little, quick checks. Coherent states are eigenstates of the Hamiltonians of a or a dagger? For the easy question, I will just give you 10 seconds. All right. Stop. Display. Yes. OK. That's good. Next question. Which states realize the minimum quantum uncertainty with delta p delta q equals h-bar over 2? I give you five different states. And in order to use the clicker, you should just answer how many of them are minimum uncertainty states. Ready, set, go. So three or two. Let's just go through them. The vacuum, definitely. The coherent state, definitely. The thermal state, definitely not. Now, number states, they are eigenstates. But let's just take a big number state. It's a big lean in the quasi-probability. It has a huge delta x and a huge delta p. It's definitely-- it's a very quantum like state, very nonclassical. But it doesn't have minimum uncertainty. Final question. x and p, are those minimum uncertainty states? x and p are a little bit cryptic, because one is infinite, the other one is 0. And infinite times 0 is h-bar over 2, yes. So it's a minimum uncertainty state, because we discussed it. It is the limit of squeezing, when we perform the limit that the squeezing parameter goes to infinity. You can send me your lawyer, and I may have to step back from it. But the way I present it, this is one correct answer. It's a limit of an infinitely squeezed state. So if the squeezed vacuum-- which I forgot actually to ask-- if the squeezed vacuum is a minimum uncertainty state, then the x and p state are the limiting case. By the way, your clicks are getting recorded, and the clicks are responsible for half of your semester grade. This was a joke. You can actually use clickers and record them, but then each student would have a clicker with an already set number. But as you see, you just take a random clicker out of the box and put it back in. So I have no idea what your clicker number is. And to use the clicker is solely for our joint entertainment here. OK, next question. What features are unique to nonclassical light? And nonclassical light means-- well, we have discussed nonclassical lights but their different aspects. One is that the roots of the two quadrature components-- the uncertainty of the two quadrature components-- the cosine and sine quadrature components are unequal; negative cosine probabilities; a correlation function, which is smaller than 1; and sub-Poissonian statistics. So I've given you four criteria. How many of those four are features which we will not associate with classical light, only with nonclassical light? OK. Stop. Display. So there's a wide distribution. Let's quickly discuss the answer. Negative cosine probabilities, well, if you use w distribution and p distribution, negative probabilities are always associated with nonclassical light. You can also say it's the definition of nonclassical light. g 2 function of-- maybe, I should not say smaller or equal -- smaller than 1 is only possible for nonclassical light. Sub-Poissonian statistics is also only possible for nonclassical light. Now, the fact that the two quadrature components are different, you can just use some, and you can choose classical states and prepare them, that you have more certainty in the sine quadrature than in the cosine quadrature, so definitely not. OK, now, we really need prominently to discuss quantum noise and shot noise, the coherent state. So just to make sure that we relate-- I will relate later everything to photon number. Because photon number is intuitive. You can count, you can feel the photons fluctuation of square root n and immediately tell you Poissonian, if it's less, sub-Poisson, all that. So I want to make sure we are correctly associating photon number with coherent states. So, therefore, the first question is if you have a coherent state, alpha, and we detect all the photons with a photodiode of 100% efficiency, is the number of detected photons alpha or alpha squared? Yes. Alpha is, so to speak, the field. And alpha squared is the intensity. So, therefore, form now one, I will always use units for the photocurrent in such a way that the photocurrent, or the integrated photocurrent, is the number of photons. So when I say, what is the current associated with a coherent state, it's just alpha squared. Just one side remark. If you use a laser beam, you sort of have a stream of photons, but you have a time constant in your detector, and every millisecond you do a measurement. And then, what I mean by alpha or alpha squared is the number of photons which arrive during the time constant. So that's sort of the mapping. We all usually talk about coherent state in an isolated system. But if you have a beam, you just chop the beam into pieces and associate that with your coherent state. OK. So now, we are talking about the measurement of coherent beams. We will have now lots of coherent beams, running through beam splitters, being mixed at beam splitters and such. And we will always characterize the strength of the coherent beam by alpha squared. And that means in photons. So now, the next question is-- again, just to make sure we are all on the same-- we have the same basis, what is now the variance? If I measure the fluctuating current, I measure the current, I have a coherent state, and I repeatedly measure the current, it will show fluctuations. And the question is, what is not the standard deviation of the current, what is the variance, which is the standard deviation squared? Yes, the standard deviation is square root n. The variance is n. OK. Stop. Hide. OK, now finally, now we can have fun with beam splitters. So, so far, we have really just repeated a little bit of definitions of classical and nonclassical light at coherent states. We measured current. We know now in what units we need to measure the current. And we characterized the fluctuations in variance not in standard deviation for the rest of this unit. But now, we run the coherent state through a beam splitter. And we have two photodiodes, one i1 and one i2. And we measure the current, i1. What is the variance now in the current you measure? OK. It is A. It's n/2. The way how you think about what it is, when you split a laser beam with a beam splitter, you get two laser beams. Each is half the intensity. So therefore, each creates a photocurrent, which is n/2. And since one laser beam, which is a coherent state, gives two laser beams, which are two coherent states, each of them has a variance which is equal to the photon number. And this means, we have Poissonian statistics. So we have n/2 photons in each beam. The standard deviation is the square root of n/2. And the variance is n/2. Ask me, if there are any questions. I mean, this is still the preliminaries. We are not yet bringing in squeezed light. Then, it gets really interesting. OK, we will always refer to that situation. We have a laser beam. We split it. We create-- and that's actually your homework assignment. We'll actually show wonderful operator algebra. If you split a coherent beam, you get two coherent beams. And they are simply characterized that the alpha value is now down by square root 2. The number is down by 2. And if the number is down by 2 and you have a coherent beam, the variance is down by 2. Poissonian statistics. OK, next question is we have a beam splitter. We have two photodiodes, i1 and i2, and now we measure-- it's important for balanced homodyne detection-- we can now take the two currents, and we can add them or we can subtract them. So the question is, if we add the two currents, what is now the variance in the two currents? Remember, each current had a variance of n/2. So now, you are asked, when you sum the currents of the two photodetectors, what happens to the two variances? Do they sum up? Do they--? I don't know. Do people want to-- does somebody want to defend his or her choice? What do you think is the variance? AUDIENCE: It should be the answer, you just measure before the beam splitter. PROFESSOR: You said it should be as if I measure before the beam splitter, and then it would be n, because we have n photons. Well, you can also see if you put in a mirror here, and you shine the two beams on the same photodiode. And then you would say you have n photons. And if there are Poissonian fluctuations, you would get n. But there is one thing you should keep in mind. Since we have the beam splitter here-- this is the correct answer. But I want to point out one subtlety for you, which will become important later. The beam splitter has an open part with the vacuum. And so when you said we can measure before or by sort of just combining the two beams with a big photodiode, we can measure after, that's now a question for you-- which, maybe hold the thought and ask me later, if it doesn't become clear with the following question. Hasn't the vacuum entered here, and hasn't the vacuum, which has entered here added fluctuations? So if you put in a beam splitter, and we mix in the vacuum with our coherent states, shouldn't that possibly mean that if you use a big photodiode and measure everything together, that we measure the variance we had before. But isn't that possibly a contribution of the vacuum, because we were opening the door to the vacuum? Please hold the thought. OK. The next thing is now, you can't have the solution that we take sort of a big photodiode and combine the two beams, because I'm asking you now is, what is the variance in i1 minus i2? So you subtract the two currents. So the question is, if you have technical noise, the two beams would be fluctuating, and you would form i1 minus i2. All the technical noise would cancel away. But what is shot noise? Is shot noise technical noise? Is shot noise that every time you prepare an ensemble, there is a hiccup, and you can split it and subtract it? So now, it's getting really interesting. To what extent does a beam splitter-- does a split coherent state, which has shot noise, and you now perform the difference of the measurement, can you get rid of some of the shot noise or not? You vote, please. Yes. So it doesn't matter whether we measure the sum or the difference. So it seems, we have two beams. And it seems that the noise in i1 and i2 is completely uncorrelated. And whether we perform the sum or the difference, it doesn't matter. As long as you think about classical states and coherent states, you can get away with this notion. But now I'm going to add some squeezed light. And now, it's getting more subtle. So we have exactly the same situation. We have our beam splitter. We split our coherent beam into two beams, exactly as before. But now, I'm playing with a vacuum. I'm playing with the open part of the beam splitter, and I put in squeezed vacuum. And just to be specific, I've squeezed the vacuum in such a way that what is narrow now is the quadrature component of the coherent state. In other words, the coherent state is cosine omega t. In sine omega t, we haven't been put in any photons. Our coherent state has really a classical oscillation in cosine omega t. And what I have squeezed here with the vacuum is the noise in the cosine omega t quadrature component. So in other words, the coherent state and the short axis of the ellipse are the same quadrature component, which means, they can now interfere. So the question is now the following. We split the beam. We have n/2 photons here. We have n/2 photons there. And what is now the result for the variance? I formulated a condition up there, which is the strong local oscillator condition. You should assume, if you get confused, that the local oscillator is very, very strong. And if you have any doubts about some smaller terms, just crank up the power and make sure that it's really the local oscillator which dominates, which is the biggest number in all equations. OK. So the reference is without squeezing, the noise, or the variance, in i1 was n/2. With n/2 photons, Poissonian statistics was n/2. The question is now, when we use squeezed light, does the variance stay the same? Does it go up? Does it go down? Or does it go to 0? By the way, at the end of the clicker question, if you come to me after class and honestly declare you got 100% on all questions, I buy you a lunch. OK, you're done? What have you learned? It's almost random. I will give you a little bit formalism later, but I mean, let me say one thing. I had a few discussion with students after class, and I came to the conclusion after last class that I could teach today about the Deutsch-Jozsa algorithms' wonderful applications of all that, but I came to the conclusion, you can go to the Wikipedia and read about yourself. Because this is simply using beam splitters, homodyne detection. It's just using kind of all the tools. But what I said instead is, I'd rather spend half the class today in trying to give you, with very simple examples, an intuitive understanding. I mean, everything I'm saying you can do with operator algebra. But to some extent, I feel when we learn physics, we use operators, we use equations, but in the end, we want to shape our intuition. And what I try to sort of challenge you today is, what happens to noise? So what I have to tell you now is-- OK, got to hide this. The way you should look at it very soon, I hope, is the following, that when you split the coherent beam, you also split the noise. In other words, your little uncertainty circle goes down in the amplitude by square root of 2. It is split-- the reflection coefficient of the beam spit and transmission coefficient is 1 over square root 2. But what happens is, you let in the vacuum here. And so, the vacuum comes into the transmission coefficient of 1 over square root 2. So what I want to teach you is that when we think back to this example and we had this situation with a vacuum, yes, it looked to us that we had n/2-- so this was i1 plus i2-- we had n/2 photons and the variance was n/2, it looked completely Poissonian as if nothing has happened. But what you should understand is, that the beam splitter has taken out-- in amplitude, 1 over square root 2, when I square it-- half of the noise out of the coherent state. But half of the vacuum state was added. And half of the vacuum state and half of the intrinsic noise of the coherent state give you back this circle of unity. And you can pretty much neglect all subtleties of the quantum field. But now, by messing up with a vacuum, by squeezing the vacuum, half of the shot noise comes from the beam splitter. It's a 50-50 beam splitter. But the other half comes from the open part. But if you squeeze it strongly, it's on the order epsilon. That means, it doesn't contribute. And the correct answer is now, we have-- just a second. It is linear or quadratic? Variance is quadratic. The correct answer is here, because this is epsilon; this is small. We only get half of the noise we had before. We had n/2, and now, we have n/4. OK. Now, we take exactly the same situation. So the thought is that you measure the fluctuation here and you have only half of them, because this squeezed vacuum, you've squeezed out the noise from the vacuum part. And since you have a strong coherent state, everything which is sine omega t doesn't matter. Cosine omega t is homodyned, everything fluctuation and cosine omega t. When we measure, the intensity is the square of phi, or it's a dagger a. And so if you measure a quadratic quantity and we have a very strong local oscillator, everything is projected on cosine omega t. This is why only the cosine omega t fluctuations of discrete statement are here. It's a strong local oscillator limit. OK. Now, I hope you've learned something. Next question is now, we do exactly the same thing, but now we have the two currents, and the question is, what is the variance in i-plus? I add the two currents. What do I get? And I give you the same choices as before. Just as a reminder, when we did not squeeze, the variance was n. So the question is now, has this squeezing done nothing, or has it reduced the variance? What is happening? The correct answer is C. It's n. And I will show you-- I owe you a little bit of math now. But the way how you should feel is the following. The noise of the coherent oscillator is equally split. But the amplitude-- I mean, you can say 50% go through, 50% go through here. And if you add the two currents, you get back the full noise of the original coherent state. You would say what about the squeezing? Well, the issue is the following. The squeezed light, the squeezed vacuum, is reflected with a reflection coefficient, which is minus 1 over square root of 2. You know when you have two beams reflected, by unitarity, it's necessary that one has a negative, the other one has a positive reflection coefficient. So therefore, you have this squeezed vacuum appears in this arm with a minus sign and this arm with a plus sign. And if you sum up the two currents, you lose everything which is related to the vacuum part. So when you measure i1 plus i2, what came from the vacuum part entered here with a minus sign, here with a plus sign. And therefore, you [INAUDIBLE] what you've done at the vacuum part. This is when I told you earlier, you measure i1 plus i2, and we all agreed, the variance was n. And I said, well, this is exactly the same if you put a photodiode-- big photodiode-- behind the beam splitter or before the beam splitter. The vacuum has not contributed. That's what it seems. But even if you squeeze the vacuum, it cancels out. OK. Now, the obvious next question-- but I think I've given you the answer-- is what happens when we now measure the difference, i1 minus i2? Again, without squeezing, the variance was n. So now, we measure i1 minus i2, and what happens? OK. The correct answer is this. Well, when I say 0, it's always epsilon squared terms which I neglect. I just assume strong squeezing. I mean, I just want to give you the most conceptional problem. What happens is the following. And I will show you the equation. The noise of the coherent state is split equally. So we have a delta a-- a is the field operator, or the quadrature operator here-- and a delta a here. And therefore, the original shot noise is common mode, because the reflection and transmission coefficients are the same. And if you subtract the current, you completely get rid of all noise, which was originally associated with your coherent state. Now, what happens is the small noise we get, but now, we have to consider the vacuum part. But what comes in on the same quadrature component point on the vacuum tube is strongly split. And because the beam splitter has different signs for reflection and transmission, the noise is anti-correlated. And if you take the difference, you will actually measure the noise of the input of the open part of the beam splitter, which would mean, in terms of order epsilon-- I didn't work it out. I think it's epsilon squared. But it's a power of epsilon. So anyway, it seems your intuition fails here. So let me try to give you a few equations which explain that. In your homework and in most of the course, we really want to work with operators. We do transformation of operators and such. But I've found-- for shaping my intuition-- I found a simplified approach very helpful. And I've posted two references for you. One is the article by Schumacher, and one are some older lecture notes, where I summarize some results of this article. And this goes as follows. When you have a coherent input, you can say you have a deterministic classical field. And then, you have some noise in one quadrature component, which is cosine omega t. And the i is a reminder that there's also noise in sine omega t. So you can really say the electric field has a sharp value, but then, there is some fluctuation in the coefficient of cosine omega t and a little bit fluctuation in the coefficient of sine omega t. And what is important now is, if you shine that on a photodiode, this is sort of the field. The current is proportional to the square of it. You have to be a little bit careful where you put complex conjugation, but it is the square of it. And now, if we use the strong local oscillator assumption, we only take alpha squared, which gives us a number of photons. But then, those fluctuations are homodyned. We only take-- we don't consider that the square of the fluctuation. We only homodyne it. We use, at the cross term, where we multiply with alpha. So if I now put in photon numbers for alpha, we get n photons. Yes, OK. And now, we get some noise. Well, the noise is the current squared. The noise is-- we take the average of the current squared and subtract the average of the current squared. So if you do that-- this is the current squared-- the current squared average, and the average current squared takes that away. So we get this here as the variance. And now, I have to tell you that in the units I'm using here, this delta a1 squared for coherent state-- so if I show the standard circle-- the standard Heisenberg uncertainty circle-- this quantity is 1/4. I've just kept track of my constant. So therefore, we obtain in this picture that the variance is n. And this is Poissonian light with a coherent state. And the fluctuation squared of the current, n, this is Poissonian statistics, square root in fluctuations of the current. The variance is n. But what you should realize is one thing. The moment I have a strong coherent state, the sign quadrature component doesn't matter at all. Because if you multiply cosine with sine and average, you get 0. So we only consider the quadrature component, which is given by the coherent state. This is a principle of homodyne. You set your quadrature component with your local oscillator. OK. But now, I have a beam splitter. And the way you can think about it is, you can get everything in operator algebra, but just take my word or follow the procedure now that we are saying, we can always think about a quantum field that it has an average value and some fluctuations. And if you pass it through an optical element or a beam splitter, we have a transmission coefficient, but it multiplies the mean value of the field. But it also multiplies the fluctuations of the field. So that's why I'm saying the beam splitter is actually splitting the fluctuations of the field. And therefore, you get reduced fluctuations in either beam. But we have to consider that there may be vacuum, and the vacuum has also some fluctuations and will reflect in or transmit through parts of the vacuum. So now, we want to understand photocurrents. For photocurrents, we take those quantities and square them. So which one should I take? Well, let me take the first line. So this is i1. So if I calculate now i1, I obtain t squared alpha squared, plus the cross product between the first two terms is 2 times t alpha times t delta a1. And then, I get a second term from the vacuum, which is 2tr alpha delta b2. And let me assume that we have a balanced beam splitter, 50-50. So t squared is simply 1/2. Alpha squared is the number of photons. So this term becomes n/2. t squared is 1/2. It cancels with the 2. We are left with alpha. And this is shot noise. Here, appears our square root n. So you can say the shot noise comes because we have homodyned the quantum noise. The quantum noise, which is the same for all coherent states, has now been multiplied with square root n through the homodyne process. OK. So I lost my line. That's delta a2. And then-- and this is important-- because of the mixing at the beam splitter, the quantum noise off the vacuum gets also homodyned by a factor of square root n. So now, I think you see what I told you before. The transmitted beam has half of its noise coming from the coherent state, and half of the noise comes from the vacuum. So if I would now take the current, square it, calculate the variance, I would find that I have n/2 photons with a variance of n/2. But an identical contribution of the noise will come from the original coherent state and from the vacuum. So therefore, when I replace the vacuum by a strongly squeezed vacuum, the beam, the current i1, lost half of its shot noise. We had n/2 photons with a noise-- with a variance, which was n/4. Any questions? Yes. AUDIENCE: Why are the subscripts on the delta operator of [INAUDIBLE]? PROFESSOR: Because I should have dropped them all together, saying we are only using the quadrature component of the local oscillator now. Everything else is orthogonal to it and cancels out. But if you want me to put them back, they should all be equal. I compiled this writeup from different lecture notes, which one had the cosine, one had the sine component. I just made a [? stitching error. ?] OK. So this was i1. If I would now look at-- let me change the color-- at the current, i2, the only anything which changes is that because the quantum noise is transmitted with a positive coefficient, that I have to put a plus sign here. And now, you realize the magic. When you form i1 plus i2, whatever came through the vacuum part, or through the open part, cancels out. Therefore, when we calculated-- or when I asked you for the variance in i1 plus i2, it was independent, whether there was squeezing, whether whatever was at the vacuum part. And therefore, it doesn't matter whether I put my power meter, measuring both currents-- both beams-- before the beam splitter or after the beam splitter. But-- yes, a question? AUDIENCE: This last really depends on the convention of where we put the negative sign in reflection? Because if I had put the negative sign here and a positive there, then I would have been canceling my delta a's instead of b's? [INAUDIBLE] the process? PROFESSOR: Yes. Well, so you should [INAUDIBLE] conventions here. It's clear that you have to be careful. We're talking about [? face-sensitive ?] detection. So what happens is-- number one is-- there is really physics associated. When you have a beam splitter, one, let's assume you have a piece of glass, which has some coating on it. One beam is reflected when it goes from the vacuum to material interface. The other one is reflected from the material to vacuum interface. AUDIENCE: So then, it depends on which face of the beam that will be put which--? PROFESSOR: But one thing is absolutely clear. The two reflected beams, one of them has a minus sign. So therefore, when I do i1 minus i2, the contribution of the-- AUDIENCE: --If a's will cancel. PROFESSOR: Oh. It is clear that in i1-- OK, yes. But the question is whether we have to perform i1 minus i2 or i1 plus i2. Well, I don't know if it helps you, but what happens is we have a propagating laser beam. If you would simply move your mirror by half a wavelength, you get a minus sign. So eventually, all of the phases-- the beam splitter phase, the propagating phase-- all the phases have to be controlled. And in essence, what you do is you put an experiment together, you measure some noise, and then you move your mirror or you change your phase. And then, you really figure out which noise is now subtracting and which noise is constructively interfering. Other questions? OK. But the other thing is now obvious. When we measure i1 minus i2, everything related to the coherent input, to the original noise of the beam, cancels out. And what we are measuring is only the noise which came in through the other part. So this is the reason why, if you have interesting light you want to measure it, you use a local oscillator. But then, you do i1 minus i2. And this is what I presented to you at the end of last class, is the way how people have measured the signature of squeezing. Any questions? And finally, I think this is one of the coolest things. Now, we want to talk about the displaced squeezed vacuum. Now, we put everything together. Remember, we have learned by applying the squeezing operator, how we can squeeze the vacuum. And then, with the displacement operator, I can move the displaced vacuum, that there is an expectation value, alpha, for the field. So it now looks like a squeezed coherent state. So this is now the state we have generated. And I explained to you that such a state can be generated by using a beam splitter in the limit where the transmission coefficient goes to 1. So the squeezed vacuum is pretty much transmitted without loss. But now, we have a very strong local oscillator. Also, the reflection coefficient goes to 0. We can just crank up the power in alpha in such a way that r alpha squared-- the number of photons which get reflected-- is n. So the limit we are looking at it is small n and large alpha. But what we keep constant-- throughout actually all those questions-- is that we have n photons in that beam. And I explained to you in class-- and I showed it to you with operators-- that this setup is really creating the displeased squeezed vacuum. OK. Now, we have one laser beam, which has an average value of the field, alpha. It has certain properties. And now, I want to ask you, what is the variance of this beam? We measure the photocurrent of this beam. We measure-- of course, the photocurrent is n photons. But what is this variance? So the choices are we have n photons. Is the variance n? We learned before, with the 50-50 beam splitter, that we were able, by squeezing the vacuum, to eliminate half of the shot noise. Maybe what I've done is wrong, and we get 2 and we increase the noise? Or does the noise really go to 0, which means it's on the order of epsilon, whatever I define for the squeezing for the short axis of this squeezed vacuum. So now, we are not subtracting currents. No i1 minus i2. We have one beam hitting straight the photodiode. There are n photons per unit time. And the question is, what are the fluctuations of this beam? What is the variance? Well, the correct answer is this one. What happens is the following. In terms of fields-- you should go through that again a little bit more slowly-- but what happens is the coherent state is sort of this disk. But when we make alpha larger and larger, the disk of the coherent state is the same. But now, we have of a reflection coefficient, which goes to 0, such that the product of r alpha stays the same. But that means the reflected beam, we have divided the disk by the reflection coefficient-- multiplied with the reflection coefficient-- so therefore, the original noise of the coherent beam has been completely eliminated because of the small reflection coefficient. So the only noise which is there, is from the other part. But we have strongly squeezed the noise in this quadrature component. And therefore, the noise goes really down to 0 or to I think it's epsilon squared when epsilon is the normalized widths of the narrow part of the ellipse. I've written down the math for you here, which is the following. We have a reflection coefficient for the coherent beam and the reflection coefficient for the fluctuating fields. Whereas, the squeezed fluctuations, they are fully transmitted. And what I'm completely neglecting is-- and you'll realize it already-- the other quadrature component. Because the moment I square it to obtain the current, I'm only considering parts, which are in the right quadrature component, that they get enhanced by alpha. So if I square that and neglect all terms which do not get multiplied with alpha, I get this line. And now, you see that this term here-- the one which had the fluctuations of the coherent beam-- is what I just said. r times alpha is n. So this is n. It would be the normal variance. But now, we have another power of r. So therefore, delta a1 really got diminished. I mean, you see that actually already better in the previous line. So therefore, in the limit of r, in the limit considered, there is no contribution from the fluctuations of the coherent beam. The only fluctuations come from the squeezing. And therefore, the variance is, well, 0 or epsilon squared. So you can create a beam, which has no fluctuations-- or almost no fluctuation-- in the photon number by mixing a strong local oscillator with a squeeze vacuum. And that would mean now, if you take this beam and send it through a cesium cell and do spectroscopy of cesium, you have reduced noise, and you are more sensitive to obtain, to observe, a small quantity of cesium. I've posted it actually-- this is exactly what I described here, how the first spectroscopy with squeezed light was done. And I've posted the paper of Jeff Kimble's group on our website. So I'm wondering, by going through these examples, if I would ask you that the shot noise, which is square root n in the fields and n in the variance, after looking at the beam splitter and the modes which are involved, how would you now answer this question? Is the shot noise related to quantum measurement, is it related to quantum fields, both, or none? That's interpretation of physics; I think there is no right and wrong. I would actually very strongly-- my opinion is, it's B. It's really the quantum fields. The quantum fields which are split. If the noise were done in the measurement, you cannot have constructive and destructive interference due to the positive and negative reflection coefficient. So for me, the noise, which is responsible for shot noise, is a field which can be split where you have constructive and destructive interference even before the measurement process takes place. But this is sort of my interpretation. And let me maybe point out why I think that for those cases, the quantum measurement process is not relevant. Here, we've always talked about square root n fluctuations, where n is big. So the fluctuations of the coherent field-- all of the fluctuations we were discussing here-- had many, many photons. So to some extent that we have maybe just a single photon, and we can only get one click and not two clicks, doesn't measure it. That doesn't matter here. All the fluctuations we considered here were fluctuations in the field, where many, many photons were involved. So, therefore, I think the quantum character of the measurement process does not really show up. If I would construct something-- other examples-- where we wouldn't always have this strong local oscillator, this strong homodyning. Then I think we would find quantum noise of a single photon, and then I would say, we're getting closer, that's really quantum measurement. But here, I would say the short noise related to a strong local oscillator-- at least, I find it easier for go through those questions if I think about it as a quantum field, which can be split, added up. Constructive and destructive interference. Good. Questions? Yes? AUDIENCE: So, when we're talking about [INAUDIBLE], can we actually talk about it without referring to i, because that makes me think of measurements. PROFESSOR: You mean i, as the current? AUDIENCE: Yeah. PROFESSOR: But we have to do something. OK, what happens is i was nothing else than the expectation value of a dagger a. In other words, whenever we had a photodiode-- I mean, we learn how to transform modes, and i was synonymous to performing a measurement, and the measurement is a dagger a. And when we talked about i squared, in order to get the noise, the operator involved was a a dagger, a a dagger. I think I will ask you something about the displaced. Yeah, it's too nice a problem. I should ask about the displaced squeezed vacuum in the next problem set. I really want you to calculate, maybe with a higher order term-- it's a three-liner-- in operators, what is the noise? And in your calculation, photodiode will not appear. What will appear is "Calculate the expectation value of a dagger a. If you call it a measurement process, it's [? inobservable ?] and it's an expectation value of an emission operator. Other questions? AUDIENCE: [INAUDIBLE] understand the [INAUDIBLE] for a stream of particles, which [INAUDIBLE] with a constant probability per unit time? [INAUDIBLE] The detector is also [INAUDIBLE]. I think you also get the square root. PROFESSOR: Well, a lot of classical counting statistics needs to short noise. If you have independent particles with a certain probability, then you get short noise. But I would sort of press you hard to make a classical explanation for the last example I gave you, where we had a strong local oscillator. We reflected it and superimposed it with squeezed vacuum. At that point, all the short noise has disappeared. So therefore, I would be hard-pressed to connect it with the classical probability of detecting a photon. Of course, classically speaking, the empty part of a beam splitter is empty, and squeezed vacuum doesn't exist classically. Yes, [INAUDIBLE]. AUDIENCE: So it might be kind of a silly question, but we say we have a minimum uncertainty state when delta x delta p is 0. But a lot of times, we talk about the number-phase uncertainty. So, I'm just curious, because the coherent state has the same circle of-- the area of the circle is the same, right? I know this is wrong, but can't you always make the delta m delta phi uncertainty very small by having a very large n? Delta n stays the same, but the angular spread always goes down as you make the circle farther away from the origin? PROFESSOR: So you're talking about an uncertainty relation we have not discussed. In the last homework, you looked at it. So what happens is if you have a coherent state, you would say the phase angle phi has this uncertainty. And then there's an uncertainty relation. AUDIENCE: Maybe you should have talked about this problem yesterday. I think the uncertainty and the size of the disc is constant, [INAUDIBLE] in alpha. But alpha is proportional not to n, but the square root of n. [INAUDIBLE] an arrow propagation [INAUDIBLE] square root of n. If we move the circle further away, the uncertainty alpha stays the same, but the uncertainty of n goes actually up. But we don't see this [INAUDIBLE], because we only have alpha, not [INAUDIBLE]. PROFESSOR: Yes, you have to be careful. That's another comment I wanted to make, but thanks for reminding of that. When we have coherent states, it looks that the uncertainty is always the same, because it's a disk. But we learned today, and in the last class that the uncertainty of the coherent state when you measure the photocurrent is square root n. What happens is the following: the uncertainty-- and this is what I was referring to-- the uncertainty, let's say delta a in one [? quadrature ?] component, is constant for the coherent state, independent of n. But that would mean what you really calculate, when you calculate the photocurrent, is alpha plus delta a which is constant. But then you square it. In the limit of a strong coherent state, you get alpha squared, which is the number of photons, plus 2 alpha times delta a. So the delta a, which is constant, gets multiplied with square root n here. So therefore when you go to launch a coherent state, you increase the number uncertainty, but by the same factor, you reduce the phase uncertainty. Well, I thought this would take half an hour. It took an hour, but I hope it was time well spent. So let's get started on the next unit, which is fully dedicated to single photons. So after learning about different aspects of the electromagnetic field, we now want to focus on the purest quantum states, which are single photons. And this unit it is definitely motivated by quantum information processing. We want to use control over single photos to realize interesting devices. We talk about beam splitters and phase shifters, but then we put beam splitters and phase shifters together to make Mach-Zehnder interferometers. And then eventually I can show you how Mach-Zehnder interferometers with a non-linearity can be used to realize quantum gates. But let me make maybe one general comment here. This unit here on single photons has actually two purposes. It tries to give you some fundamentals, which are really necessary to follow recently literature in quantum information science, so we introduce some concepts of gates and truth tables and such. But for those of you who are maybe less interested in quantum information science, I have to say that as a more traditional atomic physicist, I've never worked in quantum information science myself. I started with atomic molecular spectroscopy, and advanced from laser cooling to quantum gases. I feel that using the quantum logic formalism is a very elegant way of organizing your thoughts. So I do feel even for the rest of the world, who is not doing kind quantum information process, which is a pretty big fraction of AMO science these days, even for the rest of us, to use concepts of quantum logic and quantum information is very, very powerful. So in other words, I would say right now every atomic physicist should have some knowledge in what happens to beam splitters, what happens to singular photons when they're manipulating, and the way, the approach, how we formulate it using concepts of quantum information processing is extremely powerful. Like in quantum mechanics, you need to know the Schrodinger representation and the Heisenberg interpretation. You can't get away with one. Similarly, for some aspects of the propagation of quantum states, you should just use the language of quantum information processing. Just sort of to give you already an outlook where I think everything comes together, I've taught for 15 years, 20 years, about decoherence and master equation. And I know how to derive it from general quantum physics, and I will show you that later in this course. But then I learned from Professor Ike Chuang, when I co-taught the course with him, that he could actually make a model for decoherence by just passing light through many, many beams splitters. So there is a beam splitter model for decoherence, which is just beautiful. You can give derive a master equation by just sending light through many, many beams splitters. So that taught me something, that actually those was elementary devices for quantum information science can really act as conceptual models for more complicated, more general things. So I've decided now, over the last few years, to make quantum information language an essential ingredient of this course. But I should say over the last couple of years, I've de-emphasized some of the more formal algorithmic things, but I'm focusing on the concepts. And I think some concepts are really highlighted by using quantum information aspects. So with that introduction, we want to talk about single photons. And I want to introduce for you the single photon qubit. Now, [INAUDIBLE] would think, if you talk about single photons, you automatically have one qubit. 0 is no photon. One is the single photon state. We make in addition later on, this is not the most robust qubit, because if, due to unavoidable losses or limited quantum efficiency, you have one photon, and you lose it, you think the bit has been flipped. So you don't want to really use that, but we will use, in a moment, a single photon in two different quantum states. And then, if you detect zero, you know you've lost it. We will talk about one photon, which is either here or there. And this is zero of our qubit, and this would be one of our qubit. But we will come to that later. I first need the Hilbert space of zero one to introduce a few things, and then we'll use what we've learned to define the dual-rail single photon state. But for now we don't have to make this distinction. The question we want to answer is what happens when those states pass through optical components. I want to show you how beam splitters and phase shifters can be used-- very simple optical components-- they can be used to realize an arbitrary single qubit operation. And then we introduce non-linear interferometers, and we get two qubit operations. So let's start out very simple and talk about phase shifters and beam splitters. And this is pretty much how far we will get today in this chapter. Well, if you have a single photon and time passes on, the single photon gets phase shifted. If you pass it through a dispersive medium, it will have another phase shift. So this is how phase shifts can be realized. But there is one caveat, and that immediately tells us we can't get away with just a single photon in one fiber and phase shifting it, because everything phase shifts a photon. Time and everything which is in the beam path. So whenever we talk about the phase, we need a reference. So therefore, we're immediately drawn that one mode and one photon is not enough, even if you're interested in one-photon gates, we need two modes. Just think about two wave guides or two optical fibers. We call them mode A and B. If you put one photon in-- and this is now the symbol for a phase shifter, a little box with the letter phi-- then we get a phase shift of the state one. And the second mode A acts as a reference mode to detect the shift of phase, let's say by interfering the two photons. So in general, we have an input state, and we obtain an output state, which has a phase shift. We'll come back to phase shifters later. So we'll have two modes. The next thing we can do is we can combine modes, and this is done with a beam splitter. So we've talked about beam splitters in the context of homodyne detection. We've talked about beams splitters most of the first hour with a clicker question, and I was mainly referring to the beam splitter before in a rather non-formal way. If you've a 50:50 beam splitter, and you have two inputs, you get a symmetric and antisymmetric combination of the two, if it's a 50:50 beam splitter. What else can you get? Or, I used the concept of just reflection and transmission coefficients. I think that's all important to get an intuitive feel for the beam splitter. But now I use a more rigorous approach. I have a beam splitter, which is characterized by an angle theta. We will later find out what theta is. And we have an input mode A and a second input mode B, and those modes get mixed now. We have output modes B prime and output modes A prime. And we can use the Schrodinger picture and talk about the photon states in those modes, but I prefer to use the Heisenberg picture, so when I put A, A prime, B and B prime in, just think about annihilation operators which annihilate photons in those modes. So what I postulate is that this system is described by the following Hamiltonian. And I'm sort of telling you, this Hamiltonian describes this beam splitter. You will say, why and how? Well, I will tell you what this Hamiltonian does to the modes, and then we have the quantum description of the beams splitter, and you will then see in your homework that this quantum description of the beam splitter does exactly what it is expected to do to coherent states. It just splits a laser beam into two, with a reflection and transmission coefficient which are cosine squared theta and sine squared theta. And I will also show you that this beam splitter is doing the same, splitting with cosine squared and sine squared theta probabilities, a single photon. So in essence, I give you the Hamiltonian, I discuss that the transfer matrix, which is built upon this Hamiltonian, does everything you ever wanted a beam splitter to do, and therefore you can say, this defines the beam splitter. OK, if you have a Hamiltonian, that means after propagating through the Hamiltonian, we have the following transfer. You may be familiar with putting the unitary time evolution here, and the time which has elapsed, but we are always using a fixed time after the light has transformed the beam splitter. So you can see the time t is absorbed in the definition of h, the Hamiltonian, and b is really the unitary time evolution from [? being ?] before and [? being ?] after the beam splitter. So the transformation, which transforms the modes at this beam splitter, has now the following unitary operator. And we have modes a, b, mode one and two, before the beam splitter. After the beam splitter, they have been transformed to a prime and b prime. And we obtain that by taking the input modes and doing unitary time evolution. You do simply the operator transformation for unitary transformation. So we can now transform the operators a and b. I'm not going with you through the exercise how you transform an operator. I think in undergraduate quantum mechanics classes, you do the Baker-Campbell-Hausdorff formula, how to use with operators in the exponent. And if you do with that, you find-- not surprisingly, a beam splitter is a linear element-- that what you get is a linear combination of a and b. So if you don't like me postulating an operator and saying, "Believe me, this is a beam splitter," you can reverse engineer. You can say, this really looks like the modes, the operators a b get linearly superposed with cosine and sine. This is what a beam splitter does, is it mixes two modes. And then you can reverse engineer and figure out that it's exactly the Hamiltonian above which does it. So your homework will show that this happens exactly for coherent states. So if you've a coherent state with alpha, you get a coherent state with alpha cosine theta in one arm, and alpha sine theta in the other arm, and you will see that everything works beautifully. I have to use one convention now for a 50:50 beam splitter. And I use a convention that I chose theta to be pi over 2 b and 90 degrees. And we want to use symbolic language, because we want to draw diagrams of multiple beam splitters and such. So this tilted square is the symbol for beam splitter. We have input ports a and b on the left side. We have output ports, which are now b prime and a prime. And for the 50:50 beam splitter, the modes are symmetric and antisymmetric superpositions. So this is the symbol for the unitary transformation b. And since one output port has a minus sign, it's antisymmetric, and one is positive. We usually put a dot there which distinguishes the two different ports. So this is the operator B. We have already used, for the unitary transformation, the operator b dagger. So for the operator b dagger, using the equations above, we have input modes a b, but what happens now is for the operator b dagger, the output modes are interchanged. So we have a plus b over square root a, a minus b over square root 2, and therefore the symbol for b dagger has the dot here. I just want to warn you that there are other phase conventions in the literature which are equivalent. But sometimes if you read a paper and all of the minus signs appear somewhere else, well, people have a different phase convention. You can postulate that you have a beam splitter without having the i here. And everything works similarly. It's just that, for instance, in this equation, you get an i here. So imaginary units and plus-minus signs and that busyness are phase conversations. If you'd say, hey, but if you another definition, don't you have another beam splitter? Yes, it is another beam splitter, but I can always get it out of my beam splitter-- which now is our beam splitter-- by putting a phase shifter afterwards. So it doesn't limit the generality of the discussion by defining our beam splitter in this way. OK, time is over. Reminder, we have another class this week on Friday, and finally, your graded homework is here. Have a good afternoon.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

19_Bose_gases.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, good afternoon. So this week we talk about quantum gases, ultra-cold atomic gases. And sure, they're ideal Bose gases, ideal Fermi gases, but I will spend one or two minutes on each, because that's exactly solvable. It's in all the textbooks. That's simple. Quantum gases become interesting because of interactions. And in my lecture today and my last lecture I want to introduce you to Bose gases and Fermi gases with inter actions. And they both turn superfluid, and the superfluid properties are determined by the interactions between the atoms. So the purpose of those lectures is number one, to acquaint to you with important phenomena in cold gases-- superfluidity, superfluidity in lattices, superfluidity just in free gas, in a normal gas without lattice, and superfluidity of fermions. But at the same time, I also want to have sort of a theme for you how we deal with interactions. And that kind of theme is how theoretically we reduce unsolvable Hamiltonians to Hamiltonians which can be solved. And you will see that actually appearing in different situations with similarities, but important differences. So just to remind you, we started out with the interacting Bose gas in a homogeneous system. We have a very general way to describe scattering two particles with initial momentum disappear and two particles with momenta appear. This is a scattering event. Now this means we have products of four operators, which is very difficult to solve. And the Bogoliubov approximation which we discussed on Monday replaces the operator for the condensate for the zero momentum state with a Z number, saying that N 0, N 0 plus 1 is the same-- a little bit waving your hands. But it's also you can say the macroscopic limit when we have a photon field with many, many photons, we can use a Z number in our Hamiltonian to describe the electric field. So that's the same spirit. So maybe I should emphasize it. This is for atoms. What you're used to do with photons for your whole life. OK. With that we have transformed the Hamiltonian into a bilinear expression. And so at the moment your bilinear expression-- you do a Bogoliubov transformation. You simply diagonalize it by finding a new set operators where the cross term between a and b or between a of k and a of minus k disappear. And then you've solved it. And this is what we arrived at at the end of the last lecture. With those approximations, we have diagonalized the Hamiltonian. Our Hamiltonian is now, you can say harmonic oscillator Hamiltonian, or you can see it has become a gas of non-interacting quasi particles. Each of those operators creates a quasi particle, and the quasi particle energy is U of k. And I explained to you that as expected, the quasi particle energy is simply particles with momentum h-bar k for high momenta, because in high excitation of Bose quantums it is a free particle. But a low-lying excitation is affected in a major way by all of these interactions with the Bose-Einstein condensate. And that turns the quadratic dispersion relation into a linear one, or you can say that turns the free particle into a photon or something. So that's my review of the last class. Are there any questions before we go further? Colin. AUDIENCE: Does this require low density and [INAUDIBLE], or one or the other? PROFESSOR: We come a little bit later to that, but in the end, there is a small parameter. The small parameter will be N a cubed-- the gas parameter-- N, the density, a cubed, the scattering length. It's usually the dimensionless combination of the two which decides whether we are in the weakly or strongly interacting limit. There is another assumption which we have made here, but it's related to that-- as you will see a few moments-- that most of the are in the condensate. We assumed N0 is peak. The condensate depletion-- the number of atoms which are not in the condensate, t equals 0 is small. But let me just first show you-- I want to mix in theory and experiment how sound can be observed. I should actually say this experiment is sort of dear to my heart, because at some point it clicked to me how the new world of atomic physics connects with condensed matter physics. Soon after we had realized Bose-Einstein condensates, a famous condensed matter theorist said hey, Wolfgang, you should now observe sound. Sound is important. Yeah, OK, but how do you observe sound? You know, use a piezo and just kick the system and create a sound wave. And I said, oh, gosh, this guy doesn't know with that if you put anything in contact with the quantum gases-- a piezo-- the gases will just stick to it. He has no idea what our system is. But then I said wait a moment-- we have to translate. He says use a piezo. Well, I have to translate. In atomic physics how do we kick atoms? With a laser beam. And in that moment, I had the idea that we can take a Bose-Einstein condensate, suddenly switch on a blue de-tuned repulsive laser beam. This will do exactly as a piezo-- create a density perturbation, and then the density perturbation will propagate with the speed of sound. And that worked. That was just when Bose-Einstein condensations were very fresh. One of the first scientific experiments done-- we switch on the laser and you see color-coded in red the density perturbation which propagates out of it. And the slope of this line is the speed of sound. And here we determined the speed of sound as a function of density. So that's how phonons-- or at least how the speed of sound and wave packets which propagate with the speed of sound can be prepared. I come back to phonons and collective excitations in a few moments. But let me first say when we have diagonalized the Hamiltonian we know everything we want to know. I just focused on the quasi particle energies, but you also know the ground-state energy. And actually here Colin, you see that the corrections to the ground-state energy scale with N a cubed, so this is really the small parameter in this system. But we can also find out what is the ground state wave function. And let me use it to introduce an important concept to you called the quantum depletion. When you have two atoms in the condensator at zero momentum and you switch on the interactions, the delta function interaction couples zero momentum state to higher momentum states. So therefore, the effect of interactions is that the condensate is not just at zero momentum. It has some probability, or some admixture, of finite momentum states. This is the ground state. This is how the Hamiltonian is diagonalized. So from the Bogoliubov approximation, where we introduced this V and U parameter to transfer from one set of Bose operators to another set of Bose operators, those coefficients give us the population of those momentum states in the [INAUDIBLE] of the condensate. So if I now ask, what is the condensate fraction? What is the number of atoms in the zero momentum states? It's all atoms, but those who have finite momentum. And we find again the small parameter N a cubed. Now this quantum depletion allows me now to make a distinction between the cold atomic gas condensates and superfluid helium form. In cold atomic gas condensates, this correction is about 1%. So therefore, I can say with 99% probability, or with 99% weight, the condensate, the many-body wave function of the condensate is just the zero momentum state to the power N. And this very complicated admixture of correlation into the ideal gas wave function is only 1% for alkali gases. But for liquid helium, the condensate fraction-- even at zero temperature-- is only 10%. So when people use neutron scattering-- it would be a long story in itself how this is done-- but when they use neutron scattering to analyze the liquid helium at low temperature, and figure out what is the fraction of atoms which have zero momentum, they found 10%. The quantum depletion is 90%. But that's just the difference between a quantum gas and a quantum liquid. And in the quantum liquid-- in liquid helium-- N a cubed is on the order of unity. OK, I mentioned this dispersion relation. These are sort of quasi particles. And I at least showed you how you can measure the slope of the quasi particle dispersion relation by propagating sound waves. But let me now tell you how we can observe quasi particles' elementary excitations directly. And this is actually simply done by light scattering. If you scatter a photon, and the scattered photon loses energy h-bar U. And it is scattered at an angle. Therefore, it transfers momentum q. These forces can only happen if there is an elementary excitation with momentum q and energy h U. So in other words, on a photon basis, you can see photon by photon, if you scatter a photon, the photon transfers momentum and energy. The process can only happen if you can form a quasiparticle with this momentum and energy. So since this is sort of the direct way of mapping out whether the system has the possibility to absorb momentum q and energy h nu, this has sort of a name. The scattering probability is called the dynamic structure factor. And the dynamic structure factor is just an integral over all the possibilities that a many-body system can absorb momentum and energy. Now there is one nice feature which was introduced by us at MIT, and that is if you're going to measure the dynamic structure factor, often you do it with neutron scattering. You scatter neutrons or x-rays and they scatter spontaneously at an angle, and you need a detector to detect the scattered particles. But in our case, because the gas is so dilute, the distance between atoms is on the order of the optical wavelengths. We don't want neutrons. We don't want x-rays. We want photons, because the wavelengths of photons is perfectly matched to the properties, to momentum transfer and such we need for our system. But photons-- we have photon lasers. So instead of in a painstakingly way analyzing the frequency and the momentum of scattered photons, we don't do the spontaneous scattering process. We do a stimulated process. We use two laser beams, and we stimulate a photon to be scattered into the other laser beam. And by having the two laser beams at frequency difference delta omega, we're really asking the system, are you ready to absorb delta omega energy? If yes, then you have a quasiparticle. So this is how we do quasiparticle spectroscopy. And a few years later, this method was defined by the [INAUDIBLE] Institute. And what you see here is they varied the angles between the two laser beams, realized different momentum transfer, and what you see is the low dispersion, the linear dispersion relation for low momenta, and then the quadratic part at high momentum. So this is called BEC spectroscopy. It's a variant, you can see, of Raman spectroscopy where you go from a zero quasiparticle state to a one quasiparticle state through a simulated Raman process, and this is how the dispersion relation is determined. OK so I've so far dealt with aspects of a homogeneous Bose-Einstein condensate where, of course, if you're a homogeneous system, you formulate everything in momentum space. But now we want to deal with the situation that our condensates are finite in size. And in addition, they're usually in a harmonic oscillator potential, and therefore their density is inhomogeneous. Let me start the discussion with the inhomogeneous Bose-Einstein condensate by showing again a picture which brings in some memories of '90, '96. It was the first time that we could see the Bose-Einstein condensate in the trap. Before people have just seen it by time of flight when it was pretty much already destroyed or it was just flying out. But here we see the trapped Bose-Einstein condensate and we could even take multiple pictures of the same condensate. So you can see that this was the first time that a condensate was seen alive. What you see here is actually the bimodal distribution. You see the condensate and then due to speckled pattern, the Fermi cloud looks fragmented, but this is just a speckled pattern due to the signal to noise. But here you really see now-- not in ballistic expansion as I showed you on Monday-- but really insight to the size and shape of the condensate. OK, how do we describe it? Well, the message is pretty much exactly the same way as we did with the Bogoliubov approximation and such, but instead of in momentum space, we do it now in position space. Actually, I have to say this week, I go through a lot of material. But what I'm trying is to give you sort of a spirited and animated overview that you really know what is important. Where is the same concept appearing again in a different way. The details-- I've posted, actually, the original articles, references. Some of are school notes, some of them written by myself on the backside. So yes, it's a little bit different character than other lectures. I want to show you a lot of things, and put a special emphasis on the ideas. OK, so in second quantization, we are now using field operators which create and annihilate particles at position R. This is the single particle Hamiltonian-- kinetic energy and potential energy. And our interaction term has now, again, four operators-- two interrelation operators, two creation operators, but they're now formulated in position space. We do exactly the same as we did in the homogeneous gas. We assume the potential is short range. That means delta function. And that means we can get rid of one of the integrations in r and r prime. But we still have product of four operators. We can formally solve the equation by writing down Heisenberg's situation for the equation of motion for those field operators, but of course this cannot be solved. Remember products are four operators are hard. We have to reduce it to two operators. And this is actually done by the Bogoliubov approximation again. But it's done here in the following way. Remember, in the momentum space we replaced a0 and a0 dagger by the square root of a0. What we do here is we say this is sort of a quantum field. And when we have a superfluid, this field operator has an average value. It's actually the macroscopic superfluid order parameter. So we replace the operator by an average value which we assume is large, because many, many particles are in the superfluid state. And then we have fluctuations which are small. And what we will do is-- it's pretty clear-- when it comes to the fluctuations, we will neglect higher products-- products of four fluctuation operators. And we are down to C numbers and, as you will see, two operators. We can-- and this is even more dramatic-- it wouldn't make sense in momentum space, but it does make sense in position space-- we can even do a first approximation where we completely neglect those fluctuations. And then what we have is when we simply insert a [? Z ?] number into this Hamiltonian. We have then an equation for this number, for this function psi. And this is the nonlinear Schrodinger equation, or it's also called Gross-Pitaevskii equation, which is now the analogy to Schrodinger's equation, but now for the macroscopic wave function which is occupied by many, many particles. And in addition to the kinetic energy and the trapping potential in the external potential, it has a term which is proportional to the density. And this is just mean field repulsion which one atom feels exerted by the other atoms. I should say-- just to connect it to what I've said earlier about the quantum depletion-- that you can regard the function psi as the best approximation to an ideal Bose-Einstein condensate. So if you want to write your many-body wave function as a function psi to the power N, all particles are in the same quantum state, then a variational calculation would say that this function psi should obey this equation. If you want to read up what I've said, I've posted a recent paper, which I found very pedagogical, where they derive the Gross-Pitaevskii equation without even using an operator psi dagger, without using any second quantization. They just say we have an Hamiltonian for an interaction system, and we try to write the complicated, many-body wave function as a wave function to the power N. And then you put this ansatz into your Hamiltonian and do a variation optimization which single-particle wave function psi, if taken to the power N, minimizes the total energy of N particles, including interactions. And the answer is this equation. So there are many ways this equation is sort of very natural. Now you all know about single-particle physics-- kinetic energy in an external potential. Let's now learn what this density-dependent term does. Well, it's pretty clear. There is a price to be paid for density. And while the total integrated density, of course, is given by the total number of particles. That means if you lower the density somewhere, you have to increase the density. But because this is a non-linear term, if you have an average density and you lower it here and increase it here, you have actually more repulsive energy than if the density is flat. So this term has only one goal. It wants to flatten out densities. So therefore if we have a box potential and you know the ground state in a box is just half a period of a sine function. This is a non-interacting condensate. This would be the macroscopic wave function in a box potential. But if you now put in strong interactions, the strong interaction's going to flatten out the potential. They flatten out the potential, and only at the very last moment-- all right, now it's time to go down because we have to meet our boundary condition. And the length scale where you eventually go down-- it has a famous name, the healing length-- is the length scale where the kinetic energy due to the curvature of the wave function is now comparable to the interaction energy. If the system would curve down earlier, it would cause too much repulsive energy, because the density is not kept flat. If it would curve down too late, the enormous curvature would mean a lot of kinetic energy. And this is just the best compromise between those two criteria. So that's how you derive the healing length. The healing length is now the length scale over which the system is willing to meet boundary conditions, and not stay flat, as flat as possible. Yes. AUDIENCE: Just curious-- in that equation, the atom has the mass of each individual-- PROFESSOR: The mass is the mass of a single atom. AUDIENCE: Single atom, it's not-- sorry, I wasn't here last time. But the quasiparticles are the same mass [INAUDIBLE]. PROFESSOR: In a Bose-Einstein condensate they are. We haven't changed the mass. But we're not talking actually about quasiparticle. We're actually really talking about here qualitative features of the solution of the Gross-Pitaevskii equation. And the Gross-Pitaevskii equation is sort of a single particle equation for particles of mass, of the original bare mass M. It's a macroscopic wave function, but I sometimes say it describes the wave function of a single particle where all the other particles are included at the mean field level. So therefore, it is really this, the atomic mass, and not any form of collective mass here. Well, if you would now ask how do interactions transform the first excited state in the box potential? Well, then it's again flat this as much as possible. But then if you want to maintain the parity of the wave function, then it's only close to the zero crossing within a healing length that the system says OK, now I change sign. So that's sort of what is inside the Gross-Pitaevskii equation. Now once we realize that, we can take it to the next level and say, well, if you neglect-- let's say we're interested in the ground state and we neglect this boundary region where the kinetic energy becomes important-- maybe we can simplify the Gross-Pitaevskii equation by neglecting the kinetic energy. If potential energy dominates by far, we can neglect that. And then we should get a good description which will not be valid in the wings of the wave function but in most of space. But now if you look at this equation, it's no longer a differential equation. It has no derivatives. It's just something which applies to the wave function psi itself. And we can simply solve that. The solution of that is that psi squared-- the density-- is nothing else than there is a constant minus the inverted trapping potential. So therefore in this Thomas-Fermi approximation where we neglect the kinetic energy completely, you just take your trapping potential, turn it upside down, and then you fill it up with density until you have accommodated the number of atoms you want to accommodate. And this is, of course-- the constraint in the number of atoms is determined by the chemical potential. Or if you have a more complicated W-shaped potential, the same construction. Flip it over, fill it up, gives you the condensate wave function, the density distribution of the condensate in this potential. I don't want to belabor it, but coming back to the question of the small parameter, if you look at those equations, you can identify a parameter-- this is now the small parameter, or the parameter in the system which is the important dimensionless parameter. It depends on the number of atoms, and it depends now on the ratio of the scattering length-- which characterizes the interaction-- and the harmonic oscillator length. You can say the harmonic oscillator length is the ideal wave function harmonic oscillator. So this parametrizes the importance of kinetic energy, whereas the scattering length parametrizes the importance of interactions. And all those solutions can be nicely written as what you would have in an ideal gas, and then this parameter X. So in typical experiments, N a, N is a million. The scattering length is smaller than the oscillator length, but N is a million, and this parameter X is usually large. So therefore-- and I'm simply just talking about this solution-- we have the situation that the chemical potential is larger by a power of X than the ideal gas solution, which would just be the ground state with its zero point energy in the harmonic oscillator potential, or the width-- the size-- of the wave function is larger than the ground state of the harmonic oscillator, but only with an exponent, which is one-fifth. Well, we can see that. These are now, again, somewhat improved pictures of condensates inside the trap. For the expert, the previous picture was dark-ground imaging. This is now phase-contrast imaging. And if you take a profile, we clearly see the condensate wave function and we see the thermal wings. When we look at the size of the condensate wave function, you realize what I just meant-- that the ground-state wave function, the harmonic oscillator length is 7 micron in the axial direction, but here you have 300 micron. So this condensate is completely dominated by interactions. And it fulfils very nice, and pretty much the whole shape, except maybe some details in the wings, are quantitatively described by the simple approximation I have explained to you. So just a little bit show and tell. We have this Gross-Pitaevskii equation. The Gross-Pitaevskii equation has, as Schrodinger's equation, a time independent form to get the ground state. It has also a time dependent form. You simply replace the