Where Did I Go Wrong in Degenerate Perturbation Theory?

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SUMMARY

The discussion centers on the application of degenerate perturbation theory in quantum mechanics, specifically addressing the calculation of matrix elements \( W_{n,n} \) and \( W_{n,-n} \). The user initially finds both matrix elements equal to zero, which raises questions about potential errors in their approach. A more experienced participant highlights the importance of considering cases where the denominator of the correction term becomes zero, particularly for integer values of \( l \). The conclusion emphasizes that the non-zero contributions to the first-order correction arise from specific cases, leading to a clearer understanding of the perturbation theory application.

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mathsisu97
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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.
 
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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:
 

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