# Electronic Properties of Materials/Quantum Mechanics for Engineers/Momentum Velocity and Position

** ROUGH DRAFT**
Next, we are going to talk about momentum, position. What makes this discussion particularly useful is that it provides the basis for later parts of this course where we start talking about the velocity of electrons moving in a material; relevant to the conductivity of a material. First we must define velocity in quantum mechanics in terms of position and momentum.

## A Closer Look at the Free Particle

Looking back at our free particle from <CHAPTER>. We solved the free particle already and our resulting Hamiltonian was ${\hat {H}}=H(x)$. (Note that this is for a 1D particle. The solutions are valid for 2D and 3D, but for the purpose of this exercise we will constrain ourselves to 1D.)
Additionally, our wave function, Ψ(x, t)=X(x)T(t){\textstyle \Psi (x, t)=X(x)T(t)} where T(t){\textstyle T(t)} gives the time evolution of the system, is separable. You can prove to yourselves that substituting these in will definitely get a Schrodinger equation that separates into a time dependent and position dependent part. Furthermore, our time dependent part looks like: T(t)=exp⁡[−i E/ℏ t]{\textstyle T(t)=\exp[-i\ {E \over \hbar }\ t]}. Here −i E/ℏ t{\textstyle -i\ {E \over \hbar }\ t} must be dimensionless, which means that E/ℏ {\textstyle {E \over \hbar }} is the frequency (ω{\textstyle \omega }) with units of 1/t {\textstyle 1 \over t}, and E=ℏω{\textstyle E=\hbar \omega }.
Similarly in blackbody radiation, we use E=nhν{\textstyle E=nh\nu }, or E=n(2πℏ){\textstyle E=n(2\pi \hbar )}, where ℏ{\textstyle \hbar } is the reduced Plank constant. By multiplying this by the relationship between frequency and angular frequency, ω/2π {\textstyle \omega \over 2\pi }, we get E=ℏω{\textstyle E=\hbar \omega }.
Going back to the position dependent part of our original equation, we know that this is just another planewave, as proved in <CHAPTER>. Once again our planewave function was: X(x)=Ae$^{ikx}$+Be$^{−ikx}${\textstyle X(x)=Ae^{ikx}+Be^{-ikx}}, and our solution was: E=ℏ$^2$k$^2$/2m {\textstyle E={\hbar ^{2}k^{2} \over 2m}}. In this case, because it is a free particle, k{\textstyle k} is a continuous variable; we haven't quantized this at all.
<MATH CHECK>
We know also that momentum and energy commute: [p$^^$, H$^^$]=0{\textstyle [{{\hat {p}}, \ {\hat {H}}}]=0}
In fact we solved momentum in 1D which provided us the solution $X(x)=Ae^{ikx}$, where $k$ is still a continuum value. In this case it could be positive infinity or minus infinity. Just remember that $k$ in the case of a planewave is the wave vector, and it's telling you the direction and the wavelength of the wave.
Finally, we also found that p=ℏk{\textstyle p=\hbar k}, for the free particle, which means that if we measure a particular value of momentum, for instance, we will get a particular value for $k$ ($k'$). Once we measure this particular value, the wave function collapses and we can write the solution as:${\begin{aligned}\Psi (x,t)&=Ae^{ikx}e^{-i{E \over \hbar }t}\\&=Ae^{ikx}e^{i(kx-\omega t)}\end{aligned}}$Having the commutation of energy and momentum equal zero means that we can simultaneously measure these two properties.
<MATH CHECK^^^>

### Velocity

<FIGURE> "Classic Particle Movement" (Description)
So let's say we've got a particular value we'll call $k'$, and this free particle is going to have some sinusoid, $\sin(x)$. <FIGURE> If we wait some small amount of time, and look at it again, the wave will have propagated. This is, after all, what planewaves do. Now let's say that after that "certain amount of time" the planewave propagated by some $\Delta x$, shown in <FIGURE>, which means that this new sinusoid is now $sin(x+\epsilon )$, where $\epsilon >0$. Essentially, if there is some $f(x,t)$, it can be rewritten as $f(x-\nu _{p}t)$.
<MATH CHECK> sin wave propagation (+) or (-)
<FIGURE> "Sinewave Translation" (Propagates towards +x)
Now, if this wave is propagating, then we can talk about the velocity which also propagates. From wave mechanics, we have that the velocity is equal to the angular frequency divided by the wave vector (ω/k {\textstyle \omega \over k}). Multiplying both the top and the bottom by $\hbar$, and substituting variables from our eigenfunction solution gives us:$\nu _{p}={\omega \over k}={\hbar \omega \over \hbar k}={E \over \hbar k}={{\hbar ^{2}k^{2} \over 2m} \over hk}={\hbar k \over 2m}={p \over 2m}={v_{c}\ m \over 2m}={v_{c} \over 2}$
<ASIDE> Extra math from notes with no home:${\begin{aligned}e^{i(kx-\omega t)}&=e^{ik(x-{\omega \over k})t}\\&=e^{ik(x-v_{particle}t)}\\\ \\\end{aligned}}$<END ASIDE>

