Quantum particles on the torus

Here is a simple yet beautiful problem in quantum mechanics: find the explict form of the wavefunction in the position basis for a single free particle (or multiple, noninteracting, distinguishable free particles) confined to move on the surface of the two-dimensional torus \(\mathbb{T}^2\).

We will just be solving the time-independent Schrödinger equation at first, since we know how to go from those solutions to the time-dependent solutions. Our beginning equation is then

I’ll note now, for any physicists reading, that I’ve set all the constants (mass, reduced Planck constant, etc.) to one in the correct units. Anyway, the fun part of this problem is the torus; what coordinate system should we use? I’ll think of the torus as \(\mathbb{T} = S^1 \times S^1\), which means that we should use a product of polar coordinates \(x = (\theta, \psi) \) with the radius held fixed. (We could also see this essential linearity of the coordinates by thinking of the torus as the quotient group \(\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2 \).) Expressing the Hamiltonian in this basis gives \( \langle x | H | \Psi \rangle = \nabla^2 \Psi \), so our equation to solve is

We also need some boundary conditions. Thankfully, those aren’t hard to figure out: since we’re on the torus, we know that any admissible solution must be periodic in both variables, so \(\Psi(\theta, \phi) = \Psi(\theta + 2 \pi, \phi)\) and \(\Psi(\theta, \phi) = \Psi(\theta, \phi + 2\pi)\). Since \( \Psi^*(\theta, \phi)\Psi(\theta, \phi)\) must be a probability distribution, we also have the normalization condition

We’re ready to solve! This PDE is linear, so we write the wavefunction as a product of univariate functions \(\Psi(\theta, \phi) = \Theta(\theta)\Phi(\phi)\). Now we can write the PDE as a sum of ODEs:

We will write \(E = \alpha + \beta\) and break this sum apart into two separate ODEs as

These are easily solved; the solutions take the form of complex exponentials. Since we are in a compact space, we will discard the exponentially growning solutions to find \(\Theta(\theta) = c_1 \exp\left( -i R_{\theta}\sqrt{\alpha}\theta \right)\) and \(\Phi(\phi) = c_1 \exp\left( -i R_{\phi}\sqrt{\beta}\phi \right)\). Thus the wavefunction has the form

There is still some work to be done: we need to find the values \(\alpha\) and \(\beta\) so that we can find an explicit form for \(E\). We also need to figure out the value of the normalization constant \(c\)–let’s do that first. Using our normalization condition, we see that

so that \(c = \frac{1}{2\pi\sqrt{R_{\theta}R_{\phi}}}\) and the wavefunction is \( \Psi(\theta, \phi) = \frac{1}{2\pi\sqrt{R_{\theta}R_{\phi}}}\exp\left( -i(R_{\theta}\sqrt{\alpha}\theta + R_{\phi}\sqrt{\beta}\phi) \right) \).

Now we will deal with the boundary conditions as we figure out the energy levels. (You will note that this is a way to see that energy must be quantized in a quantum system.) For \(\theta\), we have that \(e^{-iR_{\theta}\sqrt{\alpha}\theta} = e^{-iR_{\theta}\sqrt{\alpha}(\theta + 2\pi)} \), so that \(e^{-iR_{\theta}\sqrt{\alpha}2\pi} = 1 = e^{-i 2\pi n} \). Solving, we find that \(\alpha = \left( \frac{n}{R_{\theta}} \right)^2 \). Performing an identical procedure for \(\phi\) gives \(\beta = \left( \frac{m}{R_{\phi}} \right)^2 \). Substituting into the wavefunction, we have (finally!) that

with energy levels given by

Now that we’ve done all the hard work, we can have fun. Let’s introduce time and solve the time-dependent Schrödinger equation in the position basis:

This is a very easy equation; we just integrate to find \(\Psi(t) = \exp(-i E t)\). Great! Now we can put everything together to find that

with energy levels given above.

The great part about this problem is that, for noninteracting distinguishable particles, we could just repeat this process ad infinitum if we wanted to. The stationary wavefunction just becomes a product of the particles: \(\Psi(\theta, \phi) = \prod_{i=1}^n\Psi^{(i)}(\theta, \phi) \); in this case, we’d have

Written on July 20, 2017