Periodic Systems

Wave packets

We have seen how particles behave as waves in quantum mechanics. You've seen the wave functions, and how a momentum eigenstate is a plane wave. But how do we rationalize their particle-like behavior?

In one sense quantum particles are particle-like simply because they are countable, and because the wavefunction is normalized. But how can we understand them moving in classical trajectories? How do we describe a particle that is behaving in a sense that is in-between the classical and quantum limits?

We describe such particles in terms of wave packets. A wave packet is a form of wave function that has a well-defined position as well as momentum. Thus wave packets tend to behave classically and are easy (and fun) to visualize. Naturally, neither the momentum nor the position is precisely defined, as is governed by the uncertainty principle.

A classic application of the uncertainty principle is in understanding wave packets. A wave packet with a very well-defined position will have a very uncertain momentum, and thus will quickly disperse as the faster components move on ahead of the slower ones. Conversely, if you construct a wave packet with a very definite momentum it will travel a long distance without dispersing, but it starts out being very broad already in position space.

The simples (and most commonly used) wave packet is a Gaussian wave packet. We can construct such a wave packet most easily in reciprocal space:

$$ \tilde\psi(k) = e^{-a^2(k-k_0)^2} $$

SWBQ
What are the dimensions of $a$?
BWBQ
Work out $\psi(x)$ for this wave packet.
What is the expectation value of $p$?
What is the expectation value of $x$?
extra
What is the uncertainty in $p$?
extra
What is the uncertainty in $x$?

So this is a minimal-uncertainty state with this desired momentum and position.

Speed of a wave

Let us say this particle has mass $m$.

BWBQ
What is the frequency of a particle with wave vector $k_0$?
What is its phase velocity?

The phase velocity is how fast the nodes of the real part of the wave travel. Or how fast the "phase" travels.

Generalization of velocity

For a free quantum particle, we have a dispersion relation given by $\omega \propto k^2$. As you saw last term, for a simple classical wave $\omega \propto k$. You will find that for many materials and systems, the relationship between frequency and wave vector is more complicated. We call this relationship the dispersion relation, although in the case of electrons in a solid it is usually called the band structure.

The time dependence of a plane wave (quantum or classical) is given by $$ \psi(x,t) = e^{i(kx-\omega(k)t)} $$

If we write a wave as a superposition of wave vectors, we can write that $$ \psi(x,t) = \frac1{\sqrt{2\pi}}\int \tilde\psi(k) e^{i(kx-\omega(k)t)} dk $$

BWBQ
Solve for $\langle x \rangle$ for an arbitrary state whose initial state is given by $\tilde\psi_0(k)$.

$$ \begin{align} \langle x\rangle &= \int |\psi(x,t)|^2x dx \\ &= \int dx\, x \left(\frac1{\sqrt{2\pi}}\int dk' \tilde\psi(k')^* e^{-i(k'x-\omega(k')t)}\right) \left(\frac1{\sqrt{2\pi}}\int dk \tilde\psi(k) e^{i(kx-\omega(k)t)}\right) \\ &= \frac1{2\pi}\int\! dx\int\! dk\int\! dk'\, x \tilde\psi(k')^*\tilde\psi(k) e^{-i(k'x-\omega(k')t)} e^{i(kx-\omega(k)t)} \\ &= \frac1{2\pi}\int\! dx\int\! dk\int\! dk'\, x \tilde\psi(k')^*\tilde\psi(k) e^{i(k-k')x} e^{-i(\omega(k)-\omega(k'))t} \\ &= \int\! dk\int\! dk' \tilde\psi(k')^*\tilde\psi(k) \left(\frac1{2\pi}\int\! dx\, x e^{i(k-k')x}\right) e^{-i(\omega(k)-\omega(k'))t} \end{align} $$

Here we get to a funny bit. We wish that the $x$ integral would turn in to a $\delta$-function, but it isn't quite. We can, however, relate it to a delta function:

