Finite square well, excited states

1. Oct 8, 2014

photomagnetic

1. The problem statement, all variables and given/known data
Consider a particle of mass m in the ground state of a potential well of width 2 a and depth.
the particle was in the ground state of the potential well with V0 < Vcritical, which means the well is a narrow one.
At t = 0 the bottom of the potential well is shifted down to Vo' from Vo.
The resulting well is deep enough to support two bound states with energies
-e0 and -e1 > -e0.
What is the probability that the particle will get ionized – that is, it will leave the potential well
by occupying levels with positive energy?
2. Relevant equations

3. The attempt at a solution
I've found the wave function for e0
and the 2nd wave function for e1

Now I have to integrate them (-a to a)
But this is getting outta control. Is there a trick? or I am doing something wrong because there is no way to integrate so many terms. Screw mathematica btw. I need to learn the trick. Thanks!

2. Oct 8, 2014

Orodruin

Staff Emeritus
You should not be integrating from -a to a. This would simply give you the probability of finding the particle inside the well if it is in a bound state. What you want to figure out is the amplitude (and hence probability) of the particle ending up in the bound states.

3. Oct 8, 2014

photomagnetic

I'm pretty sure it's -a to a
"The resulting well is deep enough to support two bound states with energies"

[integral ( first wavefunction) (2nd wave function) dx ]2

4. Oct 8, 2014

Orodruin

Staff Emeritus
The wave functions will have overlap also in the regions with |x| > a as long as the well is not infinite. Neglecting this contribution is going to introduce errors. This is why you need to integrate over the entire real line.

5. Oct 8, 2014

BvU

I suspect Photo is also chewing on the wrong integral. E0 changes to E0' when V changes...

Particle is in state $|\psi_{0,V_0}>$, ground state for the V0 well.

V changes, bound states are now the wave functions you claim to have found: One even $|\psi_{0,V_0'}>$, one odd $|\psi_{1,V_0'}>$, right?

For a given state $\phi$, the amplitude of $\psi_0$ is $<\phi\;|\;\psi_0>$. As Oro says: from $-\infty$ to $\infty$.

The other state is odd, so I expect that amplitude to be zero.

So there is one (admittedly hideous) integral left to do. Answer is then 1 - (that integral squared)

I would expect that finding the expression, not its numerical value, is the object of the exercise.