# Homework Help: Finite Square Well Solutions

1. Oct 22, 2012

### chris_avfc

1. The problem statement, all variables and given/known data
Already defined that for a 1D well with one finite wall the eigenvalue solutions are given by
k cot(kl) = -α

Show the eigenvalue solutions to well with both walls finite is given by

tan(kl) = 2αk / (k^2 - α^2)

Well is width L (goes from 0 to L) with height V_0

2. Relevant equations

k cot(kl) = -α

tan(kl) = 2αk / (k^2 - α^2)

Time Independent Schrödinger Equation.

3. The attempt at a solution

I will explain briefly as there is a lot of equations and it will look a mess, I can always upload a picture of my work if needs be.

General Solution:
u(x) = Ce^-αx + De^αx

For x < 0
C = 0, so exponential doesn't go to ∞.
For x > L
D = 0 so exponential doesn't go to ∞.

Using boundary conditions where the function and it's derivative must be continuous
With the solution between 0 and L being

u(x) = A sin(kx)

Substituting in L for x

A sin(kL) = Ce^-αL
Ak cos(kL) = -αCe^-αL

Dividing the second by the first

k cot(kL) = -α (As given in question)

This breaks down when substituting in 0 for x.

A sin(k0) = De^α0
Ak cos(k0) = αDe^α0
0 = αD

I thought about changing the coordinates so the well runs from -L/2 to L/2, but that just appears to give

k cot(kL/2) = -α

And

k cot(kl/2) = α

I was/am looking in one text book, and that manages to get it down to

k tan (kL/2)= α
k cot (kL/2)= -α

Still not what I need, but might be closer?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 24, 2012

### chris_avfc

Guess nobody can help then. :(