energy by the derivative of the wave function. This is the time-dependent form. Everything is very simple, and you can do many-body physics, but on your computer, you pretty much look for single-particle wave function and take them to the power N. So some areas where the Gross-Pitaevskii equation has quantitatively explained experiments-- one is the expansion of a Bose-Einstein condensate. It's this famous situation when you have an elongated condensate and let it expand. It expands faster in the radial direction. One simple argument is the pressure of the mean field is larger and leads to faster acceleration. So therefore, if you have a cigar-shaped condensate and release it, it turns into a disk. It inverts the aspect ratio, going from a cigar shape to a disk shape. And that has been beautifully and quantitatively described by the Gross-Pitaevskii equation. Here we have measured the interaction energy as a number of condensed atoms. And I mentioned to you that this X parameter comes often with power 1/5 and 2/5. And this here is a fit to the power 2/5, beautifully confirming the theory. So the Gross-Pitaevskii equation was invented in 1962, about 50 years ago by Gross and Pitaevskii to describe vortices. Actually, Lev Pitaevskii is still alive, going strong, and publishing papers. One of his latest predictions was solitons in Fermi gases, and I know in Professor Zwierlein's group, one of his lab just looked at the same physics, and compared to the theory of Lev Pitaevskii. And this year we are celebrating-- I forgot, the 85th or 90th birthday of him. AUDIENCE: [INAUDIBLE] PROFESSOR: 90? AUDIENCE: There's a poster. PROFESSOR: I know there's a poster next to my door, but I forgot which anniversary. So I mean, he's an legend. But he's still walking. He's still doing science. So if you meet him, you go back to 50 years in history. Anyway, it is this Lev Pitaevskii, and he invented his equations to describe vortices. So anyway so we have this nonlinear Schrodinger equation. Let me just show you what vortices are and how they are formed. Vortices come-- if you solve the Gross-Pitaevskii equation-- if you ever any quantum fluid or quantum gas, and you add angular momentum, the angular momentum cannot lead to rigid body rotation. This would violate the fact that the velocity field has to be irrotational to make sure that the phase of the wave function is well defined. So if you rotate the system, it can absorb angular momentum only by forming vortices. And vortices are singular points of the wave function-- singular points where the density is 0. In other words-- I don't have time to be exhaustive here-- but when something rotates, there is a dynamic phase you can say, because there are matter waves going in circles. And the integral of the phase has to be an integer number of 2 pi, otherwise you would not have a well-defined phase of the wave function. And if you now say you make a circle, which is 2 pi. You make the circle smaller, smaller, smaller. In the middle of the circle, you go around. The wave function changes by 2 pi, but what should the wave function do on one point? Which number between 0 and 2 pi should the wave function peak? Well, the wave function says I can't peak. I just go to 0. And a 0 value has no phase and I'm fine. That's exactly what the wave function tells you. So therefore if you want to describe vortices, you want to now describe the Gross-Pitaevskii equation, but with the boundary condition that the wave function goes to 0. And if you want to describe one vortex, you want the wave function to go 0 in the center of the cloud. And so you make a corresponding Ansatz. You allow the phase to wrap around by 2 pi, when the angle phi is varied. And then when you solve it, you have, of course, put in that there should be a 0 of the density at the center. And it's now nice for me to-- there's an important review paper which describes all that and more-- but it's nice to show you now the two extreme cases of the ideal condensate without interaction, and the strong interacting condensate. The ideal condensate-- well, we are in the first excited state of the harmonic oscillator potential. And this is the dashed line. And of course, the size of the whole, the 0, is pretty much on the order of the oscillator length, because there is no other length scale in the ideal harmonic oscillator. But if you have the interacting system, remember what interactions are doing. They want to keep the density as constant as possible. Well, we are not in a box potential. We're in an inverted parabola potential. And remember, our Thomas-Fermi solution, which neglects kinetic energy, is the inverted parabola up to here. And then the tail is when we can no longer neglect the kinetic energy. But if we now say, OK, fine, but now in addition, we want a vortex in the center, the wave function out there says no. We do what minimizes kinetic energy, and what minimizes repulsive energy. We follow the inverted parabola, and only at the very last moment-- on the scale of the healing length, the system meets the required boundary condition that the density goes to 0. Anyway, with this qualitative understanding, you can get a lot out of those. You can immediately understand the salient feature of the solution. Question? AUDIENCE: I know we're not talking details, but is there a simple, maybe clear reason why when you spin a Bose-Einstein condensate you get many vortices, but when you spin your coffee in your cup, you just get one big one? PROFESSOR: Yes. So the question is, if I have a wrap-around of 4 pi in phase, whether the system should have a doubly charged vortex, or two singly charged vortices? What is the difference in energy? Well, what happens is doubly charged vortices are unstable. My group, at some point, were the first to create doubly charged vortices, but we saw that they immediately decayed. AUDIENCE: [INAUDIBLE] PROFESSOR: Pardon? AUDIENCE: [INAUDIBLE]? PROFESSOR: Initially, we couldn't observe. We just saw that it was unstable. But it's a decay into two vortices. But it's easy to understand. If you put two vortices on top of each other, what is the energy of the system compared to one vortex? Well, the energy of a vortex is the rotational field around it, and if you superimpose two vortices, the velocity around the doubly charged vortices is twice the velocity of one vortex. And therefore, the kinetic energy is four times. If you have two vortices which are far separated that it has its own velocity field. It has its own velocity field here. And when the two velocity fields come together, the velocity is already so low that you don't have to consider that, then those two vortices have an energy which is two times the energy of a single vortex. So therefore, when the two vortices, they start out with four times the energy of a single vortex, and when they dissociate and repel each other, they have shed half of their energy. So this argument tells you immediately that vortices are in effect repulsive. And therefore, any multiple-charged vortex will spontaneously decay. It's also this net repulsion between vortices which makes the vortices arrange in a regular lattice. The regular lattice which you saw before is Nature's answer to how can we minimize the energy of all those vortices? And the idea is let's keep the average distance between them as large as possible, and the answer is a hexagonal lattice. Yeah. AUDIENCE: So a few slides back, when you write the G-P equation, when you plotted sort of for the box potential, the ground state and the excited state comparing the G-P equation to the typical single-particle [INAUDIBLE]. And so I guess I'm a little bit confused about how excitations manifest themselves in the system. Because in some sense, if you were to solve the G-P equation, would you arrive at a spectrum of solutions? Then I guess I'd be confused at whether excitations manifest themselves into that spectrum of solutions, or rather than they become excitations, sort of deviations from the mean field as including the [INAUDIBLE]. PROFESSOR: You are now asking about something more complicated. You're asking what are the excited states of the many-body system? And actually, we have already found one answer. And this is if you want these small excitations, it's one particle becomes a phonon, or one particle becomes a quasiparticle. What we are talking about here is what happens if the whole macroscopic wave function is in an excited state. So we are asking here in the box potential what happens if you force all N atoms to have one node in the wave function. And this is what I'm talking about here. So the excitation energy of this state here is much, much higher than of single quasiparticle excitations. I'm not sure if I'm addressing your question here, but-- AUDIENCE: So with single quasiparticle excitations, are those essentially sort of deviations from the mean field? PROFESSOR: No, they are-- you really look for many-body physics-- main field deviations-- probably. I mean, what you do is in that sense, yes. You're allowing in this Ansatz that psi operator is psi average plus fluctuations. You now look for fluctuations, and you're looking for the energy eigenspectrum of those fluctuations. And the answer are quasiparticles. So it is, actually, the Bogoliubov solution for the spectrum of the fluctuations. And the answer is the dispersion relation I presented to you. Whereas here we are asking what are excited states of the macroscopic wave function? It's a very, very different question which we've addressed here. Yes. AUDIENCE: So when you stir the condensate, do you view it as-- do you stir, sort of, at the trap frequency, so I do sort of N single excitations? Or do you stir at N times the trap frequency, so you get one sort of N particle excitation? PROFESSOR: Well, the experimental answer is you want to stir at the quadrupole frequency, which is square root 2 times the trap frequency. Then you create quadrupolar excitations. The quadrupolar excitation can be regarded as a standing sound wave, but you create a macroscopic number of those excitations. So you make, actually, the whole condensate wave function oscillate in a quadrupolar pattern, but it's a rotating quadrupole, and that eventually then rearranges itself and leads to vortices. So I think the correct answer is the most efficient way to create vortices would be to excite quasiparticles, but then create so many-- and this may relate to Matt's question-- create so many quasiparticles that you have really a time-dependent and oscillating macroscopic wave function. So eventually, you have a coherent excitation of quasiparticles. And that eventually means the condensate moves in a quadrupolar pattern, but it's a quadrupolar pattern with rotation, and that eventually turns into many vortices. But that's really a very rich question which you're asking, which has been studied-- a lot of different aspects of that have been studied. OK, so we've talked about Bose-Einstein condensates in a homogeneous system, just sort of to lay the groundwork. We've talked about Bose-Einstein condensates in traps in inhomogeneous system. Now we want to talk about Bose-Einstein condensates in optical lattices. Well, there are two reasons why we want to do that. One is we want to use the Bose-Einstein condensate to obtain deeper insight into the properties of matter. And a lot of forms of matter appear in periodic lattices. So if you put a Bose-Einstein condensate into a periodic potential, we can at least understand some of the properties of crystalline matter, or electrons which are block waves in a periodic potential. So this is one reason why. The other reason why we want to go to optical lattices is the following-- ideal Bose-Einstein condensates are trivial. Weakly interacting Bose-Einstein condensates are entertaining, and you can write a lot of papers, have a lot of fun with it, develop your methods-- also mildly intellectually interesting because how those weak interactions manifest itself in vortices and all. It's really rich and interesting. But the conceptional problems appear when you go to strong interactions. Strongly correlated matter is where mean field descriptions no longer work. This is really the frontier of our understanding of many-body systems. And when you want to be there, you want to create a strongly correlated system. And strongly correlated systems means that the interaction energy dominates over kinetic energy. In Bose-Einstein condensates you can only go so far with increasing interactions-- why are Feshbach resonances? Because you get into some inelastic collisions. So that's one knob to turn. You go to larger and larger of scattering lengths, and crank up the interactions. But if atoms strongly interact, they start to do bad chemistry. They start to spin flip. They do other things. And in some cases you keep it under control. In others, you just can't keep it under control. But another way to get to strong interactions means you reduce the kinetic energy. It's the ratio of the two which matters. And when you put particles in a lattice, well, the lattice actually reduces the kinetic energy. You may know from condensed matter physics that in a lattice there is an effective mass, which is higher than the [INAUDIBLE] mass. Therefore you've reduced the kinetic energy. Or if you want another hand-waving approximation, the kinetic energy is given by the bandwidths. And if the tunneling becomes slower and slower, the width of your band becomes narrower, and your kinetic energy is less. So anyway, I can give you many hand-waving approximations why in a lattice, kinetic energy is reduced and repulsive energy is probably enhanced, because instead of having your atoms spread out, they're bunched up at each lattice site into a higher density cloud. So anyway, maybe you're interested in parity potentials, or you're interested in quantum systems with strong correlations. For whatever reasons, you want optical lattices. Now I have here a few slides which introduce periodic potentials, but it is really just the single particle physics in a periodic potential. It has nothing do with quantum gases. This could be the first lecture of how to describe electrons in a metal. So let me just quickly go through assuming that almost all of you are familiar. But I use those slides also just to give you a few definitions and introduce a few symbols. So this is simple, boring, exactly understood physics. We have a Hamiltonian which has kinetic energy, and a periodic potential which is our lattice potential. It's rather trivial to solve, but exactly for your wave function, you use Bloch's theorem and divide it by an exponential factor with quasi momentum times a periodic function. And if you now solve Schrodinger's equation, you want to use Fourier space. You Fourier analyze the wave function. You Fourier analyze the potential. And since the potential is periodic, sine square potential has only three Fourier components at 0 plus/minus 1 times the periodicity of the lattice. You do a Fourier expansion for your periodic wave function. And if you insert that into Schrodinger's equation, well, the Fourier transform has turned the differential equation into an algebraic equation, because the second derivative simply becomes now k squared or q squared. So in other words, you have a set of linear equations. There is an index which is the band index-- how high do you want to go? And usually you truncate it. But it's the same. It is that same trivial story which is told in all textbooks of condensed matter physics. If you have no band structure, well, you have the parabola, but to prepare for band structure, I've opened it down here in the first Brillouin zone. And if you now introduce a lattice, you introduce band gaps, and you go from the left to the middle to the right for stronger and stronger lattices. So the case which I will focus on, because it is the most extreme case away from free space, is this case which is called the tight binding limit, where the potential energy of the lattice is large. And large means compared with the kinetic energy at the Brillouin Zone, which is k squared-- k is the lattice momentum-- k squared over 2 m. And that's the recoil energy. So that our dimensionless parameter here is the depth of the lattice measured in recoil energies of the photon, because that is the kinetic energy of the free gas at the Brillouin zone. OK there are few things which immediately simplify that. Once we're in the tight binding limit, our lattice is really deep, and each side forms a harmonic oscillator. And the harmonic oscillator frequency is analytically given by the depth of the lattice. And the solution here for the lowest band is that the lowest band-- the energy in the lowest band, the dispersion relation is-- well, we have a harmonic oscillator in each site at x y z. So the average energy in the lowest band is the 0 point energy in x, y, and z-- three half h-bar omega 0. And then we have a cosinusoidal band structure where q is the quasimomentum. And what appears here as the only interesting parameter is j. And j is I think this 4 should be 4j is the bandwidth. So what appears here now for the first time is j. It appears here as the bandwidth. But let me immediately give it another interpretation as a tunneling matrix element in the following way-- right now, we have formulated the physics in Bloch wave functions which are infinitely extended. The Bloch wave functions are for the lattice what plain waves are for free space. But if you have tight binding limit, there is another limit which is important. Namely, a particle is localized and hops around in the lattice. The localized particle, of course, is in free space. It would be a wave packet-- a superposition of plain waves. So let's do the same in the lattice. Let's construct superpositions of Bloch waves. And these are our wave packets. And the wave packet is now called the Wannier function. And there is a mathematical procedure how you should pick the phases here to get the maximal normalization. But the simple picture is those Wannier functions are very, very close to the Gaussian ground state solution of the harmonic oscillator on each side. The wings are different, but I don't want to go into that. You transform from an orthonormal basis of Bloch wave function to another orthonormal basis Wannier function. And the Wannier functions are as well localized as possible. That's the procedure. And now we can simply rewrite our total Hamiltonian or everything we're interested in, not in Bloch wave functions but in Wannier wave functions. And what comes out now is, is that the bandwidth j is nothing else than our Hamiltonian with kinetic energy in the periodic potential. But j becomes now a matrix element between two Wannier functions. But the Wannier functions have now-- I'll put in some indices in a moment-- connect now two different sides. So it is-- you have Wannier function. You have the Hamiltonian and connect it to another Wannier function. So it is the amplitude that, with the Hamiltonian, the particle can hop from one side to another side. So therefore j, which was the bandwidth, is now the tunneling energy divided by h bar-- the tunneling rate from on one side to the other side. For very deep lattice, everything is analytic and it can easily be solved. And I mentioned already in the tight binding approximation, you should think about your Wannier function as just localized Gaussian eigensolutions of the harmonic oscillator. Yes? AUDIENCE: So qualitatively, this question sounds sort of silly, but normally when we write down the solution for j, we're only considering nearest neighbor. But from the math, I don't immediately see why we wouldn't include i equal to j. PROFESSOR: Give me one more slide. So I should've actually-- what I should have done is that-- j has an index here. I just didn't want to overload you with indices. I mean, this is sort of just telling you what j is in its simplest form. It is loaded with indices and I well show you in a moment where those indices come in. So I just wanted to give you the idea if we want to have Wannier function. We hop from one Wannier function to the other one, and the operator is the Hamiltonian. And therefore, there should be an index j and l. And maybe-- let me just do the next step first and then come back to it, but I wanted to tell you here what I'm aiming at-- namely interpretation of j, of tunneling from one side to the next. But right now I haven't really told you which j I really mean. There should be-- based on the right-hand side, there is a j which has two indices. And I will make the indices disappear in a moment. But before I make those indices disappear, let me introduce the other relevant parameter, which will also have indices, and that is we have to bring in the interactions. We want to describe an interacting system. We describe an interacting system using the short-range approximation by assuming that two particles interact with a delta function. And if you have two particles on site, each of them has a density which is the Wannier function squared. And the product of the two densities integrated gives us the expectation value for the repulsive energy. And this is given here. So the moment we introduce interactions, we are now interested in the interaction energy between two particles which in this case occupy the same side. OK so I've tried to introduce was what sort of j is. j is a matrix element between two Wannier functions with a Hamiltonian in between. And u is the matrix element of two Wannier functions with the interaction operator in between. And now I want to use that concept to take my full Hamiltonian and transform from field operators localized at x to Wannier functions. So they B operators are now creation operators. They create an atom in a Wannier function. In other words, B dagger means you have a particle in a Wannier function at site i. And I just use that as a basis transformation and the exact transformation of this Hamiltonian is now into this form. So What I have right now is, I have the tunneling matrix element between particles at site i and j, and I have to sum over all of them. And in terms of interaction, I can calculate this matrix element by using Wannier function with four different indices and ask what happens. And this is simply an exact way of rewriting it. And here I've given you the definition. So if you want to forget everything I told you about j and u, I've done an exact transformation from field operators to Wannier function creation operators. And this introduces tunneling terms like this from site i to site j. And these here includes products of four Wannier functions which are responsible for the interaction term. Actually, if this is not complicated enough, I've suppressed band indices here. I should also now sum over all possible bands. But OK, I want to come now to the leading approximation in a tight binding model. And that is where-- I mean those Wannier functions are overlapped. Two neighboring Wannier functions barely overlap. If I go further away, the overlap becomes even smaller. So the most dominant terms are nearest neighbor interactions. And the nearest neighbor interaction is where i and j differ by 1. And this is what I call j without indices. And similarly, when it comes to the interaction term where we have products of four Wannier functions and we want to get the overlap of all four, and then multiply it with g, the prefactor of our delta function to get an interaction term, well the best overlap is if all indices are the same. And this is what I call U. So in that limit, in that tight binding limit, my Hamiltonian is now very simple. It consists of a tunneling term parametrized with j, and an on site interaction term parametrized with U. Yes. AUDIENCE: So you're saying i, j and k are all the same, so the interaction is with itself? PROFESSOR: No, two particles per site. When the four particles are the same, it turns into-- this Hamiltonian here has all indices are the same. And if you calculate that, it turns into the product of 2 b. b dagger is the occupation number at each site. But if you're careful with commutators, it becomes occupation number times occupation number minus 1. If you have only one particle per site, this term is 0. So technically U is, if you put in 2 times 1, if you have two particles per site, U is the interaction energy between two particles. Just use this expression to figure out what it is for three or four. But for one particle, you get 0. The self interaction of a particle is absolutely 0. One particle does not interact with itself. AUDIENCE: That gets back to my question with the [INAUDIBLE]. So when you have i equals j, the [INAUDIBLE] term, do you just-- PROFESSOR: No, I assume, i j is nearest neighbor. I assume that the index i and j differ by 1. AUDIENCE: Yeah, I mean your argument was that the best overlap is between nearest neighbors. PROFESSOR: Yes. AUDIENCE: So an overlap with the Wannier functions at the same site is sort of the kinetic energy term. Is it approximately 0, or you just absorb that into the chemical potential? PROFESSOR: This is a 0. AUDIENCE: [INAUDIBLE] So we're only looking at nearest neighbor, so are we essentially saying the case where i equals j is approximately 0 [INAUDIBLE]. AUDIENCE 2: Because it's the matrix element of a Hamiltonian. So p, the candidate, plus the lattice operator. So if I take my Wannier function, which is built out of eigenvalues for that equation, [INAUDIBLE] bunch of energy terms. On every site [INAUDIBLE] they're the same. Because you have a common [INAUDIBLE] energy that you can [INAUDIBLE]. AUDIENCE: Well then just multiple it by [INAUDIBLE]. PROFESSOR: My gut feeling is-- and this is why nobody considers it-- it's just a constant energy, which is probably something like the 0 point energy times the number of particles, which is not affecting any dynamics. It's pretty much a constant which can be simply dropped. Let me just go back to that slide. I hope-- let me just look up the reference. I haven't looked at it recently. When we do this exact transformation, there should be a reference whether i and j, what happens when i equals j. I think it's just a constant term. AUDIENCE: Because if you're summing over every single lattice site, so you get the number of particles times 0 [INAUDIBLE]. PROFESSOR: Yeah, but even mathematically it gives just a constant term, here, which is [INAUDIBLE]. Thanks. It was good to clarify it. OK, I think within the next 20 minutes, I can step you through the superfluid to Mott insulator transition. First, references for what I've just said are given here. But let me now come take this Bose-Hubbard model and discuss its two limiting cases. One case is where U is much larger than j. The other case is where j is much larger than U. These are the two limiting cases, and it will turn out that one is an insulator and one is a superfluid. And that makes perfect sense, of course. If U is much larger than j, I can set j equals to 0 if I can neglect it. No tunneling means no transport, and that means an insulator. It's also clear that when there is no tunneling, that the system is really described by a product of so and so many particles per site. So it could be one particle per site. And I have a product over all sites, or two particles per site. So this will be the ground state. And it's called the ground state of the Mott insulator. It's also trivial to discuss what happens if j is much, much larger than U, because then I simply neglect U, and I have a free gas. Well, a free gas of Bloch wave function, but that's the same as a free gas of planar wave function. It's just that quasimomentum replaces momentum. And if you have just an ideal Bose gas in a periodic lattice instead of Bose-Einstein condensation in the lowest momentum state, you have Bose-Einstein condensation in the lowest quasimomentum state. The lowest quasimomentum state is a superposition of all Wannier function-- I mean, the 0 momentum state in free space is a superposition of all position delta functions. The plane wave is delocalized. And the lowest quasimomentum state is just completely delocalized over all Wannier functions. But this is nothing else than the q equals 0 quasimomentum Bloch wave. The interesting question is-- and this has led to hundreds if not more papers in the literature-- how do we go from one limit to the other limit? And now you would say, well, maybe we should use what has worked so well. We should use the Bogoliubov approximation. Just assume we have a condensate, replace all those kind of a 0 operators for the lowest Bloch waves. However, this doesn't work, because the interesting thing here is to find the transition from the superfluid from the Bose-Einstein condensate to an insulating state. The nature of this approximation is that you need N 0 to be large. But we're interested when the condensate wave function turns to 0, and in the insulator it becomes 0. So we are actually interested in the opposite limit. And indeed, if you would ignore everything I've just said that doesn't make sense to make this approximation and make it nevertheless, you will find that you never get the insulating state because you've pretty much eliminated the possibility to describe an insulating state by doing this approximation. So I want to show you now that we have to do another mean field approximation, which is actually nice. It's very different from this Bogoliubov approximation, but it's also a mean field approximation which will describe our system. So the goal is now that I want to find an effective Hamiltonian which describes the transition from here to there. And the important approximation I will use was is the following-- again, I have to get rid of operators. Products of two many operators cannot be solved. And so what I will do is I will use products of operators, write them as average value plus fluctuations. And then when I multiply that out, I take the product of the average values, and I include the fluctuations in leading order. So I take delta A times B and delta B times A average, but I neglect the product of those fluctuations. You can say I neglect the correlation of fluctuations here. So this is spelled out here. But the sign is important. Just look at this equation-- A delta B plus delta B with B plus A average times B average. If I absorb A times B by upgrading delta B to B, but I do the same here, I have to subtract 1 product of the two average values. It's actually this minus sign which will play a role later, but here you see already that I will make this decoupling approximation-- that I decouple the fluctuations from each other and I write it in this way, there is an important minus sign. OK, so we want to start in the insulating state. And we're going to figure out how does the system develop superfluidity out of the insulation state? And the perturbation operator which takes me out of the insulating state is tunneling. So therefore, the operator which is responsible for breaking out of the insulating state is the operator which induces tunneling between neighboring sites. And the tunneling operator in this Hamiltonian-- remember, the Bose-Hubbard model had j times B dagger B-- this was our tunneling term-- this involves now products of operators. And I told you we want to get rid of products of operators to get something we can easily solve. So we use now this product of operators on two neighboring sites. And we use exactly this decoupling approximation. So therefore, we replace each operator by an average value, and we neglect the product of the fluctuations. And then we obtain this. And I explained to you where the minus sign came from. So now I call this average value of the operator B l, I call the superfluid order parameter psi. I think I could have chosen psi to be complex, and then B l dagger B l would have complex conjugate. But it's sufficient here to restrict the discussion on real numbers, and it makes the notation simpler. So anyway, I introduce that. And what happens now is the following-- that I have my Hamiltonian. The Hamiltonian had the interaction energy. But the tunneling term is now very simplified, because instead of tunneling from one site to a neighboring site, the other side is sort of absorbed by the mean field, by the superfluid order parameter psi. And therefore, and this here gives me a psi squared term which appears here. Trust me, this is just an identical rewrite of the previous Hamiltonian by using this decoupling approximation. So what we have gained now is something really dramatic. We had many sites and tunneling from site to site. But if you look at it now, we simply sum the effective Hamiltonian over site index l. So our effective Hamiltonian is now the sum of an identical Hamiltonian per site. The sites no longer interact with each other. Each site interacts with all the other sites described by the mean field by the superfluid order parameter psi. So therefore, our many-body problem, which is still a many-body problem, but has turned into an effective Hamiltonian for each site, because each side has the same Hamiltonian. Colin. AUDIENCE: What happened to the j psi squared term? PROFESSOR: This here? AUDIENCE: Oh, OK. PROFESSOR: OK so I can say instead of solving for the sum, I can just solve for each site individually. And this is now my effective Hamiltonian for each site. Now I want to catch the onset of superfluidity. So I want to get the system when psi is small. And therefore, I can just ask-- I don't know what psi is. It's part of my solution. But I'm interested in the moment when psi begins to take off from 0, when superfluidity emerges. So what I can therefore do is, I can regard psi as an epsilon parameter, as a small parameter. And the psi parameter comes with an operator V. And this operator V is nothing else than B l dagger plus B l. So in other words, what I'm doing is, I'm separating my Hamiltonian into a Hamiltonian which is diagonal in the quantum numbers of the isolating state-- just one, two, three particles per site. And psi squared is the Z number. Psi squared is also diagonal in that. And now the possibility of tunneling, the possibility of superfluidity is now perturbative in this term psi times V. OK, I don't want to explain, actually, this expression. It just formalizes [INAUDIBLE] intermediate step. When we have the chemical potential, and we raise the chemical potential-- we go from zero to one particle per site to two particles per site. And whatever the chemical potential is determines whether we have one or two particles per site. This is just telling me as a function of chemical potential, what is the ground state of the insulator? So now we take this ground state-- it's actually much easier described in words then by this formula-- we take this ground state and do perturbation theory in our term psi times V. Remember, the operator is B dagger B. And the epsilon is psi. So in second-order perturbation theory, we get psi squared. And then V, because B B dagger, is very simple. It only couples one occupation number N to N plus 1 and N minus 1, because B and B dagger destroy or create a particle per site. So therefore, I can immediately write down what this matrix element is in second-order perturbation theory. I mean, these are all-- sorry, it's all defined here. I know I'm losing you now. Nobody will tell me what is the difference between U bar and U, but it's trivially defined. So the idea is we have the isolating system. We do perturbation theory in tunneling. And the perturbative operator is psi times B plus B dagger. The B plus B dagger matrix elements are trivial, because they admix to N particles per site-- N plus 1 and N minus 1. And this is what we've done here. AUDIENCE: And the j is just occupation number? PROFESSOR: The j is the occupation number of our site. AUDIENCE: I have two questions. So in the bottom equation, what happened to the psi squared? PROFESSOR: Sorry, this is the sum. The psi square is missing. AUDIENCE: I guess I had the same question for the equation next to the green thing. Is there supposed to be a psi squared in there? Because originally, there's a psi squared and an h0. PROFESSOR: No, this is the ground state where psi is 0. And now we do perturbation theory in psi V. These are the unperturbed energies which appear in the energy denominator. The wave functions we are using are Fock states-- number states-- per site. And here we couple occupation number j to all possible N's. But because of B and B dagger, N has only two values-- j plus 1 and j minus 1. And this is what's given here. It's mathematically trivial. The notation is more complicated than the physics behind it. But now comes again-- I think I need five minutes and I'm done. But now comes the interesting physics. We have to ask what are we doing here? It's all the mathematical, and the math is really simple here. What we have done is we have looked at the isolating state, and we have done perturbation theory in psi times B plus B dagger. And now we get an energy correction which is psi squared. And what are we really doing here? Well, you can say the following-- we started with a hypothesis that we have a superfluid state characterized by psi and psi, at the onset of superfluidity-- is small. But now we have done the calculation assuming that there is a psi. But now we are turning around and said have we really done the system a favor by introducing superfluidity? In other words, has our perturbation theory in psi lowered the energy of the state or raised the energy of the state. So in other words, we've done a hypothetical calculation. Hey, what would the system feel like if there were a little bit of superfluidity? If the system says great, I've lowered my energy, then we know we are in the superfluid state. When the system says no, I raised my energy because of the psi, then the system has rejected our idea to introduce superfluidity. So therefore, the question we are raising now is after we have done the calculation, for what values of U and j is it favorable to introduce a psi or not? Now I was expecting the question of some of you that in second-order perturbation theory, second-order perturbation theory always lowers the energy. But remember, this is why I emphasized the minus sign-- we had a psi squared term which came from that, which came from the last term of the decoupling approximation, which had a minus sign. And therefore, we have in psi squared one term which came from this special psi-- and I emphasized in the decoupling approximation. And we have a contribution of psi squared which comes from perturbation theory. And the two together can actually change their sign. So what we have right now is if you describe the ground state as a function of psi, we have our unperturbed energy of the Mott insulating states, and then we have a term in psi squared. And we should-- and we could, but we don't-- calculate the next order in psi to the four. And it turns out in fourth-order perturbation theory, this term is always positive. So what happens now to the total energy when this term A2, which we have exactly analytically calculated-- if this term A2 is larger or smaller than 0. Well, if you have a parabolic term and a quartic term, in this case, both the quadratic and the quartic term are opening up like in a U shape, but here the total energy turns into W shape. And the interpretation is fairly simple. Under those conditions, the system prefers to have psi equal 0, whereas here, it's like the system wants to have a psi parameter which is finite, either positive or negative. And whether it's positive or negative is sometimes called spontaneous symmetry breaking between two degenerate solutions. What I'm showing here is, of course, very similar to the Ginzburg-Landau theory of phase transitions, where you have an effective potential and the phase transition takes place when the second-order coefficient changes sign, and the effective potential turns from U-shaped into W-shaped. So that's what we have done. So anyway, I think this is a nice problem where the interpretation is, I think, much more subtle than the calculation itself. But we've calculated the phase transition. Everything is analytic. Here is the analytic result, which I know with very indigestible notation. I followed exactly the paper by Vanderstraeten, [? Stouff ?] and collaborators, which is posted on the website. You can plot this phase diagram in this way. But if you use what is more common, normalize the chemical potential by U and normalize j by U, you get these wonderful lobes of the Mott insulator where you see that, if you increase the chemical potential, you have N equals 1, N equals 2, N equals 3, Mott insulator, and in between you go through superfluid regions. However if your tunneling is larger, if your tunneling is too large, you're only superfluid. So this is sort of the way how you derive this rather rich phase diagram of bosonic atoms in an optical lattice. Green is an insulating state. And white is the superfluid state. Let me just conclude by showing a few slides how this can be observed. So in one case, we have an insulating state with a definite number of particles per site. And here we have superfluid state, which has the normal fluctuations in number. Here's another cartoon picture of an condensate in the lowest Bloch wave function. And here you have localized number states. Often, you observe the transition by taking the system and releasing it in time of flight into ballistic expansion. If the superfluidity is coherent, the wave function on the different sites is coherent. And what you get is a multi-slit interference pattern, which is a diffraction pattern. And you see these characteristic diffraction patterns. It's an in-slit diffraction pattern which characterizes the coherence in the initial state. However if you have a completely isolating state, you have a Gaussian wave packet on each side, which has nothing to do with the next neighbor. And then the Gaussian wave function simply expands in a structureless way. And so the transition from here to there happens exactly when the ratio of U over j crosses the value we have derived. There's another technique of observing that which is more recent. This is a quantum gas microscopes where you do the same physics in two dimensions, so you have only one plane where this physics takes place. And you can observe atoms now with a microscope. So each green dot here is an atom. If you sort of zoom in. And if you now study the region where you go through the superfluid to the Mott insulator, you see that you have exactly one atom per site. If you increase the atom number, you have an insulator with one atom per site. And in the middle, you have state with two atoms per site, which-- for reasons I don't want to discuss-- are color-coded here in black. And that sort of goes on and goes on. So you can really resolve site per site the occupation of the number of atoms per site. So I think I'll stop here with the Bose gases. When you think the Mott insulator is the end of the story, all motion has been frozen out and you have one particle per site, well, it could be the beginning of a new story. Because if you use two different hyperfine states, spin up and spin down, we can talk about spin ordering. But this would be a whole different lecture.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

14_Solutions_of_optical_Bloch_equations_Part_1.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Back to master equations an optical Bloch equations. I hope you remember that on Monday we derived, under very general conditions, how dissipation, relaxation comes about in a quantum system, namely because we have a quantum system which interacts with a reservoir. The total system evolves, as quantum mechanics tells us, with a unitary time evolution. When we derive life what happens with the small system, due to the coupling with a bigger system, there are now suddenly-- the time evolution of the density matrix has relaxation terms, and we derived that. And the special equation we are interested in for two level system interacting with the vacuum-- interacting with a vacuum for spontaneous emission are optical Bloch equations. Today I want to discuss with you some characteristic solutions of the optical Bloch equations. But before I do that, let me first discuss with you the assumptions we made to derive a master equation. And I hope you remember the Born approximation and the Markov approximation. The Born approximation says that the reservoir will never change. The reservoir is always, for instance, in the vacuum state. And the Markov approximation said that the correlations in the reservoir are delta function related. It can absorb photons instantaneously. The vacuum has no memory time. It's like a black hole. It sucks up everything. But it's always what it was initially, namely the vacuum. Do you know examples where those assumptions are not fulfilled? I make it so natural. These are the most natural assumptions, but what are systems which cannot be described with this approach. AUDIENCE: I was thinking maybe like an atom in a cavity. If you change the vacuum somehow. PROFESSOR: An atom in a cavity. Very good. If an atom emits-- now it depends. We talk today about the master equation for photons and atoms in the cavity, but we will assume that the cavity is rapidly damped. If it's a high Q cavity, the atom emits a photon, but the cavity has memory. It stores the photon, and after one period of the vacuum Rabi oscillation, the photon goes back to the atom. So the vacuum has memory time. The vacuum has now a time constant, which is the same as the time constant for the atom, namely the period of the Rabi oscillation. Big violation of the Markov approximation. We do not have the very short correlation time of the system. We just integrate over in the much longer time scale, during which we are interested in the dynamics of the system. Maybe another example. Actually, this example violates both the Markov approximation, also the Born approximation, because the vacuum has changed when it has a photon. Do you know an example where maybe only the Born approximation is violated, namely the reservoir is changing due to interaction with the system? Well? Before we talked about the system in a big reservoir. What happens when you make the reservoir smaller? Well then when it absorbs energy. It will heat up and such. It will change. The Born approximation is actually, when it would come to energy transfer, the assumption that the reservoir has an infinite heat capacity. It can just take whatever the system wants to deliver without changing, for instance, it's temperature. OK, so at least you know now there are systems for which the treatment has to be generalized, and you understand maybe better the nature of approximations we have made. [INAUDIBLE] AUDIENCE: Just a question. For the last case which is where the Born approximation is not violated, is there any example of that? PROFESSOR: Well for instance, spontaneous emission into the vacuum, but the speed of light is so slow that it takes forever for the photon to escape. The Markov approximation is more about the time scale for the reservoir to react. And the Born approximation is if the reservoir is really big enough to simply absorb everything without changing. I don't know. You could maybe think about you have some reservoir, but it has very bad heat conduction or something. The transport is just very slow, and therefore you do not have the hierarchy of the two time scales. OK, any other questions about the master equation? Optical Bloch equations? Well, then we can learn more about the characteristic features of the optical Bloch equations by looking at solutions. And I would like to discuss three aspects of solution. I want to discuss the spectrum and the intensity of light emitted by the atoms. Secondly, we want to talk about transient and steady state solutions. And finally, actually I put it in the chapter of optical Bloch equations, but I will use a cavity for that example, which will actually go beyond optical Bloch equations. It's also nice for you to see that there is more than the optical Bloch equation, but it's the same formalism. And this is when we discuss the damping of the damped vacuum Rabi oscillations. I also picked this one example sort of as an appetizer, because I can introduce two concepts for you in this last example. One is the quantum Zeno effect. And the other one is the adiabatic elimination of coherences, which plays a major role in many, many theoretical treatments. But let's start with the first part, namely the spectrum and intensity of emitted light. And I want to start off this discussion with a clicker question. Do everybody take a clicker out of the box? So the situation is that we have an atom which is excited by laser light of frequency omega L. The atom has frequency omega 0. I'm discussing the limit of very low intensity. And the question for you is what is the spectrum of the emitted radiation? So in other words, we have our atom. We have the laser beam which comes here. And now the atom emits in all directions. And you collect and frequency analyze the light. And I want to give you four possibilities how the spectrum of the emitted light may look like. Let's assume that on the frequency axis this is the laser frequency. And this is the frequency of the atom. And the first possibility is that we observed a spectrum which is centered at the atomic frequency where the line meets gamma. The second option is that we observe a delta function, a very sharp peak at the frequency omega 0. The third possibility is the we observe a sharp peak at the laser frequency. And the fourth option is that we observe radiation centered at the laser frequency, but with gamma. So I would ask you for your vote. A, B, C, or D. Was there any questions about the choices? 40 for in spectra. The low intensity limit of light scattered emitted by an atom which is excited with a laser beam. We can assume close to resonance, but not on resonance, so we can distinguish between the laser frequency and the atomic frequency. Anybody had a chance? OK. OK, good. A, B, C, D. Yes, let's discuss it. I want to draw your attention to one thing I said, namely low intensity. At low intensity, you can assume that there's only one photon scattered at a time, because the intensity-- we work in the limit of infinitesimal intensity. So I want to as you now, why don't you consider energy conservation and think about the problem again? OK, maybe I'm indicating I'm not happy with the answer. Maybe an obvious question which sometimes people get confused about, what about Doppler shifts? We assume the atom is infinitely heavy and cannot move. Or we trap it in an ion trap. It's in the ground state of the ion trap, and no kinetic energy is exchanged. So don't get confused, and I think none of you got confused about Doppler shifts in kinetic energy here. We eliminate that. OK. Well it's moving in the right direction. Now the two most frequent answers are those two. OK, I've asked you to take energy conservation really seriously. If you start with an atom in the ground state, and the monochromatic photon at the laser frequency, and you've done a wonderful job, which many labs can do. You've stabilized your laser with a precision of 1 hertz. And 1 Hertz, this is sort of what I do as a delta function. So if you take energy conservation absolutely seriously, the photon, which can-- the atom is in the ground state after it has scattered light, so the energy of the atom has not changed. And you can only fulfill energy conservation if there is no change in the spectrum of the light. And that means delta function in, delta function out. Now I know I say it as if it would be the simplest thing in the world, but maybe you have questions. Does anybody want to maybe argue or defend why the light should be broadened? And actually discuss very soon other examples. We go to higher intensity, and at higher intensity other things end and other things are possible. But at the low intensity limit, do you have any suggestion what could cause the broadening of light? AUDIENCE: I guess this is more if [INAUDIBLE] but then look at it as pinned down, but you're going to see a Doppler shift from the recoil as the light comes out. So it accepts the light, and then it moves something, so then it has some velocity. And then when it emits, it's going to be emitting in a random direction. And so its final velocity is going to be in some basically arbitrary direction, because the two [INAUDIBLE] boosts that you get from accepting and emitting. PROFESSOR: OK now first of all, I wanted for this problem to assume that the mass of the atom is infinite. And at that point, the recoil energy has 1 over the mass term. The recoil energy is 0 and we can neglect it. So the limit I'm talking about is sound. But you're right. If you want to bring in the recoil energy, what would then happen is if an atom absorbs a photon and emits it pretty much in the same direction, there is no momentum exchange. Then it will be close to the laser frequency. But then you will find, depending on the angle now, there is a shift between 0 and 2 recoil energies. If you would, however, measure the light at a specific angle, you would still find a delta function, but you would find it shifted. But this is really a complication which involves now the mechanical effect of light and the recoil of a photon. It doesn't change the result. Energy conservation requires the spectrum to be, so to speak, a delta function in total energy. And what we have just discussed is that, depending on the scattering angle, maybe there's a kinetic energy to consider. So what else can broaden the light? Yes? AUDIENCE: Finite pulse time? PROFESSOR: Yes. So things are very different if I use a pulse laser. But here I said the incident light is monochromatic, just one frequency. And therefore you don't have any time resolution in the experiment. The moment you have time resolution, you would introduce into the whole discussion Fourier [INAUDIBLE] and such. But frankly, I ask this question every time I teach the course. And every time the vast majority of the class gives the wrong answer. Let me ask you something else which is maybe a little bit simpler, and that's the following. You have a harmonic oscillator which has an eigen frequency omega 0. You drive the harmonic oscillator at frequency omega L. What is the oscillation of the oscillator? Is it omega 0, the harmonic oscillator frequency? Or omega L, the drive frequency? AUDIENCE: [INAUDIBLE]. PROFESSOR: Of course, the drive frequency. There may be a short transient, but if we do not switch on the drive suddenly, we avoid the transient. The steady state solution is the harmonic oscillator always oscillates at the drive frequency. Now, isn't the atom a harmonic oscillator? In 8421 we discuss the oscillator strings. We discuss that the polarizability of the atom, the dipole moment, the AC stock shift can really be regarded as an oscillating dipole moment. So the atom for that purpose is nothing else than a driven harmonic oscillator. And when you drive the atom at a laser frequency, the dipole moment oscillates at which frequency? The laser frequency. Not the resonant frequency. And therefore, the light which is scattered-- think about it semi classically. This is like an antenna. It's an electron. It's a charge. It's a dipole moment which oscillates at the drive frequency. And the light, which is emitted by this accelerated charge is at the drive frequency. The atom is, for that purpose, nothing else than a driven harmonic oscillator. And I'm saying that probably several times in that course, when you have a semi classical picture and a fully quantum picture, and they do not fully agree. I've never gone wrong with the semi classical picture. But myself and many of my students have come to very wrong answers by somehow holding on just to the quantum picture, and now seeing what the semi classical limit is. You may think too much here in the atom not as a harmonic oscillator, but a two level system. The two level system gets excited, and then after it is excited, it emits at the natural frequency. But you've taken something to seriously here. It's a two level system, but you cannot read the system that you first populate a level, and then you emit, because the energy of the laser is not exactly the energy needed to populate the level. So therefore you shouldn't do first order perturbation theory where you populate a level and then you wait and wait for the emitted photon. This is light scattering. It's a second order problem. One photon in, one photon out. And you have to conserve energy. Sorry for hammering on it, but it's really an example where you can maybe realize certain misconceptions, and I really want to make sure that you fully understand what's going on in this situation. Questions? Well, then let's make it more interesting. Let's introduce higher intensity. So let me first maybe show you-- we've talked a lot about diagrams-- what the diagram for this process is. So we have the laser frequency, and then we have the photon emitted. And of course, the length of this vector, this frequency is the same as the laser frequency. So let's move on now and talk about higher intensity. The high intensity case we discuss in great detail in 8421, but you pretty much I think know also from basic quantum physics what happens if you take a two level system and drive it strongly. If you drive it strongly, stronger than spontaneous emission, that means there is a limit where you can just forget about spontaneous emission in leading order, and the system is nothing else than a two level system coupled by a strong monochromatic drive. So I assume you've all heard about this solution? Get the Rabi oscillation between ground and excited state. So the high intensity limit are Rabi oscillations at the Rabi frequency. So what happens now to the spectrum of the emitted light if you are in the limit that you have Rabi oscillation of the atomic system? And I want you to think classically or semi classically. AUDIENCE: Three peaks. PROFESSOR: Three peaks you generate side bends. Classically, you have an emitter, an oscillating charge. But now you want to throw in that the atom goes from the ground to the excited state, from the excited to the ground state. So, so to speak, when it's in the ground state, it cannot emit. When it's in the excited state, it can emit and such. So you should think about an oscillating dipole which is now intensity modulated. And you know when you take a classic light source which is monochromatic, but put on top of it an intensity modulation at frequency omega Rabi, the solution of that are three peaks. The carrier, which is the laser frequency, plus two side bends at omega Rabi. So therefore, what we now expect for the spectrum is we have the laser frequency, and this is sort of our result for low intensity. However, when we-- in the limit of strong drive, and you will see in a moment where strong drive clearly comes in. So if you have this modulation at the Rabi frequency, you obtain side bends at the Rabi frequency. And the fact that there is a structure with three peaks is actually the famous Mollow triplet. I know when I was a graduate student there was big excitement because people measured for the first time the Mollow triplet. You need high resolution lasers and such, and people were just ready to measure that. Now it seems something which we just mention in basic courses because it's such a basic phenomenon. OK, but let's work a little bit more on that, namely the Rabi frequency is the Rabi frequency of the drive, or you can say the coupling matrix element. But when we have a detuning-- and this is what you want to discuss now-- you have to add the detuning in quadrature. So this is a frequency at which the atomic population oscillates, and it is just semi classically the modulation of the atomic population which creates the side bends. Now if you take that to the limit of large detuning, or small drive, small Rabi frequency, this becomes delta. So that actually means if you go to our stick diagram here, the laser frequency is, of course, detuned by delta from the atomic frequency. But the Rabi frequency is delta, and this is the Rabi frequency. Maybe some people who pressed A and B feel now vindicated, because now you have a component of the emission spectrum, which is at the atomic resonance frequency. And you have a second peak, which is omega 0 minus 2 delta or 2 omega Rabi. Question. How would you give a simple answer if I would ask you but what about energy conservation? I was hammering so much on energy conservation. How can we conserve energy by have a laser photon and emitting one at a different frequency? AUDIENCE: [INAUDIBLE]. PROFESSOR: It's compensated with that. And now let me ask you another question. If you would scatter n photons and you do an experiment, you would expect that, due to Poisson fluctuation and such-- we've talked about fluctuations of intensity and squeezed light and all that. But unless you do something special, you would expect if you excite an atom it emits n photons. You experiment. You measure plus minus square root and you measure Poisson in statistics. But now let's expect you observe n plus photons on the upper side bend. And you measure and you observe n minus photons on the lower side bend. Would you expect that n plus and n minus both have now Poisson fluctuations, or would you expect something else? In other words, let me be very specific, if you count n plus-- we can make it a clicker question. So you have a counting experiment, and you have n plus photons in the upper side bend and n minus in the lower side bend. And if you simply look at the variance, which is the square of the standard deviation from n plus, you find that it is Poisson distributed. So the question which I have now is what is the variance between n plus minus n minus. Is it simply the variance of n plus plus the variance of n minus? The second answer I want to give you is 0. And the third answer is something else. So this is A. This is B. And this is C. Yes. Energy conservation. So there shouldn't be any fluctuation. The emission of the upper side bend and the lower side bend have to be correlated. And you would immediately verify that if I show you what is the diagram. We have ground and excited state. We excite with a laser which has detuning. But then we emit a photon on the lower side bend. The second laser photon can now resonantly reach the excited state. And so you see we have absorbed two photons from the laser beam. And what we have emitted in this diagram is one lower and one upper side bend. This is what happens in second order-- well, second order in the laser beam, but fourth order in the diagram. We emit four photons. And each time the system does it, goes through that, it fulfills energy. Or if I call the upper side bend click and the lower side bend clack. The atom only makes click clack click clack clack click clack clack click clack. It never does click click clack clack. It cannot do two clicks because it has to fulfill this diagram, and they always come in pair. The two photons always come correlated. OK, so we understand now the spectrum. We understand when the resonant photons emerge. They emerge in the high intensity limit due to Rabi oscillations. The question is now what is the widths of those peaks? And I'm want to short you that too, because no you realize some things are easy, but other things are harder. In order to get that, you have to solve the optical Bloch equations, but I want to show you in the next half hour how we can at least get the salient features of this. Colin, you have a question? AUDIENCE: The picture you just offered about the [INAUDIBLE]. That's assuming that we're not depleting the carrier [INAUDIBLE]. Because if we're depleting the carrier, wouldn't we expect to see sort of like a high order, almost like a Bessel function. PROFESSOR: All right. OK, yes. AUDIENCE: First order on to the second [INAUDIBLE]. PROFESSOR: Yes, thanks for bringing that to our attention. There is an assumption. When I said we have a laser beam, I assume we have a laser beam which delivers zillions of photons in such a way that depletion doesn't play a role. What that means technically we replace a laser beam by a c number, and a c number never changes, therefore cannot be depleted. But if you would use a very weak laser beam, we have to modify the answer. The extreme case would be if we use single photon sources, then of course we can never scatter two photons because there's only one photon in the source. But already if you would have hundreds or photons in your cavity and you put in a single atom, the scheduling of single photons would not cause major depletion effects. [INAUDIBLE]? AUDIENCE: In regards to the semi classical [? fabrication ?], we said we have a Rabi frequency, and when it's excited it's likely to emit, and when it's in the ground state it's not likely to emit. So whatever frequency you emit should be modulated by your Rabi frequency, but aren't you most likely to emit when you have a dipole moment? Which would be in the intermediate states between e and g, in which case those occur at twice the Rabi frequencies I think. PROFESSOR: OK, now you go to subtleties of a semi classical picture. Number one is-- I just want to say we're a little bit dangerous, on slippery slope here. One is I really discuss the spectrum here. When I discuss the spectrum, I don't know when the photon is emitted. So when I said we have a Rabi oscillation, I told you that the system is modulated, and this gives rise to side bends. But I can only spectrally solve the side bends if I fundamentally do not have the time resolution to measure when in the Rabi cycle is the photon emitted. The moment I would localize when the photon is emitted, whether the atom is in the ground or in the excited state, if I have a time resolution better than the Rabi oscillation, I do not have the spectral resolution to resolve the side bends. So therefore the question when the photon is emitted is not compatible with observing the spectrum. The second question, I think, can be addressed. And that's the following. I will show you in the next 45 minutes during this class that the steady states solution of the optical Bloch equations give us a rate of photon scattering which is simply gamma times the excited state fraction. So I think from that I would say the more you have atoms in the excited state, the larger is the scattering rate. So therefore, the semi classical picture that you have only a dipole moment when you are halfway between ground and excited state is overly simplistic here. Also, I just want to give you one word of warning. It's really an important comment. And this is the following. The oscillating dipole moment is a picture which uses sort of the analogy with the atom and the harmonic oscillator. And I mean yeah, I really sort of was a bit provocative a few minutes ago when I said just regard the atom as a harmonic oscillator. But you have to be careful. There is one fundamental limit between-- one fundamental difference between the harmonic oscillator and a two level atom. And the fact is the following. This is a two level atom, and this is a harmonic oscillator. So the difference is the following. An harmonic oscillator can be excited. You can pump more and more energy in it. You can go to larger amplitude coherent states, and there is no non linearity. The atom saturates. And this is the difference. Saturation. When saturation comes, when you put more than a few percent, let's say 50%, into the excited state, you have to be very, very, very careful with analogies with the harmonic oscillator. So if you have most of the population in the ground state, a little bit here, and almost nothing here, then we can describe the atom as an harmonic oscillator. And everything you glean from the model of an electron tethered with a spring is not only qualitatively, it's quantitatively correct if you throw in the oscillator strength of the atom. But the moment you pile up more population here, and you just ask about 50-50, at least I would say immediately be careful. Do not take the harmonic oscillator fully seriously at this point, because there's a fundamental difference. In the harmonic oscillator, when you're 50% here, you would have things over there. The harmonic oscillator stays linear. But the two level system starts to saturate. Good. So the question is now we understand the stick diagram of the Mollow triplet that we have always a line at the carrier. Let me just put marks here. We have the carrier and the resonance frequency, but the question is now what is the width of the spectrum. The general answer is rather difficult, but let me give it to you. And I will derive it from the optical Bloch equation. Let me discuss that one limit where the detuning is larger than anything else. And the other case is the resonant case. In general we started out by saying we have a delta function at low intensity. If you now go to higher and higher intensity, you still sort of have the delta function from the lowest order diagram, but the higher order diagram become more important. And the spectrum is in general a superposition of three broadened peaks and the delta function. At low intensity, you only have the delta function. At high intensity, the weight of the elastic scatter the delta function goes to 0. And now let's discuss the widths. And that shows you that even light scattering by a two level system, those very, very simple Jaynes Cummings model has interesting and maybe non intuitive aspect. The side bends have a width of gamma. The carrier has a width of 2 gamma. But when you are on resonance, the situation changes. The carrier has now a width of gamma and side bends have a width of 3/2 gamma. But if you look at it, the sum of the three widths is always 4 gamma. And you will understand that in a few minutes. OK, so this was more qualitative discussion. Let's now work more quantitatively and describe the Mollow triplet. The Hamiltonian for our system is the atomic Hamiltonian times the the z Pauli matrix. The z Pauli matrix is excited excited minus ground ground state. And because of the conservation of probability, it's 2 times the excited. Also just mathematically it's just that. We have the three Hamiltonian of the radiation field, which is a dagger a. And then we have the coupling of a two level system, which is the product of the atomic dipole moment written as sigma plus plus sigma minus raising and lowering operator. Just to remind us, sigma plus is e g. And sigma minus is g e. And then we have to multiply with a plus a dagger. If you want, the atomic part is the dipole moment, and this is the electric field both written as operators. So this is the electric field. And if you look at the coupling term-- we've discussed it before. You have four terms. But in the rotating wave approximation, which we want to use here, we keep only the two resonant terms, which is when we raise an excitation in the atom, we lose an excitation in the light field. Or if you go from the excited to the ground state, we emit a photon. So we have two terms which form the rotating wave approximation. The other two terms are highly of resonant and can often be neglected. So this rotating wave approximation becomes of course the better, the more we go in resonant, because in one term is fully resonant and the other one is very off resonant. And in this case, we can simplify the Hamiltonian because the number of excitations is conserved. The number of excitations is the number of excitations in the photon field plus the number of excited atoms. So this is the operator which measures the number of excitations. If in a [INAUDIBLE] counter rotating term, this operator is conserved. So if I introduce as usual the detuning-- this is what my notes say. It seems to have the opposite sign. I'm not sure if I made a sign error here or later, or if I changed the definition. Let's just move on and see how it works out. So if you now do the rotating wave approximation, our Hamiltonian has now-- by introducing the operator N, the number of [INAUDIBLE] times h bar omega. However, we have multiplied the N operator with the photon energy. If the energy is in the atom, we have made a mistake, which is delta, and we correct that by using the z Pauli matrix. And then our interaction term is a dagger times sigma minus plus a times sigma plus. Let's now discuss what are the eigen states of this Hamiltonian In the simple case when the detuning is 0. And we have sort of two levels of eigen state. One is the ground state with n photons. And one is the excited state with n photons. 