## Particles vs. Planewaves

In this instance we solved for a delocalized particle, and found the phase velocity. Notice how this equation describes the relationship between the classical velocity (v$_c${\textstyle v_{c}}) of a particle and the velocity of the propagation of a particular planewave, referred to as the phase velocity (ν$_p${\textstyle \nu _{p}}). Most importantly, these two velocities are NOT the same. As it turns out, a real particle will be localized.
When talking about the particles that we are interested in, which have a classical velocity, they simply don't travel as a particle, they travel as a wave packet. <FIGURE> Inside these wave packets there are lots of waves with different k{\textstyle k} values, and the packet as a while moves with the same group velocity $\nu _{g}$ equivalent to our classical velocity.
Particles are not single planewaves. They are a superposition of planewaves, and tend to group themselves together in these wave packets which have a group velocity of the entire group of waves in superposition. Additionally, they are within some sort of envelope function which also travels at the group velocity, equivalent to the classical velocity.

### Superposition

<MATH CHECK> phi or psi in linear wave state equation?
Imagine a superposition of plane waves. In our first example the states in superposition were discrete. They were a summation of states where $\Psi =a_{1}\phi _{1}+a_{2}\phi _{2}+...$. This form has our wave function as a linear superposition of wave states ($\phi$). Each state is a particular solution to our Schrödinger equation where each coefficient provides us with a projection of the wavefunction into these particular basis states. (Thinking back to our eigenfunctions as a basis in Hilbert space.)
This equation is equal to the infinite sum: ∑$_{j=i}$$^∞$a$_j$ϕ$_j${\textstyle \sum _{j=i}^{\infty }a_{j}\phi _{j}}. Note that most of the time, when dealing in practical matters, energy is considered finite so infinite distributions are rare.
<VOCAB> dispertized?
Alternatively, instead of thinking about energy, which is <dispertized>, we generally talk about a continuous distribution. For example, if instead of talking about energies, we express this in terms of momentum. As we saw already, a particle in free space can take any value for momentum giving us the continuous distribution:
$\Psi (x,t)={1 \over {\sqrt {2\pi \hbar }}}\ \int _{-\infty }^{+\infty }\underbrace {\exp \left[\ i\left[p_{x}x-E(p_{x})t\right]\ {1 \over \hbar }\right]} _{Basis\ Function}\ \phi (p_{x})\ \operatorname {d} \!p_{x}$
Here we simply replaced the sum from the infinite energy equation with an integral and integrated over all the allowed values of momentum. The resulting equation is that of the wave packet. Here ϕ(p$_x$){\textstyle \phi (p_{x})} is our coefficient. This is a direct analogy to the summation from earlier as now instead of summing over all these coefficients, we are integrating over them instead, but what are these coefficients?
<FIGURE>
This coefficient is simply a function of $p$, representing the probabilities of finding the particle in one of these particular states. It can be thought of as $P(p_{x})=|\phi (t)|^{2}$. Physically this describes the distribution shown in <FIGURE> where the probability of measuring the particle at a particular momentum is related to the value of our coefficient in front of our basis states.
Now let's simplify our equation and say that $\beta (p_{x})=p_{x}x-E(p_{x})t$, providing us with:
$\psi (x,t)={1 \over {\sqrt {2\pi \hbar }}}\ \int _{-\infty }^{+\infty }e^{i\beta (p_{n}) \over \hbar }$ Looking at this solution, we know that the whole wave function, and the coefficients ϕ(p$_x$){\textstyle \phi (p_{x})}, must be well-behaved. The coefficients are well-behaved as they are just some statistical distribution, going out to zero on either end and integrating to one. On the other hand, e$^{iβ(pn)/ℏ}${\textstyle e^{i\beta (p_{n}) \over \hbar }} will oscillate rapidly, so the only way that our wave function can be well-behaved on the whole is if e$^{iβ(pn)/ℏ}${\textstyle e^{i\beta (p_{n}) \over \hbar }} is a constant defined by:
$\left.{\partial \beta (p_{x}) \over \partial p_{x}}\right\vert _{\underbrace {p_{x}=p_{0}} _{centered\ on\ \phi _{max}}}=0$Solving this relationship looks like:${\begin{aligned}{\partial \beta \over \partial p_{x}}&={\partial \over \partial p_{x}}(p_{x}x-E(p_{x})\ t)\\&=x-t\ {\partial E(p_{x}) \over \partial p_{x}}=0\\x&=t\ {\partial \over \partial p_{x}}E(p_{x})\end{aligned}}$
Looking simply at the units in the final equation, we have length=time∗(length/time){\textstyle length=time*\left({length \over time}\right)}, meaning that ∂/∂p$_x$ E(p$_x$){\textstyle {\partial \over \partial p_{x}}E(p_{x})} has the units (length/time){\textstyle \left({length \over time}\right)} or $velocity$ (ν$_g${\textstyle \nu _{g}}). Going back to our definitions of energy and momentum we can further transform ν$_g${\textstyle \nu _{g}}:${\begin{aligned}\nu _{g}&={\partial E(p_{x}) \over \partial p_{x}}\quad {\begin{cases}E=\hbar \omega \\p_{x}=\hbar k\end{cases}}\\&={\hbar \partial \omega (k) \over \hbar \partial k}\\&={\partial \omega (k) \over \partial k}\end{aligned}}$ Here, $\omega (k)$ and $E(k)$ are called "dispersion relations". They are essentially the energy/velocity of a particle vs. the wave number, $k$. They are important and researchers spend huge amounts of time, money, and resources to determine them for various material systems. For example, the band structure of a material is a dispersion relation. <CHAPTER REF> The group velocity, $\nu _{g}$, is the scope of the dispersion. When we talk about electrons moving in a crystal we talk about the group velocity, the magnitude of which generally depends on $k$.
<FIGURE> "Title" (Description)

