Talk:Afshar experiment/Archive 5

message: Afshar's error is to assume, on the basis of a particle being found in a given lensed detector, that the particle must, therefore, have "passed through" the corresponding pinhole. But this is not the case. The only thing that passes through the pinhole is the probability of finding a particle there. This is not the same thing as actually finding a particle there. If you want to find a particle in the pinhole you have to measure it there. But in Afshar's experiment the particle is found (measured) at the detector - not at the pinhole.
Afshar's website says that the failure of aeroplanes without wings to fly should not be used to prove an aeroplane, with wings, can't fly. The analogy is in relation to an alternative experiment meant to discount Afhsar's. 
But an interpretation of quantum theory is the same thing. You can't just ignore one of the principles (remove the wings) and then complain that one of the other principles fails (doesn't fly). 
Particles are not things that pass through pinholes. They are more like events taking place in the measurement plane.
The Copenhagen interpretation of quantum theory is a reasonable deconstruction of the interference pattern for the purposes of a classical audience insisting on upstream causal models.  The interference pattern can be reconstructed from such deconstructions. It certainly can't be constructed from classical models. 
But there is nothing stopping neoclassical reinterpretions of quantum theory from working. The transactional theory is a good example. The "flight path" of a classical particle is theorised (but not observed!) as the result of a superpostion of classical waves, both of which propogate classically, but one of which is inverted with respect to the other in terms of it's temporal direction (it travels backwards through time!). The resulting classical image of a particle being somewhere (without being measured) correspondingly satisfys the classical mind.
And time travelling waves would appeal as well.
But the Copenhagen interpretation, while aimed at a classical audience, was not intended to entertain such an audience. It constitutes a criticism. It is an early example of critical deconstruction in which criticism (of the classical model) is done in terms of the very model being criticised. 
For example, when Heisenberg introduces the uncertainty principle, he does not do so in terms of quantum particles (ie. those that are measured in the measuring plane). He does so by analogy, ie. in terms of colliding classical particles, in which the desired classical information is compromised. What he is introducing, to a classical audience, is that classical reality, on the quantum scale, can't actually be experienced (or "realised").
In and of itself this does not necessaily invalidate classical assumptions. It just prepares the classical audience for the principles they should need when dealing with what really will deconstruct their assumptions - namely the results of quantum experiments.
That said, there is always room for other ways of doing this.
Carl Looper
15 January 2006
:Dear Carl, the error of Afshar is not interpretation, but mathematics. This pseudo-professor is not aware that to claim that the photon carries the "which way" info is indeed to say that the photon arrives in a "mixed" state at the detectors. Actually the "mixed state" does not mean that the photon is in a classical state, but it maybe well entangled with its environment or agents that measure through which pinhole it passes. In contrast Afshar's result shows that the photon's state is pure state i.e. it is quantum coherent superposition. I do not want to teach Afshar about basics that he missed in the University, but I warn the readers that Afshar is incompetent fraud, than a real scientist. Danko Georgiev MD 09:49, 15 January 2006 (UTC)
thread_title:  Afshar's Error 

message: There are many things on this page which are good or not concerning Afshar's claim . 
It seems that no body agree on the definition of the other and that math or philosophy are concerned. Now beeing physicist I want just (the last time I hope ) summarize why this experiment proves nothing concerning Bohr's principle. 
Forget here entanglement: Afshar experiment does not need this concept to be disqualified.
Bohr's principle tells us that we can not with the same photon build up the statistical patterns of two non commutative observables associated with a given quantum state (wave function or density matrix ). Here the two mutually excluded cases are i) spots A' B' in the image plane (images of respectivelly pinholes A and B), and ii) the Fourier image in the back focal plane of the lens (i. e. the interference pattern).
Without wire(s) of course this is true since one can not observe or absorb a photon twice (I consider only destructing detection in agreement with Afshar set up). 
Now the introduction of the wire change nothing of really fundamental : the wires absorb (practically ) nothing so the waves and the images are the same. For this reason the claim of Afshar is necessarily wrong :)
I will stop here since after I will be technical and go again into details 
Aurelien Drezet, Graz Austria
Dear Drezet, stop repeating like parrot that you cannot measure photon twice, this is not relevant to the Afshar's experiment. There is a principle of counterfactual defitiness, that requires that if the photon's state is say A, you may imagine many such photons and measure them in your imagination. And what is more - your imaginary measurements will have to produce exactly results compatible with the photon being in state A. 
The real problem is that YOU, are ignorant in physics and you don't know whether the photon is in pure state, or mixed state at the detectors. If you say that the photon is in pure state of superposition A+B, THEN NO WHICH WAY info in there. And IF you say that the photon is in mixed state being either at A or at B, then THERE IS WHICH WAY info. 
You will not have good career in physics, simply because you don't understand basic principles, and you cannot clearly present your ideas when discuss the issue with someone. And do not be afraid to be "too technical" - I challenged all of you to write down the math, but as it seems I am the only one that has provided clear math formulas, all else posted here is just bla,bla,bla .. Danko Georgiev MD 09:27, 17 January 2006 (UTC)
replies: text: Can you guys find some other forum to have this conversation? I don't beleive WP is the correct place to have debates over topics such as this. linas 01:09, 18 January 2006 (UTC)
thread_title:  Afshar's Error(bis repetita)= I can not see a photon twice (without disturbing him at least) 