0, 1, 2. So we can label this by the photon number. And the excited state of course starts with a higher energy for 0 photon. This is the energy with one photon. And in the case of 0 detuning, the two letters are degenerate. And if we introduce the interaction, if g is now non 0, the levels split. And the splitting is given by g. There's a factor of two. But now if you have photons in the field, a dagger acting on the photon field gives n plus one. So therefore for one photon we have square root 2. In general, we get an expression which is square root n plus 1. Or to write it mathematically, if we have strong mixing between an excited state with n atoms-- with n photons. And a ground state with n plus 1 photons, for the case of delta equals 0, it's just plus minus. You metric as a metric contribution times 1 over square root 2. And these are now with the states we label plus minus. The upper and lower state of each manifold. And n is the photon number in the excited state. n plus 1 is the photon number in the ground state. And the energy of those states is the number of [INAUDIBLE] n plus 1. But then we have the resonant splitting g. And because of the matrix element of a and a dagger, we get square root n plus 1. Now if you use very strong drive-- and this addresses Colin's question about depletion-- we use a laser beam, which is described by a Rabi frequency. This is proportional to the electric field. So be introduce the Rabi frequency, which is given by that. And the laser beam has so many photons that we don't care about n plus 1 or scattering a few photons, so we simply replace n n plus 1 n minus 1, whatever contains n, by the z number, which is the Rabi frequency of the laser beam. So this eventually means that we have too many folds. Plus, minus. Plus, minus. So we have two states plus minus, and they are periodic when we add one photon to the field. Periodic in n. The splitting is the Rabi frequency defined above. And now we realize that when we have spontaneous emission between the manifold for n minus 1 and the manifold n, we have two possibilities. To emit light on resonance. And we have then one possibility to emit an upper side bend. And one possibility to emit the lower side bend. So what is called the dressed atom picture is nothing else than the solution of an atom driven by a strong monochromatic field. The atom is dressed with photons, and the eigen states are no longer just atomic eigen states. These are dressed eigenstates of the combined atomic system plus laser field. So this dressed atom picture explains the Mollow triplet. And you can immediately generalize it if you want to arbitrarily tuning when the superposition states plus minus do not have equal weight. But it's what you've seen a million times, the diagonalization of a two by two matrix. Colin? AUDIENCE: How does this explain the [INAUDIBLE]? PROFESSOR: It does not. This is what I just wanted to say. It explains the Mollow triplet, but not the line widths. The line widths cannot be obtained perturbatively. So for that we have to discuss now the Bloch vector, which is a solution of the optical Bloch equation. So now I'm going to explain you not all glorious details, but the salient feature of the line bits. So the density matrix for two level atom can be written as-- let me back up a moment. The density matrix has four matrix elements. But if the sum of the diagonal matrix element is one conservation of probability, we have actually three independent matrix elements. And those three independent matrix elements can be parametrized by what is called the Bloch vector. A lot of you have seen it. We also discuss it in great detail in 8421. But I'm giving you the definition. And sigma is the vector of Pauli matrices. To be specific, the three components of the Bloch vector are as follows. The z comportment measures, you can say, the population inversion. The difference between ground and excited state population. And the x and y component measure the coherencies. Either the sum of the coherencies or the difference of the coherencies. And here is minus imaginary unit. If you replace the density matrix or the matrix elements sigma ee, or ee, or gg by definition the optical Bloch equation turns into differential equation for r. It's now a differential equation for r. And let me just write it down. Let me just write down the Hamiltonian part. This is the equation of motion for the density matrix. I will add on in a minute the relaxation part which comes from the master equation. This is just the Hamiltonian part. And this would result into an equation of motion for the Bloch vector, which has delta here minus delta here minus g. g and the rest of the matrix elements is 0. The Hamiltonian, using rotation matrices, can be written by a z rotation matrix plus g times the x rotation. And if you actually look at the solution-- it's not a finite. It's an infinitesimal rotation. This matrix tells you that the detuning does a z rotation, and the drive is responsible for an x rotation. Let me tell you what happens. You have seen the Bloch vector many cases. If you don't drive the system, the Bloch vector in the ground state is down. In the upper state it's up. If it's in between-- that's where Timo's dipole moment comes in-- we have a superposition of ground and excited state. And it rotates at the atomic frequency omega 0. But we are in the rotating frame of the laser at omega, so in the rotating frame, when the frame rotates with omega. And omega 0 rotation becomes the rotation with delta. So therefore, the free evolution of the atom is based on this Hamiltonian that the Bloch vector rotates with delta in the rotating frame of the laser. And it's a rotation around the z-axis. If you drive the system now, we take the Bloch vector from ground to excited, there are Rabi oscillations. And this is now a rotation around the x-axis. Well, whether it's x or y is a convention of notation, and I've defined it such that the driven system is an x rotation. The free evolution is a z rotation. And that's all you have to know. This is the most general solution of two level system without dissipation that we have two rotation angles. One is the free rotation, which is the detuning. It is the z-axis. And the driven system rotates around the x-axis. But now this is well known. This is boring. But now we want to add what the master equation gives us. And the master equation uses as its [? limb ?] platform where we have the sigma plus jump operator. And I told you that we have to use the following form. We'll hear more about it, probably not today, but on Friday. You know that this gives simply the optical-- this gives us simply the optical Bloch equations. We discussed it on Monday that we have now a damping gamma term for the population. An excited state population decays with a rate of gamma. And the coherences, the off diagonal matrix elements, decay with a rate of gamma over 2, and we spent a long time discussing this vector of 2. So this would mean now that the equation for the Bloch vector has gamma 2 terms, which comes from the dampening of the coherences. And it has gamma, which is the damping of the population. Well if that would be all, everything is damped to 0, but everything is in the end leads to the ground state population. So we have to add this term. It's just a different way of the optical Bloch equations. We have discussed in great lengths this form of the optical Bloch equations. And the sigma plus sigma minus is because these are just simple matrices with one matrix element. You've seen the optical Bloch equation, I think. We did also that in 8421 21 that you have the derivative of the density matrix includes now damping of the off diagonal matrix elements by gamma over 2 damping by the diagonal matrix elements by gamma. If you simply substitute the r vector for the matrix element, you get this equation in one step. These are now the optical Bloch equations written as a differential equation for the Bloch vector. OK, sorry if that was confusing, but it's simple definition substitution no brainer. These are the optical Bloch equations written in terms of the Bloch vector and no longer in terms of the density matrix. And questions about that? OK, because now I want to draw conclusions. Our goal is to understand the spectrum and the line widths of the Mollow triplet of the emitted light. So I have to make the connection. How do I make the connection from the Bloch vector to emitted light? Well, it's done by the dipole moment, because it is the oscillating dipole which is responsible for emitting and scattering light. The dipole moment can of course be obtained from the solution of the optical Bloch equation for the density matrix. It is the trace of the density matrix times the operator we are interested in. And this involves the matrix element dge. The dipole moment has only matrix elements between excited and ground state. And ground and excited state. Trace rho other parentheses. So this is the matrix element. Let me now write down the density matrix in its matrix element rho ge plus rho eg. But now we want to go from the rotating frame back to the lab frame. And the rotating frame rotates at omega. So now I have to put back e to the i omega t and e to the minus i omega t. This is sort of going back from the rotating frame to the lab frame. And the picture of we want to use is-- the intuitive picture is to use the Bloch vector. So I'm expressing now those matrix elements by the component x and y of the Bloch vector. And since the x and y component was the sum or difference of diagram matrix element of the density the coherences. I get now cosine omega t and sine omega t. So that tells us that one component is in phase-- well in phase means with respect to the driving field we assume that the system is driven by classical field e0 cosine omega t. And now we find that the oscillating dipole moment, it's a driven harmonic oscillator. And there is a part which is driven in phase with the drive field, and one which is in quadrature. So we have those two components. And the fact is now the following. If you have an oscillating dipole, this gives rise to immediate-- or more accurately, I should say scattered light. Remember, if you assume there is photon first emitted and then a resonant photon absorbed, this picture is wrong. You should rather think of scattering light, but the light scattering is induced by the oscillating dipole. And what happens now is-- and this is why the Bloch vector and the Bloch sphere is such a wonderful picture. The dipole moment are the transverse component of the Bloch vector. So therefore, you can say that if you're interested in the spectrum of the emitted light, what you should find out is what is the spectrum of r. Let me back up one step. If you want to get the spectrum of the emitted light from first principles-- I'm cheating a little bit here. I'm saying let's get it from the dipole moment. You should actually get it from the correlation function of the dipole moment. It's sort of the two time correlation function. But you can show-- and this takes another 10, 20, 30 pages in API. The temporal correlation function of the dipole moment fulfils the same differential equation as the Bloch vector. So I've given you here sort of the intuitive link that would oscillates is the Bloch vector it emits it. But technically, because you may not control the phase of each atom, you should rather say the spectrum is a correlation function of the dipole, and we should look at the spectrum of the correlation function. But instead, I'm looking now at the spectrum of the optical Bloch vector. And you can show mathematically exactly that they obey the same differential equation, but it's a little bit tedious to do that. OK, so therefore if you take this little inaccuracy-- forgive me this inaccuracy. Once we know what the optical Bloch vector is doing spectrally, we know the spectrum of the emitted light. And for that, for the optical Bloch vector, we simply look at this matrix, and we ask what are the eigenvalues of this matrix? This matrix has three by three matrices, three eigenvalues. The real part of those eigenvalues gives us the position of the Mollow triplet. And the imaginary part gives us the widths of those peaks. So if you look at this matrix-- and I'm not doing the complicated cases here. If you look at this matrix for delta equals 0 and g equals 0. Well it is already in diagonal form, so it has three imaginary eigenvalues minus gamma over 2 minus gamma 2 minus gamma. So this matrix has three complex eigenvalues. And I want to discuss two cases. For this case we have minus gamma over 2 for the x and y component of the optical Bloch vector. And the z component is minus gamma. And this is the situation. Apart from a factor of two that we have gamma, 2 gamma, and gamma. The fact of 2 comes of course. Are you asking what is the spectrum of the electric field or what is the spectrum of the power? If you go from e to e squared, then if you have an exponential decay with gamma of e, e squared decays with two gamma. So that's where factors of 2 come from. AUDIENCE: Can you label the top of the diagram with [INAUDIBLE] just the delta equals 0. PROFESSOR: Yes. Give me a second. AUDIENCE: [INAUDIBLE] g equals 0, right? PROFESSOR: The fact is here we're looking at limiting cases. And the more important thing is that this is the case of weak drive. So what is more important is that the drive frequency is 0. I'm not giving you the full picture here. The full glorious derivation is in API. And just tried to sort of entertain you by giving you a few appetizers. If you want, read it up there. But what I want to show you is they're two simple cases, but it looks already intriguing because here this is wider than the side bends, and here it's the opposite. And I want to give you a taste of it. And I'm not doing it rigorously, and you're right, I have to think about it. But the important part is that the drive frequency is 0. We are not driving, we are not rotating in x. The system is mainly rotating around the z-axis. And at that moment, the matrix for the differential equation for the optical Bloch vector has three imaginary parts. It's gamma, 2 gamma, and gamma or half of it if you look at the amplitude. And now I want to just show you in a second how we can get the second part. But yeah, I'm guilty as charged. There are small gaps in my argument. But all I want to show you is how you find-- I want to just sort of convince you that the optical Bloch equations with this matrix have all the ingredients to explain it. OK, so what I've done here is I've shown you that we have those three ingredients. And actually the moment-- I should have said it before answering your question. When we now crank up delta, what we're physically doing is we're doing a rapid rotation of the Bloch vector around the z-axis. So when we rapidly around the z-axis, we're not doing anything to the z component, but we're strongly mixing the x and y component, but since [? half ?] the imaginary part-- I'm waving my hands now. We're not changing the imaginary part for x and y. And we still have two imaginary parts, which are gamma over 2. One imaginary part which is gamma. I know I'm running out of time, but I only need a few more minutes. So if you do a rapid rotation around the z-axis, we obtain three eigenvalues. Since we rotate around the z-axis, we have minus gamma. Sorry, my notes are better than what I remember. I looked through the notes. But what I wanted to say was the following. I first show you what are the eigenvalues of this matrix in one case, which is just a warm up. I'm telling you there are those three values. And now I'm saying we are interested in the physical situation of rapid rotation. The rapid rotation modifies the three eigenvalues. The three eigenvalues are no longer real. The rapid rotation around the z-axis adds an imaginary i delta to the x and y-axis. Because now we are rotating, x and y are getting mixed. And this means now that we have three peaks. They are located in the rotating frame at 0. This is the carrier at plus minus delta. These are the two side bends. And this part are the widths of it. This is how you have to interpret the result. So we have to know three peaks. And the full widths of half maximum for the intensity, which is the amplitude squared is gamma, 2 gamma, and gamma. And we can now immediately proceed to the next case. If we are on resonance, and we drive The system strongly, now we are driving the system strongly around, not the z-axis, around the x-axis. So now if we add the strong drive around the x-axis, we have now eigenvalues which is plus minus ig. This is the rotation we add, and this gives e to the i-- in rotation ge to the i omega t gives an ig. And what happens is the following. The rapid rotation around the x-axis-- well, if you rotate around the x-axis, you're not touching x. And without any rotation, we had an eigenvalue for x which was minus gamma over 2. And if you take the matrix of whatever the system and rotate it around the x-axis, this is preserved. AUDIENCE: Excuse me. PROFESSOR: Yep, I need one more minute. Sorry. But the rotation around x strongly mixes y and z. So therefore the two other eigenvalues appear at rotation ig. And it is now the average of gamma and gamma over 2, which is 3/4 gamma. And if you go to the amplitude squared, it is 3/2. So therefore we have the situation that we have those three eigenvalues. And this is responsible for the three peaks at plus minus g and 0. And the widths of it is I have to multiply by a factor of 2 gamma and 3/2 gamma. And this explains the two limiting cases I presented to you earlier. No time for question, the other people are waiting. Reminder we have class on Friday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

7_Metrology_shot_noise_and_Heisenberg_limit_Part_2.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Good afternoon. So it has been a little bit more than a week when we met the last time. Before I continue our discussion of metrology and interferometry. I just want to share something I saw on my visit to the Netherlands. When I visited the University of Delft and the Kavli Institute, they had just accomplished the entanglement of two NV centers. And we had just talked in class about the entanglement of two ions. So I'm sort of excited to show you that people have just taken the next step. And what often the next step means is that in atomic physics, we use pristine systems-- ions, neutral atoms in [INAUDIBLE] vacuum chamber. And we create new forms of entanglement. Or with quantum gases, new forms of quantum matter. But ultimately, we hope that those concepts, those methods, and this knowledge translates to some room temperature materials or solid state materials which can be handled more easily, and are therefore much closer to applications. So NV centers are kind of nature's natural ion trap, or nature's natural neutral atom trap. Let's not discuss whether this is neutral or ionized. It's a vacancy in nitrogen. And it has a spectrum which looks like an atom. So you can say once you have such a defect in nitrogen, you have an atom, a single atom in an atom trap or an ion in an ion trap. And you don't have to create the vacuum. It's there. Every time you look at it, it's there. And you can excite it with the laser. It has a spectrum similar to atoms. It has spin structure So you have these vacancies in diamond. So these are little quantum dots. But now you have two problems. One is you want to collect the light emitted by them. And what is best is to mill a lens right into the material. So this is the diamond material. A lens is milled. And that already gives some [? culmination ?] of the light emitted by it. But the bit problem until recently has been when you create those some people call it artificial atoms. Every atom is a little bit different, because it experience a slightly different environment. A crystal has strain, so if you have seemingly two identical defects in a diamond crystal, the two defects will have a resonance line, which is a few gigahertz difference. You would say, well, maybe it's just [? apart ?] in 10 to the 5, but it means the photons are distinguishable. So if you want to do entanglement by having two such artificial atoms emitting a photon onto a beam splitter, and then by performing a measurement we project the atoms into Bell state-- I hope you all remember what we discussed for the trapped ions-- then you have to make sure that fundamentally, those photons cannot be distinguished. And the trick here is that they put on some electrodes. And by adding an electric field, they can change the relative frequency. And therefore, within the frequency uncertainty given by Heisenberg's Uncertainty Relation, they can make the two photons identical. And when, let's say, experiment. Well, you have two NV centers, defects in diamond. You can manipulate with microwaves coherently the spin. You need lasers for initialization and readout. But then closer to what we want to discuss, you need laser beams which excite the NV center. And then the two NV centers emit photons. And what you see now is exactly what we discussed schematically and in context of the ions, that the two NV centers now emit photons. And by using polarization tricks and the beam splitter, you do a measurement after the beam splitter. And based on the outcome of the measurement, you have successfully projected the two NV centers into a [INAUDIBLE] state. OK, good. All right. Let me just summarize what our current discussion is about. This section is called Quantum Metrology. It is a section where we want to apply the concepts we have learned to precision measurements. It's actually a chapter which I find nice. We're not really introducing new concepts. We're using concepts previously introduced. And now you see how powerful those concepts are, what they can be used for. So we want to discuss the precision we can obtain in quantum measurement. We apply it here to an atom in the interferometer, to an optical interferometer. We could also discuss the precision of spectroscopic measurements. A lot of precision measurements have many things in common. So what we discuss here is as a generic example for precision measurement that we send light from our center interferometer. Here is a phase shift. And the question is, how accurately can we measure the phase shift when we use n photons as a resource? And of course, you are all familiar with the fundamental limit of standard measurements, which is short noise. And sort of as a warm up in our last class Wednesday, week ago, I showed you that when we use coherent states of light at the input of the interferometer, we obtain the short noise. Well, it may not be surprising, because coherent light is as close as possible, has similar to classical light. But then we discussed single mode, single photon input. And by using the formalism of the Mach-Zehnder interferometer, we've found that the phase uncertainty is again 1 over square root n. So then the question is, how can we go to the Heisenberg limit, where we have an uncertainty in the phase of 1 over n? And just as a reminder, I find it very helpful. I told you that you can always envision if you have n photons and you multiply it, put the n photons together by multiplying the frequency by n, then you have one photon which [INAUDIBLE] with n times the frequency. And it's clear if you do a measurement at n times the frequency, your precision in phase is n times better. So what we have to do is we have to sort of put the n photons together. And then we can get a precision of the measurement, which is not square root n, but n times better. So this is what we want to continue today. This is the outline I gave you in the last class. If you have this optical interferometer, this Mach-Zehnder interferometer. And if you use coherent sets or single photons as the input state, we obtain the short noise. Now we have to change something. And we can change the input state, we can change the beam splitter, or we can change the readout. So we have to change something where we entangle the n photons, make sure in some sense they act as one giant photons, with either n times the frequency, but definitely with n times the sensitivity to phase shift. Any questions? Good. The first example, which we pretty much covered in the last class, was an entangled state interferometer. So instead of having the Mach-Zehnder interferometer as we had before, we have sort of a Bell state creation device. Then we provide the phase shift. And then we have a Bell analysis device. And I want to use here the formalism and the symbols we had introduced earlier. So just as a reminder, what we need is I need two gates. One is the Hadamard gate, which in matrix representation has this matrix form. And that means if you have a qubit which is either up or down and you apply the matrix to it, you put it in a superposition state of up and down. It's a single qubit rotation. You can say for the bloch on the Bloch sphere, it's a 90 degree rotation. The second gate we need is the controlled NOT. And the controlled NOT we discussed previously can be implemented by having an interferometer and using a non-linear Kerr medium coupling another photon to the interferometer in such a way that if you have a photon in mode C, it creates a phase shift with the interferometer. If there is no photon in mode C, it does not create an additional phase shift. And as we have discussed, this can implement the controlled NOT. So these are the ingredients. And at the end of the last class, I just showed you what those quantum gates can do for us. If you have two qubits at the input, we apply the-- and I will assume they are both in logical 0-- the Hadamard gate makes the coherent superposition. And then, we have a controlled NOT where this is the target gate. Well, if this is 0, if the control beat is 0, the target beat stays 0. So we get 0, 0. If the control beat is 1, the target beat is flipped to 1. So we get 1, 1. So the result is that we have now created a state 0, 0 plus 1, 1. And if we apply a phase shift to all the photons coming out on the right hand side, we get a phase shift which is too fine. So we already get the idea. If we take advantage of a state 1, 1, it has twice the [? face ?] sensitivity as a single photon. And we may therefore get the full benefit of the factor of 2, and not just square root 2. And this is what this discussion is about. Questions? OK, so that's where we ended. We can now use another controlled NOT and bring in the third qubit. So what we create here. Just make a reference, this is where we start today. So we create here the state, which is either all beats are 0, all beats are 1, and then the phase shift gives us three times the phase shift phi. And therefore, by bringing in more and more qubits, I've shown you n equals 1, n equals 2, n equals 3. So now you can go to n. We obtain states which have n times the phase sensitivity. Let me just mention in passing that for n equals 3, the superposition of 0, 0, 0 and 1, 1, 1 goes by the name not gigahertz, GHZ. Greenberger. This third one is Zeilinger. The second one is-- AUDIENCE: Horne. PROFESSOR: Horne. Thank you. Greenberger-Horne-Zeilinger state. And those states play an important role in test of Bell's inequality. Or more generally, states which are macroscopically distinct are also regarded cat states or Schrodinger cat states. If we apply a phase shift and go now through the entangler in reverse, just a reverse sequence of CNOT gates and Hadamard gates, then we have all the other n minus, have the n minus 1 qubits. All the qubits except for the first reset to 0. But the first qubit is now in a superposition state where we have the phase into the power n. And now, we can make a measurement. And P is now the probability to find a single photon in the first qubit. So in other words, our measurement is exactly the same as we had before where we had the normal Mach-Zehnder interferometer. We put in one photon at a time. And we determined the probability. What is the photon at one of the outputs of the interferometer? But the only difference is that we have now a factor of n in the exponent for the phase shift. And I want to show you what is caused by this factor of n. Let me first remind you how we analyzed the sensitivity of an interferometer for single photon input before. So for a moment, set n equals 1 now. Then you get what we discussed two weeks ago. The probability is [INAUDIBLE] as an interferometer. Sine also, cosine also infringes with cosine phi. If you do measurements with a probability P, that's a binomial distribution. Then the deviation or the square root of the variance in the binomial distribution is P times 1 minus P. And by inserting p here, we get sine phi over 2. The derivative of P, which is our signal, with a phase is given by that. And then just using error propagation, the uncertainty in the phase is the uncertainty in P divided by dp dt. So then we repeat the whole experiment n times, and if your binomial distribution and we do m [INAUDIBLE] of our [INAUDIBLE] with probability P, we get an m under the square root. And then we get 1 over square root m. And this was what I showed you two weeks ago, the standard short noise limit. But now we have this additional factor of n here, which means that our probability has a cosine which goes with n phi. When we take the derivative, we get a factor of n here. And then dp d phi has the derivative. But then we have to use a chain rule, which has another factor of n. And therefore, we obtain now that the phase sensitivity goes as 1 over n. And therefore, we reach the Heisenberg limit. So if you have now your n photon entangler, the one I just-- the entanglement operation with these n cubits is special beam spreader replaced by the entangler, we have now a sensitivity, which goes 1 over n. And then of course, if you want, you can repeat the experiment n times. And whenever you repeat an experiment, you gain an addition with the square root of m. Just give me one second. We have this. OK, so this is an example where we had n qubits entangled, like here. And the sensitivity of this interferometer scales now as the Heisenberg limit with 1 over n. OK, I want to give you, because they're all nice and they're all special, I want to give you three more example how you can reach the Heisenberg limit. The next one goes by the name super beam splitter method. It's actually a very fancy beam splitter where you have two prods of your beam splitter. One has 0 and one has n photons. And behind the beam splitter, you have two options. Either all the n photons are in one state, or the n photons are in the other state. You never have any other mixture. Of course, you know your standard half silvered mirror will not do that for you, because it will split the n photon state with some Poissonian statistics or whatnot into the two arms and such. And you can calculate exactly what a normal beam splitter does to that. It's very, very different. This here is a very, very special beam splitter. Just to demystify this beam splitter a little bit, I want to show you that, at least conceptually, there is an easy implementation with atoms. I really like in this section to grab an example from atoms and to grab an example for photons, because it really brings out that the language may develop equally applies to atoms and photon. So if I take now an easy example, if I take an example of the Bose-Einstein condensate with n atoms, let's assume we have a double well. But now we have attractive interactions. This is not your standard BEC. Your standard Bose-Einstein condensate has repulsive interaction, with attractive interaction. If you go beyond a certain size, certain a atom number, the condensate will collapse. So let's assume we have a Bose-Einstein condensate with n atoms. There are some attractive interactions, but we stay within the stability diagram that those atoms do not-- that those atoms do not collapse. So now, if the atoms are attractive, they all want to be together. Because then they lower the energy of each other. So if you have something with attractive interaction, you want to have n atoms together. But if your double well potential is absolutely symmetric, there is no symmetry breaking, and you have an equal amplitude for all the atoms to be in the other well. So therefore, under the very idealized situation I described here, you will actually create a superposition state of n atoms in one well with n atoms in the other well. And this is exactly what we tried to accomplish with this beam splitter. OK, so if we have this special beam splitter, which creates that state, then when we add a phase shift to one arm of the interferometer, we obtain-- which looks very promising-- a phase shift phi multiplied by n. And if we read it out, we have now if you pass it through the other super beam splitter, we create again a superposition of n0 and 0n. But now, because of the phase shift with cosine and sine factors, which involve n times the phase shift. For obvious reasons, those states also go by the name noon states. That's the name which everybody uses for those states, because if you just read n00n, it gives noon. So this is the famous noon state. So it seems it smells already, right, because it has a phase shift of n phi. Let me just convince you that this is indeed the case. So the probability of reading out 0 photons in one arm of the interferometer scales now with the cosine square. If you do one single measurement, the binomial distribution is the same as before, but with a factor of n. The derivative has those factors of n. And that means that the variance in the phase measurement is now 1 over n. If you want a reference, I put it down here. As far as I know, the experiment has been done with three photons, but not with larger photon number. Questions? Collin. AUDIENCE: This picture with the double well, isn't there going to be-- all right, some people would say you spontaneously break the symmetry and [INAUDIBLE] either ends up in one side or the other. Like isn't this [INAUDIBLE]? PROFESSOR: Well, Bose-Einstein condensation with attractive interactions was observed in 1995. And now, 18 years later, nobody has done this seemingly simple experiment. And what happens is really that you have to be very, very careful against any experimental imperfection. If the two double well potentials are not exactly identical, the bosons always want to go to the lowest quantum state. Well, that's their job, so to speak. That's their job description. And if you have a minuscule difference between the two, the dates of the two wells, you will not get a superposition state. You will simply populate one state. And how to make it experimentally [INAUDIBLE], this is a really big challenge. [? Tino ?]. AUDIENCE: I have a question. Let's say we're somehow able to make the double well potential perfect. But if we didn't have attractive interactions, then wouldn't he just get a big product state of each atom being in either well? PROFESSOR: OK, if you're a non-interacting system, the ground state of a double well potential is just your metric state 1 plus 2 over square root 2. And for non-interactive BEC, you figure out what is the ground state for one particle and then take it to the power n. This is the non-interactive BEC. STUDENT: So then entanglement is because of the interaction, right? PROFESSOR: If you have strong repulsive interaction, you have something which should remind you of the [INAUDIBLE] insulator. You have n atoms. And n over 2 atoms go to one well. n over 2 go to the other well. Because any form of number fluctuations would be costly. It will cost you additional repulsive interaction. So therefore, the condensate wants to break up into two equal parts. So that's actually a way how you would create another non-classical state, the dual flux state of n over 2, n over 2 particle. And for attractive interactions, well you should create the known state. OK, so this was now a state n0 and 0n. There is another state, which you have encountered in your homework. And this is a superposition not of n0 and 0n and n0. It's a superposition of n minus 1n and n n minus 1. This state goes by the person I think who invented it, the [? Yerkey ?] state. And you showed in your homework that with that, you also reach Heisenberg Limited Interferometry, where the phase scales is 1 over n. What I want to add here to it is how one can create such a [? Yerkey ?] state. And again, I want to use the example with an atomic Bose-Einstein condensate. And here's the reference where this was very nicely discussed. So let's assume we can create two Bose-Einstein condensates. And they have exactly n atoms. And I would actually refer to [? Tino's ?] questions how to make them have two n atoms in a trap and then make a double well potential, deform the harmonic oscillator potential to a double well potential. Then for strong repulsive interactions, the condensate will symmetrically split into two flux states, each of which has n atoms. So now how can we create the [? Yerkey ?] state from that? Well, we simply have-- we leak out atoms. We leak atoms out of the trap. My group demonstrated an RF beam splitter, how you can just split in very controlled way start rotating the cloud on non-trapped state. And then atoms slowly leak out. Well, when you can measure, of course, you can take an atom detector and measure that an atom has been out-coupled, that an atom has leaked out of the trap. If you don't like RF rotation, you can also think that there is a tunneling barrier, and atoms slowly leak out by whatever mechanism. And the moment you detect that an atom, you then project the state in the trap to n minus 1 atom, because you've measured that one atom has come out. But now you use a beam splitter. And a beam splitter could simply be a focused laser beam. And the atoms have a 50% probability of being reflected or end up tunneling through. So therefore, if you have now a detector which measures the atoms on one side and the atoms on the other side, then you don't know anymore when the detector makes click form which atom trap the atom came. Or more formally, a beam splitter transforms the two input modes ab into a plus b and a minus b normalized by square root 2. So therefore, if this detector clicks, when you project the remaining atoms into the symmetric state, here you detect one atom in the mode a plus b. And that means that the remaining atoms have been projected into the [? Yerkey ?] state. If the lower atom detector would click, well, you get a minus sign here. So that's one way how at least in a conceptually simple situation, you can prepare this highly non-classical state by starting with a dual flux state of Bose-Einstein condensates, and then using-- and this is an ongoing theme here-- by using a measurement, and then the post-measurement state is the non-classical state you wanted to prepare. Question? Yes. AUDIENCE: How can you ensure that only one atom leaves the [INAUDIBLE]? PROFESSOR: The idea is that we leak atoms out very, very slowly. And then we have a detector, which we assume has 100% quantum efficiency. So therefore, we simply wait until the detector tells us that one atom has leaked out. And in an idealized experiment, we know the atoms either have been measured by the detector, or they are still in the trap. AUDIENCE: And also, is there a property that holds the [INAUDIBLE] you detect atoms [INAUDIBLE]? PROFESSOR: In principle yes, but the idea here is if you have a very slow leakage process, the probability that you detect two atoms at the same time is really zero. You leak them continuously and slowly, but then quantum mechanically, that means for most of the time, you measure nothing. That means the leakage hasn't taken place. The quantum mechanical system has not developed yet. But the moment you perform a measurement, you project-- it's really the same if you say you have n radioactive atoms. You have a detector. And when the detector makes click, you know you have n minus 1 radioactive atoms left. It's just applied here to two atom types. Other questions? AUDIENCE: [INAUDIBLE]? PROFESSOR: No. And maybe I'll tell you now why not. We have discussed the noon state. And we have discussed here a highly non-classical superposition state. Let's just go back to the noon state. We have n atoms here, zero here, or n atoms, or the reverse. But now, assume a single atom is lost, is lost from your trap by some background gas collisions. And you have surrounded the trap with a detector. So if you have the noon state, the symmetric superposition state, all atoms here and all atoms there. But by your background process, by an inelastic collision, you lose one atom. And you detect it. You could set up your detectors that you know from which that you figure out from which trap was the particle lost. So therefore, a single particle lost if you localize from which trap the particle is lost would immediately project the noon state into a state where you know I have n or n minus 1 atoms in one well, and 0 in the other well. So I've already told you, with the attenuator, you can never assume an attenuator is just attenuating a beam. An attenuator can always regard it as a beam splitter. And you can do measurements at both arms of the beam splitter. Or you have to consider the vacuum noise which enters through the other part of the beam splitter. And if you are now add that those atoms, n atoms, in a trap have some natural loss by inelastic collision or background gas scattering, the loss is actually like a beam splitter that particles don't stay in the trap, but go out through the other part. And then you can measure it. So in other words, every loss process should be regarded as a possible measurement. And it doesn't matter whether you perform the measurement or not. And I think it's just obvious that the noon state, the moment one particle is lost and you reduce this particle to figure out if n particles are here or n particles are there, the whole superposition state is lost if you just lose one single atom out of n. So the lifetime of a noon state is then not your usual trap lifetime, where you lose half of the atoms, 1 over e of the atoms. It is n times faster because it is the first atom lost, which is already completely removing the entanglement of your state. So more quantitatively, to say it more specifically, the limitation is loss. When you have a fully entangled state, maximally entangled state, usually a loss of one particle immediately removes the entanglement. We had the situa-- no, that's not a good example. But for the most entangled state, usually-- and for the noon state, it's trivial to see-- a single particle lost allows you to measure on which side of the potential barrier all the atoms are, and all the beauty of the non-classical state is lost. So if you assume that in a time window, you have an infinitesimal loss, a loss of epsilon, what usually happens is if you have an entanglement of n particles, then you lose your entanglement. If you expand 1 minus epsilon to n, it becomes 1 minus n times epsilon. So that's one reason why people have not scaled up those schemes to a large number of photons, or a large number of atoms. Because the larger n is, the more sensitive you are to even very, very small losses. Questions? AUDIENCE: [INAUDIBLE] PROFESSOR: Yes. AUDIENCE: Or is that physically? PROFESSOR: The super beam splitter would create the noon state. I've not explained to you what it physically would be for photons, but I gave you the example for atoms to BECs in a double well potential with attractive interaction. So to start with the condensate in-- AUDIENCE: Start with a bigger [INAUDIBLE]. PROFESSOR: If you start with a double well potential, and you put in n atoms initially in, but then you switch on tunnel coupling, then you would create the noon state if the interactions are attractive, and if everything is idealized, that you have a completely symmetric double well potential. OK, the last example is the squeeze light Interferometer. I just want to mention it briefly, because we've talked so much about squeezed light. So now I want to show you that squeezed light can also be used to realize Heisenberg Limited Interferometery. So the idea is, when we plot the electric field versus time, and if you do squeezing in one quadrature, then for certain times, the electric field has lower noise and higher noise. We discussed that. And the idea is that if there is no noise at the zero crossing, that this means we can determine the zero crossing of the light, and therefore the phase shift with higher accuracy. You may also argue if you have this quasi-probabilities, and with squeezing, we have squeezed the coherent light into an ellipse. And things propagate with e to the i omega t. Then you can determine a phase shift, which is sort of an angular sector in this diagram, with higher precision if you have squeezed this circle into an ellipse like that. So that's the idea. And well, it's fairly clear that squeezing, if done correctly, can provide a better phase measurement. And what I want to show you here in a few minutes is how you can think about it. So we have discussed at length the optical Interferometer, where we have just a coherence state at the input. This is your standard laser Interferometer. But of course, very importantly, the second input port has the vacuum state. And we discussed the importance of that. So the one difference we want to do now is that we replace the vacuum at the second input by the squeezed vacuum where r is the parameter of the squeezed vacuum. OK, so that's pretty much what we do. We take this state, we plug it into our equation, we use exactly the same formalism we have used for coherence states. And the question, what is the result? Well, the result will be that the squeezing factor appears. Just as a reminder, for our Interferometer, we derived the sensitivity of the Interferometer. We had the quantities x and y. And the noise is delta in this y operator squared divided by x. This was the result when we operate the Interferometer at the phase shift of 90 degrees. Just a reminder. That's what we have done. That's how we analyze the situation with a coherent state. The signal x is now-- the signal x is the number measurement a dagger a. And we have now the input of the coherent state. And b dagger b. We have an input mode a and a mode b. They get split. And then we measure it at the output. And we can now at the output have photons-- a dagger a, which come from the coherent state, and b dagger b, which come from the squeeze vacuum. So this is now using the beam splitter formalism applied to the Interferometer. So this is now the result we obtain. And in the strong local oscillator approximation, it is only the first part which contributes. And this is simply the number or photons in the coherent beam. The expectation value of y is 0, because it involves a b and b dagger operator. And the squeezed vacuum has only, if you write it down, in the n basis, in the flux basis, has only even n. So if you change n by one, you lose overlap with the squeezed vacuum. So therefore, this expectation value is 0. For the operator y squared, you take this and square it. And you get many, many terms, which I don't want to discuss. I use the strong local oscillator. Limited a dagger and a can be replaced by the eigenvalue alpha of the coherent state. So therefore, I factor out alpha squared in the strong local oscillator limit. And then what is left is b plus b dagger squared. And since we have squeezed the vacuum, this gives us a factor into the minus r. So if we put all those results together, we find that the phase uncertainty is now what we obtained when we had a coherent state with the ordinary vacuum. And in the strong local oscillator limit, the only difference to the ordinary vacuum is that in this term, we've got the exponential factor e to the minus r. And since we have taken a square root, it's e to the minus r over 2. So that result would actually suggest that the more we squeeze, that delta phi should go to 0. So it seems even better than the Heisenberg limit. However-- well, this is too good to be true-- what I've neglected here is just the following. When you squeeze more and more, the more you squeeze the vacuum, the more photons are in the squeezed vacuum, because this ellipse stitches further and further out and has overlap with flux states at higher and higher photon number. So therefore, when you go to the limit of infinite squeezing, you squeeze out of the limit where you can regard the local oscillator as strong, because the squeezed vacuum has more photons than your local oscillator. And then you have to consider additional terms. So let me just write that down. However, the squeezed vacuum has non-zero average photon number. And the photon number of the squeezed vacuum is, of course, apply b dagger b to the squeezed vacuum. This gives us a sinc function. And we can call this the number of photons in the squeezed vacuum. So we have to consider now this contribution to y square. So we have to consider the quadrator of the ellipse, the long part of the ellipse, the non-squeezed quadrator component. And we have to consider that when we calculate the expectation value of y square. And then we find additional terms, which I don't want to derive here. And the question is then, if you squeeze too much, you lose. So there's an optimal amount of squeezing. And for this optimal amount of squeezing, the phase uncertainty becomes approximately one over the number of photons in the coherent state, plus the number of photons in your squeeze vacuum. So this is, again, very close to the Heisenberg limit. So the situation with squeezed light is less elegant, because if you squeeze too much, you have to consider additional terms. This is why I gave you the example of the squeezed light and the squeezed vacuum as the last. But again, the Heisenberg limit is very fundamental as we discussed. And for an optimum arrangement of the squeezing, you can also use a squeezed vacuum input to the Interferometer to realize the Heisenberg limit. Any questions? Why is squeezing important? Well, squeezing caught the attention of the physics community when it was suggested in connection with gravitational-- with the detection of gravitational waves. As you know, the laser Interferometer, the most advanced one is LIGO, has a monumental task in detecting a very small signal. And pretty much everything which precision metrology can provide is being implemented for that purpose. So you can see, this is like precision measurement. Like maybe the trip to the moon was for aviation in several decades ago. So everything is really-- a lot of things pushing the frontier of precision measurement is motivated by the precision needed for gravitational wave detection. And what I want to show you here is a diagram for what is called advanced LIGO. LIGO is currently operating, but there is an upgrade to LIGO called advanced LIGO. And what you recognize here is we have a laser which goes into a Michelson Interferometer. And this is how you want to detect gravitational waves. But now you realize that the addition here is a squeezed source. And what you are squeezing is not, while it should be clear to you now, we're not squeezing the laser beam. This would be much, much harder, because many, many photons are involved. But it is sufficient to squeeze the vacuum and couple in squeezed vacuum into your cavitational wave detector. If you wonder, it's a little bit more complicated because people want to recycle light and have put in other bells and whistles. But in essence, a squeezed vacuum source is a major addition to advanced LIGO. Yes. AUDIENCE: Where is the squeeze actually coming into the system? I see where it's drawn, but where is it actually entering the interferometer? At that first beam splitter? PROFESSOR: OK. We have to now-- there are more things added here. Ideally, you would think you have a beam splitter here, the laser comes in here, and you simply want to enter the squeezed vacuum here. And this is how we have explained it. We have one beam splitter in our Interferometer. There is an input port and an open port. But what is important here is also that the measurement-- here you have a detector for reading out the Interferometer. And what is important is that the phase is balanced close to the point where no light is coming out. So you're measuring the zero crossing of a fringe. But that would mean most of the light would then exit the Interferometer at the other port. But high power lasers are very important for keeping the classical short noise down. So you want to work with the highest power possible. And therefore, you can't allow the light to exit. You want to recycle it. You want to use enhancement cavities. And what I can tell you is that this set up here integrates, I think, the signal recycling, the measurement at the zero fringe. And you see that kind of those different parts are copied in a way which I didn't prepare to explain it to you. All I wanted you to do is pretty much recognize that a squeeze light generator is important. And this enters the interferometer as a squeezed vacuum. What I find very interesting, and this is what I want to discuss now is, that when you have an interferometer like LIGO, cavitational wave interferometer, and now you want to squeeze. Does it really help to squeeze? Does it always help to squeeze? Or what is the situation? And this is what I want to discuss with you. So let's forget about signal recycling and enhancement cavities and things like this. Let's just discuss the basic cavitation wave detector, where we have an input, we have the two arms of the Michelson interferometer. And to have more sensitivity, the light bounces back and forth in an enhancement cavity. You can say if the light bounces back and forth 100 times, it is as if you had an arm length which is 100 times larger. And now we put in squeezed vacuum at the open part of the interferometer. And here we have our photo-diodes to perform the measurement. So the goal is to measure a small length scale. If a cavitation wave comes by, cavitational waves have quadrupolar character. So the metric will be such that there's a quadrupolar perturbation in the metric of space. And that means that, in essence, one of the mirrors is slightly moving out. The other mirror is slightly moving in. So therefore, the interferometer needs a very, very high sensitivity to displacement of one of the mirrors by an amount delta z. And if you normalize delta z to the arm length, or the arm length times the number of bounces, the task is to measure sensitivity in a length displacement of 10 to the minus 21. That's one of the smallest numbers which have ever been measured. And therefore, it is clear that this interferometer should operate as close as possible to the quantum limit of measurement. So what you want to measure here is, with the highest accuracy possible, the displacement of an object delta z. However, your object fulfills and uncertainty relationship, that if you want to measure the position very accurately, you also have to consider that it has a momentum uncertainty. And this fulfills Heisenberg's uncertainty relation. You will say, well, why should I care about the momentum uncertainty if all I want to measure is the position. Well, you should care because momentum uncertainty after a time tau turns into position uncertainty, because position uncertainty is uncertainty in velocity. And if I multiply it by the time tau it takes you to perform the measurement, you have now an uncertainty in position, which comes from the original uncertainty in momentum. So if I use the expression for Heisenberg's Uncertainty Relation, I find this. And now, what we have to minimize to get the highest precision is the total uncertainty in position, which is the original uncertainty, plus the uncertainty due to the motion of the mirror during the measurement process. So what we have here is we have delta z. We have a contribution which scores as 1 over delta z. And by just finding out what is the optimum choice of delta z, you find the result above. Or if you want to say you want that this delta z tau is comparable to delta z, just set this equal to delta z, solve for delta z, and you find the quantum limit for the interferometer, which is given up there. So this has nothing to do with squeezing. And you cannot improve on this quantum limit by squeezing. This is what you got. It only depends on the duration of the measurement. And it depends on the mass of the mirror. Now-- just get my notes ready-- there is a very influential and seminal paper by Caves-- the reference is given here-- which was really laying out the concepts and the theory for quantum limited measurements with such an interferometer, and the use of squeezed light. Let me just summarize the most important findings. So this paper explains that you have two contributions to the noise, which depend on the laser power you use for your measurement. The first one is the photon counting noise. If you use more and more laser power, you have a better and better signal, and your short noise is reduced. So therefore, you have a better read out of the interferometer. And this is given here. Alpha is the eigenvalue of the coherent state. But there is a second aspect which you may not have thought about it, and this is the following. If you split a laser beam into two parts, you have fluctuations. The number of photons left and right are not identical. You have a coherent bean and you split it into two coherent beams, and then you have Poissonian fluctuations on either side. But if you have now Poissonian fluctuations in the photon number, if those photons are reflected off a mirror, they transfer photon recoil to the mirror. And the mirror is pushed by the radiation pressure. And it's pushed, and it has-- there is a differential motion of the two mirrors relative to each other due to the fluctuations in the photon number in the two arms of the interferometer. So therefore, what happens is you have a delta z deviation or variance in the measurement of the mirror, which comes from radiation pressure. It's a differential radiation pressure between the two arms. And what Caves showed in this paper is that the two effects which contribute to the precision of the measurement come from two different quadrature component. For the photon counting, we always want to squeeze the light in such a way that we have the narrow part of the ellipse in the quadrature component of our coherent beam. We've discussed it several times. So therefore, you want to squeeze it by e to the minus r. However, when what has a good effect for the photon counting has a bad effect for the fluctuations due to radiation pressure. So therefore, what happens is-- let's forget squeezing for a moment. If you have two contributions, one goes to the noise, one goes with alpha squared of the number of photons. One goes inverse with the square root of the number of photons. You will find out that even in the interferometer without squeezing, there is an optimum laser power, which you want to use. Because if you use two lower power, you lose in photon counting. If you use two higher power, you lose in the fluctuations of the radiation pressure. So even without squeezing, there is an optimum laser power. And for typical parameters, so there is an optimum power. And what we're shown in this paper is whenever you choose the optimum power which keeps a balance between photon counting and radiation pressure, then you reach the standard quantum limit of your interferometer. But it turns out that for typical parameters, this optimum is 8,000 watt. So that's why people at LIGO work harder and harder to develop more and more powerful lasers, because more laser power brings them closer and closer to the optimum power. But once they had the optimum power, additional squeezing will not help them because they are already at the fundamental quantum limit. So the one thing which squeezing does for you, it changes the optimum power in your input beam by a squeezing factor. So therefore, if you have lasers which have maybe 100 watt and not 8,000 watt, then squeezing helps you to reach the fundamental quantum limit of your interferometer. So that's pretty much all I wanted to say about precision measurements. I hope the last example-- it's too complex to go through the whole analysis-- but it gives you at least a feel that you have to keep your eye on both quadrature components. You can squeeze, you can get an improvement, in one physical effect, but you have to be careful to consider what happens in the other quadrature component. And in the end, you have to keep the two of them balanced. Any questions? Oops. OK. Well, we can get started with a very short chapter, which is about g2. The g2 measurement for light and atoms. I don't think you will find the discussion I want to present to you in any textbook. It is about whether g2 is 1 or 2, whether we have fluctuations or not. And the discussion will be whether g2 of 2 and g2 of 1 are quantum effect or classical effects. So I want to give you here in this discussion four different derivations of whether g2 is 1 or g2 is 2. And they look very, very different. Some are based on classical physics. Some are based on the concept of interference. And some are based on the quantum indistinguishability of particles. And once you see you all those four different explanations, I think you'll see the whole picture. And I hope you understand something. So again, it's a long story about factors of 2. But there are some factors of 2 which are purely calculational, and there are other factors of 2 which involve a hell of a lot of physics. I mean, there are people who say the g2 factor of 2 is really the difference between ordinary light and laser light. For light from a light bulb, g2 is 2. For light from a laser, g2 is 1. And this is the only fundamental difference between laser light and ordinary light. So this factor of 2 is important. And I want to therefore have this additional discussion of the g2 function. So let me remind you that g2 of 0 is the normalized probability to detect two photons or two particles simultaneously. And so far, we have discussed it for light. And the result we obtained by using our quantum formulation of light with creation annihilation operator, we found that g2 is 0. In the situation that we had black body radiation, which we can call thermal light, it's sometimes goes by the name chaotic light. Sometimes it's called classical light, but that may be a misnomer, because I regard the laser beam as a very classical form of light. This is sometimes called bunching because 2 is larger than 1. So pairs of photons appear bunched up. You have a higher probability than you would naively expect of detecting two photons simultaneously. And then we had the situation of laser light and coherent light where the g2 function was 1. And I want to shed some light on those two cases. We have discussed the extreme case of a single photon where the probability of detecting two photons is 0 for trivial reasons. So you have a g2 function of 0. But this is not what I want to discuss here. I want to shed some light on when do we get a g2 function of 1. When do we get a g2 function of 2. And one question we want to address, when we have a g2 function of 2, is this a classical or quantum effect? Do you have an opinion? Who thinks-- question. AUDIENCE: When we did the homework problem where we had the linear's position, like alpha minus alpha [INAUDIBLE] plus alpha, we found that one can have a g2 greater than 1, and one can have a g2 less than 1. So maybe g2 isn't a great discriminator, whether it's very quantum or very classic. PROFESSOR: Let's hold this thought, yeah. You may be right. Let's come back to that. I think that's one opinion. The g2 function may not be a discriminator, because we can have g2 of 1 and g2 of 2 purely classical. But why classical light behaves classically, maybe that's what we can understand there. And maybe what I want to tell you is that a lot of classical properties of classical light can be traced down to the indistinguishability of bosons, which are photons. So in other words, we shouldn't be surprised that something which seems purely classical is deeply rooted in quantum physics. But I'm ahead of my agenda. So let me start now. I want to offer you four different views. And the first one is that we have random intensity fluctuations. Think of a classical light source. And we assume that if things are really random, they are described by a Gaussian distribution. And if you switch on a light bulb, what you get if you measure the intensity when you measure what is the probability that the momentary intensity is I, well, you have to normalize it to the average intensity. But what you get is pretty much an exponential distribution. And this exponential distribution has a maximum at I equals 0. So the most probable intensity of all intensities when you switch on a light bulb is that you have 0 intensity at a given moment. But the average intensity is I average. So you can easily, for such a distribution, for such an exponential distribution, figure out what is the average of I to the power n. It is related to I average to the power n, but it has an n factorial. And what is important is the case for our discussion of n equals 2, where the square of the intensity averaged is two times the average intensity squared. And classically, g2 is the probability of detecting two photons simultaneously, which is proportional to I square. We have to normalize it. And we normalize it by I average square. And this gives 2. So simply light with Gaussian fluctuations would give rise to a g2 function of 2. Since it's random fluctuations, it's also called chaotic light. And the physical pictures is the following. If you detect a photon, the light is fluctuating. But whenever you detect the first photon, it is more probable that you detect take the first photon when the intensity happened to be high. And then since the intensity is high, the probability for the next photon is higher than the average probability. So therefore, you get necessarily a g2 function of 2. So this is the physics of it. So let me just write that down. The first photon is more likely to be detected when the intensity fluctuation-- when intensity fluctuations give high intensity. And then we get this result. Yes. Let us discuss a second classical view, which I can call wave interference. This is really important. A lot of people get confused about it. If you have light in only one mode, this would be the laser of, for atoms, the Bose-Einstein condensate. One mode means a single wave. So if we have plain waves, we can describe all the photons or all the atoms by this wave function. So what is the g2 function for an object like this? AUDIENCE: 1. PROFESSOR: Trivially 1, because if something is a clean wave, a single wave, all correlation functions factorize. You have the situation that I to the power n average is I average to the power n. And that means that gn is 1 for all n. OK, but let's now assume that we have two. We can also use more, but I want to restrict to two. That we have only two modes. And two modes can interfere. So let me apply to those two modes a simple model. And whether it's simple or not is relative. So it goes like as follows. If you have two modes, both of Unity intensity, then the average intensity is 2. But if you have interference, then the normalized intensity will vary between 0 and 4. Constructive interference means you get twice as much as average. Destructive means you get nothing. So therefore, the [? nt ?] squared will vary between 0 and 16. So if I just use the two extremes, it works out well. You have an I square, which is 8, the average of constructive interference and destructive interference. And this is two times the average intensity, which is 2 squared. So therefore, if you simply allow fluctuations due to the interference of two modes, we find that the g2 function is 2. So this demonstrates that g2 of 2 has its deep origin in wave interference. And indeed, if you take a light bulb which emits photons, you have many, many atoms in your tungsten filament which can emit, re-emit waves. And since they have different positions, the waves arrive at your detector with random phases. And if you really write that down in a model-- this is nicely done in the book by Loudon-- you realize that random interference between waves results in an exponential distribution of intensity. So most people wouldn't make the connection. But there is a deep and fundamental relationship between random interference and the most random distribution, the exponential distribution, intensity which characterises thermal light or chaotic light. So let me just write that down. So the Gaussian intensity distribution-- actually it's an exponential intensity distribution. But if you write the intensity distribution as a distribution in the electric field, intensity becomes e square. Then it becomes a Gaussian [INAUDIBLE]. So Gaussian or exponential intensity distribution in view number one is the result, is indeed the result, of interference. Any questions? So these are the two classical views. I think we should stop now. But on Wednesday, I will present you alternative derivations, which are completely focused on quantum operators and quantum counting statistics. Just a reminder, we have class today and Friday. And we have class this week on Friday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