## The Momentum Space Representation

Looking closer at these wave packets, let's begin by rewriting our planewave equation, putting time dependence into the general coefficient function and setting $t=0$ to get rid of the energy variable. This results in the $\psi (x,t)$ equation:
$\psi (x,t)={1 \over {\sqrt {2\pi \hbar }}}\ \int _{-\infty }^{+\infty }\operatorname {d} p_{x}\ \phi (p_{x},t)\ \exp[i\ p_{x}\ x{1 \over \hbar }]$
Now let's apply a Fourier Transform to our equation: ${\begin{aligned}{\mathfrak {F}}[\psi (x,t)]&=\phi (p_{x},t)\\{\mathfrak {F}}^{-1}[\phi (p_{x},t)]&=\psi (x,t)\end{aligned}}$
Putting this transformation into the above planewave equation results in:
$\phi (p_{x},t)={1 \over {\sqrt {2\pi \hbar }}}\int _{-\infty }^{+\infty }\operatorname {d} x\ \psi (x,t)\ \exp[-i\ p_{x}\ x{1 \over \hbar }]$
If the set $\left\{\psi _{n}(x,t)\right\}$ are orthogonal to one another and normalized, then $\left\{\phi _{n}(p_{x},t)\right\}$ are ${\mathfrak {F}}\left\{\psi _{n}(x,t)\right\}$ also. We refer to this as the momentum-space representation of the wavefunction and Fourier space has certain properties which makes this representation extremely useful. Truthfully, there is only one wavefunction, (it is a state function!!) but here it is projected on to momentum representation where as $\psi (x,t)$ is projected onto position representation.
Let's consider a physically meaningful distribution. In this case, the equation for gaussian momentum is:$\phi (p_{x})=c\exp \left[-{(p_{x}-p_{o})^{2} \over 2(\Delta p_{x})^{2}}\right]$
<FIGURE> "Gaussian Momentum" (Description)
<MATH CHECK> Everything below...
To define $c$, let's use a well-known relationship: $\int _{-\infty }^{+\infty }\phi ^{*}(p_{x})\ \phi (p_{x})\ \operatorname {d} \!p_{x}=1$
$\int _{-\infty }^{+\infty }|c|^{2}\exp \left[{-1 \over (\Delta p_{x})^{2}}(p_{x}-p_{o})^{2}\right]\operatorname {d} \!p_{x}$
Using a "well-known" relationship to find $c$:
$\int _{-\infty }^{+\infty }e^{-\alpha u^{2}}e^{-\beta u}\operatorname {d} \!u=\left({\pi \over \alpha }\right)^{(1/2)}\exp \left[{\beta ^{2} \over 4\alpha }\right]$
${\begin{aligned}|c|^{2}(\Delta {p_{x}}^{2}\pi )^{1 \over 2}&=1\\c&=(\Delta {p_{x}}^{2}\pi )^{-1 \over 4}\end{aligned}}$
thus...
$\phi (p_{x})=(\Delta {p_{x}}^{2}\pi )^{-1 \over 4}\exp \left[{-(p_{x}-p_{o})^{2} \over 2(\Delta p_{x})^{2}}\right]$
Substituting this into $\psi (x,t)$ and solving...
