# Particle in a box Part II

1. Apr 9, 2006

### eku_girl83

O.k. I posted earlier concerning this question. I think I know have a better understanding of what's going on.

Particle in a box with boundaries at x=L and x=3L.
U(x) = A cos kx + B sin kx
Applying boundary conditions, I get that
0 = A cos kL + B sin kL
and that
0 = A cos 3kL + B sin 3kL

Then I normalize U(x).

I also had a question concerning the ground state wave function. In my first post, I was informed the ground state was a sine function for the situation x=0 to x=L. So for the case of x=L to x=3L, it should be a linear combination of sine and cosine. So is psi(x,0) = A cos kx + B cos kx? If not, then what?

I really thought I understood the problem for x=0 to x=L, but the x=L to x=3L situation is giving me fits!

Thanks for the help!

2. Apr 9, 2006

### Euclid

The value of k will come from the (time-independent) S.E. A and B come from the boundary conditions.
This is exaclty the same as the PIB for [0,L], except the boundary conditions are slightly different. Find the solution for the first case ([0,L]), and follow it through step by step. The solution to the present case will look almost identical.