Perturbation Theory

  • #1
1,444
0
a particle moves in one dimension in the potential

[itex]V(x)=\infty \forall |x|>a, V(x)=V_0 \cos{\frac{\pi x}{2a}} \forall |x| \leq a[/itex]

now the unperturbed state that i use is just a standard infinite square well.

anyway the solution says that perturbation theory is only valid provided that the energy scale of the "bump" (set by [itex]V_0[/itex]) is less than the difference in energy between square well states.
Q1: is this just a fact: perturbation theory is only applicable providing the perturbation is less than the energy difference of the states of teh unperturbed system?

it then explains the above mathematically by saying:

[itex]V_0 << \frac{{\hbar}^2 \pi^2}{8ma^2}(2n-1)[/itex]
i can't for the life of me see where the RHS of that comes from. arent the energy levels dependent on [itex]n^2[/itex] and not [itex]n[/itex]?
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
i think if you look at the difference in energy is given by
[tex] n^2- (n-1)^2 = n^2 - (n^2 - 2n+1) = 2n-1[/tex]

i think less than is probably not strong enough, i think the perturbation has to be "small" relative to the energy difference. Looking in Ballentine, this can be seen if you look at the first order contribution to the eigenvector. It effectively contains the ratio of the perturbation to the energy level difference. Higher order terms carry the ratio at higher powers, so for the perturbation sum to converge (and quickly) the ratio must be small
 

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