${\begin{aligned}\psi (x,t)&={1 \over {\sqrt {2\pi \hbar }}}\int _{-\infty }^{+\infty }\operatorname {d} \!p_{x}(\Delta {p_{x}}^{2}\pi )^{-{1 \over 4}}\exp \left[{-(p_{x}-p_{o}) \over 2(\Delta p_{x})^{2}}\right]\exp \left[i\ p_{x}\ x{1 \over \hbar }\right]\\&={1 \over {\sqrt {2\pi \hbar \Delta p_{x}{\sqrt {\pi }}}}}\int _{-\infty }^{+\infty }\operatorname {d} \!p_{x}\exp \left[{-(p_{x}-p_{o}) \over 2(\Delta p_{x})^{2}}\right]\exp \left[i\ p_{x}\ x{1 \over \hbar }\right]\underbrace {\exp \left[i\ x{1 \over \hbar }(p_{o}-p_{o})\right]} _{p_{o}=1}\\&={\exp \left[{ix \over \hbar }p_{o}\right] \over {\sqrt {2\pi \hbar \Delta p_{x}{\sqrt {\pi }}}}}\int _{-\infty }^{+\infty }\operatorname {d} \!p_{x}\exp \left[{-1 \over 2\Delta {p_{x}}^{2}}{(p_{x}-p_{o})^{2}}\right]\exp \left[{ix \over \hbar }{(p_{x}-p_{o})}\right]\\&={\exp \left[{ixp_{o} \over \hbar }\right] \over k}\left({\pi 2\Delta {p_{x}}^{2} \over 2\pi \hbar \Delta p_{x}{\sqrt {\pi }}}\right)^{1 \over 2}\exp \left[{ixp_{o} \over \hbar }\right]\exp \left[{-x^{2} \over 2\left({\hbar \over \Delta p_{x}}\right)^{2}}\right]\\&=\left({\Delta p_{x} \over \hbar {\sqrt {\pi }}}\right)^{1 \over 2}\exp \left[{ixp_{o} \over \hbar }\right]\exp \left[{-x^{2} \over 2\left({\hbar \over \Delta p_{x}}\right)^{2}}\right]\end{aligned}}$
$\psi (x,t)$ is itself a gaussian centered on $x=0$.
Width of a Gaussian: ${\hbar \over \Delta p_{x}}$
$\Delta x\Delta p={\hbar \over \Delta p}\Delta p=\hbar$
As $\Delta p$ becomes large, $\Delta x$ becomes small and vice versa.
In the limit, $\Delta p\longrightarrow 0$
${\begin{aligned}\phi (p_{x})&\rightarrow p_{o}\\\psi (x)&\rightarrow \exp \left[{ixp_{o} \over \hbar }\right]\end{aligned}}$
Watch the evolution of the $\psi _{in}$ over time...
Substitute $\psi (p_{x})=(\Delta {p_{x}}^{2}\pi )^{-{1 \over 4}}\exp \left[{-(p_{x}-p_{o}) \over 2(\Delta p_{x})^{2}}\right]$ into $\Psi (x,t)=(2\pi \hbar )^{-{1 \over 2}}\int _{-\infty }^{+\infty }\exp \left[{i(p_{x}x-E(p_{x})t) \over \hbar }\right]\phi (p_{x})\ \operatorname {d} \!p_{x}$
If you remember, $E(p_{x})={{p_{x}}^{2} \over 2m}$ is our plane wave solution from <LINK>.
Solving the integral gives us:
$\Psi (x,t)={\pi }^{-{1 \over 4}}\left[{{\Delta p_{x} \over \hbar } \over 1+{i\Delta {p_{x}}^{2}t \over m\hbar }}\right]^{1 \over 2}\exp \left[{{{ip_{o}x \over \hbar }-\left({\Delta p_{x} \over \hbar }\right)^{2}{x^{2} \over 2}}-{i\ {p_{o}}^{2}t \over 2m\hbar } \over 1+{i\Delta {p_{x}}^{2} \over m\hbar }}\right]$
Plot $P(x,t)=\Psi ^{*}\Psi$
Just because you're a theorist doesn't mean you shouldn't learn by experimentation. Let's put some numbers in and see how this wave function behaves.
<FIGURE> "Example Graph 1" (t=0)
<FIGURE> "Example Graph 2" (t=5000)
<FIGURE> "Example Graph 3" (t=10000)