I have read similar threads about this problem but I wasn't able to make progress using them.(adsbygoogle = window.adsbygoogle || []).push({});

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=[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}

2. Relevant equations

Time Independent Schrodinger Equation (V(x)=0 within the well)

-\hbar^{2}[itex]\frac{1}{2m}[/itex] ([itex]\frac{∂}{∂x}[/itex])^{2}[itex]\psi[/itex]=E[itex]\psi[/itex]

3. 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.

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

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