# Infinite well potential - changed bottom

## Homework Statement

The bottom of an infinite well is changed to have the shape

$$V(x) = \epsilon \sin {\dfrac{\pi x}{b}}, 0 \le x \le b$$

Calculate the energy shifts for all the excited states to first order in ##\epsilon##. Note that the well originally had ##V(x) = 0## for ##0 \le x \le b## and ##V = \infty ## elsewhere.

## Homework Equations

/Attempt at a solution[/B]

I know I should use ##<\Psi _n | H_1 | \Psi_n>##, and that ##H_1 = \epsilon \sin {\dfrac{\pi x}{b}}##. If I had ##\Psi _n## all I to do was integrate between ##0## and ##b##. However, I don't have ##\Psi _n##. For the "normal" potential well, I know that ##\Psi _ n = \sqrt{2/b} \sin {\dfrac{n \pi x}{b}}##. However, that is not the case in this exercise.

Any suggestions?

DEvens
Gold Member
You say: "However, that is not the case in this exercise." Why is it not? That is, what would the wave functions be if epsilon was zero?

You say: "However, that is not the case in this exercise." Why is it not? That is, what would the wave functions be if epsilon was zero?

For epsilon = 0, we'd have no potential inside the box, so the solution would be ##\Psi _ n = \sqrt{2/b} \sin {\dfrac{n \pi x}{b}}##. I say that it's not the same because there's a space (below the potential line) in which the particle can not be. Is it the same though?

DEvens
Gold Member
I say that it's not the same because there's a space (below the potential line) in which the particle can not be.

Um... What? It's a finite potential value inside an infinite square well. What do you mean "in which the particle can not be"? Where can't it be? And why not?

Um... What? It's a finite potential value inside an infinite square well. What do you mean "in which the particle can not be"? Where can't it be? And why not?

I'm imagining something like this: http://sketchtoy.com/64403013

I said that the particle couldn't be in C, only in A. But now that I think about it, I don't see why it could not be in C.

DEvens