Finite square well, excited states

[photomagnetic]

Homework Statement

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?

Homework Equations

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!

 

[Orodruin]

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.

 

[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

 

[Orodruin]

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.

 

[BvU]

I suspect Photo is also chewing on the wrong integral. E₀ changes to E_0' 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.