Where Did I Go Wrong in Degenerate Perturbation Theory?

In summary: 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.That sounds great. Thanks!
  • #1
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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  • #2
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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  • #3
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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  • #4
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:
 

FAQ: Where Did I Go Wrong in Degenerate Perturbation Theory?

What is degenerate perturbation theory?

Degenerate perturbation theory is a mathematical method used in quantum mechanics to calculate the energy levels and wavefunctions of a quantum system that has multiple degenerate states. Degenerate states are states with the same energy but different wavefunctions.

How does degenerate perturbation theory differ from regular perturbation theory?

In regular perturbation theory, the energy levels and wavefunctions of a system are calculated by treating the perturbation (such as an external field) as a small disturbance to the system. In degenerate perturbation theory, the perturbation is treated as a large disturbance, and the energy levels and wavefunctions are calculated by diagonalizing the perturbation matrix.

What are some applications of degenerate perturbation theory?

Degenerate perturbation theory is commonly used in quantum mechanics to study systems with multiple degenerate states, such as atoms, molecules, and solid-state materials. It is also used in other fields, such as nuclear physics and particle physics, to study the behavior of particles with multiple energy states.

What are the limitations of degenerate perturbation theory?

Degenerate perturbation theory is only applicable to systems with a small perturbation. If the perturbation is too large, the calculations become unreliable. Additionally, it assumes that the degenerate states are well-separated in energy, which may not always be the case.

How is degenerate perturbation theory used in practice?

In practice, degenerate perturbation theory involves solving a set of coupled equations, known as the secular equations, to determine the energy levels and wavefunctions of the system. This can be done analytically for simple systems, but for more complex systems, numerical methods are often used. Computer programs, such as the popular quantum chemistry software package Gaussian, use degenerate perturbation theory to calculate the properties of molecules and materials.

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