Intro Quantum: Expanding infinite square well

[camron_m21]

Homework Statement

Griffiths Intro to Quantum, problem 2.38:
A particle of mass m is in the ground state of the infinite square well. Suddenly the well expands to twice its original size: the right wall moving from a to 2a, leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured.

Homework Equations

c_{n} = \sqrt{\frac{2}{a}}\int^{a}_{0} sin(\frac{n\pi}{a} x) \Psi (x,0) dx

The Attempt at a Solution

Since the particle starts out in the ground state in a well of length a, at t=0 (right when the well expands) the wave function should be

\Psi (x,0) = \sqrt{\frac{2}{a}} sin\left( \frac{\pi x}{a} \right)

I know this can be written as a sum of the new wave functions,

\Psi(x,0) = \sum c_{n} \psi_{n} (x)

The problem wants the most probable result of measuring the energy, as well as the next most probable. For this, I was thinking of using |c_{n}|^{2} as the probability of getting an energy. However, when I do this, I get pi times an integer as the argument for the sine in the answer, which gives me zero.

I'm mostly just at a loss on how to start on this, so any help would be appreciated. I'm not sure what I need to calculate to find the most and next most probable energies.

 

[camron_m21]

Scratch that, I figured it out. For anyone else stuck on this, calculate c_{n} the regular way, but use [0,a] for the integration bounds rather than [0,2a], since the initial function at t=0 is 0 for x > a.

 

[vela]

Are you using your expression for c_n above? It's not quite correct for this particular problem because it's using the eigenstates for the original well, not the expanded well.

 

[camron_m21]

Yeah, sorry about that, that was just the formula for the general infinite square well. The formula I used replaces a with 2a, except in the t=0 function.