Infinite well potential - changed bottom

1. Feb 5, 2015

Kidiz

1. The problem statement, all variables and given/known data

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.

2. Relevant equations/Attempt at a solution

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?

2. Feb 5, 2015

DEvens

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?

3. Feb 5, 2015

Kidiz

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?

4. Feb 5, 2015

DEvens

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?

5. Feb 5, 2015

Kidiz

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.

6. Feb 5, 2015

DEvens

Um... You do realize that your graph is plotting the value of the potential, right? It does not tell you where the particle is. It does not make sense what you said. The particle is neither in A nor C. It's not on that graph.

The particle has a wave function. You have to work out that wave function. Or at least, you are asked to work it out to first order in epsilon. You are being asked to do a perturbation calculation. You had most of it in your first post.