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

1. Oct 25, 2011

### chrisd

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

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

Solve the Schrodinger equation for this case to calculate the normalized wave function $\psi$n(x) and the corresponding energies En

2. Relevant equations

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

3. The attempt at a solution

Skipping ahead to the general solution for $\psi$(x) , I get:

$\psi$(x) = Aeikx + Be-ikx, k = $\frac{\sqrt{2mE}}{\hbar}$

Using the boundary conditions,

$\psi$($\frac{a}{2}$)=Aeika/2 + Be-ika/2=0
$\psi$($\frac{-a}{2}$)=Ae-ika/2 + Beika/2=0,

together with some substitution I am able to prove that,

k =$\frac{\pi}{a}$n

En =($\frac{n\pi\hbar}{a}$)2$\frac{1}{2m}$

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.

Last edited: Oct 25, 2011
2. Oct 26, 2011

### vela

Staff Emeritus
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.

3. Oct 26, 2011

### 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!