13_Derivation_of_optical_Bloch_equations.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Good afternoon. The title here is the topic for our class today. We want to discuss formally the derivation Optical Bloch Equations, but usually when I teach you something, I have a general concept in mind and the concept right now here is, how can we get from unitary time evolution of a quantum system to rate equations anticipation. And this is sort of the subject of master equation open system dynamics, and this cartoon sort of tells you what you want to do. We have a total system, which is one part we are interested in, and the other one, often it has many, many degrees of freedom, and we don't want to keep track of them. But they have one Hamiltonian. But we are only interested in how our atomic system evolves, and we want to find an equation, which is no longer a Schrodinger equation. How does an initial density matrix describing our system develop with time? And as we will see, in general, it follows a master equation, and we want to discuss what are general principles of such equations. But it's very important here is that we are not keeping track of what happens in the environment, and that's associated here with a little bucket or trash can. Every result photons or such are measured, the environment is constantly projected on a measurement basis, and therefore, re-introduce probabilistic element into the part we don't observe mainly the part which characterizes the atoms. So for that we need the formalism of density matrix, and that's where we want to start. At the end of last lecture, I reminded you that the density matrix can always be written as an ensemble where you say you have a certain probability for certain rate function. This is always possible, but it is not unique. So each of you could actually create the same density matrix by preparing a number of quantum states with a certain probability and saying this is my density matrix. And also each of you has prepared different quantum states. If you sum them up in that same way, you get the same density matrix and therefore, all observables or measurements you will do on your ensemble in the future will be identical because the density matrix is a full description of the system. So at that level, it may look sort of very trivial that we have different unraveling. Unraveling means you look microscopically what is behind the density matrix, but in your homework assignment, you will all show that you can have a system which have very, very different dissipation mechanisms, but they're described by the same equation. So therefore, it's physically not possible to distinguish by just measuring the density matrix what causes a dissipation. Of course, if you know what causes a dissipation, fluctuating fields or collisions, you know more than the density matrix knows. Any questions about density matrix or the agenda we want to go through today? Just a quick reminder and since this was covered in 8421, and most of you know about it, I pre-wrote the slides. The density matrix is a time evolution. There is one which is sort trivial and covered in more elementary takes, and this is the Hamiltonian evolution, and I'm sure you've seen it. The time evolution, the unitary time evolution involves the commutative of the Hamiltonian with a density matrix, and later on, we want to specialize including dissipation, including the environment to the evolution of a tool evolutionary system driven by a monochromatic field And this the famous Jaynes-Cummings Hamiltonian, and we can characterize this system by a density matrix, but will be very important is that we distinguish between populations, the diagonal parts, and the coherences. And if you simply put this density matrix into this equation, you find the time evolution of the system, and we will refer to that result later on. Many of you have seen it in 8421. we can parametrize the density matrix for the two-level system. So we can parametrize the densities of the two-level system by a local vector, r, which is defined by this equation. And then the equation of motion is simply the rotation of a Bloch vector on the Bloch sphere, and it has a rotational axis, which is given by-- the undriven system rotates around the z-axis. This is just e to the i omega naught t, the normal evolution of the free system. But if you divide it with a monochromatic field rotation over the x-axis and the x-axis, of course, can take your two-level system, and flip it from the ground to the excited state. So it's just a reminder of the simple unitary time evolution, but now we want to add dissipation on top of it. And what I've decided that before I discuss with you Optical Bloch Equation and master equation in general, I want to give you a very simple model. I really like simple models which capture the essence of what you are going to discuss. So what I want to use is I want to use a beam splitter model I formulated for photons, but it would also immediately apply to atoms. And this model, what I like about it, it has all the ingredients of integration of the master equation we'll do later on without the kind of many indices and summations and integration, but it captures every single bit of what is important. And I usually like to present exactly solvable, simple models where you get it, and then I can go a little bit faster for the general derivation because you know exactly what the more complicated equations, what they are doing. So in other words, what I want to derive for you is we want to have the following situation. We have a beam splitter, and we know everything about beam splitters because we talked about them in the first part of the course, and we have a wave function, which is the input, and this is a photon. And we want to understand after the beam splitter, how has the system evolved. In general, it will be a density matrix, and what you want to find out is, what is the equation for the density matrix. Maybe this density matrix goes through the next beam splitter and then we want to know what comes out of it. And all we have to apply is the formalism we developed for the beam splitter earlier in the course. Of course, the beam splitter is not as harmless as it looks like. There is another part and another part one here brings in the environment, and for the environment, we will use the vacuum. That's the simplest environment. It's actually important environment because it is the environment we will use all the time when we discuss spontaneous emission. We send photons into the nirvana, into the vacuum and they disappear, and this is our modified. But the other one which is often not so explicit. If you send your photons away, you're not keeping track of what happens, but you could as well perform a measurement, and this is what we put in here. We say those photons hit a bucket or detector and measure them, we observe them. There's nothing else we do with them, so we can as well measure them, and this being immediately lead to the equation for the density matrix. So this is what you want to discuss, and it will have all the ingredients later on in a mathematically simple form for the derivation of master equation. So let's consider a similar photon for that. So the wave function, this is superposition of no photon and 1 photon, and the coefficients are alpha and beta. And just for simplifying notation, I pick alpha and beta to be real. So what do we expect to happen at the beam splitter? Well, there's a probability that the photon gets reflected. Probability to reflect the photon and therefore to observe the photon. This probability is, of course, beta square-- the probability that we have a photon to begin with-- and then the beam splitter, remember we categorized the beam splitter with angle sine theta cosine theta. Sine theta was the reflection amplitude, cosine theta, the transmission amplitude. So this probability, which I call P1 is the probability for reflection and for measurement. And now naively you would think what happens after the system has passed through the beam splitter, with a probability of P1, we've measured the photon. We know for sure there is no photon left. The system is in the vacuum state, but then you would say, well, maybe with probability 1 minus P1, we have not measured anything. Nothing has happened to the wave function, and that means the wave function just continuous. Well as we will see, this is wrong. We are missing something. What we are actually missing is that if you measure nothing, the wave function is not sine. The possibility that we could have measured something, changes the wave function. I will comment on that in much more detail in a few lectures down the road when I derive for you quantum Monte Carlo wave function. I will have a wonderful discussions with you about how does non-observation change a wave function. So we will talk about the physics behind it in some more detail. Right now, I don't want to get into this discussion. I simply want to use our beam splitter equation, so we can just take the beam splitter equation and apply it. So our output state is obtained by taking the operator for our beam splitter, and maybe you remember that the propagation for beam splitter was discovered by an operator, which had a dagger b dagger a in the exponent. a and b are the two input nodes. And the angle of the beam splitter, which interpolates between 0% and 100% reflection transmission is theta. And we're now looking for the output state of the total system. We're not performing the measurement yet, and this is now acting on the total system, which is the cross product of our photon system, of our system of interest. And the other input, which we call the environment or the vacuum, is 0. Well, look a few weeks back, we have done that all. The output state is, well, there was a probability, alpha, that we had no photon in the state psi. And if we have no photon in the state psi and no photon in the vacuum, this is the state 0, 0. What I denote here with this second place is the environment. And now we have one photon. We have exactly one photon with the amplitude beta, and this photon is split with cosine theta transmitted and with phi theta reflected. If you transmit it, we have 1, 0. If we reflect it, we have 0, 1. And again, this is the environment, and here is a photon in the environment. So let me just be clear that this is where the environment comes in. It is a vacuum state, and here, this is the output part for the environment. This is where we do the measurement. And I don't think it matters. I haven't really told you which is mode A, which is mode B. It doesn't matter, but one, let's say the environment is mode B, and the system evolves in mode A. As you can see, I'm using a new program, which has some nicer features in terms of handwriting, but it is a little bit rough in scrolling, so I sometimes have to scroll back and forth. So what is our output? Now, we have two possibilities. The environment is 0, or the environment is 1, and we perform a measurement. So we have to now go into a probabilistic description. So with probability P1, we have done a measurement, and our output state is now the vacuum state. With probability 1 minus P1, we have not detected anything in the vacuum, and therefore, our state of the system is alpha 0 plus beta cosine theta 1. Is alpha 0, so it is not beta 1 as naively would have assumed. It's not the original state. There is a cosine theta factor, which we got exactly from the beam splitter from the unitary evolution provided by the beam splitter. And since these state is no longer normalized, I have to normalize it by alpha squared plus beta square cosine square theta. So now we have done our measurement probability P1 to detect the photon. This projects the system into the vacuum state with the probability 1 minus P1. We have that state. Just one second. Scroll in the pictures. Write that down. In fact, millions that our system is now described by a density matrix with probability P1 and 1 minus 1 minus P1. With probability P1, we are in the vacuum state, and with probability 1 minus P1, we are in that state, the denormalized state psi naught, which I just hold down. Question? AUDIENCE: Have you considered theta to be some like a dynamical phase evolution system. It's very low order like when you expand it the first time it looks almost identical to [INAUDIBLE] quantum effect maybe. The environment is measuring the state in some way, and I mean, it's the lowest order now. PROFESSOR: Yeah, I Quite agree that random 0 is just an example of it. Pretty much, it's all the same. Yeah. What we do here is, I like the beam splitter because the beam splitter provides an exact formulation of the measurement process. You really can use a beam splitter to discuss what happens fundamentally when you perform a measurement. And the beams splitter is one typical implementation of that, but it has all the features you'll find in any measurement system. And especially what you observe here, let me just emphasizes is, the fact that we you do not make a measurement is changing the way function from the initial wave function psi to psi naught, we have a factor of cosine theta here. And that's also very general. A measurement perturbs, modifies your wave function no matter what the outcome of the measurement is. So let me write it down because we want to take it to the next level. So we have now found in terms of the beam splitter, angle theta, and the parameters of the initial state, alpha beta. We found the density matrix after the beam splitter. Yes. So what is the next step? Our goal is to derive the master equation for the density matrix, the time evolution of the density matrix. So since we want to discuss the time evolution, we want to find a differential equation. So what we want to figure out is, what is the difference between the output density matrix and the input density matrix. The input was, of course, pure state characterized by the matrix population alpha squared and beta squared of diagonal matrix element of alpha beta. The difference between the density matrix is can just calculate the difference. You can simplify things by applying some trigonometric identities, so this is an exact result, cosine theta minus 1. Here we have alpha beta cosine theta minus 1, and on the diagonal, we have cosine 2 theta minus 1 divided by 2. Anyway this is an intermediate result. We're interested in the differential equation. We want to sort of find out what happens when we observe, when we have the density matrix interacting with the environment all the time. And this can be simulated by beam splitters by using many beam splitters with a small degree of reflection. So we want to simplify this result now for the case of many beam splitters, and each of them has a small tipping angle, theta, and for later convenience, I defined theta to be gamma times delta t over 2. That's just my definition of theta. So what we have in mind now is that we start with the system psi, and we have many such beam splitters with an infinitesimal tipping angle. Each beam splitter has the vacuum at its input state. And we always perform the measurement. If I take the equation above, which I know you can't see anymore, we find a differential equation for the density matrix, which is we find an infinitesimal change delta over the density matrix, which looks like this. So all I've done is, I've used the equation above, and I've done a Taylor expansion in the small angle theta. And the reason why I brought in the square root, well, we get cosine theta. The first order Taylor expansion or the lowest order Taylor expansion from cosine is 1 minus theta squared. So I get the square root squared. So I get gamma, which appears here, and then I divide by delta t. So this is just an exact mathematical expression, and the next step is to form a differential equation. But before I do that, I want to emphasize the two features we are using here. They sort of enter automatically, but these are the two big assumptions we make when we derive a master equation. The first one is that we always have a vacuum state as the input. So in other words, the environment is always in the same state, which is a vacuum state, and this is sort of called a Bohr approximation. What it means is that we do a measurement here, but the vacuum is not changing. In other words, we are not overloading the vacuum with so many photons that suddenly the vacuum Is no longer in the vacuum state. Or in the case of spontaneous emission, the vacuum can just take as many photons as you dumping into it. They disappear so quickly that for all practical purposes, the environment stays in the vacuum state. So this is called the Bohr approximation. The environment is not changing. It has enough capacity you to be modified by the measurement process. And the second thing which is related is, the vacuum is always in the same state, and there are no correlations from here to here to here, there is no memory effect. Everything is completely uncorrelated. So the environment is uncorrelated. It has no memory. It's correlation function is a delta function, and this is called Markov approximation. So these are the two effects which are important. One is no memory for the environment, Delta function correlation, Markov approximation, and the two, of course, are related in the environment is all of this in the same state. So remember this is a change for the density matrix, and alpha beta where the original parameters of the density matrix for the input state. So I can now rewrite everything as a differential equation. The density matrix has a derivative for the diagonal matrix elements and for the coherences. Here we have plus gamma. Here we have minus gamma 0, 1, 1, makes sense because we conserve the trace. We have unity probability that we have a stellar system. And therefore the two diagonal matrix elements, the population, the sum of them cannot change with time. And for the coherence, we have gamma over 2. And if you're familiar with Optical Bloch Equations, which we derive next, we can say that these means if 0 is the ground state that the ground state changes because-- call it spontaneous emission from the excited state. This equation would say that the excited state decays with the rate gamma, and sometimes you may have wondered about that there are factors of two appearing, which also appears here that when the excited state decays with a rate gamma, we have a factor here for the coherences, which is gamma over 2. So what we have accomplished in contrast to let's say Einstein's equation with the Einstein a and b coefficient, which lead to rate equations for the population, we have now a new feature. We have an the equation for the coherences, and we find a decay of the coherences with half the rate as a decay of the population. Questions? So if you want you could rewrite this model for photon, which goes through beam splitters, undergoes measurement. You can rewrite it from atomic wave function and you measure whether the atomics in the excited and ground state and the equation for the measurement performed on the atom is exactly as the equation by which the beam splitter acts on the photon state. So what I've shown here it's very specific for a single photon because I could use simple equations, but everything is what you find in a much more general situation. So before I give you the general derivation of the master equation, let me talk about what we have learned from this example and what the general procedure is. The first thing is our goal is to find a differential equation for the density matrix of the system. I just remember. There was one thing I wanted to mention. In the previous derivation with the beam splitter, I started with a purer state, and the purer state developed into a statistical mixture, and this statistical mixture would then transform the next beam splitter into another statistical mixture. I derived the differential equation for you for the first state from the pure state to the statistical mixture. But if you would spend a few minutes, you could immediately show that you can start with an elementary density matrix, look how it evolves through the beam splitter, and you get exactly the same differential equation. So the general procedure is, we want a differential equation how the density matrix evolves with time. And this will be obtained by finding an operator which acts on the initial density matrix. This operator is not a unitary operator because we are performing measurements through the environment, even if you don't actually perform them. Once we dump something into the environment, it's out of our control and anybody could go and perform a measurement, and so we should assume that this measurement has been taken. It's one of those quantum mechanical things that you don't even have to care whether somebody does it. The environment does it for you. So this operator is called a Liouvillian operator. It's sometimes called-- and I haven't really traced down why-- it's called sometimes super operator. I know what superconductivity is, but I don't know what the super powers of this operator are, but that's just a name which you will find. The second thing which we have used is that the evolution of this system can be obtained from the time evolution of the total system by performing a trace over the degrees of freedom of the environment. This was exactly what we actually did when we said the system continues with probability P naught in one state and probability P1 in the other state. The operation which lead to this density matrix was exactly the partial trace. So this is the second general feature which we have to implement. Thirdly, if we could do one and two exactly, we would have an exact formulation for a small part of a quantum system no matter how complicated the environment is. In practice, we can solve the equations only when we make simplifying assumptions about the environment. One is, it is large, and more important, therefore, it's unchanging, and this is the Bohr approximation. And the second feature is, it has a short correlation time, tau c. In the beam splitter, I have made the assumption that there is no correlation between different beam splitters in the derivation, which I want to walk you through. Right now, you will see explicitly where the correlation times enter. And this is called the Markov approximation. And finally, this is number four, the whole possibility to derive a master equation hinges on the fact that we have different time scales which are very different. We are interested in the evolution of our system. We want to know how it relaxes, and this is on a time scale 1 over gamma. So we call this slow. We are interested in the variation of our system, the atomic system or the photon state, which passes through the beam splitter, and this time scale has to be much slower than the fluctuations of the environment. So therefore, if the environment has fluctuations, which in the beam splitter model where assumed to be 0, it was delta function of time, if that correlation time is much smaller than the time it takes for the system to relax and to evolve, that opens a window delta t, and this is the time scale of the master equation. Just to give you one example for the spontaneous emission, the correlation time, tau c, would be the time it takes the photon to disappear from the atom. And the photon has disappeared from the atom when it is one wavelengths away. So typically, the correlation time f the vacuum for spontaneous emission is one cycle of the optical frequency. It's very, very fast. Whereas typical decay times of the excited states, a nano second. It's six orders of magnitude slower, and this is what we describe. But on a time scale of a femtosecond of one optical cycle, the photon has not detached from the atom and it could go actually back to the atom. During that time, we talked a little bit about it when we did this diagrammatic discussions of resonance scattering for very, very early times. You don't have exponential decay because you cannot do the approximations where we approximated the kernel by something which was completely energy dependent. And so what happens at such short times, we encounter here again in such short times, we will not have a simple description of the system. So the last point, let me summarize. Our goal is that we describe the density matrix of the system, and we want to find the Liouville operator or some matrix which acts on it. And because we will integrate over time steps, which are larger than the correlation time of the system, we can also call it, it will be a coarse-grained evolution. Any questions about that? I really like the discussion, the derivation of the master equation, how it is presented in atom-photon indirection. But it is presented on more than 50 pages with many, many equations. So after giving you all of the principles, all of the concepts, I want to go with you now over those equations and point out how the principles, which we encountered with the beam splitter, how they are now implemented in a very general context. I will not be able to give you all of the mathematical aspects of it, but I think by now you know that the book atom-photon interactions are actually wonderful. You can get a lot of conception information out of it by looking at the equations without understanding every technical detail. So I would really encourage you, if there is something which piques your interest and I hope there will be things which you'll find very interesting, that you go to the book and read it. So I'm exactly following that actually I used copies of the book. So we have a Hamiltonian, which is describing the atomic system. It describes the reservoir and then there is an interaction. We keep it very general here, but you may always think well, the atom is your favorite two-level system. The environment is maybe the vacuum with all its possible modes, and the interaction is the dipole interaction or the a dot p interaction. So we start out with an equation, which is nothing else than Schrodinger's equation for the density matrix. The time derivation of the density matrix is commutative with the Hamilton. But it is often useful and you've seen it many times, to go to the interaction representation, that the time dependence due to the unperturbed part of the operator is absorbed in a unitary transformation, so therefore, this density matrix in the interaction representation evolves not with, h, because h naught is taken care of, it only involves due to the coupling between the two systems, between the system and the environment. So now this equation, we are interested in a time step delta t. And this time step, delta t remember, we want to coarse grain, will be larger than the correlation time of the reservoir, and you will see exactly where it comes about. So we want to now do one of those coarse-grain steps. We take this equation and we integrate from time t to time t plus delta t. So this is exact here. But now we want to iterate, and that means the following. We have expressed the time step in the density matrix by having the density matrix there. But now we can do in a first-order perturbation theory, we can do one step and we get the second order result by plugging the first order result into this equation. It's the same we have seen with our diagrams and such. We have an exact equation. It's useless unless we do something, and what we do is, we realize that we can iterate it because the part we don't know involves one more occurrence of the interaction potential. And when you plug the nth order solution in here, you get the n plus first order solution. And this is exactly what is done here. And I skipped a few equations here. This is what he's done here, number one. And number two is, we are interested in the system, not in the reservoir, so therefore we perform the trace over the reservoir. And the trace of the reservoir for the photon beam splitter mend, we say we have two possible states, we detect a photon or not. And for the system, we have now a density matrix which is probability P naught in one state, probability P1 in the other state. And this is exactly what the operator partial trace does. Remember also I want to really make sure that you recognize all the structures. The time evolution of a density matrix was a commutator with h, but in the interaction picture, it's a commutator with v. But since we are putting the first order result in here, the second order result is now the commutator of v with the commutator of v and o. It's just we have iterated one more time. So the sigma tilde, the density matrix for our system, tilde means in the interaction picture is now the partial trace over the reservoir of the total density matrix. Tilde means in the interaction picture. And the important part here is that it is exact. We have not done any approximation here. Any questions? Of course, now we have to make approximations because we cannot solve an interactive problem exactly. The first one is-- and what do we want to do in the end? We want to keep the first non-trivial term but to the extent possible, we want to factorize everything. We want to get rid of the entanglement of the environment in the system and only get sort of the minimum which is provided by the coupling. So this evolves as follows. The interaction we assume is a product of two operators. One operator acts on the system, one operator acts on the environment. So this could be the dipole acting on the atom, the vacuum field, e, interacting with the environment or it could be p dot a. Or maybe your system has a magnetic moment, m, and the environment consists of fluctuating magnetic fields. So we'll pretty much find in every kind of measurement that a measurement involves the product of two operators. One is an operator for your system and one is an operator for the reservoir of the environment. And so this is one thing we want to use, and now there is one thing which the moment we will set in our equations, there is one thing which will naturally appear. Let me scroll back. What we have here is the interaction operator, v, at two different times. So this means if something happens at different times and we integrate over times. This is a correlation function, a correlation function between v at the time t prime and the time t double prime and since the reservoir part of this interaction is the operator, r, so what we have here now is, we have a correlation between the operator, r, at two times, which characterizes the environment. And now comes an important approximation. You remember I said we want to assume that the environment has a very short correlation time. Whenever a photon is emitted, it appears dramatically fast. It disappears in one optical cycle, and the environment is sort of reset, it's back in the vacuum state. So this is now expressed here that this product over which we take the partial trace has a very short coherence time. And the fact is now the following. We are integrating over a coarse-grained step delta t, but this correlation function goes to 0 in a very short time. So therefore it will not contribute a lot. Let me write that down. What we are going to approximate is our total density matrix is now approximately factorizing in a density matrix describing the atomic system. Well, we describe the atomic system when we trace out the environment. We describe the environment when we trace out the atomic part. And if we now form the direct product, we are back to the total system, but we have factorized the total density matrix into two parts. What we neglect here is a part which cannot be factorized which is the correlated part of it. But what happens is, since we are integrating over time steps delta t and the correlation decay in a very, very short time, the result is that this complicated part, which we could never calculate, is smaller than the first part by the ratio of the time where the correlations contribute over the time delta t, the time step we are going to take. So this is a very critical assumption. There is a whole page or two in the book where an photon-atom interaction they discuss the validity of this assumption, but I've given you the physical motivation that we indicate over much larger time, and if this time is large and the correlation is lost for short time, they only contribute with this small parameter to the result. So in other words, this means after-- we have an interaction between the environment and the system. We write it down in second order, but the second order result is now we evaluated by factorizing the density matrix into our system in the reservoir. So that means in that sense if you factorize something, it looks as if it's not interacting, but the trick is the same. You write down something to first and second order, and once you have factored out the important physics, now you can evaluate the expression by using an approximation, which is now the approximation that the density matrix factorizes. So with that, this is the approximation that we have made that the correlation time is very short. And now we have a differential equation for the density matrix sigma, which describes or atomic system. We have traced out the degrees of the reservoir. And now we want to insert B.17. You probably don't remember what B.17 is. It says that the interaction operator is a product of a, the operator a for the atoms and r for the reservoir. So the reservoir part, tau prime tau double prime, gives a correlation function. This is the correlation function between the operator, r, at two different times and the part which acts on our system, the a part, is explicitly kept here. So this is now a general master equation. It tells us the time evolution of the density matrix in this form. It looks very complicated, but this is because it's very general. In order to bring it into an easier form, we want to now introduce a basis of states, energy eigenstates of the unperturbed system, and write down all of these operator into such a basis of states. But anyway you saw here how we had an exact equation, and the main approximation we made is that the operator acting on the environment has a very short correlation time. Any questions? Well, you're only a few minutes away from producing this result to Fermi's golden rule, which you have known for a long, long time. It's just we have made very general assumptions. You see sort of how the assumptions propagate, but now if you write it down for an energy eigenbasis, you will immediately see results you have probably known since your childhood. So we want to have energy eigenstates of the atomic operators, so this is sort of ground and excited state if you think about a two-level system. The previous equation, I have to go back to it. Our previous equation is a differential equation for the density matrix here, and here is the density matrix. So now we formulate this equation into an energy eigenbasis, and what do we get? Well, we get an equation for the matrix elements, and what matrix elements are important? diagonal matrix elements which are population of diagonal matrix elements which are coherences. So we pretty much take this equation, use the energy eigenbasis and look, what do we get for the populations and what do we get for the coherences. So the structure is now the following. That we have our matrix elements ab. There is one part which looks like a unitary time evolution. This is what comes from the Hamilton operator. This is sort of the-- we'll see that in a moment-- but this is the time evolution without relaxation and now we have something here which are generalized relaxation coefficients. And you will find if you go further above that those relaxation coefficients are directly related to the correlation function of the reservoir. So we can now specify what happens between populations. Population means that we have a differential equation, let's say between sigma aa the puller, and sigma cc, so we have a rate coefficient which connects the population in state A with the population in state C. And if you take this expression, you find several things. Well, you find Fermi's golden rule, in a generalized way-- that's always nice-- you find Fermi's golden rule. When you integrate over time, you often get data function, and you expect to get a delta function because of energy conservation. So you get that, of course, naturally. Secondly, we have second order matrix element, which you know from Fermi's golden rule, but now we have the following situation that the matrix element in Fermi's golden rule may actually depend on the state, mu, of the environment. So you have maybe 10 different possibilities for the environment, and Fermi's golden rule gives you spontaneous emission, which is different for those 10 states. And naturally, since we have performed the partial trace over the environment, we have all those rates weighted with the probability that the environment is in one of those states. So what you find here is a simple generalization of Fermi's golden rule. And if you look at the off-diagonal matrix elements, for instance, you want to know what is this rate, what is the rate coefficient, which gives you the time derivative of the coherence, and it's multiplied with the coherence. You'll find now that in general, this rate coefficient has a damping term, but it may also have an imaginary term. And I hope you remember when we played with diagrams, that we had something similar. There was something which we called the radiative shift. I called it the AC stock effect of a single photon, and here it is a level shift which comes because the environment interacts with your system and it shifts the levels a little bit. So in addition to just relaxation, spontaneous emission, and damping, there is also a dispersive part, a level shift, and it has exactly the same structure. Let me add that delta ab is the difference between the shift of state b and the shift of state a. And those shifts have exactly the same structure. You have to take the principle part of something which has 1 over the difference of energies, and we discussed that this has to be understood by somewhere adding an infinitesimal imaginary part and doing the right thing with complex function. It's actually related to Laplace time difference between Laplace transformation and Fourier transformation. So anyway what I find sort of beautiful is that we started with a most general situation. We perform the partial trace. We made one assumption of short correlation times, and a lot of things we have known about quantum system just pops out in a very general form here. Any questions about so far? Well, the coherences are, of course, more interesting than the population. Coherence is always something physicists get excited about it because it captures something which goes often beyond classical system that we have quantum mechanical coherences. And what happens is the coefficient here, which provides the damping of the coherence, just comes out of the formalism, has two parts. And one part is an adiabatic part and the other one is a non-adiabatic part. Well, and that makes sense. If you have two quantum states and there is a coherence, some phase between the two, the phase can get lost if you do a transition between the two states or one state undergoes a collision and is quenched. So you definitely have one part which is due to the fact that the quantum states or the population changes, and you find that there is this state-changing part, which is pretty much the sum of all the rate coefficient leading out of state, leading to the decay of state a and leading to the decay of state b. In other words, if you have a two-level system, which has a coherence and you have decay of the excited state and decay of the ground state, you would expect that those decay terms appear also in the decay of the coherence between the two levels and they do, and they appear with the correct factor of 1/2. But there is another possibility and this is the following. You can have no [INAUDIBLE] of the population of the state, but you can still lose the coherence. The model you should maybe make is that you have spin up, spin down. You are not perturbing the populations in spin up and spin down, but the environment provides fluctuating magnetic field. Then due to the fluctuating magnetic field, you no longer can keep track of the phase, and that means in your identity matrix the off-diagonal matrix elements decay. And we find that here this is the second part which in this book is called the adiabatic part, and the physics behind it is now pure de-phasing. So it's an independent way for coherences to decay independent of the decay of the population. Questions? Collin. AUDIENCE: Where does the Markov approximation come in? PROFESSOR: The Markov approximation is, so to speak, the delta function approximation, which would say that-- I mean, I introduced the correlation function between the reservoir operator and said the correlation time tau c is very, very short. The Markov approximation would actually state that it would actually say in a more radical way the correlation time is 0. And the Bohr approximation, the fact that the reservoir is unchanged came in when we said the total density matrix for the second order expression just factorizes. It factorizes into the environment, which is just this density matrix of the environment. It's not changed by the interaction with the system. And this is the Bohr approximation. We just use the same expression for the reservoir independent of the measurements the reservoir has done. Other questions? Yes, Nancy. AUDIENCE: [INAUDIBLE]. PROFESSOR: This is something very general. Thank you actually for the question. Whenever we have some damping of the population, the coherence is only damped with a factor of 1/2. One way to explain it in a very simple way is, that if you have an amplitude alpha excited and alpha ground, the population in the excited state is this squared. So you sometimes make the model that the amplitude decays with gamma over 2, but the population is-- because you take the product, decays with gamma. So you would say alpha e and alpha g both decay with half the rate, but the probability is for the total rate, and the coherence is the product of the amplitudes. So therefore when you look at the coherence, this decays with 1/2 gamma e. This decays with 1/2 gamma g, and this is what you get here, 1/2 gamma in state a, 1/2 gamma in state b. Whereas the probability to be in this state decays with twice that because the probability is squared. So you find that pretty much in any quantum mechanically corrects derivation, which you do about the decay of population coherences. Other questions? So we have done two things. We have done the very, very simple derivation using the beam splitter model where you may not even notice where I did the Markov approximation because I jumped from beam splitter to beam splitter and left all the correlations behind. Here in the most general calculation, you have seen exactly where it enters, but maybe now you have the full forest in front of you, and you don't recognize the trees anymore. So let me wrap up this lecture by now focusing on the system we want to discuss further on, namely a two-level system interacting with a vacuum through spontaneous emission. But I also want to make some generalizations. I want to give you some generalizations about what kind of environments are possible in quantum physics. Let me just see how I do that. So this part I actually owe to Professor [? Ikschuan ?] who wonderfully compiled that. So what I want to do now is, I want to call your attention to the operator form which, is rather unique. Remember when we did second order perturbation theory, we had sort of the commutator of v with the commutator of v and rho. This came from iterating the exact equation of motion for the density matrix. And you want to specialize that now to Jaynes-Cummings model. I mean, in the end, at least in this course, we always come back to the Jaynes-Cummings model because it captures a lot of what we want to explore. So the Jaynes-Cummings model in the rotating wave approximation is very simple. It raises the atom from ground to excited state and destroys a photon, or it does the opposite. So this is our simple interaction between our system, the two-level atom and our reservoir which is just the vacuum of all the modes. And you, again, recognize what I said in general. You usually always find it by linear form, an operator which acts on the modes, on the reservoir on the vacuum, and an operator which acts on the vacuum. Now, we want to make the explicit assumption that the initial state of the reservoir is the vacuum state. It's empty. And I want to show you what is the structure of the operators we obtain. And so if you put v into here, you have first the commutator with rho, which I write down here, and then we have to take another commutator with v. And the result of that is the following, that when it comes to relaxation pauses, based on the general structure of the time evolution of quantum mechanics, we have this double commutator. And the operator which, couples our system to the environment is erasing and lower an operator. I mean, the atom because it interacts with the environment either absolves the photon or emits the photon. But those operators, sigma plus and sigma minus, appear now always as products because we have two occurrences of the interaction, v. But if you look at the double commutator structure, the operator sigma plus sigma minus appears. This is just the general structure of this double commutator appears to the left side of the atomic density matrix to the right side of the atomic density matrix and then there is the coarse term where the atomic density matrix is in the middle of the two. So this is actually something which is very general and very important in the theory of open quantum system. What I'm discussing with you now is this famous Lindblad form. And the story goes like that. You want to know what are possible environments, not just empty vacuum. You can have fluctuating fields. You can have, you name it. But if you are saying that your environment interacts with your system through an operator, and our operator is now the operator sigma minus, which is a spontaneous emission, you need heavy t. The mathematical structure of a valid quantum field remains that if your system interacts with an environment by emitting a photon sigma minus, this is now the structure of the master equation. This is the structure of the time evolution of the density matrix. So the operator sigma minus and its and its Hermitian conjugate sigma plus, all of this have to appear in this combination. Yes, Collin. AUDIENCE: This is still in my interaction picture. Right? There's no dynamical phase evolution that we put back in there. PROFESSOR: Yeah. OK. What we do in general if the system is driven by a laser beam, for instance, Rabi oscillation, we simply add up the dynamics of the Rabi oscillation of the unitary time evolution to the time evolution done by the reservoir. AUDIENCE: So this form is always in the interaction picture? PROFESSOR: Well, this is, you would say, this form is what is the relaxation provided by the environment. And if you drive the system in addition with the coherent field, unitary time evolution, you would add it to it. So in other words, what I'm telling you here is that this is the general structure, and if you have a system which interacts with an environment in five different ways, with a dipole wand, with a magnetic moment and such, you have maybe five interaction terms and then you have to perform the sum over five operators, and here one of them is a sigma minus operator. So in other words, if you want to know what is the whole world of possibilities for quantum system to relax and dissipate with an environment, you can pretty much take any operator which acts on your system, but then put it into this so-called Lindblad form and you have a possible environment. And I mentioned I think last week that people are now in our field actively working on environmental engineering. They want to expose a system to an artificial environment and hope the system is not relaxing, let's say, to a broken ground state, but to a fancy correlated state. So what this Lindblad form, if the operators appears in this way, what it insures is the following. Just imagine if we have an equation, a derivative of the density matrix, which depends on the density matrix, you could write down a differential equation and say is it possible? Well, it has to be consistent with quantum physics. You have certain requirements. One requirement is that rho, the density matrix, always has to be the density matrix. The trace equal 1 has to be conserved. A density matrix always must have non-negative eigenvalues, otherwise, what you write down, it might be a nice differential equation, but quantum mechanically, it's nonsense. But now there is one more thing, which is also necessary. This time evolution of the system's density matrix must come from a unitary time evolution of a bigger system. So you must be able to extend your system into a bigger system, which is now the environment, and this whole system must follow a Schrodinger equation with a Hamilton as a unitary time evolution. And this is where it's restrictive. You cannot just write down a differential equation and hope that this will fill some requirement, and what people have shown is under very general assumptions, it is the Lindblad form which allows for it. So some operator always has to appear in this form. So often in this Lindblad form, you have an operator, which is called a jump operator, which is responsible for the measurement, which the environment does on your system. The jump operator is here, the operator which takes the atom from the excited to the ground state. With that you need a photon, and the photon can be measured. So often you can describe a system by a jump operator, and if the jump operator is put into this Lindblad form, then you have a valid master equation for your system. So let me wrap up. If you take now the definition of the raising and lowering operator, and you take the form I showed you, the Lindblad form, you'll find now this differential equation for your two-level system. And this is one part of the Optical Bloch Equation. Now, coming to Collin's question, if you include the time evolution of the classical field, this is a coherent evolution of the Bloch vector, which I showed at the beginning of the class, and we add this and what I wrote down at the beginning of the class. Then we find the famous Optical Bloch Equations in the Jaynes-Cummings model. So these are now the Optical Bloch Equations, and I hope you enjoy now after this complicated discussion, how simple they are. And it is this simple set of equations, which will be used in the rest of the course to describe the time evolution of the system. Just because I did some generalizations about the Lindblad equation, I copied that into the lecture notes from Wikipedia, and probably now you sort of understand what is the most general Lindblad equation. It has a Hamiltonian part and then it has jump operators like you sigma minus operator, but it has to come in the form that the jump operator and its complex conjugate, emission. conjugate, is on the left side, on the right side, and left and right of your density matrix. So this is the generalization I've mentioned. Yeah. with that I think with that we've derived the master equation, and on Wednesday, we will look at rather simple solutions, transient and steady-state solution of the Optical Bloch Equation. Any questions? One reminder about the schedule, this week we have a lecture on Friday because I will not be on town next week on Wednesday. And of course, you know today in a week, next week on Monday. It's [INAUDIBLE] day. So we have three classes this week, no class the following week, and then the normal schedule for the rest of the semester.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

15_Unraveling_Open_System_Quantum_Dynamics.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: I apologize for the delay. But we are ready for multimedia show with three different applications. One of them are the clickers. I hope you all took a clicker. It's just two questions today, but really good questions. [LAUGHTER] So today's actually very nice topic. I think almost every lecture we have a nice topic, but today's especially nice. It's called Unraveling Open Quantum System. That may not tell you so much. Well unravelling means we want to sort of untangle it, We want to understand what is inside. And so we want to develop a better understanding what is inside a density matrix. How can we think about the quantum system in an even more microscopic way than the density matrix tell us? And again, this finds applications in atomic physics, because one application of it is spontaneous emission optical block equations. On the other hand, actually, this class will also teach you something very general about quantum physics. So today's lecture is this part. We'll have three topics. One is I want to provide a motivation for single quantum systems. And you should appreciate this. A single quantum system is not what Schrodinger and Heisenberg told us about. Quantum mechanics was formulated as an ensemble average. But today we want to talk about, is there something in addition? About single quantum systems. Part two is-- addresses the derivation of the quantum Monte Carlo wave function method. And this is a formalism, which is really simulating in quite general terms, quantum physics by following-- using quantum Monte Carlo simulations-- trajectories of individual quantum systems. So here the individual quantum system-- even if you have one, even if you don't have one-- becomes a computational tool. And eventually in part two I want to give you another example for unraveling open quantum system, which differ from the models for dephasing. And if you remember that this was on one of your homework assignment that's correct. I sort of want to here take your homework into the classroom discussion. Well I think it's nice to start this lecture with the quote by Erwin Schrodinger. Where he said, "We never experiment with just one electron, or atom, or molecule. In thought experiments, we sometimes assume that we do; this inevitably entails ridiculous consequences." Well today we want to talk about those ridiculous consequences, because many people in our field trap, address, and observe single electrons and single atoms. Now this is related to quantum jumps. Quantum jumps has even made it into the popular realm. You can find lots and lots of references on quantum jumping, but I warn you this is about quantum jumping into different worlds. It's about broadening your horizon. And those authors, those people actually, sell you something by claiming that people like Einstein and Stephen Hawking would support the idea that you can jump into different universe, experience another version of yourself-- probably a better version of yourself-- and this will inspire you to change your life in the world you live. But-- So this is quantum jumping. I'm not talking about quantum jumping, I will only talk about quantum chunks. So what you see here is really a classic experiment. One of three experiments which were done almost simultaneously. There have always been races for, you know, the first observation of something. And there was the [? Hamburg ?] group by [? Toshak ?], the Seattle group by Hunt [INAUDIBLE], and the Boulder Group by Dave Wylen, who observed the phenomenon of quantum jumps for the first time. Let me just tell you what it is, because it's exactly what you're going to describe. Here in this situation you have a system. It's a trapped barium ion with a p and s state. And there is very fast transition where you can rapidly scatter light, and get lots of fluorescence, at 493 nanometer. But what happened is, every once in a while, the particle has a branching ratio to decay to d state, and, well, d to s 2 units of angular momentum. The transition here down to the ground state is very slow. It's slow because it's forbidden. It's even slower because it happens at longer wavelengths of 1,600 nanometer. And this is slow. And it has a lifetime of about 50 second. So you-- If you think about it naively and that's probably the correct approach-- you would say the particles in the s state, it's like a light bulb. It scatters many, many photons, but then it branches to the d state. And for 50 seconds the light bulb is switched off, until the particle spontaneously goes down to the s state, and then the light bulb is switched on. And this is exactly what people observed. But I mean I was-- This was '86. That was the time I was a graduate student, and I talked with theorists in the field, and respectable theorists told me that there was a controversy-- probably not among experimentalists, who use the naive approach-- but theorists who believed too much in what they calculate. Because if you solve it quantum mechanically, and look what you get. It's a steady state solution. Atom in steady state. Laser in steady state. And if you calculate the quantum mechanical expectation value, the expectation value is constant as a function of time. So a simple quantum calculation would say the fluorescence is on average. Well just the time average and his constant, and his time independent. So there were some serious doubts whether you would observe this, or something which looks more like the ensemble average. Well made all of the theorists soon realize you cannot calculate the steady state average. You have to calculate the correlation function in steady state. And the steady state correlation function has structure exactly at 50 seconds. So in a steady state system, if you're interested in fluctuations and temporal behavior, you can't just calculate quantities, you have to calculate correlation functions of quantities. So anyway-- But then of course, if you now average over many, many realizations, you find exactly an exponential decay of 50 seconds. And this is sort of what you would get for an ensemble average. So this is what quantum jumps are. This is what you should always imagine when we talk about quantum jumps in the rest of this lecture. But I also want to show you more recent example from Serge Haroche, in Paris, where they observed quantum jumps. Recording a single photon in a cavity. So what happens is there's a single photon in this cavity, and they can read out whether there is a photon or not by sending Rydberg atoms through. And a single photon is never observed, but it causes a phase shift, an AC-Stark shift, on the Rydberg atom. And then they can read out with a Ramsey interferometer the AC-Stark shift. So every atom which passes through the cavity will tell you whether there is a photon or not. And-- And this is shown here. Blue means no, there is no photon. Red means there is a photon. So here you repeatedly figure out again there is no photon. And suddenly there is a quantum jump when the cavity has absorbed a photon, or when a thermal photon was released. You may say, what happened here? Well no measurement is perfect. There is some noise. But by seeing that these are outliers, you clearly see that here is a photon in the cavity, and here where the red read out of the atom overwhelms the blue readout here you have no photon. And then as time goes by, you can observe the death and birth of micro-cavity photons. So these are now quantum jumps in the-- not in atomic population, but quantum jumps in the photon number in a cavity. AUDIENCE:What's the time scale on that? Seconds? PROFESSOR: Yeah, time is second. So what is shown here is-- I think what is shown is here are 300 milliseconds, and this is 2.5 second element. AUDIENCE: What was the density of each line mean? PROFESSOR: Each line is a Rydberg atom, which has been readout with a Ramsey interferometer. And-- AUDIENCE: So that means that they're leaving no, like between-- PROFESSOR: I think that means there was a Rydberg atom in state e, which means there is a photon. And blue means, the blue bar means, there was a Rydberg atom in state g. These are two highlighting states of the rubidium atom. Sort of each spike here is one readout of the microchannel plate. So these are detector clicks. You can say, yes. You can sort of say, blue is yes, and red is no. And these are the two clicks. AUDIENCE: Yeah, but, like the separation between two clicks is not uniform. PROFESSOR: No, because if you're an atomic beam the atoms come not equally spaced out. There's a randomness when the atom comes. I think it's posed only in statistics in this experiment. It's a real experiment. It's amazing. Other questions? Team One. AUDIENCE: So one more question. So the, for example, the first bit of blue is all measurements for the same state of the cavity. I know blue means photon or no photon, but if it does mean photon, that means it's the same photon interacting with-- PROFESSOR: Yes. AUDIENCE: OK. PROFESSOR: Yes, because, I mean, this is something people know that the photon number in this cavity only changes over very long time scales. And you sort of see, kind of, exactly hear how-- You see here two quantum jumps. One from blue to red, and one from red to blue in this cavity. Yes? AUDIENCE: I don't if this is maybe just my accent's not very good, but in the bottom left graph-- Right? Or on the bottom graph, in sort of the left corner, it looks like there's some overlap between the dark blue and the dark red. Yeah. PROFESSOR: You mean here? AUDIENCE: Yeah, like right there. Is that like sort of there is a photon in that [INAUDIBLE] simultaneously? PROFESSOR: I think, to the best of my knowledge, I haven't really looked-- I've heard wonderful talks about it.You can go back and read the paper, but this probably means in the atom number in the cavity, and the algorithm they have used to analyze the data says that here the photon number was 0, jumped to 1, and jumped back again. So you-- If you see here, for instance, there is some overlap, but you would still say, since either there is a photon or not a photon, as long as you have a-- as you have a higher number of red clicks than blue clicks, you would say there is a photon. And the little overlap, there is maybe a little bit of time where you don't know whether there is a photon or not. But this really-- This is real, because every measurement takes a certain amount of time. Since you have a few random clicks, if they are a few random clicks, you do not know yet if the system has jumped, you have to wait a little bit. But this is really also showing how our knowledge about the quantum system comes with time. And therefore, the wave function, or the statistical operator changes with time, because of wave function and statistical operator are simply a description of our knowledge about the quantum system. AUDIENCE: [INAUDIBLE] is too dense PROFESSOR: You could have a situation, for instance, you could prepare the cavity, that it has an equal probability foreseeable in one. And then what you would read out may just be completely random. So that may be possible, but in their experiment, I think, the nature here is that you have most of the time stable photons, because the cavity is cooled to very low temperature, and only occasionally does the thermal distribution, the Bose-Einstein distribution of photons in the cavity have a one. And what we observe are the thermal fluctuations between mainly 0 and occasionally a 1. If you would make the cavity and little bit hotter, it would have maybe half of the time a 1 and half of the time 0. But that wouldn't mean that you have completely randomness in your results, because there is a thermal relaxation time over which a photon is created, and a photon goes back. The photon would not jump back and forth from the cavity to the walls of the cavity in microseconds. It would probably have an exchange time, and you could sort of measure the thermal correlation time of photons in a cavity. It's a really good experiment. OK. So in that sense I think I showed you that what Schrodinger said is ridiculous is real, is the subject of current research, and since for quantum information processing in quantum computers, we need single quantum system is at the heart of quantum information processing. So therefore, we want to now describe single quantum systems. But when I do that, I-- come very -- It will become clear at the end of this unit, but I want to give you the warning right away. The way how I describe now a single quantum system may not be unique. There will be, there is actually, an infinite number of ways. And I've already mentioned that to you, that there is an infinite number of ways to unravel a density matrix. I mentioned that you can write every density matrix as a probability of pure states. Weighted probability of being in pure states, but there's always different choices. You can get an infinite number of different combinations of pure states, and construct your density matrix. So, so this was sort of in at the level of the density matrix, but similarly-- and this will come out of this discussion today-- there is also an infinite number of ways to unravel the dynamics-- the dynamics involved in the evolution of an open quantum system. So-- so with that I want to describe the quantum Monte Carlo wave function method. So in the quantum Monte Carlo wave function method, we perform-- We don't need an experiment. We don't need sophisticated experiments to observe a single quantum system. It's a theoretical method, but we perform thought, or gedanken experiments. So our gedanken experiments is that we perform measurements. So we assume-- we have-- When we want to describe an ensemble of atoms, we assume we have a single atom. We assume it emits photon, and then we assume that the photon is detected. And the mom-- So the moment the photon is detected we know, for instance, that the atom is in the ground state. And so our quantum Monte Carlo simulation follows now the trajectory of the single quantum system by saying, I toss a coin, there's a certain probability for detecting the photon. If I detect the photon, the system I know is in the ground state. If I don't detect the photon, it's not in the ground state, and such. And these probabilistic approach is just simulated in the quantum Monte Carlo sense. So-- So what is in essence-- what is the essence of the quantum Monte Carlo method is that we allow a small time step delta t. It's a real time simulation. And there is a small probability that the system has decayed in that time. So we may now toss a coin-- use a random function generator-- and ask, has spontaneous emission taken place? And then the two possibilities are yes, and we continue whatever the outcome, whatever the quantum state is. Or we say no, and then with the probability-- with a-- so with a probability, p, we say yes. With a probability 1 minus p we say no, and then we continue the simulation accordingly. So this will become much, much clearer when I write down the formalism for you, but there is one important aspect, which I really want to discuss with you. And this is the essence of the method. Namely, how does the wave function change by your measurement? So-- so this is the essential, the only conceptional part for the quantum Monte Carlo wave function. Afterwards I just show you. Once you have this, once we have cleared up this conceptional part, the rest is just a few equations. And we immediately show that the quantum Monte Carlo wave function method is equivalent to the optical block equation. But let's really feature and take some time for discussing the concept, namely, the change over wave function by a quantum measurement. And maybe what this carries home to you is that the nature of an open system-- we have a quantum system, which interacts with the environment-- is, actually, that the environment permanently performs measurements. And you will actually see that the idea of performing measurements, formalized in the quantum Monte Carlo sense, gives us an equation for the density matrix, which is identical to the optical block equation. So in other words, you can say I am re-deriving for you master equation now, but from a very, very different perspective. From the perspective of an environment which constantly performs measurement. OK. So let's go through that. How-- What happens in a measurement? Well if our initial state is the ground state-- starting very in trivial ways-- well there is no measurement, because the particle will stay in the ground state forever. OK. Now let's go to the next situation that our atom starts out in the excited state. Now in a time step delta t, we have two possibilities. A photon can be observed, or nor photon is observed. When the photon is observed, we know for sure that the system is now projected into the ground state. If no photon is observed, well the system is still in the excited state. OK. So we've discussed what happens when the system is initially in the ground, and when the system is initially in the excited state. But now comes the following question. We prepare the system in a superposition of ground and excited state. And we want to discuss what happens when the photon is observed? What happens when no photon is observed? I think if you observe the photon it's pretty clear we have to assume after photon emission the system is in the ground state. But if no photon is observed, I want to give you three choices, and you should address them with the clicker. One choice is that the particle stays in the same state. The second choice is that the particle stays in a superposition state, but the probability to be in the excited state decays. Also no photon has being emitted, or no photon has been observed. And the third possibility is something else. So in other words, assume you're an observer, and you know with 100% detection probability-- let's not discuss technical issues-- you have completely surrounded your system with perfect detectors. And you know in the time step delta t, in the time t, no photon was observed. And now you should make a prediction what happens next to the quantum system. And you make the prediction by assuming that the system is described by a wave function or density matrix. And I'm asking you, which of the three choices correctly describes what the system will do afterwards? What the system will do next? Is the question very clear? Good. [TAPPING] OK. All right. Pretty good. A lot of people when I taught the class before said it's A, the wave function is not changing, because no photon has been emitted. Now in-- For those people, but there only few among those, I would have now asked the next question. If you have-- If you have a 50-50 superposition, of ground and excited state-- And let's say the spot time for spontaneous emission is 10 nanoseconds-- But now you wait one second, and after 1 second, no photon has been observed. What is your prediction? In what state is the system at this point? Is the system in the ground state? Or is the system in, still in, a superposition of ground and excited state? Well since we have to clicker, you can tell me. Is it-- Is it still a superposition state, or is the system in the ground state? OK. So now you have to help me out, because the people who said B, if they would go back to here-- And most people said it is B. If you put in gamma 10 nanoseconds, the inverse of 10 nanosecond, and the time is 1 second, the excited state amplitude, the excited stated mixture, is 10 to the minus nothing. So-- OK. If you want to be a mathematician and say-- [LAUGHING] --there is epsilon 10 to the minus 10, yes. Then you can say there is still a small admixture to the excited state. But by making it so extreme-- 10 nanoseconds versus a second-- at least in all practical terms, the system is in the ground state. And I think this should teach you what it means if you have a system which is in the superposition of ground and excited state, and after 1 second it has not emitted a photon. I mean-- Then you would say, it had its chance. It had plenty of chances to emit, but it decided it doesn't want to emit. And that would mean your 50/50 superposition of ground and excited state means the atom decided that it's in the ground state. You should also realize a 50% superposition of ground and excited state means in half, in 50% of the cases there will never be a photon emitted. So in other words, what you should realize is if your 50-50% superposition of ground and excited state, that the quantum evolution of this system is-- after a long time the system is definitely positively, absolutely in the ground state. It's always in the ground state after a long time. In 50% of the cases with emission of a photon, in 50% of the cases without emission of a photon. If you have a radioactive-- ensemble of radioactive nuclei-- and let's say half of the atoms are in the excited state and [INAUDIBLE] and half are in the ground state-- and you send the sample to your friend, and your friend waits many, many, many half times, common sense tells you that all the particles are now in the ground state. Also only half of them have decayed, because the other half were in the ground state to begin with. I mean this is what quantum physics tells us. This is what a 50/50 superposition is. But quantitatively this is, of course, included in this result-- which magically that almost all of you got right-- namely there is a superposition of ground and excited state, but the excited state amplitude decays. And to the limit, in the limit that the time is much longer than the decay time, no matter what the coefficient initially was in front of the excited state, this coefficient has decayed to 0. Any question about that? Let me just emphasize that, again in giving you also an example in the reference, a 0 measurement. A 0 measurement modifies the wave function. You can also put it like this, what the wave function is is the best knowledge you would have about this system. The wave function is a way to predict what happens next, and it's the most precise way to predict it. More accurate predictions than the wave function always cannot be made, because of Heisenberg's uncertainty relation. But what happens is, if you have a 50/50% superposition of ground and excited state, and the system has not emitted, then you would say initially my guess was half of the atoms were excited, half of the atoms are in the ground state. But if none of them has decayed for awhile, you would say I revise my estimate now. Now I have to assume that more are in the ground state. Because having not decayed means there is a higher probability that the system has actually-- is in the ground state to begin with. So this is how this way of dealing with a wave function automatically adjusts for the knowledge you have gained about the system. Let me give you another example which carries home the same message, and I take it from a pedagogical paper from Dickey, from American Journal of Physics, 49, 926, 1981. And what he discussed is the following. You have a box and originally your atomic wave function is completely localized. Your Bose-Einstein condensate if you want in this box, or let's just assume one atom to be more precise, and it's completely delocalized. But now you focus a laser beam into the system. And the laser beam would ionize your atom, and you could count the ion with 100% probability. And for a short moment, you flash on the laser and the result is yes, you count. You get a count. In that moment you would say my wave function is more localized. I revise my estimate. Well maybe I shouldn't say ionized. If it's ionized you destroy the atom. Let's just say you observe fluorescence so the atom is still alive. But now you have to say that the atom is localized in that region. However, if you don't observe anything, you would say, well now it's actually more probable that the atom is outside. Because if it had been in the laser beam, with a certain probability, I would have detected a photon. But the non-detection of a photon means that I revise my estimate. And I say it's much more probable that the atom is outside the laser beam than inside the laser beam. So this here is exactly how a non-observation of anything, a 0 measurement, modifies your wave function. And the formalism to incorporate that is what we have discussed before. Yes? Cody? AUDIENCE: So your arguments here so far work the same way as if it's in a wave function, you have just an entire Pascal probability, and we're updating our understanding of probability. Like you haven't included anything about the relative phase between these two-- PROFESSOR: I want to do that now. I mean, I-- This is sort of just to address the basic concepts about what does a measurement mean, what does a 0 measurement mean. But now I want to write down everything for you in amplitudes. We want to do exactly the time evolution of the system. Nancy? AUDIENCE: Is this, like, inconsistent with [INAUDIBLE] quantum equation? Like especially the [INAUDIBLE], because when you are saying that it is admixture of the ground and excited state, and we did not observe-- PROFESSOR: Then the superposition say that ground and excited state. We are not starting out in an energy heightened state, so therefore, we-- what we have is not a sharp value of the energy, but an expectation value. And also if you repeat the measurement many, many times, the average energy is conserved, and this is exactly what energy conservation in quantum physics says. AUDIENCE: [INAUDIBLE] PROFESSOR: Well if your particle, if your system, is not in an eigenstate of the Hamiltonian. It doesn't have a sharp energy. And that would mean that when you measure the energy now, you have fluctuations. Sometimes you measure higher, sometimes you measure lower. It's only when you're in eigenstate that you measure a sharp value. And therefore you know from the beginning that you will measure an energy distribution. And sometimes you measure higher, and sometimes you measure lower than the average value. There's nothing wrong about it, but your question is a very good one. You should-- I mean those examples are really-- On the one hand they are trivial, but on the other hand, it's very profound what they tell us about quantum physics and how to apply conservation laws and such. OK. I think-- Let's-- Let me now formalize exactly how this is done with all the bells and whistles. We assume we have an initial wave function, which is now an arbitrary superposition state of ground and excited state. And we have the environment, which, in this case, we assume is the 0 photon state. It's a vacuum. And we want to exactly solve the Schrodinger equation for that system. But we restrict ourselves too much, to a very small time step. The small time step is smaller than, you know, anything else. Than the natural decay time. Maybe the inverse [INAUDIBLE] frequency if you drive the system. You're not doing it here, but I will later show how you can add a laser beam and drive the system. Or it should also be smaller than the inverse tuning. So what is very important here is that we want to deal only with simple cases, namely one photon has been emitted or not photon. So you want to make sure that there is, at most, one spontaneous emission event during the time delta t. So what I'm writing down now is the total wave function of the system plus the environment. And now we can evolve it. We can ask what happens? What is the wave function time later? This can be exactly done by time-dependent perturbation theory. And the result is that the system will be in a superposition of ground and excited state. And we want to calculate the coefficients alpha prime and beta prime. And we still have the vacuum state. But then we have the possibility that a photon has been emitted. In that case we are in the ground state. And the direct product with the environment involves now the photon emitted into a certain direction with wave vector k, with the polarization epsilon. And we have coefficients beta, k epsilon, and we have to sum over all possibilities for the photon to be emitted. So we call this-- So the wave function of the total system has now two parts. It's a wave function, which I call psi 0. This is a wave function which involves no photon in the environment. And the wave function psi 1 involves one photon in the environment. So this is the [INAUDIBLE], which seems very natural, but it's also-- you could immediately prove that these are the only possibilities, how the system can evolve. And we can verify this by using time-dependent perturbation theory. It's actually almost everyone of you has seen it in either 8.21 or in a more basic course on quantum physics. It's time-dependent perturbation theory for the emission of a photon. But