QM: Infinite Square Well -a/2 to a/2

[chrisd]

I have read similar threads about this problem but I wasn't able to make progress using them.

Homework Statement

Consider an infinite square-well potential of width a, but with the coordinate system shifted so that the infinite potential barriers lie at x=[itex]\frac{-a}{2}[/itex] and x=[itex]\frac{a}{2}[/itex].

Solve the Schrodinger equation for this case to calculate the normalized wave function [itex]\psi[/itex]_n(x) and the corresponding energies E_n

Homework Equations

Time Independent Schrodinger Equation (V(x)=0 within the well)
-\hbar²[itex]\frac{1}{2m}[/itex] ([itex]\frac{∂}{∂x}[/itex])²[itex]\psi[/itex]=E[itex]\psi[/itex]

The Attempt at a Solution

Skipping ahead to the general solution for [itex]\psi[/itex](x) , I get:

[itex]\psi[/itex](x) = Ae^ikx + Be^-ikx, k = [itex]\frac{\sqrt{2mE}}{\hbar}[/itex]

Using the boundary conditions,

[itex]\psi[/itex]([itex]\frac{a}{2}[/itex])=Ae^ika/2 + Be^-ika/2=0
[itex]\psi[/itex]([itex]\frac{-a}{2}[/itex])=Ae^-ika/2 + Be^ika/2=0,

together with some substitution I am able to prove that,

k =[itex]\frac{\pi}{a}[/itex]n

E_n =([itex]\frac{n\pi\hbar}{a}[/itex])²[itex]\frac{1}{2m}[/itex]

which I believe to be correct for infinite square well.

I run into trouble trying to solve for the constants A and B. My approach was to try and normalize the wavefunction and then use substitution to solve for A or B, but I can't seem to get anywhere after a certain point.
I would expect to get cosines for n=1,3,5... and sines for n=2,4,6... based on what I know about the shape of the wavefunction, but I am unsure how to prove this mathemetically.
Any tips would be appreciated.

 

[vela]

Try plugging in your allowed values of k into the boundary condition equations. You'll get a different relationship between A and B depending on whether n is odd or even.

 

[chrisd]

Ah, that's the ticket. It's probably easier without using exponential form, but I was able to prove that A=B for odd n (giving me a cosine) and A=-B for even n (giving me a sine). Thanks!