# 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?