usually when you see those treatments there reservoir, the environment, is not treated as explicitly as we do it here. The theory where this is treated exactly in this way, how we need it, is the Viegener Biscoff theory. Which is nothing else in the perturbative approach, but it's really a perturbation theory. Not just for the atomic system-- how we sometimes present it in a simplified version-- it's perturbation theory for the total wave function of the complete system. Philmore? AUDIENCE: So you mentioned the time interval [INAUDIBLE] is much smaller than a lot of things. But you were ignoring, I take it, the counter rotating terms? In which case we're still at a large enough time step that's average out, so to speak? I simply say this, because I expected to be an e with a photon term. From the g going up, so to speak. PROFESSOR: Yes. Well-- The counter rotating term-- The first answer I wanted to give you, no this is exact perturbation theory. But I think I get myself in trouble if I would allow-- I would get myself in trouble if I would allow the time step to be extremely short, because then I'm in shorter than 1 over omega. Because during the time 1 over omega, we have-- you can say we have counter rotating terms. Or in other words, this particle in the ground state during a time 1 over omega, particle in the ground state can emit a photon and reabsorb it. These are where those weird diagrams with virtual states, which we discussed earlier in this course. So I think I want to assume here, which I haven't done, that the time step is larger than omega to the minus 1. And as you know the counter rotating term has a detuning delta, which is 2 omega, so therefore it is excluded. Or in other words, when we talk about photons sent into the environment, we want to talk about real photons, and not virtual photons. Yes, good point. OK. So if you do simple lowest order time dependent perturbation theory, in the Viegener Biscoff approach, you get an exact result for beta prime. You find that beta prime in this superposition is the original amplitude beta, but it has decayed with e to the minus gamma over 2dt. And since we are only interested in small time steps, we can do a linear expansion of that. The probability, dp, that a photon has been emitted is the norm of this wave function psi 1. And this, using Viegener Biscoff perturbation theory, is gamma dt times the amplitude squared, that the system was excited to begin with. And because of the conservation of the norm, the wave function, the norm of the wave function psi 0-- which is a wave function without a photon being emitted-- is 1 minus dp. So what I'm telling you here is that this is an exact result of time-dependent perturbation theory. Often when this-- when textbooks treat spontaneous emission, they're more interested in getting Fermi's gold rule at this rate. But the same approach-- if you just write it down-- tells you how the amplitude in the excited state evolves. So therefore, what we learn from perturbation theory, that alpha prime and beta prime, the wave, the coefficients of the wave function, without emission of a photon, this evolution of the wave function occurs with a non-emission Hamiltonian. And this non-emission Hamiltonian is our Hamiltonian for the atomic system. But then we have to account for spontaneous decay, and this is done by this non-emission part. So occasionally you may have heard that people wave their hands and say, your system is described by a Hamiltonian, which has an imaginary part for decay. And this is sort of phenomenological. This is definitely not the case, because, even if you have a Hamiltonian like this, a pure state would simply decay and remain a pure state. What we are doing here explicitly is we deal with probability in the correct way. Every time the system can branch out into different possibilities, two different dimensions of the density matrix-- one has probability delta p, one has probability 1 minus delta p-- and it is only the wave function associated with probability 1 minus delta p, which evolves with this Hamiltonian. So what I'm telling you is the exact solution for the evolution of the total system, in terms of a density matrix for the atomic system. And this is exact. So that means we can now-- write down the-- Just get some extra space. Oops. How does it do it? Nope So we have to insert new page. Yep. OK. OK. So the procedure, how we implement this exact solution of perturbation theory, is the following. We have a time step delta t. We compute what is the probability that a photon will be emitted. Then we need a ran-- Then we need a random number generator. So we need a number epsilon, which is a random number chosen in the interval 0 and 1. If this random number turns out to be smaller than delta p, then we continue our time evolution on the computer. That psi is now in the ground state, and maybe there is a laser beam which excites it again. And such I will add a few bells and whistles later. Otherwise psi, our wave function, is now the wave function which is the time evolution with a non-emission operator of the original wave function psi. And since we have a real wave function now with probability 1, we have to re-normalize the wave function by this denominator. And then we execute the next time step. This means go to 2 and do the next time step. And then you have, so to speak, if you do it many, many times, you get a time direct trajectory of, so to speak, one experiment. And then you start again with your system in a wave function psi, and you accumulate a second experiment. And maybe you do it 10,000 times to get enough statistics. And then you sum up, you know, everything you want and calculate all the expectation values you want to know about your quantum system. So the claim is that this method, called quantum Monte Carlo wave function method, is fully equivalent to the optical block equations. And I want to prove it to you by showing that, if I take a density matrix, which is an ensemble average, over many realizations of those quantum Monte Carlo trajectories. That then this density matrix follows the differential equation, which is your optical block equation. And the proof is shown here. So what I told you is the density matrix after time delta t, has now two matrix elements. One with probability delta p, and this is-- photon has been emitted. System is in the ground state, with probability 1 minus delta p. We evolve the quantum state with a non-emission Hamiltonian. And then all what is done in the next few steps we assume that delta t is small. We do a Taylor expansion of the exponent. We neglect quadratic terms in delta t, and then we come to this line. And if we write this as rho of t plus delta t, minus the original rho of t, we find that this follows differential equation, which is exactly the optical block equations. So therefore what we sort of implemented as a form of doing many, many quantum measurements in the environment is a procedure, which is rigorously the same as the optical block equations, which we derived using a master equation approach. OK. This method is very powerful for the following reasons. If you simulate a density matrix with, you know, an internal states, or external, internal states-- I should've said with in quantum states, you need n times n matrix elements, which can become quite a memory hog for your computer. The wave function, at any given moment, as only n components. So therefore there is a computational advantage in using a stochastic wave function approach over simulation of the density matrix. The second thing is that a lot of people, especially experimentalists, like sort of this approach, because it reflects directly what they do in the experiment. And I think it's pretty clear that with this approach you can deal with a great variety of situations. Let me just mention two obvious extensions. One is polarization. A photon is detected. That means your random number produced a number epsilon, which was smaller than delta p. At that point you can create a second random number, which determines if your pull-- if the photon has been detected with sigma plus, sigma minus, or pi polarization. Or you can discuss recoil. If the photon is-- If the photon is detected, you use-- you throw on another random function generator, which determines what the k vector of the photon is. So what direction each part k the photon has taken, and this determines now what is the recoil, the change in momentum of your atom. So you see, kind of, you can start with an atom at 0 momentum. You can see it emits a photon, and then your computer always makes a choice based on the random number. And then you say, OK my photon has now received a recoil kick at 45 degrees. Now your wave function of the photon is such and such, and if you add a laser to the Hamiltonian, the recoil may have kicked-- may have Doppler shifted the resonance, but everything is easily taken into account. So you see this quantum Monte Carlo wave function method can easily be extended to describe external degrees of freedom, multiple laser fields, and all that. Any questions? OK. Let me now just generalize the thought, but I think you know already everything about it. We talked about the master equation. We talked about the most general master equation in the [INAUDIBLE] platform, where those operators L are the jump operators. And the prominent example for jump operator was the signal minus operator, which takes a particle from the excited state to the ground state. And this is the jump operator for spontaneous emission. But you may have many jump operators. May be spontaneous emission of different photons with different polarization and such. Or the cavity may lose a photon, and then we have a jump operator for the cavity. We went through that. Well that means now in an exact way that the non-emission Hamiltonian is the original Hamiltonian with an imaginary part, which comes because of those jump operators. And the general quantum Monte Carlo wave function procedure is that first you ask, has something happened? Has any jump happened? And this gives you the probability delta p. And if-- If no jump has happened, you just do a time evolution with a non-emission Hamiltonian. But if a jump has happened, then the jump is now-- the wave function is projected by the jump operator. If you have a sigma minus operator, it takes a particle to the ground state. If you have several operators, a different jump operator may take your particle to another state. And you-- And you then have a branching ratio that you know first a jump has happened. And then the question is, which jump has happened? And you just play the probabilities game. So that you see that the quantum Monte Carlo wave function method can be immediately be generalized. So let me come back to the special case of spontaneous emission. In that case, we have only one jump operator, which is sigma minus. And the normalization is the square root of gamma. And that actually means that our quantum trajectory ground state, excited state, is very simple. If you start with the system 100% in the excited state, it's very trivial what happens. Nothing happens. But if then the jump occurs, the particle is in the ground state. Then nothing happens then until infinite time. You then have to start-- The next trajectory you start in the excited state. And this time, by chance of the random number generator, the jump happens later. Another time the jump happens earlier. And what you then have to do is, you have to average, over all of those realizations, what is the probability for the particle to be in the excited state. And this is now your estimator for the excited state diagonal matrix element of the density matrix. And if you-- If you've written your code correctly, you will find that you get a wonderful exponential decay exactly as you have expected. Questions? Now-- Yes? Let-- Let me now talk about dephasing. What I've shown you so far is how we can unravel the density matrix in many microscopic realizations. And I've given you a specific example. We assume probability for spontaneous emission, and with that we propagate our wave function. But now I come to, sort of, the nitty gritty details, or the dirty truth, that what we have assumed is by no means unique. And I want to explain it to you first by reminding you of a very, very nice homework assignment you have solved. And this was about we have optical block equations for the density matrix, and the dense -- and the solution of the optical block equation is that we have population damping with a damping time t1. And the off diagonal matrix elements are damped with a time t2. Remember if you've only spontaneous emission t2 is 2 times t1. But we can have a lot of other processes, which can lead to much, much faster defacing time than the population, than the time of population changes. And in your homework, you have discussed three possibilities. One is, well, spontaneous emission is energy loss. But you have discussed three different possibilities for t2. One-- Three different possibilities how dephasing, loss of coherence, and phase damping can happen. And in this homework assignment, you assumed either that there is a random phase, that the elastic collision which project onto the excited state, and then you got the most crazy and artificial model. That every time, randomly, the phase of the excited state flips form plus to minus. So if I would implement that now, with a quantum Monte Carlo method, it would have the following affect. And I hope you enjoy, sort of, the graphical representation. You can really think about it, that this is what happens. That this is what inside-- what is inside the density matrices described by this process. So let's assume we have a system, which would be in a superposition of ground and excited state. And that would mean that in the left frame the dipole moment would just rotate at the resonance frequency. It rotates, and it is a rotating dipole moment, which emits coherent light. But now you assume that you have a random phase. So if you add random phases to it, maybe because you're fluctuating magnetic fields, then the line becomes sort of-- That doesn't look random, but you know what I mean. It becomes sort of jagged. The phase distribution is still sort of there, but there is a jitter on top of it. And as a result, the light emitted by this dipole is spectrally broadened Well in your second model you assumed that-- with a certain randomness, with a certain time constant-- there's an elastic collision in the excited state, which projects the system into the excited state. At this moment there is no superposition state anymore, and the dipole moment is 0. So that's what you assume. Or when you have random phase flips, you assumed that suddenly this kind of data to data ministic sine function suddenly jumps corresponding to a minus sign in the excited state. And depending where you are in the cycle, it creates jumps at random places. And the question of course is, if all those three processes-- and that's what you showed in your homework-- lead to the same density matrix, which one is correct? Which one is real-- is really realized in an experiment on a system which is described by this kind of damping, or by the optical block equation? So you can say all or none. Well you can assume what you want, it doesn't make a difference. And the reason is actually profound. The reason-- The reason is profound in that sense that the way-- our goal was to have an open quantum system, which is the atomic system, interacting with a reservoir. But the way, how we phrase the question, we are only interested in what the atomic system does. We didn't do extra measurements on the environment. The environment was just a dump for photons; a dump for energy; a dump for whatever we assumed in our dephasing mechanism. So let me just take the example of putting photons into the environment. And this is described by a certain damping term, but there is an ambiguity. And this is shown here. If you evolve the system-- our, our atomic system, the density matrix evolves, but the environment sort of also evolves. And for instance here, the unitary time evolution has taken the excitation from the atom, and we have emitted a photon into the environment. And you remember we got the optical block equation by doing the partial trace here, and just focusing on the atomic part. But now wait a moment. If we trace out the environment, we can trace it out in a different basis set. Remember, I showed you the quantum Monte Carlo wave function, we emit a photon. But we can, for instance, detect the photon in our gedanken experiment. We can detect it with linear polarization or with circular polarization. If you have a situation where you have a-- an s state, which decays to a p state, sigma plus takes you to m equals plus 1, sigma minus takes you to m equals minus 1. So therefore, if photon is emitted in you quantum Monte Carlo procedure, you would say, now the atom mean is in the plus 1 state, or now the atom is in the minus 1 state. But if you detect linearly polarized light, by just putting a polarizer here-- and this is a unitary transformation in front of your detector-- you would no longer project the atomic system on plus 1 at minus 1, you would project it on m equals 0 or whatever is connected to the linear polarization of the measurement. So therefore, if you simply assume that something has been dumped in the environment, and can be used for measurement, there are many ways, many unitary transformations, what you can do to the information, to the energy, to the photons, which have been dumped into the environment. And each of them will lead to a very different trajectory in your quantum Monte Carlo system. So therefore if you just dump the photons and not measure them, you have equal rights to assume that your quantum Monte Carlo wave function jumps to states which correspond to a circular basis, or jump to states which correspond to linear basis. And there's many, many possibilities. But each of those possibilities is 100% correct unravelling of the density matrix. And if you just do it right, assume that the measurement is done in a certain basis, and you're consistent with it, you will 100% correctly describe the time evolution of your atomic system. Any questions? Philmore? AUDIENCE: Again with the dt. I'm just curious that we take a [INAUDIBLE] very short, but why don't you run into something like the quantum Zeno effect? Where if every evolution is initially quadratic, like the [INAUDIBLE] frequency, you take very short dt-- short measuring system-- Shouldn't for a choice of sufficiently short dt you, you know the simulation would give you strange results, because of-- PROFESSOR: [SIGHS] AUDIENCE: --some sort of quantum Zeno effect there? PROFESSOR: That's a very deep question, Philmore. And I, I want to think about it more, but I think I've excluded that by saying that the time step, dt, is larger than the correlation time of the environment. So I'm doing some kind of Markov approximation with the environment, where s-- the quadratic part of the behavior is a coherent time evolution, for very short time steps a wave function evolves quadratically, but this part is a coherent evolution. And that's related to the fact that an atom can emit a photon, but in the very first moment, before the vacuum has transported it away, it can take the photon back. And the result of that is that at very short times the exponential decay doesn't start out exponentially, it starts out a little bit flatter. We get the exponential decay, we get Fermi's golden rule, and we get optical blocks equations only if your time step is larger than the correlation time of the reservoir. And the fact, you remember when we derived the master equation we had to say we do a step, which is sufficiently small for the dynamics of the atomic system, but sufficiently large, that we are not getting into any memory affects of the reservoir. And we've done the same assumption here. Other questions? Nicky? AUDIENCE: Just confused. We are not always say the environment measures the system in range at time intervals, delta t, which are large compared to the correlation time? What's more [INAUDIBLE] the evolution of the atom. But now everyone isn't that defected the Markov approximation we've made when we derived the master equation? And I was just wondering for just-- [INAUDIBLE] did you derive it by making the master equation? Maybe we don't actually need-- Maybe we don't need the idea of constantly measuring. Maybe a physical interpretation of the Markov approximation. PROFESSOR: Oh yes, exactly, Nicky. I still think you said it really very, very nicely. We derived a master equation just by assuming that the system, you know, takes a photon. And past the correlation time, the photon is taken for good by the environment. And with a Markov approximation, we formalized that there is no memory affect, the photon is not stored like in a cavity. The photon has disappeared. And by saying that this time is very short, this was a Markov approximation in the derivation of the master equation. But that also means, if the photon has disappeared, has separated from the atomic system, we are now free to make a measurement. And of course it shouldn't matter whether we make the measurement or not. But what I was able to do today is, I was making now the assumption, let's assume we make a measurement. We measure all the photons which have been emitted, and the measurement is probabilistic. And we fold this probabilistic evolution into our quantum system. And what we obtained was exactly the same time evolution for the density matrix as we got from the master equation. So in other words, this should you that interactions with an open quantum system, where irreversibly energy, photons, angular momentum, or whatever-- flows into the environment. Once it has flown into the environment, you can measure it. And you get all the stochastics from the measurement. But even if you don't measure it, you get exactly the same stochastics as if you had measured it. It's all the same. Jenny? AUDIENCE: Now that I think about it, it seems sort of weird to me that we've-- that we're making these measurements, or lack of measurements, these interactions, at regular intervals. And not-- PROFESSOR: It doesn't really matter. All we-- All we have to make is-- You could actually make delta t a random variable. It wouldn't change anything. The only thing we have to make sure is we have to make the time interval short enough that we don't have, maybe, two photons emitted in that time. We just want to make sure that it's a simple probabilistic branching, yes or no. And that's the only requirement here. AUDIENCE: Is this a formula or is it based on time? You say that would be the wave function evolving and then collapse. And that's-- This is what-- Where the measurement comes in. The equivalent, so now say, the equivalent is a wave function evolved in a collapse, when you make a measurement you say that after it evolves. So is it possible that it could also have-- like evolves some kind of formalism? Like this? Where there is an operator? Because still like even though the density matrix operator, maybe it is kind of like-- [INAUDIBLE] describes a state. PROFESSOR: I haven't seen it, Mark, but I'm absolutely certain that you could describe the same physics in the Heisenberg picture, where the time evolution is with operators. Because what we have here is, we have a time step, delta t, where the system evolves as an isolated quantum system, and then we measure again. And so, I think if you would use operator equation, you would have an evolution of the operator with the same Hamiltonian, which is non-emission Hamiltonian. And then the measurement would do some form of projection. I have to think about it, what the projection would be in terms of operators. But my gut feeling is, you can always take the transformation where you put the time dependence into the operators, and not in the wave function. On the other hand, I have some little bit misgivings about that, because the quantum Monte Carlo wave function was really developed in order not to deal with matrices, not to deal with something which is dimension n times in, if n is the number of components of the wave function. It was specifically formulated to have the simpler description with n coefficients for the wave function. But conceptually, I think, it is evolution with a Hamiltonian measurement. Evolution with a Hamiltonian measurement. And I assume this could also be done with operators. Nancy? AUDIENCE: Suddenly like [INAUDIBLE] but when we've-- these simulations are actively done, how important is the randomness? Like up to epsilon [INAUDIBLE] because no computer is actually emitting random numbers. PROFESSOR: Well I think how random a random number has to be is really dealt with in computer science and mathematics. AUDIENCE: But due to our reserves we always get this exponential behavior? PROFESSOR: I don't know the answer. I mean here, conceptionally, it should be completely random number. And I think that algorithms, which even which produce pseudo random numbers, but if the pseudo is close enough, if there is a recurrence time of your random number-- if your random number-- your series of random number repeat itself after several billion random numbers, this gives you a small error bar. And if you don't-- If you'd only do a limited sampling, a limited number of time trajectories, I don't think it matters. Of course if you want to have ultimate precision, then also the precision of the random number may enter through the backdoor at some point. AUDIENCE: [INAUDIBLE] lack of simulation of the photons emitted? But if we know the photons emitted, if we have that information, then we can decide [INAUDIBLE]. PROFESSOR: That's correct. If we-- If we would know-- If we would know more about the environment, if we would say we measured the photons in a certain basis, or when energy is dumped in the environment by a dephasing mechanism, we look at the environment. Then, of course, we would add extra information to it, and then certain quantum trajectories would reflect the extra knowledge we have. And of course, we would then describe everything in this basis. On the other hand, what you should learn from this is once the information, once the photon has escaped, it doesn't matter for the time evolution of your atomic system, in which basis you measure the photon. So the evolution of the atomic system itself-- at least when you averaged-- is unaffected by the basis. However if you make a coincidence measurement, that you say your atoms which has emitted, is still flying through your vacuum chamber, and now you figure out the photon is emitted as circularly polarized light. Then of course you know that this atom, which is still available, has emitted circularly polarized light. And then you have actually obtained extra information about your atom. In other words, the photon in the atom was entangled, like a Bell pair. And if you now do a measurement on one part of the Bell pair, you know more about the other part. But this is just how quantum physics works. If you don't use this information for anything, then you could have-- you could have as well not measured the polarization. And it would have no effect on the atomic density matrix. OK that's it for today. We have class at the usual time on Wednesday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

3_Quantum_description_of_light_Part_2.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So good afternoon. Let's have an on-time departure. Last week, we talked about quantum states of light. So we're talking about the photon, the quantum description of the electromagnetic field. And I introduced for you the most classical quantum state of the electromagnetic field, namely the coherent state. I don't want to go through the formalism again, but the coherent state has a simple definition, simple but subtle. It's an eigenstate of the annihilation operator, and it has a complex eigenvalue alpha. And from that, you can pretty much derive everything you want to know about coherent states. Colin. AUDIENCE: What was the purpose of-- whoever discovered eigenstates-- what was the purpose of trying to find the eigenvalues of destruction? PROFESSOR: Yeah, isn't that subtle? Isn't that weird? How does somebody want to find the eigenfunctions of an annihilation operator, because the annihilation operator is non-Hermitian? And why the annihilation operator? Why not the creation operator? Well, if you try to find the eigenstates of the creation operator, you run into problems. The eigenstates are non-normalizable. It doesn't make any sense. So here, creation and annihilation are not symmetric. You have to pick the annihilation operator. Colin, my gut feeling is that somebody found it in a different way. The person who popularized those states was Glauber, and he got amply rewarded for that. And I think you define coherent state as some form of superposition state. Maybe people have figured out if you take a superposition state of number states that this has special properties, that this is a minimum uncertainty state, which is as close as we can go in quantum physics to obtain a classical electric field. So I think maybe people found it like that. And then they say, hey, but there is another operator definition eigenstate of the annihilation operator. I'm 99 percent certain. But maybe you can ask [INAUDIBLE] when he comes to a CUA seminar if he discovered coherent state as eigenstates of the annihilation operator. I don't think so. Other questions? So what I want to do is-- what the theme of the end of last lecture and today is that the coherent state is a complex number. The coherent state has a time evolution. It moves in a circle. And this is really the phasor of the electric field associated with it. So coherent state, the alpha value is directly related to an electric field. That's why this state is closely related to the classical limit of the electromagnetic field. We discussed some properties of the coherent state. We looked at the fluctuations and showed that it's a Poissonian statistic. And then we use now-- once you define something, you can use it as a tool. We are now using the coherent states to look at any other quantum state of the electromagnetic field, any statistical operator which describes photons by forming the diagram matrix element of the statistical operator with alpha. If you just want to sort of think about it intuitively, alpha is like the best description of an electromagnetic field. So if you write down this, we ask, What is the probability for an arbitrary quantum state described by an arbitrary statistical operator-- what is the probability that the electric field is alpha in the complex description? So that's what those quasi-probabilities are-- Q of alpha. And we immediately looked at some examples, of course, of probabilities. We showed that the vacuum state is sort of an area. It's a Gaussian. And the area is-- it's an area. I don't know. I forgot the [? prefect. ?] It's on the order of 1 or 1/2. We realize that thermal states are also a Gaussian centered at the origin. But it is a much wider distribution. So thermal states have much more uncertainty in the electric field than the vacuum state. And then we looked at the coherent state. And now, of course, when we looked at the coherent state, we realize that coherent states are maybe not as wonderful as I tried to make you believe. They have some nasty properties. And that is when you ask, What is the probability? The quasi-probability of the coherent state would be beta. Now, if you would expect a delta function, because when you ask, What is the probability of a momentum state to be p naught? If you have a momentum state p naught, it has a delta function at p naught. But here, it's not a delta function. It's a Gaussian of the standard deviation. I think the standard deviation here is-- what is it, square root 1 over square root 2? Because the Gaussian has usually 2 times sigma squared. And since the denominator is unity-- so the standard deviation of 1 over square root 2. And that showed us that the coherent states are not orthogonal. You have to be a little bit careful. It's not your standard basis set. It's an overcomplete basis. So therefore the coherent state is not a delta function in the causal probability. It's a little bit blurred circle with an area on the order of unity. And you have some homework assignment to look at it. So in some way, how you should look at it is that if you see some quasi-probability distribution, the distance from the origin-- this is the absolute value of the electric field. The phase here is almost 45 degrees. But if you look at this uncertainty, you would also say, Well, the phase is uncertain to within this angle. So you can pretty much read the uncertainty in the total electric field, the uncertainty in the phase. You can read pretty much everything you want from this diagram. Now, when we look at a number state-- Well, you often know in quantum mechanics, number and phase are complimentary. If the number of photons is fixed, you know nothing about the phase. And indeed the quasi-probability of a number state is a ring. It has no phase. It has completely random phase over the 2 pi circle. The energy is sharp of a number state, since the energy is e squared. You may have expected a delta function in the radial coordinate. But what you get is also something blurred on the order of unity. And I want to say something about that in a second. Finally, we discuss the time dependence. And the time dependence is very easy. After all, we're dealing with an harmonic oscillator. And in an harmonic oscillator, if you have a plane of x and p, symmetric and anti-symmetric combination of a and a dagger, in this plane in an harmonic oscillator, the quantum state is just rotating circle, a rotation with omega. And indeed we showed that when we apply the time evolution operator-- and some of you were right, of course, with a minus sign-- it moves with e to the minus i omega t. And therefore everything rotates in a clockwise way. Now, we discussed the operator of the electric field. And in this quasi-probability-- Sorry, we discussed the operator of the electric field. And I hope you remember that in the analogy with the harmonic oscillator, the electric field was a minus a dagger. And this is the momentum operator. In those quasi-probabilities-- and we will see more about it-- something which is sharp in momentum is a sliver parallel to the x-axis. So therefore, you can regard the vertical axis, which is the imaginary part of alpha, as the momentum axis. And you can regard this as-- the horizontal axis-- as the x-axis. So therefore, since momentum is electric field, you always get the electric field by projecting onto the vertical axis. And if you project this fuzzy ball, you get a value 0 with some uncertainty. And if this quasi-probability starts to rotate due to the time evolution, we get an oscillating electromagnetic field, almost classical except for that fuzziness. So that's where we want to continue. Any questions about that? Yes? AUDIENCE: Why is there no phase fuzziness in the E t? PROFESSOR: No phase fuzziness. There is a phrase fuzziness. For instance, if you would say the phase is determined by the 0 crossing, you don't know exactly when the 0 crossing happened, and that corresponds to an uncertainty in the phase. Trust me, everything is in this diagram. Now, there are two things we want to continue. One is I want to show you that the coherent state is a minimum uncertainty state. The product of delta x delta p is just-- is it h-bar or h-bar over two? One of the two. So it's a minimum uncertainty state. And therefore, you can never have a quantum state which is less fuzzy than the coherent state. So this fuzziness here is the intrinsic uncertainty of quantum physics. So that's what we want to discuss today. But then we will immediately start with non-classical states. And that is, well, if this area is determined by Heisenberg's uncertainty relation, what can be maybe deform the circle into an ellipse, and these are three states of light. So that's an outlook. That's what you're going to do in the second half of the class. But before I do that, I want to be a little bit more accurate about quasi-probabilities. And this is almost like a disclaimer now. Just to give you the bigger picture. What we want to achieve with those quasi-probabilities, we want to do what phase space densities in classical physics. We have a coordinate which is x, a coordinate which is p. And you can often describe a classical system if you know the phase space distribution, if you know the probability that a particle has position x and momentum p. So all this is about phase space densities. Probability of x and p. Of course, writing it down, you immediately see the problem. In quantum mechanics, you cannot measure x and p simultaneously. These are non-commuting variables. So therefore, what happens is, if you now define a phase space function, which is done in quantum mechanics textbooks, you can actually do it in three different ways. And the three different ways are Q, P, and W. The definition-- I don't want to go through the mathematical subtleties-- the definition of those functions involves the operator definition, a and a dagger, or, which is equivalent, the x and p operator. If you define something in units of x and p or a and a dagger, you can have a product which is fully symmetric i in the ordering of x p, which is anti-normal or normal. So in other words, the order matters. And you have three choices. In an operator product, you can have symmetric ordering, which means not x p, but x p plus p x, then it's over 2. That's symmetric. You can have an ordering which is called normal and one which is called anti-normal. Ordering of operators in the operator base definition. Of course, if you have three choices, you would say, Which one is the best? Which one is the winner? But the fact is all three have their advantages and disadvantages. So they all have pluses and minuses. The reason why I picked for the course Q of alpha is that it's a real probability, it's always positive. It is a diagonal matrix element of a statistical operator, and this has to be positive. So it's a real probability. The other guys, P of alpha, can be positive or negative. And also, W of alpha can be positive or negative. So if you use a P alpha distribution statistical operator, it can be written like this. And as a result, the coherent state is now not this Gaussian. It doesn't have this Gaussian distribution as a course of probability. It's what you want-- what maybe some of you wanted to see-- oh, by the way, it's a delta function. The probability of the coherent state alpha has a delta function peak at alpha, which is sort of nice. And the number state is not a ring of a finite radius. I just mentioned to you the number state. You would naively expect the energy is sharp. The square root of the energy's electric field, shouldn't it be sharp? And indeed, it is sharp. It's actually worse than a delta function. It's a derivative of a delta function. But at least here, in the probability P, which is also called the Glauber-Sudarshan P representation, you get the delta function, which may be very natural for certain purposes. AUDIENCE: Since the [INAUDIBLE] and you can express one in terms of the other, is that unique? PROFESSOR: It is unique by some symmetry choice here. So that's the advantage of it, that by-- I'm not going into the mathematics. I'm not giving you the definition but the way how it's defined. It's unique. It can be written in such a way. And the way how the quasi-probabilities P-- the P representation-- is defined, you get a delta function at the coherent state. And finally, W stands for Wigner distribution. And the Wigner distribution is something you actually find in most quantum mechanics textbooks. The Q and P distribution are more common in quantum optics. But the Wigner distribution has the advantage that the projection on the x- and y-axes are indeed psi of x squared, psi of p squared. So you get actually the x wave function and the p wave function. So now, as a full disclaimer, if you want the electric field, which is the momentum of the harmonic oscillator, which is the electromagnetic field, what you really want to project is the W function, because for the W function, the Wigner function, the projection is exactly the momentum distribution in the electric field. So the Wigner distribution is closest to the classic phase space distribution, as close as you come without violating commutators. But of course it has a disadvantage that it has negative values. And try to explain to your next neighbor what is a negative probability. Some people actually see negative probabilities spring out the non-classical character. And I know in our field, in AMO physics, a few years ago, in my lifetime, there were really attempts to use quantum state homography and measure for the first time a negative Wigner distribution and show that to the world. So that meant something to a lot of people. Anyway, you can read about it in quantum physics textbooks. All I want you to know is to know about it, but then also just relax. In the bigger picture, all the three distributions are the same. It's more sort of on the level of whether something is a delta function or has widths unity. So on the small scale, it matters. But if you map out something on a bigger scale, they are all related to each other. And for the rest off today and the next class, when I show you those phase space distribution, when I say I project onto the vertical axis to get the electric field, I'm not completely rigorous which of the three functions I've really chosen. Any questions? OK. So I promised you-- and this is what we're going to do now is we're want to understand in more depth the fluctuations. And in particular, I want to show you that coherent states are minimum uncertainty states. So by identifying the vertical axis with p, the horizontal axis with x, we immediately expect that we find a result related the Heisenberg uncertainty principle, which says sit the widths in P and the widths in Q has to be larger than h-bar over 2. And we are mainly talking with light about the photon creation and annihilation operator. But, just as a reminder, the momentum and position operator are symmetric and anti-symmetric combinations. So let me just call those [? prefecters ?] Q naught and this one P naught. Well, we can then immediately, by just using elementary commutator, calculate what are the expectation values in a coherent state for P, P squared, Q, and Q squared. For P, it is-- the P operator is a dagger minus a. If we act with a on alpha, we get alpha, because alpha-- the coherent state-- is an eigenstate of alpha. Now, with a dagger we act on the left hand side. And we get alpha star. Well, with P squared and Q squared, you have to use one or two more steps to get rid of the products. But ultimately, you can express space all that just in powers of alpha, alpha squared, alpha star, and such. So what we want is we want to find out what are the fluctuations delta Q, delta P. Just as a reminder, delta Q squared is of course Q squared average minus Q average squared. And Q squared average and Q squared is what we have just calculated. Actually, let me leave it in as a reminder for you-- Q squared minus Q average squared, the square root of it. And what we find is that it's square root h-bar over 2 omega and, for delta P, it is square root h-bar over omega 2. And what we verify, that indeed, for a coherent state, the product of the 2 is Heisenberg uncertainty limit. So therefore, coherent states are minimum uncertainty states. And maybe, Colin, to address your question, I could imagine that coherent states were maybe invented by simply saying, We have an harmonic oscillator data. We want to find the minimum uncertainty states for which the uncertainty in x and p, when expressed in natural units, is the same. In other words, the coherent state is the solution to the following question. If you plot the quasi-probability distribution, you give yourself an uncertainty area. One horizontal uncertainty is delta Q. The vertical uncertainty is delta P. And the minimum uncertainty state means that the area of the 2 uncertainties, the area of the shaded region, delta P times delta Q is h-bar over 2. Or if I want to rewrite that in-- Yes, in the real part of alpha times the uncertainty in the imaginary part of alpha for the quasi-probability, then this is 1/4. OK. So what we have learned is that-- just one second. We have learned about one way to characterize uncertainties-- the quantumness of the electromagnetic field-- by looking at the uncertainties in the x and p variable. And that led us to minimum uncertainty states. Now, I want to now introduce to you two other ways of characterizing the uncertainty of quantum states of light. One is we can ask, What are the fluctuations in the photon number? Right now, we have asked, What are the fluctuations in the electric field? But the next question is, What are the fluctuations in the photon number? Or we can ask, what are fluctuations in the intensity when we measure the intensity of the electromagnetic field? So let's do that. The fluctuations-- It seems I've lost my lines. OK. So the fluctuations of the intensity are usually expressed by the second order temporal coherence function. That's what we want to introduce now. Yes, this is the second order temporal correlation or coherence function. What I'm always encountering in this course is I would just like to immediately tell you how it is defined in terms of a's and a daggers. It's simple. It's quantum mechanical. It's exact. But I always feel that if you want to really appreciate the quantum character, you have to know the classic description first. So I want to first tell you what is the second order coherence function for classical light, which has a classic description. And then you'll see what is the difference for quantum states of light. So the classical description is you measure the intensity of light. You're just sitting here. You receive light from a light bulb. You measure the intensity at time t. You measure the intensity at time tau later. And then you form the product. And you normalize it by the average intensity squared. So if tau equals 0, it's nothing else than the intensity squared average divided by the average of the intensity squared. So this is the classical definition, g2 of tau. And I've left the proof to the homework to show that the classical g2 of tau is always larger than 1. But it's pretty much what you show is for tau equal 0, it means I squared average is larger than I average squared, which is, of course, always the case. But you can show that this is also the case for finite tau. So quantum mechanically, we will see that the g2 function is not necessarily larger than 1, it can be smaller than 1. And that's actually an interesting-- you can see-- litmus test for the quantumness. If you generate states of the electromagnetic field-- Fock states, photon number states-- we see that in a moment. And they have a g2 of tau which is smaller than 1. You know it is not possible in any way to associate an intensity of the electromagnetic field with that photon state, because whenever you can associate a classical intensity with it and use a classical intensity to calculate the second order correlation function, you get something which is larger than 1. So often therefore again, when people want to show we really have now non-classical photon states, they show g2 of tau is smaller than 1. This is similar to what I said before when you want to show that you have non-classical light, you do quantum state homography, measure the Wigner distribution, and show that the Wigner distribution has negative quasi-probabilities. That's again something which is classically not possible. It's only possible if you have a truly non-classical state. AUDIENCE: So in the definition of g2 classical, the expectation values would need averaging over time, not over ensembles of I of t? PROFESSOR: OK. You can average over ensembles. So you can have 1,000 light bulbs, switch them all on, and then measure at a certain time z. But what we assume here is that the light bulb is on continuously. So things don't really depend on t. I've implicitly taken care of it by defining the g2 function only of tau and not of t and tau. You assume any time is the same, because it's a distribution, it's this ensemble in steady state. But it's sort of the same story again and again. In classical physics, you can determine an ensemble average by taking an ergotic system and observing it at many, many times. And then the idea is that one system as time goes by will sample all possible states. Or you can prepare many, many identical systems and do more of what is an ensemble average. So in other words, you would actually think, if you switch on a light bulb with a stable power supply, that the light emitted by the light bulb will go through all possible quantum states as time evolves. And therefore, the temporal average is equal to the ensemble average. So how do we generalize that to quantum mechanics? Well, one possibility would be that-- OK. First of all, for quantum mechanics, we have to use operators. And one choice would be that, well, you say the intensity of the electromagnetic wave is proportional to the electric field squared. And you want to use the operator for the electric field squared. Now, there is a little bit of a problem, because what we really mean quantum mechanically by g2 of tau is we measure the intensity now and a little bit later. But measuring the intensity really means absorbing photons, because the only way how you can measure the intensity of light is with a photomultiplier. It makes click, the photon is absorbed. And this is not fully described by the electric field. Just assume you have no photon, you are in the zero-point state of your harmonic oscillator, and your E squared has zero-point fluctuations. So what is more closely related to an experiment how you measure the correlation function is you want to look at something else, namely at the probability of absorbing 2 photons. So you start with an initial state. You annihilate 2 photons. And then you have a final state. But if you're only interested in what is the probability you want to characterize your initial state, you may want to sum over all final state. And this is your total probability that you can absorb 2 photons out of an initial state. But since you're in sum now over all final state, this turns into-- Now, I could put for you as in the final state is a complete basis. But I can also take it out. So this is what we get. So this suggests that experiments where we look at two subsequent clicks of a photomultiplier where we determine the photon correlation, that this is measuring a correlation function, which, for quantum states of light, should be defined as the expectation value of a dagger, a dagger, a, a. And now we have to normalize by the probability squared of absorbing 1 photon, which is a dagger a expectation value squared. I just tried to be a little bit too motivated. In many, many textbooks, the discussion would start with this expression. You ask, Where does it come from? And you realize yes, for some measurement with photomultipliers, that's what you measure. But I wanted to show you how it is related to the intensity correlation function defined classically. Cody. AUDIENCE: But this looks like it's constant in tau, which doesn't really make sense intuitively to me, because at least classically, it shouldn't be constant in tau. PROFESSOR: OK. That's the next thing I wanted to say is I've swept here-- I just wanted to give you the structure of the operators and not get distracted by discussing time. I've dropped the time argument here. But the fact is that as long as we limit ourselves to a single mode of the electromagnetic field, a single harmonic oscillator, things are independent of tau. In other words, g2 of tau equals g2 of 0, as long as we're dealing with single mode light. You can actually say-- and that sort of tells you where the fluctuations come-- if g2 of tau changes, it comes because you have several modes of the electromagnetic field, which, as a function of time, constructively and destructively interfere. But if you have a single mode, in a single mode-- and what is a single mode? It's just a sine wave. And nothing happens as a function of time. It's constant. AUDIENCE: So both modes go inside. PROFESSOR: I'm squeezing a textbook of quantum optics into two classes. And I want to give you the ideas and the concepts. What I've sort of mixed up here deliberately is I've given you the classic definition of the intensity correlation function, which is the famous Hanbury Brown Twiss experiment, and used correlations as a function of time. Then I've motivated how this should be defined for quantum states of light. But when I transitioned to quantum states of light, I decided to deal with only one mode of the light. We should now sum over-- I should put double or triple indices on all the alphas for polarization, for spatial modes, for different frequencies, and we sum over all of them. But instead what I did is I wanted to just show you the simple case-- and I think you will be thankful for that in your homework, that you only have to deal with the simple case-- that you only have to look at a's and a daggers in the operator algebra for single mode, for single harmonic oscillator. But what we lose, so to speak here, is there is nothing interesting going on in time. I've already told you that, for a single mode, all of these quasi-probabilities, they just rotate in a circle. So the time evolution of the system you're describing right now is completely boring. It's really a rotation. And if you would rotate your head at omega, nothing will happen. And this is exactly what you see here. So you find everything you want to for coherence time. Coherence time is the time for 2 modes to get out of phase. But if you have 1 mode, there is no coherence time. And when you find, for classical light, that the g2 function has a peak which decays with the time, it is the time for modes to get out of phase. But in a single mode picture, this is absent. Anyway, what is important now for the discussion of quantum character of light is really, we find that in a single mode picture. So I want to show you now-- or at least give you the summary of the results, which can be very easily derived, because the math is very simple of those operators, that we are now with that definition. We have g2 functions, which are no longer following the classic constraint that g2 has to be larger than 1. We find g2 functions which are smaller than 1. And this sort of tells us now where do we find the most non-classical behavior, namely when g2 of tau is as small as possible. OK. In your homework, you will show immediately that the g2 function is related to number fluctuations. It's related to an average and an n-squared average. Let me just write that down. It's independent of tau. And the reason is we have now limited ourselves to just one mode of the electromagnetic field. Fano factor. Just give me one-- OK So we are back to-- We started with intensity fluctuations. But for a single mode of the electromagnetic field, we are back to photon numbers. So what we are now expressing with the g2 function are, in other words, just fluctuations of photon numbers. I want to in a minute draw you a table. What are the fluctuations in the photon number for the states we have encountered-- the number state, the Fock state, the coherent state, and the thermal state? And we want to characterize those quantum states of light by the g2 function. And actually, I can drop to time argument or set it to 0. But then we first make a reference that g2 can now be smaller than 1. So g2 for a single mode light is nothing else than a function. When you know what is an average and n squared average, you know your g2 function. There is another quantity, which we often use to characterize the fluctuations in the photon number. And this is called the Fano factor. The Fano factor is-- wants to compare the fluctuations. n square average minus n average squared, these are the fluctuations. The classical fluctuation-- well, I say classical, classical in the simplest case-- are Poissonian fluctuations. So maybe we want to normalize the fluctuations by Poissonian fluctuation. So for Poissonian statistics, what I just wrote down would be 1. And well, if you now subtract 1, we have the situation that the final factor, which is positive, is super-Poissonian-- more fluctuation than Poissonian-- and the negative final factor is sub-Poissonian. So with those definitions, we can now compare the different states of light we have introduced. We started out with black-body radiation, thermal radiation. We defined coherent states. And the harmonic oscillator description naturally gave us harmonic oscillator eigenstates, the number of states, or Fock states. So there are three ways to characterize it. They are all useful. One is we can look at n squared. We can calculate the Fano factor. Or we can calculate g2 of 0. For the coherent state-- remember, the coherent state is as close as we can come quantum mechanically to the ideal of a pure electromagnetic wave. It has a Fano factor of 0. This means it's Poissonian. The g2 function is simply 1, which is the lowest classical limit. So those two tell you that a coherent state is sometimes what you think-- what comes out of a laser is an ideal electromagnetic wave, which has no temporal fluctuations in the intensity. Therefore, g2 is 1. And the photon number is Poissonian distributed. And that means that n squared is an average times 1 plus an average. We actually discussed it earlier. The thermal state is quite different. If you use-- kind of pluck together the results we have obtained, it has a Fano factor of n-bar. So this is super-Poissonian. If the occupation number n is large, you have fluctuations which are much, much larger than Poissonian fluctuation. The g2 function is 2. It's sometimes called thermal light. Chaotic light has a g2 function of 2. Laser light or coherent state has a g2 function of one. And n square average is n-bar 1 plus 2 n-bar. OK. But now finally, maybe the most interesting state from the perspective of non-classical light-- of quantum light-- is the photon number state. Well, for the photon number state, the number of photons is an eigenvalue. Therefore, n squared average is n average squared. The Fano factor is minus 1. Sub-Poissonian distribution. And the g2 function, which classically cannot go below 1, is now n minus 1 over n. It is smaller than 1. And you see immediately that the biggest violation for g2 is to go to minus 1 for the case of a single photon state. Any questions? AUDIENCE: Shouldn't it be 0 for [INAUDIBLE]?? PROFESSOR: The g2 function? No, if you put in-- wait. Yes. Gosh. I'll double-check. I'm a little bit confused. But I-- At least it's consistent now. I'm not sure if it's right, but it's consistent. By the way, if there is a question, I sometimes make a question mark in my notes. And when I post the notes-- maybe not when I post the notes on the next day, but a few days later-- the question marks are eliminated. Just as an example, last class, there was some relation for which I didn't remember how the derivation was done. When I post those notes today, the derivation is now in the notes. I'm not announcing it separately. But whenever I have a question mark in class and I don't think it's worth an extra announcement, it is fixed in the posted notes and you find the information there. OK. So this is now drawing our attention to the single photon. And this is our next subsection. This shouldn't come unexpected. If you want to emphasize the difference between quantum light and classic light, where does it come from? Well, it comes from the quantization of light. It comes from the photon character. It comes from the effect that light is not a continuous stream of energy, it comes quantized in photons. So the granularity of light due to the photon character is, of course, most pronounced for a single photon. For instance, when we define the g2 function as a correlation of detecting a photon and detecting another photon, well, if you've only one photon, you find one photon, and for the next photon, there is no photon to be detected. So the probability of detecting two photons is 0. And that only happens when you go down to similar photons. So this is when certain fluctuations are most pronounced, because the energy is dependant on a singular photon. Let me first address one misconception. You can say, Well, let's just use a coherent state. And we talk about the attenuator-- the quantum attenuator with all its operator beauty-- probably not today but it the next lecture. But let me already sort of prepare you for that. When you have a coherent state, when you have a laser beam, you can put an attenuator. And your laser beam gets weaker and weaker and weaker. But it stays a coherent state. And I will prove that to you very soon. So you can now say that you take your coherent state and you attenuate it down that there is only one photon left. Is that a single photon state? The answer is no. It is an attenuated coherent state. Coherent states, as I've just shown you, are very classical. They've always Poissonian distribution. They've always a g2 function of 1. And attenuation is not changing it. Attenuation is preserving that. So now I want to show you explicitly why an attenuated coherent state may have an average photon number of 1 but it shares nothing else with a quantum state of 1 photon of n equals 1 Fock state. So a coherent state with an expectation value of 1 photon is not a single photon. And this can be, for instance, expressed by looking what is actually the probability, if you have such a coherent state, to find 1, 2, or 3 photons. Well, the probability to find 0 photons-- no photon at all-- is actually 1/3. So the probability to find 1 photon or to find 0 photons is the same. The probability to find 2 photons is 0.18. So here you have a probability of finding a photon and correlating it with the next photon click. And you have even, on first sight, a surprisingly large probability to find 3 and 4 photons. So in 2 percent of the cases, you will find 4 photons. Whereas in contrast, the Fock state with quantum number n equals 1 is an eigenstate of the number operator beta with eigenvalue n equals 1. So that tells you, if you want to get n equals 1, if you want to get a single photon state, you cannot just use a strong laser beam or a strong light source and attenuate it. You have to work with something which genuinely creates only 1 photon without any ambiguity, without any fluctuations, without any possibility of creating 2 photons. And I would actually say over the last 10 or 15 years, the creation of single photons has been sort of a small cottage industry, because single photons are often needed for protocols in quantum computation-- for experiments which really require accurate quantum state preparation of light-- and in particular, non-classical light. But of course, there are ways how you can get single photons. And this is, well, you start with single atoms. If you have a single atom in the excited state, it can emit only one photon. So in other words, you cannot-- Usually, we don't have the tools to prepare a single photon-- to take a single photon out of many, many photons-- and store it separately. But what we can control is single atoms. We can prepare single atoms. And then we can make sure that single atoms create single photons. It's a little bit a way that we cannot control the bullets which are fired. But we can control the guns. And we make sure that each gun can emit exactly one bullet. So that's a way how we can create non-classical states of light. Yes. So let's now look for those single photons at the quasi-probability distribution. So we obtain the quasi-probability distribution by taking the single photon state and projecting it on a coherent state. You remember-- it's higher up in the notes-- what the coherent state alpha is in the Fock state basis, in a number basis. And so we're just picking out n equals 1. And this was nothing else than alpha squared times e to the minus alpha squared. So therefore, in these diagrams with the real part and imaginary part, if I plot the quasi-probability distribution, we get a ring. The ring immediately tells us that there is no phase defined. All the phases are equally probable. And that also means, if you don't have a phase, the average value of the electric field this 0. Since the equation of single photons is essential for studying non-classical light, but also since it's a very active frontier of our field-- actually, one of the leaders in this field is Professor Vladan Vuletic here at MIT-- let me at least give you a taste of how to create single photons. I already gave you the major ingredient. Namely, it's about-- it involves single atoms. But it's a little bit more demanding like that. So you want to take 1 atom home or 1 ion. But the problem is if the atom or ion emits a light, it can emit the light into all directions. And therefore, you have a single photon afterwards. But you have many, many different spacial modes. And in any given mode, it will not have a single photon. So therefore, what you have to add to the single atom or single ion is-- you have to put it in a cavity. And then, in the limit of a very, very high finesse cavity, the probability will be very, very high that your photon is emitted into the mode of the cavity and it is not emitted perpendicular into other modes. But that itself is not yet sufficient, because you have a single photon, but you also want to know when does the single photon arrive. You want to do experiments. You want single photons on demand. And so one option is that you prepare your atoms in the ground state. You take a pi pulse, which, with 100 percent probability, excites the atom to the excited state. And then within an actual lifetime, or maybe even in resonator-enhanced inverse natural line widths, the single photon is emitted. So therefore, you pump the system and then you know within the next few nanoseconds your cavity mode will have a single photon. The problem here is that you're dealing-- You have to prepare single atoms, which is difficult. You have to couple them to a high finesse cavity. There is technically another approach, where you use many atoms. So you could say that many atoms, you no longer have a single gun. So therefore, you can get several photons. But the singleness in photons comes because the photons are now heralded-- they are announced. And the idea is the following. If you start with the state 1, and you have an excited state, and you have a pump pulse. Now, if you have many atoms, you have a much higher efficiency of pumping atoms comes to the excited state, because if you have n atoms, it's n times more efficient. But now you take the following situation. You wait until there is a Raman transition-- until you detect the photon for Raman transition, where the excited state decays to the state 2. At this moment, you know I have now 1 atom in state 2. So in other words, you're not starting with single atoms, which is sometimes more demanding. How can you prepare the system? It has other disadvantages that n atoms have. I'm not sure if I should mention it, but with n atoms, they have a super-radiant factor n with n atoms. You can get an n-times enhancement of emission into a single mode. So there are real, massive advantages in working with many atoms. But now you know that one atom is in state 2 when you detect the first photon. And then you have the same situation which we had above. You can now take your system-- let me just redraw the level diagram. Excited state, state 1, state 2. You know now that you have 1 single atom here. And by using a laser pulse, you can excite it to the excited state. And then you observe the single photon 2. So in other words, the observation off the first photon tells you that your system is prepared with 1 atom in state 2. And then you can get a single photon out of it, which is the photon for the inverse Raman transition when you pump the system back to state 1. AUDIENCE: Doesn't this rely on you being able to reliably prepare a single atom in another quantum state? Won't there always be some uncertainty as to how many atoms you can prepare in one sitting? PROFESSOR: There are some uncertainties. And what you're saying is correct. We don't have perfect single photon sources. And people characterize the fidelity of the single photon source. For instance, if you detect the single photon, you would say, Now my system is ready to emit a single photon triggered by this pump pulse. And if you can get now a single photon in 90 percent of the cases, you publish a wonderful paper, because you've set a new record for the fidelity of a single photon source. So people are really struggling with some of those uncertainties. But to involve, for instance, three levels is sort of an advantage, because you're not limited by the preparation of a single atom, for instance. You can have many atoms. The atoms are always there. And the moment one atom is prepared in state 2, this atom announces itself with a single photon. So it takes a lot of uncertainty out that the system says, with the first photon, I'm ready now. I can emit a single photon. And then you get your single photon. Then you can gate your whole experiment to a time following the detection of the first photon. And for your gated time afterwards, you have a very, very high probability of finding this photon. Or if you want, you can now do experiments with 2 photons. If you keep this laser on and pump the atom immediately back, you have actually a single photon here, followed by a single photon here. It's sort of click clack. And now you can do correlations between two different single photon states and such. It's a very rich frontier of our field. You have another question? AUDIENCE: Yeah. Is this also within a cavity now, like the single photon [? reflection ?] in the first test? PROFESSOR: Yes. Actually, these photons-- the wavy lines-- are emitted spontaneously. Spontaneous emission, pretty much by definition, goes into all possible spatial modes. And the only way to control the spatial mode is with a cavity. What I don't want to go into details-- but those who have an understanding of that-- is the following. If you n atoms and you prepare one atom here in state 2, you do not know which of your n atoms is prepared. So you have n indistinguishable possibilities. And if you have n indistinguishable probabilities which atom you have prepared, the emission of the photon back is n times enhanced. So therefore, you have actually a system which has an n times stronger coupling to the cavity. So having n atoms makes it much, much easier to construct a high finesse cavity. You get this super radiance increase of the strong coupling for free. And this is why there are many reasons why you do not want to work with the single ion or a single atom. It's possible. For very fundamental experiments, people have shown that it can be done. But technologically, it's much easier and much more robust to work with many atoms. But then you need the many atoms have to tell you when 1 atom is prepared to emit now a single photon. If you're interested in this subject, talk to the students in [INAUDIBLE] school. They're really the world experts in that. OK. Finally, we're not really getting to squeeze light. We start with squeezing the light tomorrow. The last thing in this major section is the famous Hanbury Brown Twiss experiment. This was a landmark experiment done in the 1950s. And it was the first experiment which really looked at g2 functions correlations, which one could say was the beginning of quantum optics and modern experiments with light. Probably until then, light was just an electromagnetic wave people regarded as boring. But by measuring now correlations, people realize that it's an interesting object to study. I want to just go over the basic scheme. You will look at it a little bit more closely in your homework assignment. So the idea behind an Hanbury Brown Twiss experiment is that you have a light source and you want to characterize it. So the light source emits light. And now, this is important. Whenever you want to detect 2 photons, you want to figure out it's 1 photon followed by another photon. Technically, you cannot do it with a photomultiplier, because when a photomultiplier clicks, it needs many, many nanoseconds for the photomultiplier to read out the signal and recover to recharge up its electrodes and be ready to detect the next photon. So therefore, when you want to find click click-- double clicks in the stream of light-- you have to involve a beam splitter and involve two photodetectors. Sure, you now need an adequate description. But in principle, you can now find 2 photons which are only a few picoseconds apart, because the first photon is observed by the first detector, and the second photon is detected by the second detector. Sure, you would say, What happens if you're in the beam splitter, both photons go to detector one? Well, that's tough luck. You take 1 chance in 2 that the 2 photons go to the 2 different detectors. But the beam splitter has to be part of your experiment, has to become part of your description. And this is what you do in homework number one. So what we are asking here when we measure the g2 function. Let's assume that tau equals 0. We are reading out a signal. And what we are using is a coincidence detector. We are looking that in a very small temporal window, 2 photons are detected simultaneously. Well, this is the quantum version. The classical version is the g2 function is the product of the intensity at time t times the product of the intensity at time t plus tau. So what you would do here in this circuit, you would take the signal from this detector, the signal from that, and multiply the two. This is how you determine I of t times I of t plus tau. So whether you do it in the classical domain or whether you do it in the quantum domain, this is the way how you experimentally measure the second order correlation function. In the classic limit, you don't have to worry about it. The classical limit is always a limit of high intensity. So at any given time, you have a ton of photons. And then, at beam splitter, half of the photons go left, half of the photons go right. You have an equal splitting. You have exactly half the intensity in the left arm and the right arm after the beam splitter. So there is no problem at all with the quantization, because this is the classical limit. So classically you can say-- actually, I shouldn't say photon, I should actually say intensity. The intensity splits equally. And if you do the measurement with the coincidence detector, you find the difference between the g2 function, which I discussed earlier. That the g2 function is 1 for coherent light or laser light, and the g2 function is 2 for thermal light. That's actually the only way how you can distinguish a light bulb from a laser beam. If you put a light bulb into a cavity or couple the light from a light bulb into a fiber, the light becomes spatially a single mode. If you put a very narrow spectral filter or Fabry-Perot cavity, the light becomes single mode in frequency domain. So when you take a light bulb and spatially and spectrally filter it that it is single mode, in terms of the mode, it is single mode as a single mode laser. But it is the correlation experimental which shows you that you started with a thermal source. It has a g2 function of 2, and you can never get rid of it, whereas the laser beam has a g2 function of 1. OK. So this is the classical version. In the quantum version, especially when we have a single photon, the photon can go to only one detector. And that means, for this extreme case of a single photon, the g2 function is 0. Anyway, you will look at those situations in more detail in you homework assignment. Yes? AUDIENCE: Sir, if you know that different for tau non-zero, then even the single mode in the function would have a final g2 [INAUDIBLE],, because there would be some probability [INAUDIBLE].. PROFESSOR: We have to be now a little bit careful. If you have a single mode Fock state, there's only 1 photon. And once this photon is detected, we have vacuum, we have no photon left. So what may be confusing here is, on first sight, is that the way how the experiment is done and has been historically done, it's not done we Fock states. It's done with the light source in a continuous stream of photons. So Hanbury Brown Twiss experiment, he looked at in light bulb, or he looked at star light and such, and determined the intensity correlation. He couldn't do the experiment with lasers, but now we would with a laser beam. And we find that g2 is 1. But in those situations, we actually do not have the way-- we have actually a beam, which is a stream of photons. And this requires a little bit different description. In other words, when I have a coherent state, a laser beam is always replenishing the coherent state. The laser beam preserves the electric field, whereas, strictly speaking, the way how I described it for pedagogical reasons is you have a cavity, which is filled with a coherent state alpha. And then you start analyzing that. But in other words, what we have focused in the simple description is that we have a quantum state which is prepared-- it's a closed system. And now we do our detection. The experiment how it is done is often done as in an open system, where you couple the system to a light source, which is always replenishing your experiment. So to be careful, especially with a single photon. I think, in essence, the experiment would be done with a single photon light source. But the single photon light source has a high repetition rate. I showed you how to generate single photons. You have an heralded single photon. You know the single photon comes now. You do the experiment. And then you repeat it again. And then what you have is you only look at one single photon bunch at a time. And then you find indeed, that during that temporal window, you will never find a second photon. So this may happen in a few nanoseconds. Then you wait a microsecond, and the next photon arrives. Of course, if you would now describe your light source, what you is the g2 of tau, and tau is a microsecond, and the microsecond is the time between the first single photon burst and the next single photon burst, now you will find that your g2 function is not 0. But this is then related to the repetition rate of a single photon and not to the single photon itself. I think you've got the taste. Your homework is a really simple. You just deal with a closed system. You look at a quantum state at a time. But if you map it onto a beam experiment, you have to think about what does it mean to have a replenishment of the quantum state. Any question? I know I have to stop, but. AUDIENCE: In the beam splitter, the other port of the beam splitter should do some stuff, right? PROFESSOR: Yes. Actually, this will keep us busy for a while, that even if you do not put in any light, we put in the vacuum state. And you will find that the description of the beam splitter in the quantum state is incomplete, unless you specify that your beam splitter is coupling in from a different mode the vacuum state. Yes, this will be very important. And we will cover it in its full glory. AUDIENCE: I just have a quick question. The way we can distinguish from a thermal state and a coherent state is through the g2. So from an experimental perspective, could we do most of experiments if we just had thermal sources that were very single mode, so to speak? PROFESSOR: Well, yes. All the laser cooling, all the [INAUDIBLE],, all the absorption imaging-- all that would work if you had a single mode thermal source. Let's face it, we say we use lasers for everything we're doing. But the only property which distinguishes a laser from the thermal light source-- we're not taking advantage of it. Of course practically, if you take a thermal source and filter it down to a single mode, you will be left with only a few photons. You cannot create an intense enough single mode light source unless you use stimulated emission into a single mode. And that's a laser. Colin. AUDIENCE: The correlation function for, for example, an LED light-- that's not really a thermal source. It's not described by a Maxwell-Boltzmann distribution. And that would be closer to something for a laser, even though there is no cavity? PROFESSOR: Well, LEDs in some limit are quantum objects. Actually, do you know what the g2 function of LED is? Does it become laser-like, or does it become even antibunched? Because if it's a relaxation mechanism-- So anyway, what Colin says is there are actually more different light sources that just the laser and the thermal light source. There are LEDs or semiconductor devices, which provide photons with interesting statistical properties. I've heard about it-- that LED light sources have some special properties, but I don't remember which was the one. OK. We have to stop. I'll see you on Wednesday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

