# 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.

Staff Emeritus
Homework Helper
Are you using your expression for cn 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.