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[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]?