# Homework Help: Particle in infinite well which is suddenly expanded

1. Apr 10, 2010

### Phyisab****

1. The problem statement, all variables and given/known data
Well I thought this problem was easy, turned in the homework and got it wrong. My prof is hard to get help from, so hopefully someone here can help me out. A particle is in a one dimensional infinite square well with walls at x=0 and x=L. At time t=0 the well is expanded to width 2L. What is the probability the particle will be in the nth stationary state of the expanded well? I bet a lot of people have seen this before apparently it's a pretty popular problem.

2. Relevant equations

$$\left\langle\Psi_{f}\left|\Psi_{i}\right\rangle = d_{n}$$

3. The attempt at a solution

$$=\frac{\sqrt{2}}{L}\int sin(\frac{n \pi x}{2 L})sin(\frac{\pi x}{L})dx$$

So I'm pretty sure I'm right so far, and the problem is just to evaluate the integral. I was sure the solution was

$$d_{n}=\frac{\sqrt{2}}{2}$$ if n=2

$$d_{n}= 0$$ if n not = 2

$$d_{n}= \frac{\sqrt{2}}{2}$$ if n=2
$$d_{n}= \frac{4}{\pi}\frac{1}{(n^{2}-4}$$ if n is odd
$$d_{n}= 0$$ otherwise

I can plug the integral into mathematica and of course it gives me the right answer. But I have seen integrals like this hundreds of times in solving PDE's, and they always go like I first thought. What is the difference here? This is causing me a ridiculous amount of cognitive dissonance.

2. Apr 10, 2010

### kuruman

One of the sines has a 2L in the denominator while the other has just L. That's the difference.

3. Apr 10, 2010

### morphemera

The integral is the Fourier transform of the $$sin(\frac{\pi x}{L})$$ in the new well.
From your result, I guess the initial wave function may not be the ground state of the of the old well ($$sin(\frac{\pi x}{L})$$)
or your integral use a different domain (note the integration from 0 to L is different from 0 to 2L) because the wavefunction is 0 within L to 2L
I have not tried it though.

4. Apr 10, 2010

### Phyisab****

I realize that. I just don't see why it makes any difference. For n=8 there might as well be a 4 in the numerator and just L in the denominator. How the heck do I explicitly evaluate this so I can prove the answer to myself?

5. Apr 10, 2010

### kuruman

If your problem is evaluating the integral, first substitute θ = πx/L and convert the sines to exponentials. Then

$$sin(n\theta/2)sin(\theta)=-\frac{(e^{i n\theta/2}-e^{-i n\theta/2})(e^{i \theta}-e^{-i \theta})}{4}$$

Multiply out the terms and you will get four integrals that look like

$$\int^{\pi}_0 e^{i(n/2+1)\theta}d\theta=\left[ \frac{e^{i(n/2+1)\theta}}{i(n/2+1)}\right]^{\pi}_0$$

or, after you multiply out the terms you should be able to recognize sines and/or cosines and integrate them.

In either case, eventually, you should be able to combine the answer to a real number and the role of the 2 in the denominator should become obvious.

Last edited: Apr 10, 2010