Where Did I Go Wrong in Degenerate Perturbation Theory?

mathsisu97
Messages
6
Reaction score
0
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?
 
Physics news on Phys.org
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?
 
Last edited by a moderator:
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:
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top