8_g2_for_atoms_and_light.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: We're going to continue our discussion of correlation functions, and in particular, g2 of 0. If you think I'm focusing a lot on one quantity, g2 of 0, that's correct. But I have to tell you, I have been involved in so many discussions with people who looked at g2 of 0 with both Einstein condensates when Bose-Einstein condensates in [INAUDIBLE] were discovered here at MIT by [INAUDIBLE] and collaborators. There was some mystery about, in essence, the g2 of 0 function. So a lot of physics actually goes into it. And I've learned a lot from my own research and through discussions. And in this unit, I want to summarize it. So the confusion sometimes comes because people use classical pictures, quantum pictures, and they may not see the relation. So let's continue the discussion. I want to go through four different views of the same physics that you will recognize that they're interrelated. The first one, just to repeat that, is if you have light which is Gaussian intensity fluctuations, light from random sources. I think if you would look at the twinkle of starlight, if you look at the light from a light bulb, because of the central limit theorem, you will find that the intensity that the electric field has a Gaussian distribution, and the intensity in exponential distribution. And if you then ask, what is g2 of 0, you'll find it's 2. Because for an exponential distribution, the average of the intensity squared is two times the intensity, the average of the-- what did I say. This average of the squared intensity is two times the average of the intensity squared. Before I forget it, one student asked me after class, I'm discussing g2 of 0, but what about g2 of tau? Well, my understanding is that almost all of the physics is in g2 of 0. And there is a simple number, 1 or 2, in those special cases. What happens is, if you have completely independent sources, if you completely independent statistics, then the probability of finding a second particle, a second photon, is independent of the first. So it means g2 of tau would be unity. This is the case for an uncoordinated system. But if you have g2 of 0 which is higher, this means correlation. This means some form of coherence. And usually, the correlated value simply decays to the uncorrelated value with a characteristic coherence time. And this characteristic coherence time can be inhomogeneous, technical. It can be limited by the Fourier transform of a light pulse, the monochromaticity, the bend which delta omega is 1 over tau. So it has all of the usual suspects for what limits the coherence time. So therefore in general, we're not learning something fundamental in looking at g2 of tau. We're just learning the usual spiel about correlation times and coherence times. The second angle I gave you with the brief interference is something which is deeply insightful. It shows you the following. If you have a plane wave, the intensity is constant. If you have two plane waves which interfere, you create density fluctuations, because the interference pattern can be constructive and can be destructive. So therefore, if the intensity fluctuates, then i square average is no longer i average square. And what happens is because of the nature of interference, because the average of cosine square kx is-- well, there are factors of 1/2 and 2 which just come out of cosine and sine square function. You get exactly g2 of 2. But that should already tell you something, which will be important for both quantum particles, and light, and the classic limit. The moment if you have light or particles in one single mode, you can't have any form of interference. Only two modes can interfere. And you will always get a g2 function of 1. But we'll come to that in a second. Any questions at that point? Yes. AUDIENCE: I thought that when we talked about the g2 function several weeks ago, you said that if we had a single mode thermal light, it would have the g2 function of 2. And that's therefore, all we were using for lasers when we used them in experiments is the fact that it's single mode, not the coherence in the g2 function. PROFESSOR: You have a good-- that's a good point, Cory. Let's just try to connect the two pictures. Slowly, slowly. What did we discuss? This question is great for me to repeat something. When we looked at black body radiation, I wrote down the partition function. And I said I'm only considering a single mode. And then, we have of course in thermal equilibrium a Boltzmann type, or Bose-Einstein type, distribution of finding certain numbers, number of photons populating the single mode. And that resulted in a g2 function of 2. AUDIENCE: It's not really single mode though, right? Because you're summing over many frequencies. PROFESSOR: No, it was single frequency. I have to think about it. That's a really good question. Right now, I would quickly say, in thermal equilibrium-- I'll give you partial answer, and when I think about it, I'll give you full answer on Friday at the beginning of the next class-- I think would be calculated where the thermal fluctuations in the intensity. And the intensity fluctuations, and those intensity fluctuations. Those intensity fluctuations result in a g2 function of 2. That's correct. Maybe what I want you to say here is, if you have two modes which have all maybe single occupation or constant occupation, and they interfere, then they create a spatial interference pattern if the phase between the two modes is random, which gives rise to additional fluctuation. So most likely, the two fluctuations are independent. One is the fluctuations because of the interference of two modes. But when I say the single mode has a g2 function of 1, I meant actually a single mode with constant amplitude. I'll double check and give you a more complete answer on Friday. Other questions? OK. The third view is classical versus quantum statistics. And I think this really shows that we need at least some quantumness to find correlations. Let's assume we have n particles. And we have N possible states. And if it's a classical system, let's assume we want to find one particle in a certain state. We have a big box and want to-- we have a big box, have a small sub-volume, and say what is the probability of finding one particle here? So this probability to find one particle is P1 is small n over big N. The probability to find two particles. If you have non-interacting classical particles, well, it's simply P1 squared. Independent classical particles, they don't care what they are doing. You grab into a sub-volume, which is maybe phase space cell. And therefore, your quantum state, you have probability P of finding one. But since each particle moves around independently, the probability to find two particles is just P squared. It's a little bit like if you toss two coins, one with your left and one with your right, and you ask what is the probability to find a head or tail, and you simply multiply the probabilities. But now, following reasonings which Bose and Einstein introduced, you want to use counting statistics, which takes into account the indistinguishability of particles. So if we go from the classical distribution to the distribution of indistinguishable particles, then the classical probability P 1 is reduced by n factorial, because we are not-- let me put it this way. The number-- I don't want to go through a mathematical argument. If you grab one particle, it can be the probability to find one particle is represented by a microconfiguration classically. And each configuration when you permute particles counts as independent. But if you have quantum indistinguishability, you're not counting permutations as an independent configuration. And therefore, you have a reduction by n factorial. If you look at the probability of finding two particles, the reduction is-- this is just the counting statistics n factorial by 2. So therefore, you will find that classically, that quantum mechanically for bosons, is two times. The probability to find two bosons in one quantum state is two times the probability squared to find one boson. So this is just counting statistics. This can be applied to conditions in a Bose-Einstein in a Bose gas. When you have inelastic collisions, spin relaxation, dipolar collisions, two body collisions, two-body collisions have a rate gamma 2, which is proportional to the probability of finding two particles at the same time at the same location. And that brings in the g2 function. Whereas three-body collisions, gamma 3, reflects the third order correlation function g3, which is defined in an analogous way. So if you now compare at the same density the two-body rate coefficient, like the same density means we are looking at the rate of two-body collisions. And we compare between two-body collision in a thermal cloud, which is a g2 factor of two, a thermal cloud of bosons. Whereas a Bose-Einstein condensate is a constant occupation of particles in one mode and has therefore a g2 factor of 1, exactly as [INAUDIBLE] of the Bose-Einstein condensate is for particles for meta waves, but the laser is for light. Whereas, if you look at three-body collisions, those who do experiments with Bose-Einstein condensates will know that usually the lifetime of Bose-Einstein condensates is limited by three-body collisions. Well, I didn't derive it, but actually yes, we have the Gaussian statistics for randomness. For Gaussian statistics, the n-body collision function has a factor of n factorial. So therefore, three-body collisions scale with 3 factorial, and two-body collisions with 2 factorial. So in other words, you have a thermal cloud, you have a Bose-Einstein condensate at the same density. But what matters for two-body collision is not the average density. It is the average of n square. And the fluctuations in the thermal cloud because of the g2 function are two times enhanced compared to Bose-Einstein condensate. And therefore, you find that at the same density, you have more loss. You have a higher rate of two and three-body collisions. Some of this was not-- became better understood soon after Bose-Einstein condensates were realized. The Boulder group studied three-body collisions. And for two-body collisions, myself in the post-doc clarified the situation that even the mean field energy, which everybody had measured before, requires two particles to interact with short range interaction is therefore proportional to g2. And therefore, when people had determined the mean field energy of a condensate without really knowing they had already determined the g2 function of Bose-Einstein condensates. OK, so this is pretty much the counting statistic which you do in undergraduate class when you derive the three different statistics-- Boltzmann statistics, Bose-Einstein statistics, Fermi statistics. And it gives you the result of that is also for bosons, you have the factor of 2. For fermions, of course, you get 0. The average between bosons 2 and fermions 0 is 1. And this is the classical statistics. Any questions? Then let me finish this discussion with a quantum mechanical description. If you describe the detection or collision of two particles in two modes. So you have mode. A mode operator, annihilation operator a1 you and a2. But if you are asking what happens if I measure two particles, you annihilate two particles with an operator a1 and a2. But then, because of the indistinguishability of particles, you have to consider an exchange term, which is a2a1. So whenever you detect two particles in two different modes, your signal is proportional to something related to a1a2 plus a2a1. This extra exchange term gives you an extra factor of 2 for bosons, which is exactly what appears in the g2 function. And, of course, if you have an exchange term for fermions, you get 0. And this, of course, leads to the antibunching, or the g2 value of 0 for fermions. However, and this is obvious, if you have a single mode, then the only operator to detect two particles is a1 times a1. And this has no exchange term. And this is the situation of the laser and the Bose-Einstein condensate. And that means they have a g2 function of 1. So Cory, this is another argument why single mode occupation has a g2 function of 1. But I have to reconcile it with your question. Right now, I think there may be difference between a canonical ensemble and grand canonical ensemble. In a canonical ensemble, we have [INAUDIBLE] the total particle number is fixed. But in a grand canonical it fluctuates. And maybe this leads to additional fluctuations for the case of a single mode chaotic light. But I will have to think about it, your question. But here is the same argument why if you occupy only a single mode-- if you look, for instance, at a Bose-Einstein condensate, you do not have the second exchange term. And the missing exchange term just propagates through the equations and gives you the g2 factor, the g2 function of 1. OK so let me sort of wrap it up. I've given you different angles at the g2 function. Some are simply classical intensity distribution. Some is simply interference. Another aspect is Poissonian counting statistics. My understanding and my interpretation is they look as different as they can possibly look, but they reflect the same physics. Because interference of to light only happens because light consists of photons. And photons are bosons. So the classical interference and the quantum mechanical counting of bosons lead not for random reasons, but lead for profound reasons to the same result. And chaotic light, which seems to be determined by just random fluctuation, well, the random fluctuations if you have single mode light, a thermal distribution of single mode photons, comes from random phases. And these random phases lead to random interference. And so we are back to interference. Finally, let me give you my view on the measurements of g2. When you look through the literature, you find the famous Henry [INAUDIBLE] experiment. In one of your homework assignments, you looked at a seminal experiment where people dropped atoms out of cold atom clouds and measured the g2 function for cold atoms. Sometimes it's confusing when you directly compare the two experiments. But let me try to give you a common description, or the common denominator, between all these experiments. You can say that all experiments to measure the two particle correlation function is about comparing the probability of finding two particles with a probability of finding one particle. And well, you restrict your measurement, your detection of particles, to either one quantum state, one mode of the electromagnetic field, one coherence length, or coherence volume, of light. Or if you use the semi-classical argument for the description of particles, you take one phase space cell. This is one phase space cell, one mode, one coherence volume, one quantum state, is sort of the different definitions of what a quantum state is, or what a wave packet is with Heisenberg limited uncertainty. You cannot define a particle in phase space to within better than the phase space volume of a quantum state, which is h square. Or is it h bar? I don't think there is a bar-- h square. And so this is how you can relate wave packets to quantum states to phase space volume. It's pretty much mathematically I could say in different basis sets, in different basis sets, it is one quantum state. But you can use wave packets. You can use time dependent description. But what I said here-- one quantum state, one mode, one coherence volume-- this is as well as you can define a particle or photon to be by fundamental reasons by Heisenberg's uncertainty relation. AUDIENCE: What do you mean by one coherence volume? PROFESSOR: What? AUDIENCE: What do we mean by one coherence volume? PROFESSOR: Well, if you have a laser beam which propagates, and it is TEM00, you would say the transverse coherence volume is the size of the laser beam but then the longitudinal coherence volume is given by the coherence length. Just envision you have a laser beam here. And the coherence volume where all of the photons are coherent is the area of the laser beam times the coherence length. And whenever you find a photon in this kind of volume, then it is in a Heisenberg Uncertainty limited or Fourier-limited state. It is in, if it's a wave packet and pulse laser, it's in a time dependent quantum state. But it is as coherent. It is fully coherent. So you have to figure out for your system what is this minimum-- what is the coherence volume, what defines the fundamental phase space cell of your system? And now you do the following. You are asking, what is the probability to find two particles? And what is the probability to find one particle? So if p is the probability to find one particle, then you have three options. p square, 2 p square or 0 to find two particles. One is the classical case of distinguishable particle. This is the case of bosons, bosonic atoms or photons. Collins Therefore includes the Hanbury Brown-Twiss experiment. And 0 is fermions. And the question is now different experiments, how do you define, with spatial resolution or temporal resolution, the coherence volume? And that can involve transverse coordination, temporal resolution, and whatever you can use to define the phase space cell or the coherence links, the coherence volume of a laser beam. Let me give you one example, which I think illustrates it. And this is the example of an atom cloud. If you have an atom cloud, all the particles which are in a volume of a thermal de Broglie wavelengths are coherent. The momentum uncertainty of particles in a thermal cloud is 1 over the thermal-- is the thermal momentum. And according to Heisenberg, the position uncertainty related with this momentum spread is just the thermal de Broglie wavelengths. So you can say that all the atoms in a cubic de Broglie wavelengths are coherent. All the atoms are in one semi-classical quantum state. So therefore, this is sort of atoms in one single mode. So the Hanbury Brown-Twiss experiment with atoms, or the measurement of the g2 function, could be defined as follows. You have an atom cloud. And if you had an electron microscope or some high resolution device, you grab into your cloud and ask, what is the probability for one particle. What is the probability for two particles? And you will find that p2 is 2 times p1 squared. If your volume is too big, you lose the factor of 2, because you average over uncorrelated volumes. Now, in your homework, you were looking at the question, how can I really grab into a cloud and just pick out atoms out of one phase space cell of effective size, lambda de Broglie cubed? And the way how it was done is that you take an atom cloud and you drop it and expand it. When the cloud expands, there is a mapping from momentum space into position space. Then you use some form of pinhole, which provides transverse coordination. You use a detection laser, which gives you temporal resolution. And well, this was part of your homework, but I just wanted to give you the bird's view on it. By controlling the transverse correlation and the temporal resolution of the detection, you create a situation that what you count only atoms which originated from one phase space cell in your cloud. So in other words, let's say experiment where without electron microscope, without submicron spatial resolution, you can literally grab into a cloud, capture a volume of the thermal de Broglie wavelengths, open your hand, and figure out how is the probability for two particles related to the probability of finding one particle. Any questions? All right. OK, good. So it's time for new chapter. So in this chapter, we want to look at interactions between light and atoms using Feynman diagrams. And I know this part of the course is a little bit formal. We are using exact solutions for the time evolution operator in quantum physics. But I'm not doing it to teach you sophisticated formalism. Well, it's interesting to learn it anyway. But it is this picture of really writing down the exact solution of the time evolution operator, which helped me to understand much better what virtual states are, what certain virtual photons are. So there are things we talk about it all the time. And the question is, how do you define a virtual state? What is the virtual emission of a photon? And the only way how I can really explain it to you is by showing you the equation and say look, the virtual photon is just a term in this equation. So that's the goal of this chapter. I don't want to overemphasize the mathematical rigor, but I want to really show you what it means to have virtual photons and what exactly virtual states are. So let me first motivate that by reminding you of two diagrams. In physics, we always like to draw something-- a few lines, a few doodles. And when we have a two level system, we have the two processes which are emission and absorption. Well, so this is easy as long as the light is in resonance. But if the light is not in resonance, we may ask what about this process? And since somehow the weekly line of the photon ends here, I may even put in a dashed line and say, this is a virtual state. What does it mean? Or we can even ask, is it possible-- I'm just playing with straight lines which are quantum state and wiggly lines-- is it possible that this process happens. That would actually mean that an atom in the ground state, just out of the blue, emits a photon. Can that happen? And I can again draw a dashed line and say, here's a virtual state. Or, we can say we start in the excited state. The atom is in the excited state. But can the atom-- its a two level system, so we don't include higher states-- but can a two level atom in the excited state absorb another photon? So let me just change the color of these photons to red, because I want to-- and this would remain black. All right. Good. So question is, are those processes possible? Yes or no? AUDIENCE: Yes. PROFESSOR: Yes, they are. And they have experimental, observable consequences. However, they look funny because something seems to be strange with energy conservation. And what we will see is that in the end, at the end of the day, that means if you let the system evolve for a long time, after a long time, energy has to be conserved. And therefore, we will always need a second photon to conserve energy. So those weirder processes, where photons are emitted by ground states out of the blue or excited states absorb other photons without-- also, there is no higher lying state. That looks a little bit strange. So what you will see is-- and I want to show you that mathematically-- if an atom emits a photon out of the ground state, you would say, where does the energy come from. But quantum mechanic allows us to violate energy conservation by an amount delta e for a time which is h bar over delta e. So for short movement, energy can be violated. But then, you need the second photon to reconcile energy conservation. You can violate energy only for short times. And if you would say, what does it really mean to violate energy conservation for short times? Well, I want to show you the equation, which exactly explains what it means. Next question, just to see. If you have an excited state, and you just say it's possible to absorb another atom, you are in the excited state, and you absorb another photon. And at least you said, yes this is possible. My question is, when you are in the excited state and absorb a photon, in which quantum state is the atom after absorbing a photon? AUDIENCE: Ground state. PROFESSOR: Pardon? AUDIENCE: Ground state. PROFESSOR: In the ground state. Yes. There is no other state. And the operator-- and I want to show you that-- when photons are exchanged, always transforms the ground to the excited state, because a dipole operator connects the ground to the excited state. So therefore-- but we'll see that all in the time evolution operator in the formal solution that this dash line is actually the ground state. And while it's not obvious, what is this dash line? AUDIENCE: The excited state. PROFESSOR: It is the excited state. The atom in the ground state emits a photon. Therefore, the atomic system has now lower energy, because a photon has been emitted. But the character of this state is now the excited state. So we will later see that this means that in a perturbation analysis, we are violating energy. We are violating energy, because the real excited state is omega photon plus omega atom, the resonant energy for the atom. The real excited state is omega photon plus omega atom higher. So you will actually find that in a perturbation analysis, this term has an energy denominator, which is of resonant by exactly this separation. And this term, because the real count state is here, the virtual state here which is the count state is here, has actually the same energy denominator as that state. And for this state, of course, this is something you've seen of resonant light scattering. The energy [? denominator ?] is there. So, OK. So I'll explain to you that we involve virtual states. So when we have virtual states, we want to clarify what is their energy. But we discussed that already in the discussion. And what is the wave function, ground or excited state. Yes. So just to repeat it, when we draw those diagrams, those diagrams show the energy of the atom, but taking into account the energy of the photon. In other words, if you have a ground state, and we absorb a photon, the photon has disappeared, and the energy of the atom is now the ground state and the photon energy. And this is a dashed line. Similarly, when an atom emits a photon in the ground state, the photon is emitted. Therefore, the atomic energy is lower than the ground state by the photon energy. And we draw the dashed line here. This is what we mean when we say that is the energy of the virtual state. And this is what we draw in those diagrams. By the way, yesterday we had the CUA seminar. And the speaker was actually talking about trapping atoms with quantum fluctuations. And he conceded explicitly Casimir forces, forces of the vacuum. And diagrammatically, forces of the vacuum come, because the ground state atom emits a virtual photon and reabsorbs. This is actually, I will mention it later, the same diagram is actually also the diagram which leads to the Lamb shift. These are all sort of when you have atoms in the lowest state, and they interact with the electromagnetic field. Well, if they interact with the vacuum, they cannot absorb photon. All we can do is emit a photon, and this leads to this diagram. But just to tell you that you have to be careful, yesterday's speakers actually used this process. But he drew this diagram. He drew it in the opposite way, which I think confused some people in the audience. The correct way is to draw it like this. The virtual state for vacuum fluctuations for the Lamb shift for the Casimir force is below the current state. OK, so we have-- let me just redraw those three diagrams. I want to now introduce the time evolution. So I want to use those. Seems I'm running out of space. So we want to introduce the time axis for-- so we had this diagram. We had this diagram. And we had-- this is a virtual state. OK. So those three diagrams correspond to the following situation. If time evolves from the bottom to the top, we can now draw the first process when atom in the ground state first absorbs and then emits a photon. And let's label the second photon in the purple color. So an atom starts out in the ground state. Then it reaches the time where it interacts the electromagnetic field. There is a photon which propagates from earlier times to this time, t equals t prime. At the time t prime, the photon disappears. And the atom, which was in the ground state, goes to the excited state. It may propagate in the excited state for a while. And then it emits the photon and is back into the ground state. So this is the temporal diagram for this process. For the next situation, we also start with an atom in the ground state. But now we have the situation that a photon is emitted by the ground state. As a result, the atom switches to the excited state. Strong violation of energy, but possible for short times. Then there is a real photon, which is now absorbed, absorbed by an atom in the excited state. And that takes us back down to the ground state. And finally, in the third scenario, we have an atom in the excited state. There is a real photon which is absorbed that switches the atom into the ground state. And then the atom in the ground state emits a photon. And as a result, it is back in the excited state. So let me just make a note that when a ground state atom absorbs a photon, this is the co-rotating term in the quantum description, which a lot of you have seen. The opposite process of emitting a virtual photon-- I shouldn't say virtual photon. It's a photon. What makes it virtual we will see later when we have the more accurate mathematical formulation. So this is a counter-rotating term. And we have used the rule that every photon, which with a real or virtual-- every photon which is interchanged with the atom absorbs or emitted-- changes the atomic state from ground to excited. So an atom goes from the ground to the excited state either by photon absorption or by photon emission. Both is possible. So whenever photon appears or disappears, it changes-- let's say we start in the ground state. We go to the excited state. And this is possible by photon absorption-- this is a co-rotating term-- or emission. What I've just said reflects that when we derived the dipole approximation, that the essential term of the dipole operator is of diagonal. The dipole operator is an operator between calm and excited state. Therefore, it has those two matrix elements. Whereas electric field operator is the equation and annihilation operator a and a dagger. Or which is very elegant if I use for the two level system a spin one half description. And I use sigma plus and sigma minus, raising and lowering operator. Sigma plus takes the ground state to the excited state. Sigma minus takes the exciting state to the ground state. Then in the fundamental atom light interaction, we have those four terms. And now you see that sigma plus takes the atom from the calm to the excited state. But sigma plus appears both with photon absorption and with photon emission. So therefore, the atom can-- this is what the quantum mechanical operator tells us-- go from the ground to the excited state either by photon absorption or by photon emission. Any questions? Yes. AUDIENCE: So if we were to look beyond the dipole approximation, would we see transitions that leave the [INAUDIBLE] state? PROFESSOR: Yes or no. What we need here is a bilinear interaction, which has an raising and lowering operator for the atom. And here it has a raising and lowering atom operator for the photon number. And it has the co-rotating term where you raise here and you lower there. You lower here and you raise there. But it has also terms where you raise the atomic excitation, and you raise the photon number by one, which of course violates energy. But the message I want to give you is this is possible for short times. And this is exactly what the quantum mechanical equations tell us. So the short answer to your question is, as long as we describe the system by the dipole interaction, or what we have even beyond-- I'm not using a perturbation argument. I'm writing down the operator. And this operator can now effect the system to all possible orders. So even the non-perturbative strong coupling limit will always have products of atomic raising operators, lowering operators with photon raising and lowering operators. The little bit longer answer is, some of what I'm telling you may be gauge dependence. If you work in the-- if you use, not the dipole description, but you use the p dot a Hamiltonian, this Hamiltonian has a e square term. The e square term allows the atoms to exchange two photons at the same time. So in other words, in the a square interaction, we may have a vertex point in time where the atom interact with the photon field where two photons are exchanged. And then the rules are different. So in that sense-- but this is quite often when we give a quantum mechanical pictures, which has terms, which nicely shows the time evolution. And you think wow, yeah, this is what happens. One photon at a time, the atom goes ground, excited, excited, ground state. And you have your picture. And it is fully consistent with the exact quantum mechanical result. You may be able to get the same result out of a different picture. That point will actually occur again when we talk, for instance, when we talk about density operators. A density operator is an average of our quantum states. But you can get the same density operator by averaging quantum states in different basis sets. So if you get one picture out of one specific basis set, it gives you the correct intuition, it's a correct description, but it is not the only possible description. [? Collin ?]. AUDIENCE: Maybe the thing, I don't know. AUDIENCE: I think what you're trying to get at is that picture works as long as you can limit yourself to a two level system. Because that's correct on a whole. Or is it like strong coupling? But as long as you can identify the two level system, or I think the dipole versus particle approximations [INAUDIBLE] an atom may have transition probabilities to others, like from an [INAUDIBLE] state. But-- AUDIENCE: Actually, I was just wondering if it was possible to have it like a transition matrix element to [INAUDIBLE] coupling the ground state to the ground state. PROFESSOR: Yes. The e square operator does that, because in the other gauge, which is pa, pa. p is in dipole operator. p is the momentum operator for the atoms. And the p operator will switch the atom from ground to excited state. a, the vector potential, involves a and a daggers. So the p dot a term does pretty much the same as the d dot e term, because it's a bilinear product of an operator which has an effect on the atoms and has an effect on the photons. And it means you always switch the quantum state in your atom and create or annihilate the photon. But the a squared term-- it's just a square, it doesn't have a p in front of it-- will allow an atom in without changing the state to create two photons, emit two photons. And then there's probably a cross term which means to absorb and emit a photon. You will actually, in one of your next homework assignments, you will actually look at the a square term. And I think it's a very educational problem. So you really see that you have two very different descriptions, but the results fully agree. But this is already an expert discussion. Why don't we for now just hang onto the dipole approximation. We have d dot e, and we just learn what is inside the dipole approximation, and what is an exact quantum mechanical description using the dipole approximation. But and then we can come back to the discussion, is everything we describe really real. And while the answer is, what is real in quantum physics is the final result, the intermediate results, you have to say that this is only one possibility to go to the final result. OK, so we want to-- let me just put in one more page. So what we want to do now is we want to do a calculation. It's a perturbative calculation, but we can take it to all orders. So therefore it's exact and general. We want to do a perturbative calculation of transition amplitudes. And what I am following here the discussion in atom photon interaction on those pages. And the idea is the following. We have the Hamiltonian for the atom. And then we have a Hamiltonian which describes the transverse field. And the interaction of the transverse field with the atoms. What we want to use is we want to use the interaction picture, which is often the nicest picture to describe the effect of the interaction between two systems. In the interaction picture, you are transforming from you Schrodinger type e function to wave function psi tilde. And those wave function psi tilde include already the dynamics of the Hamiltonian H naught. So that means that all operators in the normal Schrodinger picture become now operators in the interaction picture by canonical transformation, which involves the dynamics due to the unperturbed Hamiltonian. And so that's just a reminder what the interaction picture is. And we are now interested what happens in the interaction picture to an initial wave function psi of ti. This wave function already includes the time evolution due to H naught. So the only change now comes because we have interactions with the electromagnetic field. So this allows us now to focus only on the relevant interactions, the interactions we want to understand. And the formalism which is used in atom photon interaction focuses on the time evolution operator, which I'm sure all of you have seen. So the time evolution of our system, how the wave function involves from the initial to the final time, is described by the time evolution operator. I will come back to the wave function later, but in order to derive the time evolution of the wave function, it's at least convenient to focus first on the time evolution operator. So in other words, I want to show you a formal solution for the time evolution operator. And once we know the time evolution operator, we apply it to the wave function. And then we are talking again about the wave function. OK, since the derivation involves many equations and they are all printed in the full beauty in atom photon interaction, I decided to use pre-written slides here. Also, most of it is fairly elementary. The equations are complicated, but the concepts behind them are very, very simple. It's actually the beauty of using a kind of a time evolution operator. It's a little bit mathematically formal, but it allows us to express what is really going on in the solution in very, very simple terms. So the goal is to do a perturbative expansion for the time evolution operator. If we have no interaction potential v, the interaction picture, nothing happens in the interaction picture, and therefore, the time evolution operator is a unity matrix. But if you have a coupling term v, there is time evolution. And the time evolution can be treated in lowest or in higher orders in the perturbation. And we want to use now perturbation expansion, where we have correction terms, which is the first order, second order, and nth order correction for time evolution operator. So what we simply want to do is we want to do an iterative solution. We want to find an iterative solution for the time evolution operator. The next two lines is just a reminder. What is the differential equation for the time evolution operator? Well, we are really just talking about Schrodinger's equation. Schrodinger's equation for the wave function turns into differential equation for the time evolution operator. Schrodinger's equation in the interaction picture says that the time derivative of the wave function in the interaction picture has nothing to do H naught. H naught has already been absorbed in the definition of the wave function. So the time evolution in the interaction picture only comes from the interaction term. And now, if you write the wave function at time tf as the time evolved wave function from the initial time to the final time, then you take the derivative here. You actually take a derivative of the time evolution operator. And therefore, in one step, from Schrodinger's equation, you find a differential equation for the time evolution operator. In other words, this is now the operator equation for the time evolution operator. And it's nothing else than a 100% rewrite of the Schrodinger equation. Interrupt me if you have questions. This part should be-- we'll go through that for solely pedagogical reasons. And that means you should really understand it. So this equation can now be formally solved in the following way. Take this expression for the time evolution operator and insert it in this equation, and you find an identity. Of course, you haven't really solved it, because you have the time evolution operator expressed by the time evolution operator. This looks like a circular conclusion, which is just nonsense. But if you inspect it more closely, you observe that the time evolution operator here is expressed by the time evolution operator multiplied with v. And if you now think in an iterative solution that we want to express. We want to expand the time evolution operator in the different orders in the interaction parameter v. Then you find actually that if you are only interested in the first order of the interaction operator, you can use the [INAUDIBLE] order here, because you get one more power of v. So that's how you get the first order. If you want to know the second order of the time evolution operator, you can plug-in the first order on the right hand side. So therefore, you get an iterative solution. The first order solution is one. The [INAUDIBLE] order solution is the unity operator. The first order solution is by just putting the unity operator here. The second order solution is by taking the first order solution, plugging it in in here. And then you get two integrals over the operator v. So in that sense, we have formally solved the time evolution of the system in all orders. So this is what we get out of it. Yes, the equations are getting longer. But the structure is fairly obvious. So you saw that the-- let me just flashback. The first order term had a temporal integral over v. The second order term has two temporal integrals over v. And there's a time order in between tau 1 and tau 2. And the nth order term now involves n temporal integrals where the times are ordered in such a way. Now we want to go back to the Schrodinger equation. So we want to use this already fairly complicated expression for the time evolution operator and apply it to the initial wave function. And we want to calculate matrix elements between the initial wave function and some other phi f, some other wave function in a given basis. So we are also going back. I don't want to go through too many steps. Here, we're also going back to the original Schrodinger picture. So I've taken all the tildes off. But you know the step form the Schrodinger picture to the interaction picture was only taking out the Hamiltonian H naught. And the Hamiltonian H naught for eigenstates phi i and phi f, is simply a time evolution, e to the e ei with a certain time. So all what this transformation to the interaction picture gave us, it eliminated, at least for a short while, all of those phase factors, which are simply the evolution of the eigenfunction in the Schrodinger picture. So let me look at a typical term which we have right now. And you can't expect something simpler than this, because this is the general solution. It's an exact solution to all orders. It's a summation over terms to all orders. And eventually the summation of an infinite number of diagrams is an exact solution. So what we've got here is we have an integral over n times. And the n times are time ordered. This is the formal solution for the time evolution operator. And that means that we integrate overall times, but what it is under the integral sign is a product of the operator v at times tau 1, tau 2, tau 3, tau 4. Since we got back to the Schrodinger picture, what happens is between times tau 1, and tau 2, the particle simply evolves with the phase factor given by the Hamiltonian H naught. So the picture is the following. The time evolution operator in the interaction picture is a product over the operator v tilde at different times. If you simply plug it back into the above equation-- and you have to look at it for a little while to see that everything works out correctly. But what that means is then that if you go to the Schrodinger picture and write down the operator in the Schrodinger in the basis of unperturbed eigenfunction, that you have products of the operator v at times t1, tau 2, tau 3, they're time ordered. And because we have left the interaction picture, you also get the propagation with the phase factors given by the eigenenergies of the unperturbed Hamiltonian H naught. So therefore, you get just an infinite number of terms which have all the following structure, that the particle propagates in its eigenstate. It reaches a vertex where the interaction switches the particle from one state to the next. Then in the next state, which may now be the excited state, we have propagation in the excited state. Then the excited state is, again, exchanging a photon, has an interaction term, and is switched to-- well, for the two level system, it has to go back to the ground state. In a more general situation, it goes to another quantum state. So therefore, what we have exactly derived is that the propagation, or the time evolution, of the wave function is nothing else then many, many of those factors. And each of those factors can be represented by a diagram like this. So let me just write down what this diagram here means mathematically. It means that initially, we have a photon of energy epsilon, of polarization epsilon, wave factor k, and a certain energy. And we have particles in state a. That means that we get from the initial time to the time tau 1. And tau 1 is the time of the first vertex of the first photon exchange, that the wave function evolves. And it evolves with an energy of the unperturbed Hamiltonian, which is the atomic energy. And we have one photon h bar omega. At the time tau 1-- and this is when the diagram on the left hand side has a vertex-- we are now bringing in the interaction operator, which acts on the atom. It takes the atom from state A to state B. And it changes the state of the photon. In this case, a photon with a certain wave factor and polarization simply disappears. So this is now the time tau 1. The next Vertex is reached at time tau 2. And between the time tau 2 and tau 1, the system evolves according to the Hamiltonian H naught. And the energy is now, because we have no photon, is simply the energy of the atom in state B. Then we reach the next vertex. We start with a state B and no photon. The interaction switches, as now I assumed we have more than two levels. Two are state C. And we have now a photon, which may be different, k prime epsilon prime. Since we are now, for this given example, I've selected n equals 2, something which is second order in the perturbation. But that means now that between the time tau 2 and the final tie-in for the time propagation, that's now everything is done. Things propagate. And the unperturbed Hamiltonian H naught gives us now a phase factor, which reflects the energy of the atomic state, and the energy of the photon in mode k prime epsilon prime with energy omega h per omega prime. So in other words, I hope you see that at the end of this-- I mean, this is why people have Feynman diagrams. It's a complicated, involved mathematical formulation with multiple integrations. But at the end of the day, what we have derived is that the most general solutions to all orders in the probation is nothing else than a sum and integral over diagrams like this. So when I depict it here, the second order diagram, of course this has an exact mathematical meaning. And what we have to do is when we solve a quantum mechanical equation, we have to now allow those interactions, which happen at time tau 1 and tau 2, to happen at arbitrary times. So therefore, we have to sum over amplitudes by integrating over times tau 1 and tau 2. So on Friday-- I have to stop now. I think there's a seminar right now. On Friday, I will show you that with this description, we have actually captured everything I explained to you earlier-- those virtual states, the emission of virtual or real photons. So everything that which was maybe qualitative at the beginning of this section has now a precise, mathematical meaning. But that's what we do on Friday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

10_van_der_Waals_and_Casimir_interactions.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So the subject here is about if you want to understand interactions between neutral objects. So let's start in a very basic way by saying we have an atome A, we have an atom B, they are separated by a distance R, and we want to know what is the force between them. In a semiclassical picture, we would say that everything is-- the force which is responsible for that must be the Coulomb force. So the atom A and B consists of charges, so we should find the charge density in our system A, the charge density of B, we take the coulomb energy and we integrate over the volume. And this is how in e and m, in the most general way, you would to write down the electrostatic energy of a system. You take all charges, each charge has a coulomb potential with the other charge [INAUDIBLE] indicate over everything. But for objects which are localized and we can do multiple expansion, our objects are neutral, neutral atom. So therefore there is no coulomb term, and the next term is the dipole. So therefore, starting with classical physics, we would find that the interaction of those charge density distributions one for atom A, one for atom B, should actually interact with this Hamiltonian h prime. OK, but now, if you look at it classically, we find that classically, the expectation value for the dipole operator of each atom is zero. But you will also find-- if you said, well, OK, that's what I expect, because an atom is in isotopic space. How can it have a dipole moment? But maybe, what could happen is, if there is a fluctuation of a dipole moment here and a fluctuation there and the fluctuations are correlated, then the two systems may repel or attract each other. So therefore, when we are talking about fluctuations, we are talking about expectation values for d squared. But also in classical physics, this is exactly zero. So what classical physics tells us, it says, in classical physics, it costs more energy to create a dipole moment, or even a fluctuating dipole moment, than one gets back from the dipole interaction. So the system will never spontaneously form some form of dipole moment. And in classical physics, two neutral atoms will have no attraction at all. So it costs more energy. So now, we have to do quantum mechanics. And every so often, I have to go through a classical argument to make you wonder about the quantum mechanical result. Because the quantum mechanical result, you're used to it. You've heard about the van der Waals interaction. But now you know that there is really something quantum at work. In quantum mechanics, the expectation value of the square of the dipole operator is not zero. I mean, you can see that if you have an electron, which is in an harmonic oscillator, and if it is at the bottom of the oscillator-- well, the dipole moment-- what I want you to say is the dipole moment for a charged system is related to-- between electron and proton-- to the displacement operator, R. And even if you have an equilibrium position-- like in an harmonic oscillator, where R is equal to zero, R squared-- the expectation value-- is not zero, because whatever your stable equilibrium is, you will have zero point fluctuations around it. And so therefore, now in quantum physics, it's not that we have to provide energy to create a d squared. We have it. We have it by necessity, because of non-commuting operators, because of the nature of the quantum physics. So now, the fact that we have fluctuating dipole moments already, that means if now with the fluctuating dipole moments between the two atoms are synchronized, this will then lead to a non-vanishing average force between the atoms. So that's the idea. But the good thing is, you don't even have to know about it-- about fluctuations and all that. You just take-- and this is your homework-- you just take the operator H prime. And look what happens when you have an atom in the ground and excited state and what happens when you have both atoms in the ground state. In the first case, you get the leading result in first order perturbation theory. For two ground state atoms, you have to go to second order. In first order, the potential is 1 over R cubed. And you will calculate the C3 coefficient. In second order, you get a van der Waals potential, which is 1 over R to the sixth. And it's fairly straightforward to calculate the C6 coefficient. Any questions? So let me reemphasize, the beauty of this perturbation result that it's so simple. But the downside is, you don't really understand what you're doing. I mean, you are solving the Schrodinger equation. You are finding an expectation value. But the nature of the effect-- how it is related to fluctuation, what is really behind it-- is sometimes obscured. So let me give you one interpretation, which I really like. And I learned that actually from Dan Kleppner. When he taught the atomic physics course at MIT 20 years ago, that was one element of it. And I told you at the beginning that I'm actually proud that our atomic physics course is really kind of-- has hopefully kept-- the best of the decade long tradition of teachers like Norman Ramsey, Dan Kleppner, and Dave Pritchard. So this is now really due to Dan Kleppner. And I will later today actually post Dan Kleppner's original lecture notes about this effect. So we want to understand the van der Waals force in a completely different system. And I think that helps. We have two LC circuits. There's L, C, C, L. And what happens is, the two capacitors are closed. So there is this stray field of one capacitor reaching the other capacitor. So I won't go through the detailed math here, because it's not necessary. But what we have here is, we have two coupled systems-- two coupled capacitors. And the only thing I want to use is that the stray field of a capacitor has a diopolic characteristic. And it decays with 1 over R cubed. So what we have right now is, each LC circuit, when it's isolated, has a resonant frequency of omega naught. But now, when we have two pendulums-- two oscillators-- which are coupled, we get two modes, which are omega plus and minus. And if we simply solve for two modes, and they have some coupling, well, we get an upshift and downshift by the coupling. But if you look more carefully at the solution for two coupled modes, it has also a quadratic effect, which is sometimes neglected, but it's there. So this is, in general, what you will find, if you couple two modes. Now in quantum mechanics, when we ask for the oscillator in the ground state, we can find the zero point energy. And so the energy in the ground state is 1/2 the zero point energy in the mode plus, in the mode minus. The first order coupling term, the 1 over R cubed cancels out. But what remains is a contribution, which is 1 over R to the sixth. So we need quantum physics. Because in classical physics, the ground state of the LC circuit is nothing happens. No charge, no current, no nothing. I mean, this is a classical ground state, and you would not get any effect. So realize that this van der Waals potential requires quantum mechanics, and it is due to the zero point energy of the atomic oscillators. Just hold the thought, when we talk about atoms in the electromagnetic field, we have two oscillators. We have an harmonic oscillator, which is a fluctuating dipole moment on each side of the atom. But we have many, many harmonic oscillators in between, which is the electromagnetic field. And during the course of this lecture, I will refer to both the zero point energy of the atomic oscillator and the zero point energy of the oscillator, which is the electromagnetic field. And to give you the result at the end, I will sort of explain different effects by the zero point oscillation of the atoms, other effects by the zero point oscillation of the electromagnetic field. But at the end of this section, which is probably after spring break on Monday, I will actually tell you that you have to be careful. If you have quantum physics, you have to use a consistent description. And you always need that both the field and the atoms are quantized and give zero point fluctuations. And when I'm telling you now that the van der Waals potential is due only to zero point fluctuations of the atomic oscillator, there may be another way of getting the same result by looking at the zero point fluctuations of the electromagnetic field. So I just want you to take what I'm saying to you-- I'm not saying to you anything wrong. But it may not be the only way of expressing it. And I want you just to pay attention to it. But certain effects can be simply described by just simply looking at the atomic oscillators. Why do we need the quantized electromagnetic field when all we want to do is have a stray field coupling two capacitors. So we want to always learn about the simplest way to understanding an effect. But then in the end, there are maybe certain subtleties. OK. But let's keep it simple. So it's a zero point energy of the two oscillators. And maybe let me emphasize this by translating this description-- what we learned from the capacitors. Let me give you the Coulomb description of the atom. So an atom really never have in free space a spontaneous dipole moment. But because of zero point oscillation, it will have a dipole moment squared. So what happens now is, let's say at some moment, we have a fluctuation. One atom happens to have a fluctuation of the dipole moment. This creates an electric field at the position of the atom, b. So this is a fluctuating field. The fluctuating field, because there's a finite polarizability, induces now a dipole moment of the atom, db. I will later introduced polarizabilities. But here, it's just proportional to Eb. So that's proportional to da over R cubed. So now, we have no longer randomly fluctuating dipole moments with da db, which are out of phase. Positive and minus cancels out. We have a dipole moment, da, which creates-- through polarizability-- an aligned dipole moment, db. And for those, we have now the dipole-dipole interaction, which is db da over R cubed. And when db changes sign, da changes signs. So the product of them will always stay the same. And now, we find that we have an interaction. If you look, db was 1 over R cubed. And if you multiply that with the R cubed of the dipole-dipole interaction, we obtain, in a very different way, the van der Waals potential, which is 1 over R to the sixth. So therefore, what we can take away from that is that the 1 over R to the sixth potential is caused by the zero point fluctuations of the atomic dipole moments. OK. So now you have already three different ways to look at the van der Waals potential. One is just doing perturbation theory and not understanding anything. The second one is the ground state of two coupled oscillators, a la two capacitors. And finally, the fact that spontaneous fluctuations create stimulated fluctuations, and then there is a quadratic term, which doesn't vanish. OK. Well, let's now consider that the electromagnetic field-- that there is an electromagnetic field. Until now, we have not really used the fact that we have an electromagnetic field with photons. And the electromagnetic fields, each mode, is an harmonic oscillator. And each mode of the electromagnetic field has zero point fluctuation of itself. So let us now discuss what can the vacuum fluctuations of the electromagnetic field do for us. Well, if we have fluctuations-- momentary fluctuations-- of the electric field of the vacuum, the fluctuation at position, a, and position, b, will induce now dipole moments of the atoms, a and b, by multiplying with the atomic polarizability. And if you assume-- I'll say a little bit more about it-- that we're talking about the long wavelengths fluctuation of the electric field. Then the two dipole moments, which are induced by those vacuum fluctuations, are actually in phase. So the key word is now that the vacuum fluctuations create correlated dipole moments. So the dipole moments are no longer independently fluctuating by themselves. So if you write down now the dipole-dipole interaction, we will find a result which is proportional to the product of the two polarizabilities. The dipole interaction has an intrinsic spatial dependence, which is 1 over R cubed. But now, the important term is that we have to calculate the correlation function of the electric field at position, a, and position, b. And these are the vacuum fluctuations. I'm not doing the calculation, because it would be somewhat messy. I really post on the web a fairly easy-to-read paper by Larry Spruch in Physics Today. And he gives you a little bit more details. But you have already the physics in this equation. What you should use now is, you have to use the density of modes. And for each mode, the zero point energy, 1/2 h-bar omega. And then, what you get is, you get an interaction, which is 1 over R to the seventh. Yes. AUDIENCE: When you're accounting here only the dipole moment, which are [INAUDIBLE]. So they're coming from the fluctuation of long wavelengths in your EM. And once they [INAUDIBLE] those but they are EM field fluctuation from vacuum. So what are the effect of those shorter waves? Are they just higher energies, or just-- PROFESSOR: Excellent question. Actually, I need that-- in 10 minutes-- I need exactly the answer to your question. So let me give you the answer right now. If you look at this expression, there are two things which come into place. One is we have now two atoms separated by a distance, R, which is ra minus rb . If you take, now, high frequency modes, which are very short wavelengths, and you integrate and sum over all of them, it's pretty clear that you will get plus and minus, which will completely average out. So it seems very clear that the modes with wavelengths, lambda, shorter-- definitely much shorter-- than R average out. So therefore, the bulk of the contribution will actually come from the modes with wavelengths smaller than R. But now, there is something else. The density of states for the electromagnetic field dramatically increases with frequency, omega. So we have many more modes at short wavelengths. So therefore, the argument says, well, the long wavelength modes, up to wavelengths, R, are the dominant ones. They don't average out. But because the density of modes increases, the shorter the wavelength is, by far the dominant contribution, will come from the modes, which are centered around lambda equals R. So the density of modes implies now that those modes will dominate. So that's the answer. But ultimately, you have to put everything into an equation, have an integral, solve the integral, do some approximation. And you will exactly see from the density of modes that the very long wavelengths don't contribute. And you will also see that because of rapid averaging of E to the ikr phase factors, the high frequency modes will not contribute. OK. So we have now two different power laws. One is 1 over R to the sixth. And one is 1 over R to the seventh. So let's explain that now. So we obtain the 1 over R sixth by using the uncertainty principle for atoms. And here, we have used the uncertainty principle, the zero point fluctuations for electromagnetic waves. When we derive the 1 over R sixth potential, we actually used simply the electrostatic instantaneous Coulomb field dipole field. And this is only valid for short range when-- you know, I said we have a fluctuation and the fluctuation created another one, which was in phase. But if one fluctuation has to send out an electromagnetic wave, the second fluctuation may not be in. When propagation effects come into play, there's a time lag between the two oscillators and may actually be out of phase. So we really assumed in the derivation, without ever saying it explicitly, that there are no propagation effects. And this is called the short range potential. Whereas, for the second argument that we said, we have two distant atoms, which get synchronized by being driven by fluctuations of the electromagnetic wave at wavelengths, lambda equals R. This is what happens at long range. So this is a famous result, that we have a van der Waals force, which is the instantaneous force. And when propagation effects come into play, this goes by the name Casimir-Polder potential. The moment when propagation effects come into play-- when radiation plays a role-- we get a different power law. So again, I'm just playing with ideas. I'm taking a train of thought, atoms fluctuate; another train of thought, electromagnetic fields propagate; and just see, what are the ramifications? What are the consequences? So now, I want to eventually give you a treatment, which has both aspects in one. And this is our diagrammatic treatment. First, before I do with that, do you have any idea what distinguishes the short range from the long range potential? So you have two atoms. You pull them apart. And what distance do we have the 1 over R to the sixth physics? At what distance do you have the 1 over R to the seventh physics? AUDIENCE: When [INAUDIBLE] transitions? PROFESSOR: When the distance is the wavelengths of the atomic transition. That's the only thing which matters. You don't get it immediately from here. Because-- I mean, I used an instantaneous potential, which didn't have any scale. And here, I said we have vacuum fluctuations of all wavelengths. So here, I had a picture, which clearly works at long separation. And propagation effects are built in, because it's a vacuum fluctuation, which sort of act at position, a and b. And the wavelengths-- their propagation-- is part of the formulation. But yes, Bohr's conjecture that it is the wavelengths of the atom-- the wavelengths of the resonant radiation-- is correct. But I want to show you now how we can obtain this result in a, I think, really elegant and beautiful way by looking at the diagrams we have just learned. OK. What did we just have? Time to do that, yes. Any questions here? The reason why I want to show it diagrammatically is actually two fold. One is, it's really beautiful how it comes out. But the second thing is, when I introduced the Feynman diagrams and all of these formulation in terms of propagators, I gave you a wonderful picture. But in the end, I reduced it to first and second order perturbation theory. So I want to show you at least one nontrivial example where you would not get the result easily without this formalism. So enjoy one nontrivial application of diagrams. So if you follow the systematic way, how we have set up our formulation of quantum electrodynamic with a photon field-- the quantized photon field-- then the two neutral atoms have no direct Coulomb interaction. I just want to point that out. Because a few minutes ago, when I said a classical system is just Coulomb field-- each charge element has a Coulomb potential with the other charge element-- this was just the opposite approach. But in our QED formulation, we have sold our soul. We have said we only really got the near field, the longitudinal field, as belonging to the atoms. Everything which happens at longer range is included in our atom-photon