Where Did I Go Wrong in Degenerate Perturbation Theory?

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Homework Help Overview

The discussion revolves around degenerate perturbation theory in quantum mechanics, specifically focusing on the calculations of matrix elements and first-order corrections. Participants are examining the validity of certain integrals and the implications of specific values leading to zero denominators.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute matrix elements and questions the correctness of their results. Some participants share their own calculations of first-order corrections, noting that they also arrive at zero, while others raise concerns about potential issues with the denominator in their expressions.

Discussion Status

The discussion is active, with participants exploring different aspects of the problem. Some guidance has been offered regarding the need to consider specific cases where the denominator may be zero, and there is an acknowledgment of the importance of handling these cases separately.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the depth of exploration. There is an emphasis on ensuring that all relevant cases are considered in the calculations.

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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