# Finite square well potential energy

1. Jun 27, 2013

### Jalo

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

Hello.

Imagine a particle bound in a square well potential of potential energy
V0 if |x| > a
0 if |x| < a

The wave function of the particle is: (ignoring the time dependency)
-A*exp(kx) if x<-a
B*sin(3*pi*x/4a) if |x|<a
A*exp(-kx) if x>a

where k = sqrt(2mE)/ħ

Find the energy of the particle.

2. Relevant equations

3. The attempt at a solution

First of all I determined the value of A through the condition of continuity at the boundaries:
A*exp(-k*a) = B*sin(3*pi*a/4a)
A*exp(-k*a) = B/sqrt(2)
A = B*exp(ka)/sqrt(2)

Rewriting the wave function:

-B*exp(k[x+a])/sqrt(2) if x<-a
B*sin(3*pi*x/4a) if |x|<a
B*exp(-k[x-a])/sqrt(2) if x>a

After that I decided to use the normalization condition to find the value of k.

$\int$dx <ψ|ψ> = 1

Separating the integral into three, one for each region, I concluded that:

∫dx A2 exp(2kx) = B2/4k
∫dx A2 exp(-2kx) = B2/4k
∫dx A2 sin2(3*pi*x/4a) = B2(a+1/2a)

Therefore:
B2/2k + B2(a+1/2a) = 1
B2(1/2k + a + 1/2a) = 1

However this doesn't help me much... I don't know what I should do next, or if what I'm doing is getting me closer to the answer.. The correct answer is:

E = (9/32) π2ħ2/ma2

Thanks for taking the time to read my problem.

2. Jun 27, 2013

### TSny

To answer the question, you don't need to worry about the normalization.

Think about how E shows up in the Schrodinger equation.

3. Jun 28, 2013

### Jalo

Thank you very much! I completely forgot about that!

Cheers!