Where Did I Go Wrong in Degenerate Perturbation Theory?

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The discussion revolves around issues in calculating the first-order correction in degenerate perturbation theory, specifically regarding the integrals involving the potential term. The user initially finds both the matrix elements W_{n,n} and W_{n,-n} to be zero, leading to confusion about the correctness of their approach. A key point raised is the potential for the denominator in the correction formula to be zero for specific integer values of l, which necessitates careful consideration of these cases. It is suggested that separating the cases where the denominator is zero and evaluating the integrals of e^{\pm ix} cos(x) will yield non-zero contributions. The conversation emphasizes the importance of addressing these specific scenarios to resolve the apparent contradictions in the calculations.
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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?
 
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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?
 
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etotheipi said:
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.
 
nrqed said:
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 :smile:
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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