Degenerate perturbation theory matrix elements with cosine potential

[mathsisu97]

Homework Statement

For a free particle confined to a ring with radius $a$ with pertubation $H' = V_0 \cos(x)$, what is the first order correct energy and correction to the wavefunction?

Relevant Equations

$H' = V_0 \cos(x)$
$\psi_n = \frac{1}{\sqrt{2 \pi}} e^{inx}$
$E_n = \frac{n^2 \hbar^2}{2 m a^2}$
$n = 0, \pm 1, \pm 2 \ldots$

$$ W_{n,n} = \int_0^{2 \pi} \frac{1}{\sqrt{2 \pi}} e^{-inx} V_0 \cos(x) \frac{1}{\sqrt{2 \pi}} e^{inx} dx $$
$$ = 0 $$

$$ W_{n, -n} = \int_0^{2 \pi} \frac{1}{\sqrt{2 \pi}} e^{-inx} V_0 \cos(x) \frac{1}{\sqrt{2 \pi}} e^{-inx} dx $$
$$ = \frac{a n ( \sin(4 \pi n) + i \cos( 4 \pi n) - i )}{\pi (4 n^2 -1) } $$
$$ = 0 $$

This doesn't seem right? Where have I gone wrong?

 

[etotheipi]

I basically know nothing about this subject, but fwiw when I tried it I got the first order correction$$\psi_n^{(1)} = \frac{ma^2}{\pi \hbar^2} \sum_{l\neq n} \frac{(n+l)(-i + i)}{n^2 - 2ln + l^2 -1} = 0$$to be zero as well. Perhaps a more experienced member can advise?

 

[nrqed]

I basically know nothing about this subject, but fwiw when I tried it I got the first order correction$$\psi_n^{(1)} = \frac{ma^2}{\pi \hbar^2} \sum_{l\neq m} \frac{(n+l)(-i + i)}{n^2 - 2ln + l^2 -1} = 0$$to be zero as well. Perhaps a more experienced member can advise?

Watch out. Are there integer values of $l$ for which the denominator is zero?
You see that you must be careful with these specific cases. Go back to the integral and consider these two cases separately.

 

[etotheipi]

Watch out. Are there integer values of $l$ for which the denominator is zero?
You see that you must be careful with these specific cases. Go back to the integral and consider these two cases separately.

Ah, neat! Yeah, $l = n \pm 1$ both give zero denominator. Considering these separately means doing the integrals of $e^{\pm ix} \cos{x}$ between $0$ and $2\pi$, each of which gives $\pi$. So I think, these two terms will be the only two non-zero contributions to the first-order correction.

Does that sound okay to you? I won't write down further details because I'm not the OP, after all