message: In case it is not obvious how Afshar's results can still be obtained if, everything else being equal, we were permitted to respecify the upstream constraints, the following is an example.
  1. Remove the pinhole construction.
  2. Leave everything else as is, including Afshar's results.
  3. Place pin size obstacles at those locations formerly occupied by the pinholes.
  4. Place adjacent to such (at least a few pinhole widths away) a hologram of the pinholes.
  5. Project laser light through the apparatus.
If properly constructed the downstream data will be the same as Afshar's. 
Yet a "which way" inference (from the downstream data) would be incorrect since there are, in fact, obstacles at the inferred upstream locations.
In situ particle measurements made in the shadow of the pinsize obstacles would substantiate the invalidity of the inference. No particles would be detected there. But of course, Afshar's particles are measured ex situ. 
Well in this experiment, we can also measure such particles, ie. ex situ. And we'd find they'd be doing exactly the same thing here as they do in Afshar's experiment:
ie. indicating "which way" they went, despite actual obstacles to such!
Should we conclude our pin size obstacles are not really there? 
I guess we could ...
Carl Looper
19 January 2006
Dear Carl, please upload a diagram of your proposed experiment, otherwise I think noone can understand your idea. Use the upload file option in Wikipedia, and then read the instructions how you insert the image in your post. best, Danko Georgiev MD 07:00, 19 January 2006 (UTC)
thread_title:  Virtual Pin Holes 

message: There are problems with the theory section. I tried to clean it up, but promptly got tangled. I'll try to fix it in time, but any discussion here would help clarify what the intent of this section was.  linas 02:06, 7 December 2005 (UTC)
replies: text: The theory section was apparently added by anonymous user User:143.50.61.66 around 28 June 2005, but no matter how I try to fix it, it just seems broken. I'm punting for now. linas 04:03, 7 December 2005 (UTC)
replies: text: So... anyone care to help fix, or at least discuss, the theory section of this article? linas 15:25, 16 January 2006 (UTC)
thread_title:  Problems with the theory section 

message: Dear Carl, I will NOT agree that which way is not in the data - but in the decision. See again the right mathematical formulation of complementarity that nobody wants to see. All this available in basic QM course held by prof. Bob Eisenstein.
All the math below not only captures naive complementarity, but has more than this. In this paper[] is presented the original idea of prof. Eisenstein that clearly explains HOW the seemingly wave-particle duality appears. ALL IS COMPLEX ALGEBRA - usually a complex number can be written as e^{ix} or cos (x) + i sin (x). Now depending on the way you compute probability which way, or not, you may obtain "clumping" distribution with P depending on e^{x} or "wave-like" distribution depending on cos (x). Seeing these deep roots of complementarity in the nature of complex numbers per se is more than simple wording done by various pseudo-physicists.
No which way is charachterized by probability distribution of pure state - so P = |\psi_{1} + \psi_{2}|^{2} where \psi are complex numbers and denote the quantum amplitude. THIS IS CONTINUOUS FUNCTION IN SPACE!!!
Which way is characterized by probability distribution of mixed state - so P = |\psi1|^{2} + |\psi2|^{2}. THIS IS CONTINUOUS FUNCTION IN SPACE!!!
So the which way or not is OBJECTIVELY THERE in space, and Afshar uses exactly this - he puts grid in the space before the lens absorbing some photons.
Below I present two scenarios with polarization filters [which way] and without filters [no which way] and you will see clear difference of the probability distribution [i.e. light intensity in space]. NOTE: the two tiny lines accompanying the central Airy disc in the first "which way" image stand for the DIFFRACTION single slit pattern, sorry for the ugly hand drawn images, but they are very useful for direct seeing of complementarity!
Now as I have said in my previous post, polarization filters produce mixed state, but you can do it by first closing on slit, record some light, then close the second slit, record some light, and then take the average of the two single slit experiments. In this time you have mixed state because the distinguisable photons have different time arrival/detection.
UNLESS there is clear link with density matrix formalism, distinguishability of q-particles, and probability distribution you do not speak of complementarity but some WRONG popular wording. Exactly the OBJECTIVE math criteria are forcing you in your holographic experiment to project from special position behind the "virtual pinhole".  I do not want to repeat myself, there is pdf of mine in the web also [].Danko Georgiev MD 07:19, 22 January 2006 (UTC)
thread_title:  Why one cares about math of complementarity? 

