# Symmetric infinite well

1. Oct 4, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
The time-independent Schrodinger equation solutions for an infinite well from 0 to a are of the form:

$$\psi_n(x) = \sqrt{2/a} \sin (n \pi x/ a)$$

If you move the well over so it is now from -a/2 to a/2, then you can replace x with x-a/2 and get the new equations right?

If I try to actually apply these new boundary conditions to the sin function, I get something different, however:

$$\psi_n(x) = \sqrt{2/a} \sin (2 \pi n x/a)$$

How can these two techniques give different results?

2. Relevant equations

3. The attempt at a solution

2. Oct 4, 2007

### Dick

Probably because you did it wrong. You managed to halve the period of a wavefunction by translating it. That's not supposed to happen, is it?

3. Oct 5, 2007

### ehrenfest

I just applied the boundary conditions to sin(k_n * x) by plugging in (a/2, 0) or (-a/2, 0).

This yields k_n = 2 n pi/a?

4. Oct 5, 2007

### Meir Achuz

For the symmetric well, there are also solutions cos(n\pi x), with n odd.

Last edited: Oct 5, 2007
5. Oct 5, 2007

### ehrenfest

But I get A cos (k_n a/2) + B sin(k_n a/2) = 0 and A cos (k_n -a/2) + B sin (-k_n a/2) = 0 implies that B/A = tan (k_n a/2) = -B/A which implies that B is 0?

Last edited: Oct 5, 2007
6. Oct 5, 2007

### ehrenfest

Maybe the problem is that I am assuming that the k_n is the same for both sine and cosine?

7. Oct 5, 2007

### Dick

sin(n*pi*(x-a/2))=sin(n*pi*x-n*pi/2). Clearly this is +/- cos(n*pi*x).

8. Oct 5, 2007

### ehrenfest

OK. The problem was that in post number 5 I divided by A which could have been 0.