Particle on a ring with perturbation

1. Jul 10, 2014

"pi"mp

So I'm trying to solve old qualifying exam problems, one of which is a particle on a ring with a constant electric field perturbation. The un-perturbed problem is straightforward, and we then add a constant electric field in the x-direction (the ring lies in the xy-plane) of magnitude E. Therefore, we add $-ReE \cos \phi$ as a perturbing term to the Hamiltonian. R is the radius of the ring and phi is the angular coordinate on the ring.

We then ask for the first (non-zero) energy correction to the original ground state. Because of the cosine term, the first order correction vanishes, I believe, so we need to look to second order or beyond.

However, the second order correction involved a sum of terms of the form:

$$\langle \psi_{k} | H_{1} | \psi_{0} \rangle =\alpha \int_{0}^{2 \pi} d \phi e^{ik \phi} \cos \phi$$

Unless I'm screwing up badly, I believe the above terms are also zero. So am I making a mistake here? Or am I correct and I need to look for third order corrections to get a non-zero term?

Last edited: Jul 10, 2014
2. Jul 10, 2014

The_Duck

I don't think this integral is zero for all $k$. Write $\cos \phi = (e^{i\phi} + e^{-i\phi})/2$.

3. Jul 10, 2014

"pi"mp

No? I agree it certainly isn't zero in general, but k must be an integer thanks to the periodic boundary conditions on the un-perturbed eigenfunctions. Won't it always vanish when k is an integer? Thanks a lot.

4. Jul 10, 2014

"pi"mp

Oh wait...it doesn't vanish for k=1, right? It will vanish for all other integral values of k?

5. Jul 10, 2014

Delta²

It doesnt vanish for k=-1 as well but negative integers are out of play anyway.

6. Jul 10, 2014

"pi"mp

Awesome, thank you. So the point is that the infinite sum which is, in general, required for the 2nd order perturbation correction, simply contains one term?

7. Jul 11, 2014

The_Duck

No, you have to consider negative integers too; $e^{ik\phi}$ and $e^{-ik\phi}$ are both energy eigenstates of the unperturbed system.

Two terms; see above.

Last edited: Jul 11, 2014
8. Jul 11, 2014

"pi"mp

But since those two distinct eigenfunctions have the same energy, I would need to use the degenerate version of 2nd order perturbation?