message: By the way - untill there is ONLY ONE density matrix - the experiment will be either WHICH WAY or NOT WHICH WAY.
If Afshar wants to disprove complementarity he MUST prove that the photon HAS 2 DIFFERENT DENSITY MATRICES AT THE SAME TIME i.e. being in mixed state and pure state AT ONCE. What is this - new math, or new physics? Danko Georgiev MD 07:43, 22 January 2006 (UTC)
thread_title:  Density matrix 

message: Forward wave functions.
The wave functions used to predict downstream data are constructed from information regarding the upstream constraints one has imposed on the lightwaves/photons/electromagentic radiation/electron waves/carbon molecules ...
To the extent that one might introduce polarisers, pinholes, mirrors, lenses, holograms (etc.) one will accordingly adjust the way the wave function is computed.
At the end of the day there is a real life pattern on the screen with which to compare both one's mathematical model and/or the accuracy of one's measurements ...
Retrospective wave functions.
These waves are used to "predict" the upstream constraints that were in place, eg. whether one pinhole was open or two. Of course, if you know that then there's no point predicting it. You have to sort of, turn off any information, that you know about the pinholes and determine it just from the detector data. Therefore you are not permitted to use any forward wave functions in determining what is represented in the detector data. Otherwise you will be cheating! 
Now obviously, if we only choose one detector's data, we could only determine if one pinhole was open, (or not). Think about it. You are not permitted to know what lies in the pinhole plane, or further upstream. So you are not allowed to use the wave functions that you might have otherwise industrially computed, eg. the interference patterns (in front of the apertures). 
Remember, waves pass through one another, and other than the region where they interfere, don't affect each other.
If you project the data back, from BOTH detectors, they WILL interfere with each other. They will also give you the status of both pinholes - telling you whether they were both open, one closed, or the other, or both closed.
Now I don't know what it really means to superimpose the forward wave function from two pinholes, with the backward wave function from just one detector.
It's not that it's "incorrect" - it's just that it's not clear what it would demonstrate.
Carl
----------
Dear Carl, there are some important issues, and actually if you see the work of Lev Vaidman  you will see that this superimposing of forward and backward wavefunctions is very essential to properly define/understand the pure state, that quantum coherent superposition. Actually if you have quantum interference [NOTE: THIS IS NOT CLASSICAL ONE!] then this is expressed by NON-ZERO off-diagonal elements in the density matrix. This means that \psi_{1} \psi_{2}^{*} and \psi_{2} \psi_{1}^{*} are not zeroes. The forward wave could have passed through one pinhole and then returned through the other, or could have passed by half through both pinholes, and returned by half through both pinholes - it does not matter. If you have mixed state density matrix however YOU HAVE ZERO off-diagonal elements in the density matrix. In this case \psi_{1} \psi_{2}^{*} and \psi_{2} \psi_{1}^{*} are ZEROs, so you cannot have the situation when the wave have passed through pinhole 1 forward, and then returned back through pinhole 2. Actually the transactional interpretation makes the picture more intuitive, and more imaginable, yet this is not something new and you have it in the matrix formalism of standard QM - ket is forward function |\psi> or just \psi, while bra is <\psi| or \psi*, where it is transposed and complex conjugated ket - J.G. Cramer is not the first one to notice that bra is time inverse of ket, Wigner has already shown this. So there the superimposition that you wonder about is the ESSENCE of quantum coherence/pure state/interference.
Note - in quantum interference you add quantum amplitudes and then square the result to produce the observable probability distribution e.g. light intensity, in classical interference you add classical amplitudes that are observables, so you add the squared probabilities [i.e. mixed state density matrix, see the complementarity principle in which way case where P = |\psi_{1}|^{2} + |\psi_{2}|^{2}. Danko Georgiev MD 03:31, 24 January 2006 (UTC)
thread_title:  Retrospective Wave functions