Hamiltonian. So therefore, there is no direct Coulomb interaction for the longitudinal field. Everything, everything now, has to come from the interaction-- the perturbation operator-- for the quantized radiation field. And therefore, we must get it out. We have one atom; it's d1 at d prime. I think that's how it's-- I'm using now the nomenclature, which is used in atom-photon interaction, pages 121 to 126. And so, there is one atom at the unprimed location, one atom at the time primed location. And everything has to come out now of the formulation of the quantized radiation field. And we have discussed that lengths at the transverse electric field become operators, a plus a dagger-- the symbol for the dipole operators are acting on the atoms. We've done all that. So then the only thing which happens is-- and this is described by a and a dagger-- is the exchange of photons. So the only way how those two neutral objects can interact is through the radiation field. And through the radiation field means they have to emit photons. So I want to show you now that the van der Waals interaction can be thought of as one atom emitting a photon. But this virtual photon is absorbed by the other atom and vice versa. And you may have heard quite often-- but I'm not sure if you've seen it explicitly-- that a lot of interactions in physics are actually mediated by the exchange of virtual particles. The famous example is that the nuclear force comes from a virtual exchange of pions. But then, people say, the Coulomb force comes from a virtual exchange of photons. But usually, we don't really show you how this virtual exchange of photons work. So in that sense, I'm proud that I can show you, at least the basic outline, how virtual exchange of photons between two neutral atoms leads to the van der Waals potential. So that's our agenda. OK. Now the fact that this takes six pages in the book-- and the book is not solving all equations-- means if you really want to do it quantitatively, it has a certain complexity. But what I've done is, I've sort of distilled out the grand idea and which also shows you what happens when you go from short range to long range. And this is what I want to present you. And therefore, I take [? Benji's ?] question and will not discuss all the modes of the electromagnetic field. I will immediately use the fact that the dominant contribution comes from modes. And since we don't have real photons, we don't have energy to create real photons when everything is in the current state. We call them virtual photons. Virtual photons are photons which are immediate. But a short time later, because of Heisenberg uncertainty, have to disappear. So the dominant contribution comes from virtual photons at the wavelengths, which is D, the distance between atoms. So the frequency, which will play an important role, is simply the frequency of those virtual photons. And I gave you already the reason-- the density of states which favor higher frequencies. But higher frequencies cancel, because of the E to the ikr terms. So we only focus on those. The second thing I need is that when we do perturbation theory, the energy shift in perturbation theory-- well, if you do second order, you go from the initial state to an intermediate state. And then, you go from the intermediate state, one, to the final state. And what you have to do is, you have to divide by the energy denominator. If you do higher and higher order perturbation theory, in our diagrammatic presentation, that means we have propagators at intermediate energies. And I told you that the time an atom can spend in an intermediate state is 1 over the energy defect. And if you integrate the E to the i energy defect factor over time, you get something which is proportional to 1 over delta E. So therefore, if you perform the time integration of the perturbation analysis, you find by integrating over the short time particles spend in the intermediate state, you actually generate those energy denominators. That's actually the relationship between the time propagation I showed you earlier. And when you integrate over the time, you get for this intermediate state, a contribution, which is 1 over the energy defect over the intermediate state. So therefore, if you go now to higher order perturbation theory, we simply have a product of matrix elements in the numerator. And in the denominator, we have a product of energy defects. You are using that all the time for second order perturbation theory, but the structure is that you can just sort of daisy chain the expression. And you get higher order perturbation theory. So therefore, all we will have to understand to figure out what happens when virtual photons are exchanged, I will just show you two relevant diagrams. And we just look-- we stare at the diagrams-- and figure out what are these denominators? What are those energy defects? And we will find that there's a difference at long range and short range. So that's the agenda. So we will need two energies. One energy is the energy of the virtual photon, which is exchanged. And the second energy, or frequency, is the resonant frequency of the atom. So let's look at two relevant diagrams. So one is atom, a; one is atom, b. We are interested in interactions between atoms in the ground state. And after a sufficiently long time, both atoms have to be back in the ground state. Let me just get this organized, one, two, three, four. And the result is obtained in fourth order perturbation theory. It has to be fourth order, because we need photon exchange. To emit a photon here, absorb it here, is second order. But then, one atom is in the excited state and this can't be. So a second photon is required to undo the effect. So we need exchange of photon pairs. Each exchange means absorption and emission. That means two vertices. And photon pairs means another two. So we need fourth order perturbation theory. OK. But it's much, much easier than you think. So we have a ground state atom. And now the action starts. This is the first vertex. It emits a photon. The photon is absorbed by the other atom, which puts the atom into the excited state. This atom is also in the excited state, of course. And then, we know the atom cannot stay in the excited state for a long time. It can now emit a second photon, which brings it back to the ground state. And the second photon is absorbed by atom, b, and puts the atom back into the ground state. So this is one relevant diagram. And let me analyze it in the energy defects. Here, in the first intermediate step, the energy defect is-- well, we have ground-ground. This is sort our reference. But now, we have put one atom into the excited state. That causes energy, omega naught. And we have created one photon, omega. Here in that time step, there are no photons, but there are two excited atoms. And in the third time step, we have still one atom in the excited state and one photon. Now, this is one relevant class. I mean, you can now use permutations at atom, b, excites photons first and such. But the structure of many of these diagrams will be the same that we have those three energy defects. But now, we have another possibility. And this is, one atom emits a photon, but it feels it doesn't want to be excited, because the excitation energy is so precious. It wants to immediately return to the ground state. So we could have a situation that the atom fires a second photon before the first photon is absorbed. So the diagram-- the relevant diagram-- is now this. So the atom started in the ground state, is in the excited state, but as soon as possible, goes back to the ground state. And the other atom starts out in ground state, is in the excited state, and is back in the ground state. So what we have here is, we have again three intermediate steps. In the first step, of course, we have an excited atom and a photon. And we have the same here. But now the difference is, that in the intermediate step, our energy defect is not two atomic excitations, it is two photonic excitations. And now we want to compare those two diagrams for short and for long range. And it becomes clear that short range is where the frequency of the photon-- its wavelength is lambda. Short range means high frequency. The frequency of the photon is larger than the atomic excitation energy. Or that means, the distance is smaller than the resonant wavelength of the photon. Whereas, long range is that the atomic excitation is more precious than the photonic excitation. OK. So I need three more minutes. So it just fits into the class time. So we are here in the limit that omega is larger than omega naught. So that means, if we take now the product of the three energy denominators, omega dominates over omega naught. So we'll have this structure omega, omega naught, omega. Omega, omega naught, omega. Whereas, at long range, omega naught dominates. We have omega naught, omega naught, omega naught. So the structure of the product of energy denominators is 1 over omega naught. If you look on the right hand side, we have replaced-- in the intermediate part in the propagator-- the energy defect by omega naught by omega. So therefore, in this case, we have 1 over omega cubed and 1 over omega naught squared, omega. So we know that when we sum up over diagrams, we have to sum up over all possibly diagrams. And the ones which dominate at short range are those ones and at long range are the other ones. So in other words, at short range, one photon is exchanged and disappears. And then, the next photon comes. But at long range, we have an exchange that one atom sends out two photons almost simultaneously. And the other atom absorbs them. Because it sends them out almost simultaneously. Because here, we are talking about long wavelength photons, and photons are cheaper. Rather send out two photons before you spend too much time in the excited state. OK. Let's wrap it up. Since omega is proportional to 1 over the wavelengths, which was-- and I said the photons we are concentrating, is the photons which have a wavelength of the distance, D . So what we are realizing now is that, if you just compare those diagrams at long range, we go from 1 over omega squared to 1 over omega. So at long range-- omega is-- we have an additional factor of omega, which is 1 over D. And so now, if you assume-- I can't show you that without solving the real problem-- but if you assume that at short range, we have 1 over R to the sixth-- van der Waals potential-- then at least, I've proven to you that at long range, you get one more power in the distance. And this is a transition from an instantaneous potential, where propagation effects do not matter, to what is called retarded potential. So that's all I wanted to show you about interactions between two neutral atoms. Are there any questions? Timor? AUDIENCE: Does it matter the direction of both the photons that we drew? For example, can atom, a, emit to b, b absorbs, and b emits to a, rather than-- PROFESSOR: In practice, it matters. When you solve it, you have to sum over all possibilities. Remember, quantum mechanics in the system does everything. It doesn't care whether it's allowed or not. It tries everything out, whether it violates energy or not. And you have to sum over everything. But if you then analyze all the possible diagrams, you will actually figure out that they can-- no matter what you do, what permutation of photons and atoms you have-- the fourth order diagrams will be such that they are always distinguished by in this intermediate zone, whether you have two atomic excitations or photonic excitations. And so what I do for you were the two diagrams, which are representative for a whole class of diagrams here. But absolutely, yes, you are right. OK. So that's all about neutral atoms. I know you have spring break next week-- MIT spring break. Maybe that's good news for the Harvard people, who-- when is Harvard's spring break? AUDIENCE: We just finished. PROFESSOR: OK. So you have-- at least for this class, you have a second spring break. So we are not yet finished with van der Waals and passing interactions, but we finished the neutral atoms. And on Monday after spring break, we talk about interactions between metal plate and atoms in two metal plates.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

17_Dressed_atom_Part_1.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So good afternoon. Our main topic today is to understand light forces, in particular the stimulated force using the dressed atom picture. And I hope once you understand how light forces can be described in the dressed atom picture you will love the dressed atom picture, as most atomic physicists do. It is really the language we use, the intuition we apply to atomic systems. Whenever we have a laser beam which drives the atom it is, in most situations, more appropriate and better to use dressed states as your basis states and not the naked, the bare states, which are ground and excited state. Now I know last week I quickly went though the dressed atom picture. Let me sort of go through it in a fly-over by not doing any of the derivation, but making a few comments here and there. Also, because I want that some of the results are fresh in your head when we apply it to mechanical forces of light. So I told you that in the dressed atom picture we have two kinds of interactions, the atom interact with a vacuum, with a reservoir of empty modes with spontaneous emission. And we also interact with a laser field. And the new feature about the dressed atom picture is that we first solve this exactly, we rediagonalize the Hamiltonian for that Hilbert space, and then we allow spontaneous emission to happen. Now the derivation is pretty straightforward. After a number of assumptions, we simply came to the two-by-two matrix, where we have two levels of the atoms coupled with the Rabi frequency. I derived it here for you using the quantized electromagnetic field. But in your homework you derive the same two-by-two matrix using a classical electromagnetic field, which drives the atoms. And we've talked about the similarities and differences last week. If you have any questions, please raise your hand or interrupt me. Now two-by-two matrix is fairly trivial. We can immediately diagonalize it. And we find Eigensolutions, which are just sine theta, cosine theta, amplitude, linear superpositions of the naked states. And the naked states are ground and excited states, state A and state B, with N or N plus 1 photons in the laser field. So it's simple. It's just a two-by-two matrix. But the insight we can get out of it is pretty profound and pretty subtle. So what we first discussed is-- and we will need that to understand forces and energy conservation-- we can now look at spontaneous emission. Spontaneous emission coupling to the empty modes of the vacuum can only connect an excited state B with a ground state A. And therefore, we have diachromatically those four combinations, represented here by those four arrows. And if you look for the matrix element, you collect cosine theta and sine theta factors. And that's what we did here. So we immediately get in the dressed atom picture, an expression which tells us what are the matrix elements for the carrier, for the blue and for the red detuned sideband of the Mollow triplet. But then there's something else, and I edit this with the lecture notes in green. We can also, now, if you know what the transition rates are, we can set up rate equations and can solve them. And once you solve a set of differential equations for rate equations, you know what the populations are. But now, today, I want to tell you one more subtlety. You know already that you cannot, in the most general situation, formulate an equation with just a population. You know that we have the master equation, we have the Optical Bloch Equation. This is an equation for the density matrix. And usually we get equations where the populations are mixed with coherences. In other words, you cannot just say the diagonal matrix element of the density matrix is a population, and the change of population is what goes out of this state and what arrives in this state. Coherences play a role. But, and this is what the dressed atom picture does for you, in the limit that we have a strong Rabi frequency-- and this is the limit you want to consider here-- the matrix equation decouples into an equation for populations and an equation for coherences. And if something decouples that should tell you we all on the right track, we have the right description. In this description things decouple, things are simpler. And that means this description is most appropriate for the physics we want to understand. Or, to say it more specifically-- and all the details of course are in the book Atom Photon Interaction-- you can rewrite the Optical Bloch Equations not in the bare bases but in the dressed atom basis. And in the dressed atom basis you find this wonderful separation between populations and coherences. So we have the simple rate equations. By setting the left hand side to 0 we find the steady state populations. This will be very important to understand the light forces. But you should just remember the steady state solution such as we know them. They're just exactly given by this angle theta, which tells us what kind of superposition of the bare states from the dressed states. Remember in the case of theta equals 45 degrees, the dressed states are just symmetric and antisymmetric superposition of the naked states. Now what is also important is those rate equations, or those rate equations have a time constant. The time constant is here, the relaxation time for populations. And we know exactly what it is. It's a little bit more complicated to find the equivalent equation for the coherences. But just take my word, the density matrix in the dressed atom basis decouples into equation for population, which is simple, equation for coherences, little bit harder to derive but looks the same. And from this equation we have now an expression for what is the relaxation time for coherences. And just to show you one non-trivial example, if you add up cosine square plus sine square you get 1. But based on what we derived we have to take cosine to the power of 4, sine to the power of 4. And this gives us effect of 1/2, which will appear somewhere in an important place in a few minutes. So with that we can now address, we know now something about the intensities of this Mollow triplet, because the intensity is nothing else than the population, what is the population in the initial state times the matrix element squared. But what is also-- what comes out very naturally-- is that we have different widths. The features have different widths. The sidebands have a different width from the central part of the Mollow triplet. And well, this is now very naturally interpreted at one has a width, which is a relaxation rate for the population. The other one, is the sidebands have a width which is a relaxation rate for the coherences. I showed you, and this was just a reminder of a few weeks ago, that when we looked at the linear matrix equation, which are the Optical Bloch Equation, we discussed complex eigenvalues mathematically. And those complex eigenvalues, the imaginary part is the sideband frequency. And the real part was the widths. What I have now, sort of-- I can get everything from the dressed atom picture. But what I've gotten in a very natural and simple way is this case, which is the case where the dressed atom picture is particularly simple. It's called the secular case, where the population and coherences decouple. And so, what I just showed you, that the population relaxation time is gamma over 2. This explains now why, in this case, the widths of the central feature is 1/2 of what it is when we're on resonance. So everything is sort of nicely playing out here. OK, any questions about dressed level populations, rate equations, the Mollow triplet? Because we want to apply it now, in a few minutes, to the stimulated light force. But before I do that, let me show you one nice piece of physics, which can be formulated both in the bare, in the uncoupled, and in the dressed state basis. And this is the process of spontaneous emission. If an atom is in the laser field, it just emits photon, rapid fire photons. And I want to show you now, how we would describe it in the bare basis, how we describe it in the dressed basis. And we get two very different physical pictures out of it. Well, in the bare basis, maybe just think of quantum Monte Carlo simulation, a photon has been emitted, and that means we are in the ground state. Now it takes a Rabi cycle before the ground state atom is in the excited state. And when it is in the excited state, it can emit again and goes back to the ground state. And then things start over. So the picture you obtain here is the picture of antibunching, that the atom emits a photon, but then it takes the inverse Rabi frequency to load the gun again. And then it can fire again and emit the next photon. So therefore, the probability of getting another photon right after you've observed the first photon is reduced, or T equals 0, it is exactly 0. And this is antibunching, the photons are antibunched. So this is how we would describe the radiative cascade, how we go from N to N minus 1, to N minus 2, how we lose photons from the laser field and dump them into the empty modes of the vacuum. That's how we describe it in the uncoupled basis. Well in the dressed basis we have sort of diagonalized the atoms interacting with the strong laser field. And now we would have the following picture. We have this manifold here. We can emit from this manifold. And we can go to the upper or lower dressed state. And then, the next thing will happen again. But what we realize here, if we are in the upper dressed state, we can either emit on the carrier or we can emit a blue detune sideband. But once we have emitted a blue detune sideband, the next sideband has to be red detuned because we are now in the lower dressed state, and we need a lower sideband to go back to the upper dressed state. So what this picture gives us, it gives us the sidebands because we are resolving the splitting in this picture. And secondly, it tells us that there is an alternation between the blue detuned photons and the red detuned photon. Now you would say, well aren't those two pictures really different? In one case we have antibunching, and all of the photons are the same. And over here I talk about an alternation between the blue and the red sideband. Well again, the two pictures are complementary. In this case, if I want to resolve the antibunching I need a temporal resolution which is better than the inverse Rabi frequency. But if my temporal resolution is better than the inverse Rabi frequency then I don't have the spectral resolution to observe the sidebands, and vice versa. If I want to resolve the sidebands, if I want to distinguish the red from the blue detuned sideband of the Mollow triplet I cannot run the experiment with the spectral resolution, which would be required to observe the antibunching. So these are sort of two different pictures. One is sort of in terms of energy eigenstates when we have spectral resolution. And this other picture is more appropriate when we're interested in the short time domain. So in that sense, you need both. You need the bare basis. You need to dressed basis. And you have to just know for which physical process, which of them allows the more direct description. Questions? OK, so this was partly review, partly preparation for the next big chapter, namely we want to understand the dipole forces within the dressed atom picture. The book Atom Photon Interaction has a nice summary of it. But the more detailed picture, and I heavily draw from this, is in the truly seminal article by Jean Dalibard and Claude Cohen-Tannoudji. I've linked this article to our website. So I've summarized here pretty much what I just told you. This is sort of the summary from the previous section. What we have to know is to understand light forces in the dressed atom picture is number one, the energy speaking between the two. The splitting between the two is the generalized Rabi frequency. If there's anything which is not-- well I shouldn't say 100%-- but at least 90% clear to you, ask me question. I mean these are the results we really need for the following discussion. So number one is the dressed energy levels are split by the generalized Rabi frequency. Number two is with this angle theta, which came from the solution of diagonalizing the two-by-two matrix, we have now the steady state populations in the upper and in the lower dressed state. And the third thing is when we have a population when we are not in steady state, then the steady state populations are obtained with the time constant, which is given here. The time constant is not just gamma, it's gamma times these trigonometric functions. So these are simple ingredients. But by using those ingredients we can get a very nice formulation for the stimulated force. So what I've shown here is, and this is the situation we want to focus on for a while. Assume you focus a laser beam. Then plot it as a function of position. The atom is outside the laser beam, it goes into the laser beam where it experiences the electromagnetic field. And if it goes further it has crossed the laser beam and is out of the laser beam again. So what happens is the splitting between the two dressed energy levels is the generalized Rabi frequency. And this is outside the laser beam, simply the detuning. But then when the laser beam is on, the generalized Rabi frequency gets larger. We have a larger lar splitting between the dressed energy levels, and so on. What is also important is that if the laser is red detuned, which is here, the lower dressed level correlates, corresponds to the ground state. Whereas for blue detuning the ground state in this limit, the ground state correlates with the upper dressed level. So it's important. So therefore an atom in this situation-- because atoms are always more in the ground state than in the excited state-- an atom will have more population in the upper dressed level. And here it will have more population the lower dressed level. So what we want to do now is we want to rewrite the stimulated light force, or the dipole force, as we want to express it through the forces acting on the dressed atom level. Well, I told you that we have a splitting of the dressed energy levels by the generalized Rabi frequency. So therefore, if I just cover that, the energy level of the two dressed levels is h bar over 2 plus minus the generalized Rabi frequency. And the force if an atom is in one of the dressed levels is simply the gradient of that. So therefore we have the picture that the two dressed energy levels, atoms in those two dressed levels, always experience opposite forces. Then an expression for the net force is that the average dipole force is the force in dressed level one times the probability that the atom is in dressed level one. And the same for two. And if you sum it up we get exactly the expression we have derived earlier for the stimulated light force. But let's just sort of look at it in the following picture. If we are blue detuned, and let's assume we are not saturating the transition, so therefore the atom will be mainly in the ground state. So you have more population in that state. But, well, if the laser power is 0, we have 100% population here. But when the atom moves into the laser beam there is a superposition of ground and excited state. But that also means there's a superposition of upper and lower dressed level. But we have always, of course, more population in the dressed level which connects to the ground state because an atom will-- well, it also comes out of the solution. But it's natural, just assume you make the laser beam weak, you start in the ground state and everything is sort of connected. The ground state is simply dressed by a small excited state contribution. So that explains it, why for blue detuned light, the physics is mainly in the upper dressed level. And we see the repulsive dipole force of a blue detuned laser. Whereas, for red detuned light, the dominant population is in the lower dressed level, and we have an attractive force. Well that's simple. We haven't gained so much, but you'll see, we can now rapidly gain insight by take it to the next level. Let me first mention a trivial example. And this is when we are on resonance. When we are on resonance we have equal population in the two dressed states, they are resonantly mixed. And therefore if the population in the two dressed states is the same, but the atoms experience opposite forces, that means the dipole force average out to 0. In a few minutes I want to calculate for you what the heating is. The dipole force is 0. But you know the atom in this radiative cascade is making transitions between upper and lower dressed level. So what does it mean for the atom? When it is in one dressed level it experience the force from this side. If the atom is in the other dressed level, it's experience a force from that side. So the atom is actually subject to an average force which is 0. But if you take these picture literally, and you can, the atom is really whacked left, right, left, right, left, right. And there is fluctuations. And we can use those fluctuating forces to immediately calculate what is the heating. The force average is 0, but the fluctuations of the force cause heating. So that's immediately the next step of physical insight we will get out of this picture. But before I do that I want to do something else. I want to discuss what happens now to the force when the atom is slowly moving. Remember, there is still one phenomenon which I owe you an explanation for. And this is when we discussed molasses. We always had the effect in optical molasses with a spontaneous light force that we need red detuning of the laser beam to get cooling. The simple picture is the atom absorbs a red detuned photon and emits away a resonant photon. And then that energy difference is taken away from the kinetic energy, and this is cooling. But then I told you that when you go to higher power of the standing wave, the sine of alpha of the friction coefficient reverses. And that means cooling is now done by blue detuned light, heating is done by red detuned light. And that's what we want to understand. So in order to get any cooling effect, or heating effect, we have to now change the equation we had for the average dipole force. Remember also, the average dipole force, we did that earlier, can be derived from a potential. It's a gradient over potential. So therefore, when you fly through a laser beam you go down and up or up and down the potential, but nothing has happened to you. But the assumption we had for that is that the dipole force was calculated with a steady state population. So now we want to say if the atom is slowly moving it may not be fully adiabatic, and we want to calculate the first correction to the non-adiabaticity. So in other words, we have the following picture. When an atom flies through the laser beam it needs a certain time to adjust its population. And the question, what is this time? Well, it is the relaxation time for the populations among the dressed states to adjust. And this is exactly the inverse of the rate gamma pop, which we have calculated earlier. So the physical picture now is that in the dressed level one and two, the population at any given moment is almost the steady state. But there is a time lag, a time lag by the time it takes to relax. So in other words, the atom is always a little bit behind in adjusting to the laser field because it takes a finite amount of time. So let's now try to understand, in this picture which we have obtained with a laser beam the two dressed levels and the atom flying through, let's try to understand what this extra lag time means. So the atom is flying into the laser beam. It starts out in the ground state, which means 100% in the upper dressed energy level. Now, in the laser beam the higher the laser power is, the more there is a mixture of the other dressed energy level. But what happens is, when the atom is here, when it just assume it flies in, and there was not enough time to adjust the population, that would mean that the atom-- and this is indicated with the red dots-- that the atom has a little bit of more population in the upper dressed level than it would have in equilibrium But what does it mean for the force now, if there is more population here than there? Well, the more population you have in the upper state, the more you're climbing up a hill, the more you get a force to the left. So this red arrow shows due to the non-adiabaticity, due to the time lag, there is a little extra force to the left hand side. So this is on the up hill. Let's now understand what happens on the down hill. Well, we just have to keep track of signs. Here, the atom, in a fairly high laser field, is in the upper dressed level with a mixture in the lower dressed level. But now it flies out of the laser beam, but it cannot immediately adjust its population. And that means it wants to have now more population in the upper dressed level but it hasn't fully adjusted. So therefore, instead of having the blue bullet, which represents equilibrium, the atom has the red bullet. It has a little bit lower population in that, and higher population there. But lower population here and hire population here means-- if you have a little bit higher population than you should have here it means you are climbing up the hill, and there is an extra force, which is again, to the left. So we realize the n the time lag, means when an atom goes up the hill there's a little bit extra force to the left. And when it goes down the hill there is a little bit extra force to the left. And after it has flown through the steady state part of the force roller coaster up down, which can be derived from a potential, has integrated to 0. But what is left is this non-adiabatic component of the force which is slowing down the atom. In other words, I have shown you in this simple picture that the time lag in adjusting the dressed energy population means that there is a small differential force which is slowing down the atom. And what I assumed here was blue detuning. So we have cooling, we have friction for blue detuned light when it comes to the stimulated force. Now this picture, this equation that the population is a steady state at the given position minus a time lag, we can solve it. And I'm not doing it here because it involves a few mathematical equations. But by just using these answers, and the rest is straight forward, you can then arrive at an expression for the stimulated for the dipole force. Colin? AUDIENCE: So, in the spirit of last time, keeping track of the energy, the energy here would be going into the fact that now I have-- if the atom's a harmonic oscillator it's a little behind the drive field? PROFESSOR: Yes. AUDIENCE: So you imprint some fluctuation on the laser beam, and that's where the energy's being carried off, in the laser beam? PROFESSOR: Well the energy's carried off by the laser beam. I have actually prepared something which I want to show you in 10 or 15 minutes, where the energy goes. So let's maybe do that in a moment. So this is a physical picture. And I think what is also important I want to later tell you, for other laser cooling mechanisms, which are too complicated to describe, but they also have a time lag. And I think, I try to teach you basic concepts by using this simplest physical picture I can have. And what can be simpler than two levels of an atom and one laser beam? But what you understand here with this stimulated force, which is also called blue molasses, is for the first time how a time lag in adjusting to the local laser field, how that leads to cooling. I'm summarizing here, simply taking this picture, this equation, and you know, turning the crank on it. The paper by Jean Dalibard and Claude Cohen-Tannoudji calculates what is the dipole force in steady state. We get almost the same result with the Optical Bloch Equations, but we are discussing here the limit that the Rabi frequency is larger than gamma. So therefore, if you're missing occurrences of gamma, they don't appear here because we've always assumed omega is larger than gamma in this picture. Now I want to apply this picture now, not just going through one hump of the laser field, but going through a standing wave. So we have a standing wave of light, which is blue detuned. Of course, in a standing wave of light, if you ever reach over one optical wavelengths, you go up the hill, you go down the hill, the lambda average dipole force is 0. But now you find that if you put in this time lag, you find a friction force, with a friction coefficient alpha. And this friction coefficient is cooling for blue detuned light. Now I don't want to elaborate on this expression. But if you take the treatment of Jean Dalibard and Claude Cohen-Tonnoudji, which is very transparent. It reads wonderfully easily, as the book Atom Photon Interaction. In the limit of large detuning, which is the simplest possibly limit, you find the friction coefficient which looks highly nontrivial. It depends to the sixth power of the Rabi frequency and such. The only reason why I put it here, in 20 minutes I want to give you a very, very simple picture where I derive that for you. So that's just a check mark. Any questions right now about-- because I want to go on-- about the stimulated force in the dressed atom picture? And if we allowed deviations from the steady state population we get a cooling effect, and this explains the effect of blue molasses. And we've applied it to a standing wave. Cody? AUDIENCE: Just [INAUDIBLE] the upper alpha are a friction constant and the lower one is the polarizability? Or, something. PROFESSOR: No, oh, thanks for asking. Alpha, yeah, we [INAUDIBLE]. Here it's the friction coefficient. Here, remember when we had the light force-- the spontaneous light force and the stimulated light force-- one was directed along a vector alpha, and the other one along a vector beta. Alpha was pointing into the direction of the gradient of the laser intensity, and beta was the phase gradient of the laser beam. So those two have nothing to do-- this is friction coefficient. This is the direction for the stimulated light force. And yes, a little bit later today, I will use alpha again, but for the polarizability. Yes. AUDIENCE: The [INAUDIBLE] track the atoms in the [INAUDIBLE] standing wave, [INAUDIBLE]. PROFESSOR: You're talking about trap in here, that if you have a standing wave of blue detuned light-- well if the standing wave is infinitely extended then the atom is always in the standing wave. But if you use two Gaussian laser beams to form a standing wave, there is always a net force to expel the atoms away from blue detuned light, because blue detuned light has an average dipole force which is repulsive. So whenever you want to manipulate atoms with blue detuned light you need an additional magnetic trap, or an additional red detuned optical trap. Because whatever you want to do with atoms, if you want to do it with atoms in steady state you have to keep the atoms together. And what keeps the atoms together is red detuned optical traps, or magnetic traps. The blue detuned light can be used to create a blue detuned lattice. Or it can be used at least, if you want to do laser cooling with blue detuned light it can be used for laser cooling, as I've started to describe for you now. But yes, I mean, in that sense as an experimentalist, you can run experiments where you have only red detuned light-- red detuned light for trapping, red detuned light for cooling. But if you use blue detuned light for a standing wave or for blue molasses, in addition you will always use red detuned light to keep the clouds together. So pretty much every experiment usually has some red detuned light for confinement. And you have to then match blue detuned light with red detuned light to get the two things you want. We can now use the dressed atom picture to calculate the heating. Remember the final temperature you can reach was always the ratio between a heating coefficient, which is actually the momentum diffusion coefficient, over alpha. So right now we have understood the alpha part. We know the physical picture, why alpha provides cooling, why we have a friction force for blue detuned light. But let's now take it to the next step, and calculate what is the heating. And I mentioned it already in the context of simple molasses, which we discussed, I think, two weeks ago, that we describe heating by the momentum diffusion coefficient. Well the momentum diffusion coefficient tells you that P square is simply growing as a function of time. If you have a cold cloud and you heat it up, you can describe the heating by the temporal growth of P square. If you don't like momentum, divide P squared by 2m you have the kinetic energy, which increases. So it is this momentum diffusion coefficient. And if you have a cloud which moves with an average momentum, you only want to have the P square minus P average square. But if you think about a cloud which is being cooled, the second term is 0. So the momentum diffusion coefficient is the derivative of P square. But the derivative of P is simply the force. So therefore we can absorb the derivative by replacing P-- at least one of the P's-- by the force. And we can now get rid of to second P of the second occurrence of momentum by saying well momentum is derivative of force. Sorry, derivative of momentum is force, or momentum is the integral of force. So we can exactly rewrite that momentum diffusion coefficient by an integral over correlation function of forces. And then we have the nice physical picture that with increase of kinetic energy, the momentum diffusion coefficient comes because we have fluctuations of the force. Yes? AUDIENCE: With the last time, the previous page the factor [INAUDIBLE]. PROFESSOR: The factor of 2 here? AUDIENCE: No, the next slide. PROFESSOR: Here is 2. AUDIENCE: For both terms. PROFESSOR: Here is 2. And here it disappears, because we have a factor of 2 and a factor of 2, and the two cancel each other. AUDIENCE: No, I meant, there's a 2 multiplied the first average. And there's a 2 for the second one. Are those for the first term? PROFESSOR: I think this is correct. AUDIENCE: [INAUDIBLE]. PROFESSOR: What happens is P-- you use-- AUDIENCE: [INAUDIBLE]. AUDIENCE: No, it's supposed to be-- PROFESSOR: Wait, is one missing? Maybe [INAUDIBLE]. I think you're right. AUDIENCE: [INAUDIBLE]. PROFESSOR: Let's assume this is what it should be. So here is the reminder, the kinetic energy. If you have a process which causes momentum diffusion, the momentum diffusion coefficient gives you the heating rate. And the cooling rate is given by our friction coefficient alpha. And as we discussed in the context of molasses-- of red detuned molasses, but it applies to any cooling scheme-- the finite temperature is when you have detail balance between heating and cooling And therefore its ratio of the momentum diffusion coefficient over the friction coefficient. Kensie? AUDIENCE: In the previous step, what was the significance of choosing f of 0 basically? Like, the time arguments were 0 and D. So we said that f of 0 was DPDT? PROFESSOR: Oh, we assumed that the laser beam's Hamiltonian is invariant against time translation, and we simply evaluate that at T equals 0. I mean, that's of the argument you have fluctuating forces. And you could have fluctuating forces at time T and time T prime. But then you use the augment that in a steady state solution the correlation function does not depend on T and T prime separately, but only on the difference. And then you simply set one argument to 0, and the other argument becomes a time difference. Other questions? So now, by understanding that momentum diffusion is nothing else than the integral over the force fluctuation, we can now use this physical picture which I gave you earlier that an atom which cascades through upper, lower, upper, lower dressed states will experience opposite but equal forces, form the right and from the left. So I can simply evaluate that by assuming we are on resonance. On resonance the force is h bar over 2 times the derivative of the generalized Rabi frequency. And the force correlation, what is it? Well, it's the force f of 0 times f of T. But as long as the atom is in one dressed state the force doesn't change. So I simply take here the force, I take the force squared, and then when I integrate over that, well the moment the atom makes a transition there is no longer any correlation between the force. Everything becomes random and the integral averages out to 0. So in other words, that's the way how you should look at this integral. You start with the force f0, as long as the force is correlated. Because the atom hasn't done spontaneous emission, you have sort of f of 0, f of 0. And you integrate it, nominally to infinity. But you simply integrate it until some spontaneous emission randomizes the system. So therefore, this integral over the correlation function is nothing else then the force squared times the typical time scale for a transition. So if you put that in on resonance, we have the force here, we square the force. And then we multiply with the spontaneous emission rate, which for resonance system is gamma over 2. And with that we get now the heating coefficient, the momentum diffusion coefficient for an atom which is exposed to resonant light due to the stimulated light force. The paper, the references I've given you, is simply taking this augment and extending that to arbitrary detuning. For arbitrary detuning, I told you what happens on resonance. The atom spends equal amount in each dressed state. If you detune the atom spends more time in one dressed state, then quickly in the other dressed state, and then more time in the other dressed state again. So you can, by simply understanding what are the rates for cascading between the dressed levels, you could pretty much write down the expression, find it, and here is the solution taken out of the paper. So we have to correct prefactor, but then if the detuning is non-vanishing we get a contribution from the second factor. Colin? AUDIENCE: Are you assuming the exponential correlation-- exponential behavior to the correlation. Because then you might have different prefactors, if it were something like if you were in a [INAUDIBLE]. PROFESSOR: Well what happens is the following. I think if you're in one dressed state and the force pushes you in one direction and you integrate over it you have an exponential decay. But instead of integrating over the exponential decay you can just take the force squared and multiply it with the average time the atom stays. In other words, if I have an exponentially decaying function I can always approximate it by the function of T equals 0 times the time. And what you are asking me now is is the time the one over E time, or is it an average time? What is the correct time to use here? AUDIENCE: I guess when the time scale depends on how the correlations decay for [INAUDIBLE] different. PROFESSOR: Yes, OK, I agree. The shape of the decay of the correlation function would matter. I'm not going into that. But what I'm sort of assuming here is that the correlation function indicated with time is the correlation function of T equals 0 times the correlation time. And how the correlation time is related to the exponential time, or if there's a small prefactor and such, that may depend on details. Yes, you're right. I've swept this under the rug. How are we doing? Good. Let me just summarize. The dressed atom picture has given us two pieces of major insight into the stimulated force. The first one is that the atom experience is always opposite forces in the two dressed levels. And it is the population imbalance between the two dressed levels which results in a net force. And this requires either red detuning, then the force is repulsive, or blue detuning-- sorry, red detuning then the force is attractive, or blue detuning then it's a force repulsive. Second, with this time lag thing, we understand that time lag, non-adiabaticity, is the way to get cooling. And the third thing we have learned, when the atom does a radiative cascade the force-- when it goes from the upper to the lower dressed level-- the force is suddenly reversing it's sign. And this causes fluctuations, this shakes up atom, and this causes heating. And this has nothing to do with the other heating mechanism we discussed earlier, namely every time there is spontaneous emission there's a random recoil. In strong laser fields the fluctuations of the stimulated force, which has nothing to do with spontaneous emission, are much, much larger than the photon recoil. So therefore for strong laser fields, it is this alternation of the force when atoms switch dressed energy levels, which is responsible for most of the heating. I want to come now to another physical situation, which is actually wonderful. You learn about Sisyphus cooling. Everybody in cold atoms speaks about Sisyphus cooling because it's a very, very elegant cooling scheme. We usually apply in the laboratory Sisyphus cooling when we do polarization gradient cooling. I will say you a few things later. But this is much, much harder to understand than the Sisyphus cooling, which takes place for a two level atom moving through a blue detuned standing wave. So in other words, I try to explain now to you what Sisyphus cooling is. The relevant application is multi-level polarization gradient cooling, which I give you an idea, but it's hard to fully describe. But here, in our current discussion on the two level atom interacting with one laser beam, or one standing wave, we find the simplest physical realization of Sisyphus cooling. So just to make sure that you are following the argument, I have so far explained to you why blue detuned light cools in the limit of very small velocities. Very small velocity means we did a first order Taylor expansion in this lag time. But now I want to discuss with you a different regime, namely that the atom moves with a velocity such that it can go up and down several periods of the standing wave before it does spontaneous emission. So before when we discussed the friction coefficient, the velocity was infinitesimal. But now I assume that the velocity is such that the atom can surf a few waves and then it will spontaneously emit. And let me assume that we are blue detuned. So the atom, let's say here, where the splitting is minimum, well this is where we don't have light. The generalized Rabi frequency is just the detuning. So we are in a node of the standing wave. And then the atom starts in the ground state, and it experiences the periodic potential due to the standing wave. But we want to now ask, OK, when will spontaneous emission happen in this picture? Well, it cannot happen here, when the atom is in the ground state, when there is not light. But where there is light, when the atoms move to an anti-node, there is a mixture, indicated in red, of the excited state. So in other words, when an atom does this roller coaster, it will most likely emit a photon-- either the carrier or one of the sidebands-- when it is at the maximum of the potential corresponding to an anti-node where it experiences the light. OK, when the atom emits on the carrier it goes from the upper dressed level to the upper dressed level. Then sort of nothing has happened because the atom is in the same dressed level. Let me now discuss the case when something happens. And this is when the atom decays from the upper dressed level to the lower dressed level. And at least for weak excitation the lower dressed level corresponds to the excited state. In other words, when there is no light, the upper dressed level was ground state, the lower dressed level is excited state. But what happens now is actually interesting, the atom had to climb up the hill. But the spontaneous emission to the lower dressed level takes the atom to the bottom. So you can see, on average, it has climbed a hill here, but now it is at the bottom. And now we're going to ask how does the atom get out of the lower dressed level? Well, the lower dressed level is mainly excited-- is 100% excited if there is no light. But if there is light, and light means the generalized Rabi frequency is larger, the excited state has now in it mixture of the ground state because the light mixes ground and excited state. So now, of course, whenever there is red you can spontaneously emit. But there is a probability that the atom is more likely to emit when you are in a purely excited state, and not in a mixed state. So now, talking about the lower manifold, the atom does it's roller coaster. It has a velocity that it can cover several periods of the standing wave in one lifetime. But there is a higher probability for the atom now to emit when it is on top of the hill. So that's remarkable. We have the upper dressed level. We have the lower dressed level. The atom is doing the roller coaster. But whenever it emits a sideband, whenever it switches state, it mainly does it by emitting at the top of the hill. And after the spontaneous emission it finds itself at the bottom of the hill. So on average the atom is climbing more up hills than down hills. And this is called Sisyphus cooling. I think you've all heard about the Greek myth of King Sisyphus, who was condemned. He challenged the gods and he got his punishment. And his punishment was that he always has to roll a stone up the hill. And when he's done the stone falls down and he has to roll the stone up. So in other words, we have condemned the atoms to the same verdict. They always have to go up hill to work. Then they fall down hill with the help of a spontaneous photon. But then they have to climb up hill. And it is the uphill climb where the experience and net friction force, which cools the atoms down and reduces their kinetic energy. So this is Sisyphus cooling. And since I need it in a few minutes, I just wrote down for you, based on this simple and wonderfully elegant picture, how much energy does the atom lose per unit time? What is the cooling rate? Well it's clear that the cooling rate comes when the atom switches from one dressed energy level to the next, because nothing happens on the carrier. So it is the transition state between dressed level one and dressed level two. And then we multiply with mu naught the depths of the optical lattice, because it is this energy difference which is extracted from the atom. You may ask, OK what is time limiting? It's really the transition rate form the first dressed level, which is mainly ground state, to the other dressed level, which is mainly exciting state. Because once the atom is in this level, which is mainly the excited state, it will pretty much do the next transition at a rate which corresponds to the spontaneous emission rate. So in other words, if you're not fully on resonance, or if you're not situated in laser power, the rate, one, two, is slower than the rate two, one. And therefore this rate is the rate limiting step. Let me show you just a few pictures. What I'm just explaining to you was explored very early on, just have to sort of put you back that it was in the mid '80s, '82, '83, '84, that people developed [INAUDIBLE] slowing. It was in '85 that optical molasses with red detuned light was demonstrated for the first time. It was in '86 that the dipole trap was demonstrated by Steve Chu. And in '87 that Steve Chu and Dave Pritchard introduced the magneto optic trap. So the mid to late '80s were fantastic times, where one important technique was realized and demonstrated. And this was actually the first major experimental paper of a new laser cooling group, which was founded by Claude Cohen-Tonnoudji at the Ecole Normale Superieure in Paris. And it was exactly about cooling atoms with stimulated emission, and this is blue molasses. And, I mean, you exactly see here out of this paper, the Sisyphus cooling mechanism. And addressing the question about red detuned light here, this was actually beam experiment. You head a blue detuned standing wave, and you were sending an atomic beam through. And if this atomic beam was cooled, the divergence of the atomic beam was reduced. So this experiment had only one blue detuned standing wave and nothing else, because it was not a trapping experiment. It was a beam collimation experiment. And what is shown here is that you have the original atomic beam profile. And for blue detuned light you make it narrower, so this is cooling. Whereas for red detuned light you have a defocusing, which corresponds to heating. Any question? I want to wrap up the discussion of the stimulated light force by discussing pretty much everything again, but in the simplest possibly limit. That's actually something I haven't found in any text books. But when I derived it for myself it helped me to really understand it. It's sort of the perturbative result of the dressed atom picture. So everything is simple. And by telling you, look at this. You know this is just EC stock shift. But in the dressed atom picture it's a generalized Rabi frequency, you suddenly understand the trivial perturbative result and how it translates into the dressed atom picture. So what I'm presenting you now in the next 10 minutes, it provides a lot of insight because it connects simple pictures. So I'm discussing things here in the limit that the detuning is large. The detuning is larger than the Rabi frequency, and spontaneous emission rate is the smallest of the three rates. In that case, I can simply do perturbation theory, and I want to show you. Also, because this is pedagogical, I just want to show you how the effects come together. I neglect all factors on the order of unity. And I set h bar to 1. So let's assume we have a standing wave. A standing wave is cosine KX. So we have an electric field. But we describe the electric field with a Rabi frequency. So the Rabi frequency forms a standing wave. Now what are the two dressed states? The dressed states 1 and 2 are the bare states, ground and excited state. And then there a perturbative mixture. And in first order perturbation theory we take the matrix element over the detuning. So one dressed state is the ground state with a little bit of excited stated mixed. The other dressed state is the excited state, minus sign, little bit of ground stated mix. These are our two dressed states in the trivial limit that we can apply perturbation theory. What are now the transition rates between the two dressed states? Well, we want to make spontaneous emission from dressed state 1 to dressed state 2. We have to go from this small mixture, which is the excited stated mixture, to the ground state here. So therefore we get the product of the two amplitudes, and then we multiply with gamma. Of course, most of-- yeah. So this is the rate to go from dressed state 1 to dressed state 2. What is the inverse rate to go back from dressed state 2 to dressed state 1? Well, you go from the excited state to the ground state, multiply with gamma, and that's what you get. Sure we have, in perturbation theory, the coefficient is 1. It's only in higher order perturbation theory that the coefficient of the bare state becomes less than 1. So isn't that simple? We just discussed it with all the cosine to the 4, sine to the 4. But now in perturbation theory this is the rate for going from dressed state 1 to 2. And this is the rate to go the other way around. So what is our dressed state potential? Well, the dressed state potential is simply the EC stock shift, which is opposite for the ground and for the excited state. And the EC stock shift in perturbation theory is nothing else than matrix elements squared over detuning. So this is mu naught. This is the potential of the standing wave experienced by the atom. Now we want to do Sisyphus cooling. What is the cooling rate in Sisyphus cooling? Well, remember, Sisyphus cooling we assume that the atom is fast enough to go over several hills and valleys. The cooling rate is determined by the rate at which the atoms switch dressed levels times mu naught. And I can scroll back, but mu naught was the EC stock effect, and gamma one, two was this expression we had before. So therefore, in the simple picture, while everything is perturbative, we have a nice expression for the cooling rate. Well, now let's get something non-trivial out of it. What is the friction coefficient? Well, in many situations I've plotted for you, the force versus velocity. Now I want to plot for you the cooling rate versus velocity. That means its force times velocity. And as we just discussed, when the velocity is sufficiently fast that the atom can go over several hills and valleys in a spontaneous lifetime, the cooling rate will saturate because it is limited by the rate at which the atom switches between dressed levels. But now the question is, if we want to understand what happens at low velocities, how should we connect the two? Well, you can say we know, for analytical reasons, that the force versus velocity has to be linear. So the cooling rate, which is another power of velocity, should be quadratic. And since I neglect all factors on the order of unity, this is now the force is friction coefficient times V. So therefore, what I do in red here is a parabolic approximation, which is alpha the cooling rate times V squared. And now, if I want to know what is alpha, I just sort of connect this parabola with the saturated value, which I know. And I know the transition happens at the velocity when the atom moves further than one wavelength in one spontaneous emission time. So therefore, based on this very physical picture, and on the inside how low velocities and high velocities connect together, I can actually obtain the friction coefficient in this blue detuned standing wave by taking the cooling rate and dividing it by the velocity at which the atom moves one wavelength per spontaneous emission time. And if I do that, I obtain this result, which I have already quoted earlier to you from the paper of Cohen-Tonnoudji and Jean Dalibard. It looks fairly non-trivial with power to the 6, power to the 5. But this is now the simple perturbative results for the friction coefficient. Now we have the friction coefficient. Let's take it further. I want to give you, in this perturbative limit, I want to give you the momentum diffusion coefficient. And then we will find in the perturbative limit what is the lowest temperature to which we can cool. So pretty much I'll do everything as we've done before, but I use the simple perturbative limit. So again, in order to calculate the diffusion coefficient-- the momentum diffusion coefficient-- and the heating we need the fluctuations of the force. And now, just remember, we are in the perturbative limit. The atom is mainly in the ground state, which is the upper dressed level. It has only a little excited stated mixture, so it will mainly stay in this state. But then with this excited stated mixture and this, power of the Rabi frequency, it makes the transition to the other dressed state. But the other dressed state is almost 100% excited state. So it will leave the other dressed state almost immediately. So this could be a Quantum Monte Carlo wave function result that the atom experiences the force in the upper dressed level. Then it goes to the lower dressed level, which is mainly the excited state, just 4 times 1 over gamma. And then it returns. So the picture now is that the atom for almost all the time experiences the steady state force, which is the blue dashed line. And there are only, when it goes to the other dressed state, these small spikes where the force has changed sign. And if you now calculate the fluctuations of the force-- the integral of f square-- once we calculate the square the small deviations from the average force does not contribute. What contributes are the spikes. So all I do is now I take the 4 square in one of the spikes. And the force is nothing else than the derivat-- remember we have a standing wave a mu naught. The derivative of the standing wave is the force. But if you have a sinusoid lattice the force is K times the amplitude of the lattice. I square it, so this is the spike squared. Then I have to multiply with the time 1 over gamma, over which the force stays the same. So this is, you can see, the correlation time of the force. But then I have to multiply, also, when I do an average over many, sort of, you can see, trajectories, I have to average over the probability that-- I have to multiply with the probability that the atom is in the dressed state 1 and not in the dressed state 2. It may take you a few minutes just to think about the combinatorics. But in the end all I do is I take the 4 square, I multiply with correlation time when the atom is in state 2, and get the probability that the atom is in state 2. And when the atom is in state 1, in my ensemble, it doesn't contribute at all, because in state 1 you're so close to the average force that it doesn't contribute with the fluctuation. Anyway, it's simple but subtle. But it's a one-liner. And now based on this perturbative picture we have perturbatively exact an expression for the heating rate. But now it's interesting, the rate gamma one, two, the rate to switch between dressed energy levels always also appears in the expression for the friction coefficient. So therefore, if you're now asking what is the ultimate temperature to which we can cool the atom? It is the ratio of this analytic result for the heating coefficient divided by the analytic result for the friction coefficient. Pretty much everything cancels out, and what remains is mu naught. And this is a remarkable result. It's highly non-trivial, and I really enjoyed showing to you how simply it can be derived by-- it's really just perturbation theory. But you have to put in the right concepts from the dressed atom picture. So what we learn is the following. If you have a standing wave, and we cool with a stimulated force, the lowest temperature is-- with a prefactor, which I haven't calculated-- the amplitude of the standing wave. And that means that it is impossible to ever localize atoms within a standing wave. Laser cooling cannot be used to create atoms in an optical lattice. The temperature of the atoms is always comparable to the depths of the lattice. So therefore the atoms will never be localized in one well of the lattice. And this is generally valid. The assumption we have made here is that we have a two level atom. A lot of you work with atoms to optical lattices. And you often use evaporative cooling, take a Bose Einstein condensate, which is pretty much a [INAUDIBLE] temperature put it in optical lattice. But before '95, people studied laser cooling and atoms in optical lattices. And this was only possible because atoms can be cooled to lower temperatures than we have derived here because they are not two level atoms. So laser cooling in a standing wave, localization of atoms in a standing wave by laser cooling, is only possibly by physics beyond what we have discussed. Colin? AUDIENCE: If you use something like Raman's sideband cooling, you could-- like, yeah, you'd kind of need three levels for that, but you eliminated the third. Really, it's just that it's two levels, right? PROFESSOR: If you do Raman's sideband cooling, he posted-- I think, in the next 10 minutes, I cover what you want to see. Because now it involves two ground states. And I want to show you, if you have two ground states, to Hyper fine-- Raman's sideband cooling-- if you have two different ground states there is no physics. And I briefly want to touch upon that. We'll just kind of fill in one thing here, and it is, well, if you take everything I've told you seriously, you would say OK, too bad. We cannot localize atoms in an optical lattice because the temperature is always comparable to the amplitude of the lattice. But by making the lattice smaller and smaller and smaller I can reach lower and lower temperatures. Well, yes, the some extent that's possible. But what we have done here is we've done approximations. We've only focused on the stimulated force. We have not looked at the recoil which comes in spontaneous emission. And therefore, the moment we would make the optical lattice smaller than gamma then we have to include heating by the spontaneous light force. And the result will be that blue molasses cannot reach lower temperatures than red detuned molasses. And for red detuned molasses we have already discussed the famous Doppler limit, which is gamma over 2. So that's what I wanted to present to you with the stimulated force. First, a more general discussion using the real dressed atom picture. But now in the end, to fly over using sort of the baby dressed atom picture, which is simple perturbation theory. OK, so we've done the hard work. Now we want to have some fun discussing the physics. And I'm realizing I've only five, six, seven minutes left. I want to discuss three points. One is I want to give you all the different pictures to find the potential of a dipole trap, because we've encountered several. The second thing is I want to have a discussion with you if the stimulated force is due to electric or magnetic forces. And eventually, and we'll be running out of time, I have to do that on Wednesday to discuss the question of energy conservation. If you cool the atoms where does the energy go? So if you use the stimulated force, it gives rise to dipole potential. And using the Optical Bloch Equations or the dressed atom picture, we found this expression for the dipole potential. And there are at least four ways how we could have derived it. One is the Optical Bloch Equations, we did that. The dressed atom picture, we did that today. Another picture is very straightforward and simple. It just says that the potential is the electric potential of a dipole. And then all you have to know is from the EC polarizability, if you drive an atom with a laser field below resonance-- I mean that's just harmonic oscillator-- below resonance the harmonic oscillator is in phase with the force. Above resonance, if you drive a harmonic oscillator fast, the response of the harmonic oscillator is in opposite, it's out of phase. So therefore, if your potential is minus the E, you are in phase-- it is a negative attractive potential. If the harmonic oscillator is driven above resonance there is another minus sign, you get plus the E, which becomes a repulsive potential. So it's, look and say, just the physics of a harmonic oscillator. There's one thing I like, which is sometimes not taught in an atomic physics course because it is the force on a macroscopic dielectric object. So let's assume we have an atom-- no, we don't have an atom. We have something which is bigger. Let's assume a tiny polystyrenes sphere. You can assume a small sphere of glass. What happens if I shine a laser beam on a small sphere of glass? Well, let me assume that the laser beam is-- well, if the laser beam hits this sphere right on, it's symmetric, there is no force. So what we want to do is we want to have a laser beam, shown here by this arrow, which has a maximum here so the profile is like this. And we want to discuss whether this sphere is sucked into the laser beam, this would be an attractive force, or whether this sphere is repelled. And all we want to use is optics, the optics of refraction. When we have a laser beam which is weak, medium, strong, strong here, and for red detuning we have an index of refraction which is larger than 1. So the sphere acts like a lens. So that leaves those photons up and in, and those photons up and in. And when a photon is bent in, the momentum of the photon has changed. And the momentum, of course, the opposite momentum transfer, is imparted to the sphere. So in this case, we have sort of-- we have more photons in this part of the laser beam, which are bent up, than photons are bent down. And this means the net force is that the sphere is sucked into the laser beam. And if you go for blue detuned light, where the index of refraction is smaller than 1, it is the opposite. So the picture which we have discussed here was for atoms form microscopic objects. But you have, actually, similar physics of a stimulated force for mesoscopic and macroscopic objects. And then what matters whether you trap or anti-trap, is whether the index of refraction is larger than 1 or smaller than one. And you know, two level system for blue detuned light has an index of refraction which is smaller than 1. And for red detuned light our atomic clouds have an index of refraction which is larger than 1. But this picture that you can use the same set up, the same focus beam and a similar dipole force to trap larger objects is actually exploited in a different field of physics, in biology, in the form of optical tweezers. When actually in the early '80s, when Steve Chu came to Bell Labs and had discussions with Art Ashkin, they actually discussed optical trapping, both of atoms and both of biological objects, little microspheres, or cells. And it's sort of remarkable that both approaches were realized within the same year. So atoms were trapped in '86. And macroscopic objects were manipulated, I think it's exactly the same year. And as you know, the optical tweezers is now an important technology in biology and biophysics. For instance, you can focus a laser beam and manipulate objects within a living cell. So I think I should stop here, but maybe take your questions. Then let me conclude with an outlook. We have this week and next week, and then the semester is over. So this week we have one class on Wednesday. I want to finish this discussion and teach about magnetic trapping and evaporative cooling. And the last week, because of a makeup class on Friday, we have three more classes, Monday, Wednesday, Friday. By the way, I just learned that the Friday class is in these room. And on Monday, Wednesday, Friday we talk about, well, it's really featured pick about Bose gasses, lecture one, Fermi gasses, lecture two. And the third lecture is about quantum logic with ions. OK, see you Wednesday.