$$ \begin{align} \delta(k) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} dx e^{ikx} \\ \delta'(k) &= \frac{i}{2\pi}\int_{-\infty}^{\infty} dx x e^{ikx} \end{align} $$ where $\delta'(k)$ is the derivative of the delta function. So we can recognize a derivative of a delta function above. $$ \begin{align} \langle x\rangle &= \int\! dk\int\! dk' \tilde\psi(k')^*\tilde\psi(k) \left(\frac1{2\pi}\int\! dx\, x e^{i(k-k')x}\right) e^{-i(\omega(k)-\omega(k'))t} \\ &= \int\! dk\int\! dk' \tilde\psi(k')^*\tilde\psi(k) \left(-i\delta'(k-k')\right) e^{-i(\omega(k)-\omega(k'))t} \\ &= -i \int\! dk\int\! dk' \tilde\psi(k')^*\tilde\psi(k) \delta'(k-k') e^{-i(\omega(k)-\omega(k'))t} \\ &= -i \int\! dk' \tilde\psi(k')^* e^{i\omega(k')t} \int\tilde\psi(k) e^{-i\omega(k)t)} \delta'(k-k') dk \\ u &= \tilde\psi(k)e^{-i\omega(k)t} \\ dv &= \delta'(k-k')dk \\ du &= \left(-it\frac{d\omega(k)}{dk}\tilde\psi(k)e^{-i\omega(k)t} + \frac{d\tilde\psi(k)}{dk}e^{-i\omega(k)t}\right)dk \\ v &= \delta(k-k') \\ \langle x\rangle &= -i \int\! dk' \tilde\psi(k')^* e^{i\omega(k')t} \int u dv \\ &= - i \int\! dk' \tilde\psi(k')^* e^{i\omega(k')t} \left( \left.u v\right|_{-\infty}^{\infty} - \int v du \right) \\ &= i \int\! dk' \tilde\psi(k')^* e^{i\omega(k')t} \int \delta(k-k')\left(-it\frac{d\omega(k)}{dk}\tilde\psi(k)e^{-i\omega(k)t} + \frac{d\tilde\psi(k)}{dk}e^{-i\omega(k)t}\right)dk \\ &= i \int \tilde\psi(k)^* \left(-it\frac{d\omega(k)}{dk}\tilde\psi(k) + \frac{d\tilde\psi(k)}{dk}\right) dk \\ &= t \int \left|\tilde\psi(k)\right|^2 \frac{d\omega(k)}{dk} dk + i \int \tilde\psi(k)^*\frac{d\tilde\psi(k)}{dk} dk \end{align} $$

So we end up with two terms. Let's first look at the second term. It's not too hard if we switch back to a real space representation of our wave function:

$$ \begin{align} \int \tilde\psi(k)^*\frac{d\tilde\psi(k)}{dk} dk &= \int dk \left(\frac1{\sqrt{2\pi}}\int dx \psi^*(x)e^{ikx}\right) \frac{d}{dk}\left(\frac1{\sqrt{2\pi}}\int dx' \psi(x')e^{-ikx'}\right) \\ &= -i\frac1{2\pi}\int dk \left(\int dx \psi^*(x)e^{ikx}\right) \left(\int dx' x'\psi(x')e^{-ikx'}\right) \\ &= -i\int dx \int dx' \psi^*(x) x'\psi(x')\left( \frac1{2\pi}\int dk e^{ik(x-x')}\right) \\ &= -i\int dx \int dx' \psi^*(x) x'\psi(x')\delta(x-x') \\ &= -i\int dx \left|\psi(x)\right|^2 x \\ &= -i\langle x(t=0) \rangle \end{align} $$

The remaining term is proportional to time, and tells us that the speed of the packet is constant, and is equal to the expectation value of $\frac{d\omega}{dk}$. This is what we call the group velocity.

$$ \begin{align} \langle x\rangle &= \langle x(t=0) \rangle + \left( \int \left|\tilde\psi(k)\right|^2 \frac{d\omega(k)}{dk} dk\right) t \end{align} $$

This tells us that the average position (over many measurements) moves with a constant velocity given by the average value of $v_g \equiv \frac{d\omega(k)}{dk}$. Meanwhile, the width of the packet will be increasing, since the group velocity is not the same for all the portions of the wave packet.

Phase and group velocity

Any wave will in general have two velocities: phase velocity and group velocity. You just derived the group velocity, which is given by:

$$\begin{align} v_g &= \left<\frac{d\omega}{dk}\right> \end{align}$$

More commonly, we omit the expectation value and just write that

$$\begin{align} v_g &= \frac{d\omega}{dk} \end{align}$$

which is accurate when we consider a wave that has a very certain momentum. In either case, we have shown that the group velocity tells us $\frac{d<x>}{dt}$, which means how fast a packet will be moving on average.

The phase velocity is the velocity you have learned about waves (and before), which tells you how quickly the phase moves. You could think of this as how quickly the zeros of the real part (or imaginary part) of the wave function move.

BWBQ
Work out the group and phase velocity of a particle with mass $m$ that has wave vector $k$.

Wave packet animation

See python code for wave packet animation unless you are in PH 367, in which case you should write your own code to do this.

You can see the two velocities in this figure. The group velocity is the steepish slope of the packet as a whole. In contrast, the phase velocity is the slope of the blue lines (or red lines), which is rather shallower (slower).

Video of a wave packet