MIT_8422_Atomic_and_Optical_Physics_II_Spring_2013

17_Dressed_atom_Part_2.txt

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Ready to roll? OK, good afternoon. Count down four lectures and you know everything you have to know about atomic and optical physics, at least for those of you who take part two-- for those of you who've taken part one and part two-- for the other ones, well, there is one more semester which is the following spring. So we have discussed light forces, first with the optical Bloch equations, and then we discussed the stimulated light force using the dressed atom picture. So this is sort of finishing the presentation and derivation of they light forces. Today, I want to continue to discuss for the aspects. We've discussed dipole traps, different ways how we can understand why there is a dipole potential from the harmonic oscillator to the dressed atom to the refraction of light by small spheres. So in all cases, we realize red detuned light traps. Today I want to discuss what is at work here, electrical magnetic forces, and then I want to address what several of you have asked, where is energy conservation? Where does the energy go when we cool with a stimulated force? But this will only take 10 minutes. The main part of today's lecture will be techniques to go to very low temperature to ultra-low temperature. This is sub-Doppler cooling, sub-recoil cooling, evaporative cooling, and magnetic trapping as a technique. So this is the menu today. Again, a feature-picked menu. Now talking about electric and magnetic forces, you may think it's a trick question, but it's rather subtle. All optical forces were derived from the electric dipole Hamiltonian D dot E. This is what we used as a starting point for light forces. So my question is, what about the Lorentz force? The Lorentz force comes from the magnetic part of the electromagnetic field. Now, is there Lorentz force on the atoms? And, if yes, does it contribute to the light forces or is it negligibly small? Actually, the question I'm asking you, I don't think it's addressed in any textbook and I think you can go to conferences and ask some of your colleagues. You may get a wide range of answers. So but anyway, has anybody of you thought about it? What about, is there Lorentz force on neutral atoms? If no, you're done. If yes, does it contribute, is it negligible, or is it even dominant? AUDIENCE: [INAUDIBLE] temperature or is equivalent to the T minus A picture, and the T minus A picture has the Lorentz force, so [INAUDIBLE]. So. PROFESSOR: So it should be included because we are talking about two different representations. Collin? AUDIENCE: Yeah. PROFESSOR: OK, so yes. The Lorentz force is included. Also, when you ride on D dot E, you think it's the electric force of an electric dipole. We come to that in a moment. The next question is, do you have any idea if the Lorentz force is important or whether it's negligible? AUDIENCE: For neutral, or? PROFESSOR: For neutral atoms. AUDIENCE: [INAUDIBLE] PROFESSOR: OK, who of you is working on optical lattices or wants to work on optical lattices? OK, just to catch your attention, when the atom goes up and down the optical lattice, if feels the lattice potential. The force is 100% the Lorentz force. There is no contribution from the electric force. So that's what I want to show you now. So we have two fundamental forces, the electric force and-- if you have two charges separated in the field, two different electric fields-- we derive from the Coulomb force, the dipole force. And the force on the electric dipole is-- that potential is-- D dot E. The gradient of it is D times gradient E-- with correct Victorian notation-- whereas the Lorentz force is V cross B. So these are the two forces we have to consider. Let me show two configurations where we use dipole forces, the stimulated light force. One is here where we focus a laser beam, and let's just assume for the sake of the discussion that the laser beam-- it's propagating here and the linear polarization goes up and down. So now we have a dipole moment of the atom, which oscillates up and down, which is parallel to the electric field. So now we have D dot E, and indeed if you take the atom and move it into the laser beam, it will experience an attractive force and this attractive force is purely electric. But let's now come to the situation which many of you work on in the laboratory, would that we have two plane waves-- so the laser beams are infinity extended-- we have two plane waves and they form an optical lattice. Now, the dipole, the oscillating dipole, the electric dipole, is again driven by linearly polarized light. It points up, but the gradient of the electric field is in the direction of the interference of the lattice, so the gradient of the electric field is perpendicular to the dipole moment. So therefore, D dot E is exactly 0, and therefore, the electric part of the potential-- the Coulomb force which is microscopically behind it-- will not contribute anything. So I could stop here and say, well, OK, since we have only the Coulomb force and the Lorentz force, I've proven to you it's not the Coulomb force, so therefore it's a Lorentz force. But let me just tell you why. So what I'm really telling you is it is actually remarkable that if you go along the Z direction and we take the Z derivative of the AC Stock's shift potential, and the AC Stock's shift potential is one half, alpha is now the polarizability-- we use alpha in many places-- times E square. So this is the AC Stock's shift potential, and the spatial derivative of the AC Stock's shift potential is a force, but it's the Lorentz force. Actually, I was amazed. I had to work it out myself because I haven't seen it discussed anywhere, but you can simply take the electric field of a standing wave, perform this derivative, and what you find is you get the Lorentz force. If you wonder how you get it without going through all the Victorian notation, the Lorentz force is V cross B. The velocity of the charge is related to the derivative of the dipole moment. The dipole moment is charge times position, and the derivative of the dipole moment is charge times velocity. So don't let yourself fool yourself. When you have an oscillating field, there are charges which move their currents inside the atom. So our V cross B, the V is pretty much D dot. The D dot is polarizability times E dot, this is alpha polarizability-- frequency times the electric field. But, as Cody said, we can go form P dot A to D dot E-- or, to be precise-- the magnetic field is the curl of the vector potential, and the curl gives us the [INAUDIBLE]-- I'm suppressing Victorian notation, here-- and you know that E, the electric field, is the temporal derivative of A, so therefore we get that. So if you now multiply D dot with the curl of A, you get something which is K alpha E square. K is for a standing wave, the spatial derivative. So I've shown you-- at least you can see by dimensional analysis, or by just pointing to the different terms, that if you take the AC Stock's shift potential through the derivative, you get everything which you have in the Lorentz force. It would be a 10 minute, easy homework assignment, but we're running out of homework assignments-- we're at the end of the semester-- to show that, explicitly. So it is really the Lorentz force which provides the trapping potential in an optical lattice. Questions? [INAUDIBLE] AUDIENCE: So in the simple picture you have [INAUDIBLE] the lattice, you're saying that the transverse confinement is provided by the electric force, but what about the perpendicular direction? [INAUDIBLE] electric field like this, but the other direction? PROFESSOR: OK, I can discuss with you two simple geometries. In one case, in one direction it's the pure Coulomb force at work. In the other direction, it's the pure Lorentz force. But in general, when you go through an AC Stock's shift potential, you have both. So you may wonder that the Coulomb force, when you integrate it, would not be integrable. It does not give rise to potential. Also, the Lorentz force does not give rise to potential. But if you add the Coulomb force and the Lorentz force, you can integrate it up to the optical dipole potential. I find it remarkable, but it is what it is. Questions? So energy conservation, yes. That's a good thing to discuss now. Let me first discuss I want to have-- energy conservation has to be discussed with two different perspectives. One is a transient phenomenon-- which I'll do in a few minutes-- but let me first discuss something, which is CW, and this is the following. You have atoms which move over a standing wave, and I use the dressed atom picture to show you that-- for a strong, blue detuned standing wave-- we have a friction force. So the atom-- this was the Sisyphus cooling-- that the atom is climbing up hills, there's some transition, climbs up the next hill, but on average the atom climbs up more hills than it scoots down, and therefore there is work done. There is a net force, and this net force provides friction. It slows down the velocity of the atom. My question to you is, where has this energy gone? The kinetic energy lost by atom, where does this energy appear? Which fields? AUDIENCE: [INAUDIBLE] the laser beam? PROFESSOR: Well, we have two fields, the laser fields, and we have maybe a spontaneous emission? But the force is only a stimulated force due to-- yeah? Which fields? AUDIENCE: The laser field? Because when you stimulate into a laser field, your [INAUDIBLE] modulator [INAUDIBLE]. PROFESSOR: Well, I have a little bit problem with that because when the atoms-- the force comes from distributing photons from one standing wave to the other one, and these are photons of equal frequency. So in that sense, the net effect is that those two laser beams-- actually, the atom moves, here. So what should happen to the laser beams? Photons are not disappearing. I will later tell you, in the transient picture-- and this, maybe you've already done your homework assignment number 10-- there is something happening to the laser beam. But this would only happen when atoms fly in and fly out of a laser beam. Then we have a different situation. But here, if the atoms stay in the laser beam the laser beam-- so to speak-- sees a constant index of refraction medium. It just goes through and nothing has happened to the photons except that-- occasionally-- photons are spontaneously emitted. Well, OK. If it's not the laser beams, what remains now? AUDIENCE: Spontaneous [INAUDIBLE]. PROFESSOR: Spontaneous emission. Now how can spontaneous emission carry the energy? What is the spectrum of spontaneous emission? AUDIENCE: Triplets? PROFESSOR: Yes, the mono-triplet. We have the central carrier, the central carrier is not carrying away any energy because it is at the laser frequency. Now we have two side bands. Can the side bands carry away energy? AUDIENCE: Interlacing them. PROFESSOR: Huh? AUDIENCE: Interlacing with the [INAUDIBLE]. PROFESSOR: Asymmetry in what sense? AUDIENCE: If the side bands are [INAUDIBLE] escape. PROFESSOR: Well, but didn't we discuss, with a radiative cascade, that the blue detuned side bend comes when you go from the upper dressed level to the dressed level? And in the lower dressed level, you can only emit a red detuned side band, not a blue detuned side bands. What we said in the radiative cascade after one blue detuned photon, there is a red detuned photon. So the number of photons emitted on the red side band and on the blue side band are exactly equal. [INAUDIBLE]? AUDIENCE: But then you have the [INAUDIBLE] cooling picture, where the red detuned side then was emitted in the lower field region? PROFESSOR: Yes. So the situation is the following. The atom emits an equal number red, blue. Red detuned and blue detuned, red detuned and blue detuned. But when it emits the blue detuned, the side bend spacing is larger because it emits the blue detuned photon when it's on top of the hill. And the red detuned photon is emitted when the atom is more in the dark, and therefore, the side bend spacing is lower. So it alternates blue, red, blue, red. But the blue detuned photon is emitted when the generalized RB frequency is larger, when the atoms is at an anti-node And the red detuned is more preferably emitted when the photon is at a node of the standing wave. So it is the difference of these modulations, the side bands are modulated, and there is a preference to emit the blue detuned side band when the side band spacing is larger than in the red detuned case. And this is where the energy goes. AUDIENCE: You could make the argument and that would give a simple picture for the maximum cooling rate? PROFESSOR: Yes. AUDIENCE: Where you have omega times your [? scattering ?] rate? PROFESSOR: Yes. That's exactly how, last class, I calculated the cooling rate. The cooling rate, the energy removed from the system, is, well, yeah, I said it is the height of the lattice, but the height of the lattice is also the amount by which the-- the lattice is nothing else than the generalized RB frequency. This is the energy level of one crest level, and therefore the lattice is the increment by which we modulate blue and red side bands. Exactly. This can be quantitatively worked out. If I had another half hour, I could do it for you. It's in the references, but this is the physical picture for cooling. OK, but so you realize, this is one of the subtleties. We can understand the force simply from a simple picture, energy levels, Sisyphus cooling and such, but in order to understand where the energy comes from, we need the mono-triplet. We need spontaneous emission. So the force comes from the redistribution of photons. It's the stimulated redistribution of photons which is responsible for the force, but the energy balance comes from, what can be neglected for the momentum balance, namely the few spontaneous emission events, and especially those which involve the blue and red side bands. And it beautifully works together and all the equation of physics are obeyed. OK, but that tells you now that the cooling effect hinges on spontaneous emission. And yes, spontaneous emission is dissipation. And whenever you want the friction coefficient-- you want dissipation-- you need an open system and the open system is spontaneous emission into those nodes. So let's now talk about an opposite situation where we do not have any spontaneous emission. And I think almost all the people who work with optical lattices use infrared lasers for them, or [INAUDIBLE] detuned laser where spontaneous emission happens only once every 20, 30 seconds or 100 seconds per atom, so we can really neglect it. So now the question is the following. Assume you have a focused laser beam and you load a Bose-Einstein condensate at the edge of the cloud, and now the condensate is accelerated, sucked into the laser beam. So now the atom has kinetic energy. The question is, where does this kinetic energy come from? And this time, we do not have spontaneous emission as we have when we do frictional cooling. So the question is now, where does the energy of the atom come from? AUDIENCE: But if we were talking about a dielectric medium, it's coming from those light waves that are reflected or not reflected. PROFESSOR: If comes from those light rays, but-- the atom is a dielectric medium, that's good-- but the reflection and deflection of light rays-- at least in leading order-- you would say photons of the same energy are deflected-- AUDIENCE: Wouldn't the [INAUDIBLE] of the atom just get absorbed [INAUDIBLE] and the [INAUDIBLE]? PROFESSOR: I would hate to use the picture of absorption and emission. What really happens is it's just scattering. It's scattering. you scatter a photon which has one frequency into a different direction. It's really light scattering. Well, you have a homework assignment on that, and I don't need to go into details because it can easily be worked out, but what happens is the following. The atom is a dielectric medium. It has an index of refraction. And if you put an index-- if you suddenly put an index of refraction into a laser beam-- then you change the phase of the laser beam. The atoms act as an electro-optical modulator, and they change the frequency of the transmitted photons. You have a homework assignment on that. So if you increase the number of atoms in your laser beam, because they move in, this is-- the phase of the transmitted photons is shifted. If the phase of the photons is shifted during a certain time, phase over time is frequency. So therefore you will see, when the atoms move in and gain kinetic energy, the transmitted photons of the laser beam have lost some energy. They are slightly shifted to the red. So in other words, it's funny but if you have a dipole trap and the laser beam is focused and atoms slosh through the trap, if you would carefully analyze the frequency of the transmitted beam, you will find that it's a little bit blue, a little bit red, a little bit blue, a little bit red, and this compensates for the kinetic energy. Again, I don't think you'll find it in any textbook. It confused me for a while until I could work it out. But the homework assignment gives you an exact dissoluble model where you can exactly show what I told you. So these are the two situations, here. It's either spontaneous emission, which is responsible for cooling, but here we have a conservative potential. And in this conservative potential, the only player is the laser beam which is transmitted. And, indeed, it is the photons which have been transmitted, and usually you wouldn't put that into your picture. Remember, the dipole force can be explained by eliminating the coherent field of the laser field-- for canonical transformation-- and then we have a C number. We have a classical electric field. This is your Hamiltonian, and from this Hamiltonian, you derive that you have a conservative potential. But if you now ask where the energy goes, the energy-- really, in this situation-- goes into a frequency shift of the photons. Also, in your Hamiltonian, you have assumed that the electric field is an external classical field, E naught times cosine omega T. Well, if you want my advice, whenever you use the AC Stock's shift for trapping, just use the AC Stock's shift and regard it as a classic potential, and just completely forget that there are photons behind because if you want to account what really happens to the photons-- what really happens to the energy in this system-- it really gets ugly and it can get really confusing. Questions? OK, we are done with the stimulated force. We are now in preparation for next week when I want to tell you about degenerate Fermi gases and degenerate Bose gases. Those gases require nanokelvin temperatures, so today, in the next hour, I want to tell you what are the techniques to create such ultra-low temperatures, and the techniques I want to discuss is one addition to laser cooling. We have so far discussed laser cooling to the Doppler limit, but now I have to tell you that that's not where laser cooling stops. You can go sub-Doppler and sub-recoil. And, at least when the Nobel Prize was given to Bill Phillips, Steve Chu, and Claude Cohen-Tannoudji, well, if you go to the Nobel website and read the report of the Nobel Committee-- well, they should have given it to those three people for the many, many wonderful things they have done, and I've made frequent references to their work. Bill Phillips invented Zeeman slower, Claude Cohen-Tannoudji invented the dressed atom picture, they have many, many contributions, but the Nobel Committee tried to justify their choice with a coherent story, and the coherent story was Steve Chu cooled to the Doppler limit in molasses, Bill Phillips discovered sub-Doppler cooling, and Claude Cohen-Tannoudji realized sub-recoil cooling. So the storyline was cold, colder, the coldest. And that's what we want to talk today. So sub-Doppler cooling. Well, I could have spent-- or, 10 years ago, I spent a whole lecture on sub-Doppler cooling, polarization gradient cooling. Wonderful! The kind of epiphany of elegance in [INAUDIBLE] a mechanical description of an atom. But I have to say-- and then I explained, in another lecture, sub-recoil cooling, how you can even cool below the recoil limit. Well, this year, 2013, I spent 15 minute on it. The reason is the following. When I joined atomic physics and was a post-doc in the '90s and assistant professor, the conference's [INAUDIBLE] of [INAUDIBLE], [INAUDIBLE] was full on intents to find new ways to laser cool. Ideas of how to get to lower and lower temperatures. There were many, many different-- sub-Doppler, sub-recoil techniques were discussed. This was the main topic. But then suddenly, in 1995, evaporative cooling-- an intellectually boring cooling technique, just have atoms collide and evaporate-- this led to the lowest temperature ever, and it was almost a sudden transition. Within a few months there was no research, no papers anymore, on advanced methods of laser cooling. Evaporative cooling it just completely wiped out this area over atomic physics. The reason being because all cooling schemes which have been discussed had, in the end, some problems. At some point, photons heat. Heat because they excite an atom and if an excited atom collides with another atom, there is a heating mechanism and such. So even all the optimistic proposals for laser cooling reached very low temperature, but never at high density. And evaporative cooling just did everything for you. But anyway, I think it-- not just because a Nobel Prize was given for sub-Doppler and sub-recoil cooling, this is really an accomplishment to understand how can you go through conventional cooling limits, and at least every graduate student who happens to graduate in AMO physics at MIT should know what are the concepts behind sub-Doppler and sub-recoil cooling. So what I need in order to introduce those cooling methods for you is I have to remind you that optical pumping is a cooling scheme, and one could actually say, to some extent, every cooling you do with lasers is based on optical pumping. And so let me explain that. Optical pumping happens that-- let's say you have a level structure with a circularly polarized light where you can go maybe from M equals 1 to M equals zero. And then there can be spontaneous emission back to M equals 1 and back to M equals minus 1, but the laser-- because it has angular momentum-- cannot excite this state to any excited state. So what happens is, of course, pretty clear. You have whatever distribution you have in state one and two. You switch on your laser, and after a few cycles everything has fallen into the dark state. You have 100% population in stage two. So this is the simplest example of optical pumping using the three-level scheme. Well, you may ask now, what has that to do with cooling? Well, temperature is a Boltzmann factor, and if you-- we can define a temperature by saying the population between two levels is given by a Boltzmann factor. Let's just introduce an energy-splitting delta E-- which is somewhat arbitrary here-- but in any event, you see if you completely pump out a level, if you have all the population in one state, this corresponds to zero temperature. So if you optically pump the atoms, preferentially, into certain states, you have lowered the entropy of the system, the atoms are no longer distributed over as many states as before, and this corresponds to lower temperature. And the message I want to give you is that you can understand laser cooling as optical pumping in translation space in velocity space. When you laser cool, you excite the atoms at high velocity and then because of the mechanisms we discussed, spontaneous emission leads to lower velocities, and this is one form of optical pumping. I also like the word-- optical pumping is, in some sense, a spontaneous Raman process. You go up with one photon, you go down with-- you go up with the laser photon, you go down with the spontaneously emitted photon. So this is a spontaneous Raman process, and laser cooling-- even if you have just a single ground state-- can be regarded as a Raman process where the initial and final state differ in velocity or momentum. You have the same internal state, but you have a different external state. So you can see laser cooling is spontaneous Raman scattering between different momentum states. OK, now how can we achieve sub-Doppler cooling? Sub-Doppler cooling was discovered in 1988, and I remember I was at the [? IKA ?] Conference in Paris when Bill Phillips' group said, we've carefully measured the temperature and in sodium we measure a temperature which is lower than the Doppler limit, which was rigorously derived. I derived for you the Doppler limit of molasses. So what went wrong here? Well, it became clear the only loophole-- if you have a theory which predicts something and then you find a violation, you carefully have to check the assumption. And the assumption which was made-- and which we have made in this course-- is that you had a two-level system. Now, atoms have hyperfine structure-- and for pedagogical reasons, because I want to explain to you why in sub-Doppler cooling scheme I assume this hyperfine structure with F equals one half, F equals three half-- but let me first tell you what the novel feature is about a multi-level atom which has hyperfine structure. Now, the one thing which happens is that-- instead of just going between our single ground state and our single excited state-- we now have transitions. You may want to call them Raman transitions, or optical pumping, between the hyperfine atoms. So there are now transitions not only up to an excited state which rapidly decays, there are now transitions-- Raman transitions-- between ground states which have a very low widths. The width of the excited state is gamma. Well, what is the width of the ground state? Well, if you don't have a laser beam, the width is zero, but if you have laser beams the atom doesn't stay there forever. There will be a time, which is estimated here-- depending on the RB frequency and the detuning-- this is the time, this is simply nothing else than the scattering rate which we derived before. So that's the rate at which you scatter photons. And then, depending what the bunching ratio is, this is also the time in which you may optically pump to another ground state. So therefore, what we have to take into account now is that we have narrower widths in our system, and those widths correspond to another two-level system which is driven by two photons in which connects the two ground states. I will not give you any derivation how this exactly leads to a lower temperature, but I want to give you two pictures which I have already introduced to you. One is, remember, when we discussed Doppler cooling, the final temperature was proportionate to gamma, the width of the transition. So now I wave my hands and say, if you have transitions between ground states which have a much, much smaller width-- which is the rate of optical pumping-- and for low laser power this width can become very, very narrow. But you have at least one ingredient which can lead to a lower temperature. OK, in all truth in advertising, for the Sisyphus-- for the polarization gradient cooling scheme-- the final temperature is not given by gamma prime, the rate of optical pumping. It is a proportionality factor, but there is another factor. If you want to learn about it, you have to read some of the classic papers. But there is another picture which I can also use to tell you-- and that's actually related-- why optical pumping at least gives the possibility for lower temperature. I explained to you with the stimulated force in [? blue ?] molasses-- that the only reason why we have cooling-- is that the atom has a lag time. It cannot instantaneously adjust itself to the light field. It lags behind. And you can say, in Doppler cooling, the lag time is the spontaneous emission time, and the inverse of it is gamma. Here, in this case, the lag time can be the very long time to optically pump. So therefore, if the atom is in one hyperfine state and now moves into an area where the polarization of the laser beams is different, it may take a long time to adjust. And it is this lag time to which the friction coefficient was proportional. So this is a new feature, long delay times, narrow widths, and resonances between ground state levels. OK, a little bit show and tell now. Nobody thought about polarization creating cooling, nobody thought about Doppler cooling, but it was discovered when people simply did Doppler molasses and it's one of those big violations of Murphy's law where cooling worked much better than everybody had thought, and the result is the following. That-- when I drew for you the blue curve, which is force versus velocity-- it's pretty much the subtraction of two [? Lorentzian ?]. What happened is-- for sodium atoms, or for any alkali atom-- it's the red curve. For large velocity you have the Doppler cooling mechanism, but for smaller velocities you have a steepening of the slope, and the steeper the slope, the larger is your friction coefficient, alpha. And this is the mechanism of polarization gradient cooling. Let me just take this famous paper and show you one way how polarization gradient cooling works, which is the most famous form of sub-Doppler cooling, and this goes as follows. If you have molasses with two laser beams and the two laser beams are lin-perp-lin, linear polarization perpendicular to the other linear polarization. So you have two laser beams with these polarization. When these two polarizations overlap and they have the same phase, you get light at 45 degree, but if these two polarization have 90 degree out of phase, you get circularly polarized light. So as these two laser beams counter-propagate, you periodically go from linear polarized sigma minus, linear polarized sigma plus. So you have spatial-- at any given point you have a polarization, but the polarization changes. You have a-- you would say, naively, lin-perp-lin don't interfere, don't form a standing wave. Well, they do not form a standing wave in intensity. They form a standing wave in polarization. So now what happens is the following. If you have a multi-level atom and I use this simple scheme here, the different polarizations, linear polarization drives, the pi transition, sigma plus and sigma minus [INAUDIBLE] different transitions, and those transitions have different strengths. So an atom here-- when it experience this polarization-- reaches cycle, but when it experiences the other circular polarization, it will be optically pumped over here and then it cycles. So therefore, an atom which from sigma plus polarization and was here, it flies over to an area where you've sigma minus polarization, it will actually be pumped over. So the atom will constantly be pumped back and forth between those hyperfine states. And this actually gives rise to a beautiful form of Sisyphus cooling that the atom experiences sigma plus light in one ground state. It climbs up the hill, it sees the AC Stock's shift potential-- and the AC Stock's shift potential, because of the, for instance, for one hyperfine state, sigma plus drives the strongest transition-- and therefore we have sigma plus light, one ground state has the deepest potential. Where we have sigma minus light, it is the other ground state which has the deepest potential. And so what happens is that the atom is in one hyperfine state. It climbs up the hill, and then it's optically pumped to the other hyperfine state, and we have exactly the same kind of Sisyphus cooling. I just show you pictures and you sort of match it with what you know. This is more complicated because this involves optical pumping between hyperfine states. The different hyperfine states have different [INAUDIBLE] coefficients and such. It's not really complicated but more complex. What I explained to you is how Sisyphus cooling works in the dressed atom picture just for a two-level system. And here you find a more subtle form of Sisyphus cooling, but this form is more important because whenever you operate a magneto-optic trap, you get this cooling mechanism for free. OK, so that's all I want to tell you about sub-Doppler cooling. OK, sub-recoil cooling, we can quickly deal with it because I want to prove to you here that sub-recoil cooling is impossible. And unless you tell me what this wrong in my derivation, I don't need to discuss sub-recoil cooling because I've convinced you that it's impossible . OK, so let's assume our atom has an initial kinetic energy and then we absorb one photon. That would mean the momentum has now changed by the laser photon by the recoil-- and then it emits a photon, so the momentum gets reduced by the immediate photon-- and all I do is I ask, what is the difference between the initial kinetic energy and the final kinetic energy after two photons-- a laser photon and the spontaneously photon-- have been exchanged? And what you find is-- and this is just exact-- that the change in energy is two times the recoil energy. The recoil energy is h-bar square K square over 2m, plus-- and this is now the cross-term-- plus KL-- the K vector of the laser-- minus K spontaneous emission times the velocity. And now, of course, we assume-- which is correct-- that spontaneous emission goes randomly in all directions. Therefore, if we average over many cycles, this does not contribute. So therefore we find now that the average energy exchanged by an absorption and emission event is two times the recoil energy plus h-bar K laser times [? width ?], and we want to make it negative because we want to cool. Well, we make it negative by arranging the laser beam-- surprise, surprise-- counter-propagating to the velocity. This is how we can get the best cooling. Well, surprise, that's what we expected. But you find, of course, that when this velocity is a very small then K V is smaller than K squared. So therefore, once the velocity is smaller than the recoil velocity of a single photon, you cannot choose-- you cannot make this equation or this expression go negative. In other words, what I've shown to you is if there is an atom which has a velocity which is smaller than the recoil velocity of a photon, whenever this atom scatters a single photon, it will be hotter than it was before. So is it clear, what I've shown you, in energy? When an atom has a velocity which is smaller than the recoil velocity, any further photon scattering will not cool. It will lead to an energy transfer delta E, which is large, which is possible. So that shows you that sub-recoil cooling is impossible. Any idea how we can sub-recoil cool? Any idea why Claude Cohen-Tannoudji got the Nobel Prize? AUDIENCE: Further down in the lattice, you can have an effective mass that is much heavier than [? bare ?] mass? PROFESSOR: OK, great idea. We put the atom in a lattice, a lattice has band structure, in a band structure we have an effective mass, and heavier atoms can be cooled to lower and lower temperature. Actually, it's well known that, in Doppler cooling and sub-Doppler cooling-- especially in sub-Doppler cooling-- cesium an rubidium reach a microkelvin, sodium reaches only 25 microkelvin. So making the atom heavier is OK, but I have to say you're only rescaling your recoil limit, you're not breaking through the recoil limit which is now defined with a heavier effective mass. AUDIENCE: [INAUDIBLE] emission direction-- PROFESSOR: Yes? AUDIENCE: --means that the atom must have [INAUDIBLE] responding is [INAUDIBLE]? PROFESSOR: Yes. AUDIENCE: And [INAUDIBLE]? PROFESSOR: That was the assumption. AUDIENCE: [INAUDIBLE] at least for this cooling, it needs the time much larger than the Doppler cooling [INAUDIBLE] spontaneous emission time? PROFESSOR: Well, that was the assumption here. We scatter several times, a spontaneously emitted photon do not contribute the recoil of the spontaneously emitted photons cancels out because spontaneous emission-- and this is correct, this is not wrong-- spontaneous emission goes equally probable in the plus X and the minus X direction. Well, I thought I want to show you a demonstration how-- mechanical demonstration-- how you can sub-recoil cool. It's not a demonstration in velocity space, it's a demonstration in position space. What I have here is a Plexiglas tray and it has a little hole in the middle. And I have a bunch of ball bearings and I put those ball bearings in. Let me translate. I can blindfold myself and all I do is I shake the tray. So whenever I shake it, I kick the atoms randomly in position space from here to here. So the amount of position control I have over the atoms is on the order of this size. But the question is, can I-- without having any control of the transfer of position I give to them-- can I steer all the atoms into an area here into the hole in the middle, which is very, very narrow in position space? So the translation is, if I randomly scatter photons and they kick the atoms around with h-bar K, I don't have any control about momentum transfer smaller than h-bar K, is it possible to localize the atoms in momentum space to a momentum around zero, which is much, much smaller than h-bar K? Well, let's do the experiment. I just close my eyes and I shake it, and I shake it, and I just shake it for a while and I continue shaking it. And, well, zero temperature! So without controlling the motion on the scale I'm interested in, I manage to cool into a target region which is much, much narrower. So in other words, what you need is-- all you need is-- you need some dark state which is velocity selective. If you scatter light but you create a situation where once the atoms are close to zero velocity, they're not excited. You're not steering the atoms in a deterministic way, you just wait until-- by random chance-- one photon is emitted, and by the random chance the atom hits the hole. And then, it will never be re-excited again. This was the idea of this demonstration. And I could, yes, in the old days I may have spent two hours on teaching it, but by using Raman resonances between ground states-- which are terribly narrow-- or by using VSCPT-- velocity selective coherent population trapping-- you can create such narrow dark resonances which have the effect that the atoms scatter, scatter, scatter, but the moment they reach a very narrow region around zero velocity, they stop scattering. So what was wrong in my proof is that here we have a situation where we stop the laser cooling exactly at the time when the atom, by random chance, happens to be at low velocity, at very low velocity. And then you don't need any control, you just accumulate in the same way as I accumulated the ball bearings. OK, this was sub-recoil cooling in five minutes. [? Timo ?]? AUDIENCE: So I wasn't here last class. But just to summarize, the Doppler cooling rate gets us to a temperature on the order of gamma, which is natural alignment. PROFESSOR: Yes. AUDIENCE: And the recoil limit is the recoil energy which is usually tens of kilograms. But if you have a really, really narrow line, would you use the Doppler cooling limit? You could, in principle, beat the recoil limit with Doppler cooling, no? If you have a-- PROFESSOR: No, you can't, because if you do Doppler cooling-- [? Timo's ?] question is, if you simply do molasses with a very, very narrow line, we derived that the limit of Doppler cooling is Kt equals gamma, and what you are saying now, if you use narrower and narrower lines, can we reach [? arbitrary ?] low temperature? This is not the case. In our derivation of Doppler cooling, we made a continuum assumptions when we plotted the force versus velocity and we had the friction force, we assumed that an atom, when it scatters a photon, stays, let's say, within the linear part of the force versus velocity. So we had a hidden assumption which required that the recoil energy is smaller than h-bar gamma. But there is a lot of literature where people looked into it, and it turns out that when you have a very narrow line, you can go to the recoil limit, but you can't go beyond. If you want to go-- and I think this is what I showed you in the last five minutes-- if you want to go below the recoil limit, you need some velocity-selective dark state, and you would not have that in this situation. On the other hand, if you have a very narrow resonance, you can probably engineer a dark state that there is destructive interference in some excitation in a very, very narrow velocity class, and this narrow velocity class is simply selected by the Doppler effect. So you can use the narrow line to engineer sub-recoil cooling, but the simple arrangement of just having molasses with narrower and narrower line will not work. [? Collin ?]? AUDIENCE: I think the group that laser cooled to BEC just had-- their dark state was just a Stock's shift, [INAUDIBLE] and that was their-- and it shook. There were resonances in kilohertz or something. PROFESSOR: Thy used a very-- did they go below the recoil limit in laser cooling? I'm not sure. They may have had a situation where it was enough to go to the recoil limit. I have to check out the paper, but-- AUDIENCE: I remember they did have some sort of dark state. Maybe it was to increase density or-- they had the high-- PROFESSOR: Anyway, let me just be clear. Sub-recoil cooling requires that you have a dark region in velocity space. For that, you need some narrow line widths. That's necessary. But I think you just don't get it by having counter-propagating laser beams, you have to do something more. Other question? OK, so I've told you about sub-Doppler cooling, sub-recoil cooling, but what really got atomic physicists to nanokelvin temperature were two other techniques. Laser cooling was used as pre-cooling, but then the final trapping and cooling was done by magnetic trapping and evaporative cooling. So I want to give you now, in the last half hour, a quick overview of magnetic trapping and evaporative cooling. So I think any discussions of magnetic trapping starts with a theorem, a theorem which tells you that not everything is possible you would like to do. And this is the following, then if you have a region without charges and currents without-- through an empty space, and what this theorem says-- in empty space, you cannot have a local minimum-- you cannot have a local maximum of your electric and magnetic fields. So if you take the strengths of the electric field or the strengths of the magnetic field, you cannot have a maximum. You can only have a minimum. Well, and this is important when it comes to magnetic traps. Depending on the magnetic moment-- spin up, spin down-- we could create a trapping potential around a maximum of the magnetic field or the trapping potential around the minimum of the magnetic field. You want to be at the minimum of the potential, but if you have a minus sign, you may want to be at the maximum of a magnetic field. But [? Bing's ?] theorem says that only one of them is possible. The proof really goes in two lines. You really show, if you assume that there is a maximum, that would mean that if you add a field to it, you add a field-- and no matter in which direction you add a field-- the total field strengths become smaller. But that requires a violation of the Laplace equation. It's easier for you-- you can probably read it at home-- but you simply assume you have this situation, so you add-- and you show you have an incremental field, delta E. This field, delta E, you are at the maximum of an electric field, and then delta E over R is the small difference of the electric field. And you can show that this expression is only negative. That means you have a maximum of the electric field when the electric field would not fulfill the Laplace equation. So the gist of the argument is the following. Each component of the electric and magnetic field fulfills the Laplace equation. And the Laplace equation-- I don't know if you've probably heard about it-- if something fulfills the Laplace equation, it says at any given point the function which fulfills the Laplace equation is equal to the average of function, averaged over a small sphere around it. And that means-- if something fulfills the Laplace equation-- you cannot have a local maximum or a local minimum because if the value here equals the average, that would mean in one direction the value gets higher, in the other direction the value gets lower. So if something fulfills the Laplace equation, you cannot have a local maximum or local minimum. But now, we're not asking for maximum or minimum in one component of the electric or magnetic field. We ask for a maximum or minimum in the total value, or the square of the electric field. But then you show, when you assume that you have a maximum, the field points along the Z direction, that then the Z component of the field cannot fulfill the Laplace equation. So maxima are not possible. Minima are possible. Good! So the magnetic trapping potential comes because we have a magnetic dipole moment which interacts with a magnetic field. Magnetic traps are actually classical. You can have a classical model for a magnetic trap, and in that case, you say this potentially is mu times B times cosine theta. [? Crendo ?] mechanically, of course, the angle cosine theta is quantized and we have the different M F levels, the different orientations of the spin, relative to the Z-axis. OK, the fact that the magnetic field can only have minima and not maxima means that we want to make sure that the magnetic moment times the G factor times M F is negative, or cosine theta is negative in the classical picture. And if we absorb whatever we choose for cosine theta-- or what we call mu Bohr G M F, the magnetic moment-- we want that the magnetic moment of the particle is anti-parallel to the magnetic field. Then we have a magnetic trapping potential, which is-- I've taken care of the sin now-- mu times B, and if B has a minimum this potential has a minimum, and that is a magnetic trap. But so the consequence of [? Bing's ?] theorem is that we can only trap particles which are anti-parallel with a magnetic field. And they can always lower their energy by flipping to the other state. So therefore, a magnetic trap does not allow us to trap particles in the absolute ground state. There is always the possibility of spin-flip collisions which lead to an anti-trap state which is expelled from the trapping region. Now, spin-flip collisions can happen when you have particles at sufficiently high density and they collide. For the experts, there can be spin relaxation, there can be dipolar relaxation, they are two different kinds of spin-flip collisions. Fortunately, in magnetic traps, they only become relevant when you pick the wrong spin state. But for suitable choices, magnetic traps are very long-lived. Dipolar relaxation is often-- can be a limiting process for Bose-Einstein condensates, for instance, for high [? dose ?] Bose-Einstein condensation, dipolar relaxation was-- limited the number and density of atoms in the Bose-Einstein condensate. But I don't want to talk about cold collisions, here. You should just know, based on the fundamental theorem, you cannot trap in the ground state. You have to trap into a state which has more energy than other states, and, in principle, it's possible to flip the spin and go to a lower state. Now, what we have assumed here is-- when I wrote the trapping potential like this-- I assumed that the atom stays in a given hyperfine state. So we put the atom in one state and it experiences potential. Classically it means that the angle of the dipole, with respect to the magnetic field, stays constant. And this is the case due to rapid precession. Classically, the magnetic dipole precesses around the magnetic field, and when the direction of the magnetic field changes, the rapid precession keeps the dipole aligned. Or, in other words, for slow changes of the magnetic field direction, cosine theta is an adiabatic invariant. So the word adiabatic is important, either classically or you stay in a quantum state as long as you're adiabatic. Well if an atom would rapidly go through a region of very weak magnetic field, then the precession-- the [INAUDIBLE] precession-- of the atom is very slow, and this adiabatic condition can be violated. And this violation of an adiabaticity condition is called Majorana flops. So if you operate a magnetic trap at very low magnetic field, you may destabilized the trap because you violate adiabaticity. OK, so these are the consequences of the fact that we cannot create maxima, we can only create minima of the magnetic field in free space, and therefore we have to deal with those two possible loss processes, but we have learned what kind of magnetic trap, what kind of atom to pick, and in general these are not big problems. OK, so we have to-- so this shows here the typical hyperfine structure of an alkali atom. It could be rubidium 87 or sodium 23. The fact that we have only a local minimum of the magnetic field means we can only trap hyperfine states-- which I've marked here in green-- where the slope is positive, and these are the atoms where the magnetic moment is inter-parallel with the magnetic field. And, well, for stability reasons we usually pick the highest state of the lower manifold, or the highest state of the upper manifold. In principle, it would be also possible to trap here-- you have the correct slope-- but often those states suffer form spin relaxation and collision loss. [? Mickey ?]? AUDIENCE: [INAUDIBLE] the picture you showed before, is it actually possible to make a magnetic trap with state number four from the top so that they expel from the center? From [INAUDIBLE] a-- PROFESSOR: Yeah, there is a peculiarity here, you have the transition from the weak field to the strong field region. Whenever the magnetic moment is constant, you need a local minimum of the magnetic field. But if the magnetic moment would change and you have a gradient of the magnetic field, you could actually trap here. So in this case, you have a magnetic field gradient but you have a spatial variation of the magnetic moment. This has been discussed in one special paper in the literature, but it has not really found any [? lucent ?] realization. AUDIENCE: Wouldn't it help to put the Majorana flops in the center? Because they're repelled from the center. PROFESSOR: Well, yeah, but these atoms would undergo spin-flip collisions so it's not a good choice. It would help against Majorana flops, but we have many solutions against Majorana flops. And here, you would solve the Majorana flops problem but you would solve it with another problem. OK, so we need minima of the magnetic field. This provides magnetic trapping potential, and there are two kinds of possible trap configuration. One is where the minima is at zero magnetic field, and one is where the minima is at finite magnetic field. Now, the zero magnetic field minimum one can be simply created with anti-Helmholtz coil. What happens is the magnetic field, as a function of position, would actually cross through zero. It's just a field gradient which crosses through zero, but, of course, since we are interested in the absolute value of the magnetic field, we get the V-shape potential. This V-shape potential was what was used for the first demonstration for Bose-Einstein condensation because a V-shape potential is much, much more confining in this cusp than in harmonic oscillator potential. And this had advantages for tight confinement and rapid evaporative cooling. So those quadrupoled traps with the V-shape potential are the best confinement you can get, the best confinement for the buck, and the buck here is your power supply and your coils. However, they have a problem when the magnetic field is zero. We violate adiabaticity because the different spin configurations become degenerate, and in degenerates you can't have adiabaticity. So therefore, the two first demonstrations of Bose-Einstein condensation avoided this cusp in two different ways. One way was to use a rotating magnetic field, and to use a perpendicular field but which was time-dependent, and the MIT solution was to use a blue detuned laser beam, use the optical dipole trap-- optical dipole potential-- to push the atoms away from the dangerous [INAUDIBLE] field region. I should say rotating-- none of those-- most traps which are now used are the other kind of trap, the trap which has an harmonic oscillator potential, which doesn't have the cusp and where the minimum is at a finite magnetic field. Let me just make one comment. The rotating trap was very popular because it led to the first BEC and a lot of people built that, but to the best of my knowledge, it's only used in different places. And the only real application it has is-- since you have some rotating magnetic field-- by some modification you can actually make of rotating potential. And this is nice if you want to study Bose-Einstein condensation on a rotating frame, create vortices and things like that. But for simply creating a Bose-Einstein condensate, this trap is used, by far, most frequently. But there is actually a renaissance of this trap. One reason why people use this trap and why my own research group immediately switched to this trap-- and we had the idea how to built it-- is, well, harmonic potential is nice. Every physicist loves an harmonic potential. You can solve, immediately, thermodynamics in a harmonic potential. Who wants to deal with that potential? Also, if the laser beam just drifts by a micrometer, this symmetric potential becomes asymmetric. So you characterize your potential today and a few hours later you have a different potential. Whereas a magnetic potential, harmonic potential, once it's characterized-- as long as you don't change the current in your power supplies-- it's the same week after week, months after months. Well, there is a renaissance now because a lot of groups now doing experiments in optical lattices, they don't care what shape the potential is because they don't do physics in this potential. They just take the Bose-Einstein condensate or the cooled Fermi gases and transfer them to an optical lattice. So in that case, it doesn't play any role, and then the advantage of this potential is you get more bang for the buck. AUDIENCE: [INAUDIBLE] you were saying before [INAUDIBLE] more rapid evaporation in V-shape versus the [INAUDIBLE]. Where is the differences in evap time? PROFESSOR: It depends. I may run out of time today. I needed-- I can probably-- I wanted to go through evaporative cooling, and I will talk to you that, for evaporative cooling, there could be a threshold in confinement where you go into a runaway regime that evaporative cooling is speeding up. So a little bit of confinement can make the difference without never getting into a runaway regime or being in that runaway regime. So confinement, extra confinement, can make the difference between getting BEC and not getting a BEC. So it really depends. It's highly nonlinear, but I will show you later on that when you put a system together, the kind of threshold density at which the cloud can evaporate to BEC is much lower here than it is there. But-- and this is the next thing I want to explain-- it's not really that this is a linear potential and this is a [INAUDIBLE] potential because those finite magnetic field-- these traps which have a minimum at a finite magnetic field-- are usually done in the following way. Those traps are called Ioffe-Pritchard trap. Ioffe actually suggested such a magnetic field configuration for confinement of plasma, and Dave Pritchard was the first to point out that such a configuration would be a good choice for neutral atoms. So I sometimes joke and I say, well, plasma physics is the study of very hot matter. Cold atoms is the study of very cold matter. But when matter is either too hot or too cold that you cannot put it in ordinary container, you want it confined with magnetic fields. And it happens that, for plasma, you need a minimum of the magnetic field, and for neutral atoms you need a minimum of the magnetic field. So therefore here is something which ultra-cold atoms have in common with plasma physics. I don't think there is a lot the two fields have in common, but when it comes to magnetic field configurations, yes. A similar magnetic field configuration can confine a plasma and can confine neutral atoms. OK, so the generic way of how these magnetic fields are generated is you want to have two coils, called pinch coils. You can say each of them creates a magnetic field which decays, a magnetic field which decays, and now in the middle you have the parabolic minimum. But then you have to add the so-called four Ioffe bars which create a linear potential, and this is shown here. So the pinch coil is simply creating a local minimum along the Z-axis, and if you only want to trap in one dimension, you would be done, But you want to trap in 3 dimensions. And so what happens is you add now an anti-Helmholtz-- or quadrupold-- field in X and Y, and this blue field is done via the green bars, by the Ioffe bars. So you add together a harmonic quadratic field in the Z direction with the linear quadrupoled field in X and Y. And what matters for the atoms is-- and what matters for the Zeeman energy is-- the absolute value of the magnetic field. So you add those things in quadrature, and because you add it in quadrature, you have now an harmonic oscillator potential in X, Y, and Z, so you have an harmonic trap in three directions. However, if this field is not larger than this-- if these field becomes larger than this field, so then when you add it in quadrature, you actually get a linear potential. So in other words, the Ioffe-Pritchard trap is quadratic for smaller values of X and Y, but if you go out to large X and Y, you get the linear potential. So in the end-- and this is maybe in response to Matt's question-- for a high temperature, the Ioffe-Pritchard trap is actually not harmonic in three dimensions. It's linear in two dimensions and harmonic in the third dimension. So, therefore the hit you take in evaporative cooling is not as large as we'd initially assumed. And the moment we realized that, we were building a Ioffe-Pritchard trap. And this has been the traps we' have been using at MIT ever since. Well, just as a warning, this is sort of the simplified description, but if you have curvature, the curvature has to fulfill Maxwell's equation. And you cannot have a curvature only along Z. Maxwell's equations are three-dimensional, so you get sort of all sorts of curvature terms. If you really build a magnetic trap, you should understand those. If you just want to understand why magnetic traps work, the previous slide is sufficient. The different ways to build those traps, this is the design we invented at MIT. It's called a clover-leaf trap. So you have one coil package here, to another coil package here, and what you see is the pinch coils, and when you have two of them you create this parabolic minimum. We don't like Ioffe bars because if you have a vacuum chamber you either have to put the Ioffe bars into the vector vacuum chamber or you drill holes through your vacuum chamber to put the Ioffe bars back and forth. Don't laugh, some people have done it. Of course, they put little tubes around it so it was a very highly-engineered vacuum chamber where they could string Ioffe bars through the chamber. But we realized that the same field as Ioffe bars can be generated by taking the Ioffe bars and flipping them out, and after flipping them out, they had the shape of clover leaves. It's just a variant of creating the same field geometry. Yes, that's all I want to tell you about magnetic trapping. I think I should not start with evaporative cooling. This will just take 10 or 15 minutes, but I'll do that on Monday. So since we're on time, do you have any questions about magnetic trapping? Different forms of magnetic trapping? OK. Final announcement, next week is the due date for the term paper. Yes, and the term paper is due on the date of the last class, which is Friday. Any questions about that? OK. Good.