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.

 

[paranormal]

Firstly, it is important to note that when the well expands, the energy of the particle will still be quantized and will depend on the new length of the well, 2a. This means that the energy levels will now be given by E_n = \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2}.

To find the most probable energy, we need to calculate the coefficients c_n for the new wave functions. As you mentioned, these coefficients can be calculated using the formula c_{n} = \sqrt{\frac{2}{a}}\int^{2a}_{0} sin(\frac{n\pi}{2a} x) \Psi (x,0) dx. Plugging in the given wave function, we get c_{n} = \sqrt{\frac{2}{a}}\int^{2a}_{0} sin(\frac{n\pi}{2a} x) \sqrt{\frac{2}{a}} sin\left( \frac{\pi x}{a} \right) dx.

Simplifying this expression, we get c_{n} = \frac{1}{2}\int^{2a}_{0} sin(\frac{\pi x}{2a}) sin(\frac{n\pi x}{a}) dx. Using the trigonometric identity sin(a)sin(b) = \frac{1}{2}(cos(a-b) - cos(a+b)), we can rewrite this as c_{n} = \frac{1}{2}\int^{2a}_{0} \left(cos(\frac{\pi x}{2a} - \frac{n\pi x}{a}) - cos(\frac{\pi x}{2a} + \frac{n\pi x}{a}) \right) dx.

Integrating this expression, we get c_{n} = \frac{1}{2}\left(\frac{2a}{\pi}sin(\frac{\pi}{2} - n\pi) - \frac{2a}{\pi}sin(\frac{\pi}{2} + n\pi) \right). Since sin(\frac{\pi}{2} - n\pi) = (-1)^n and sin(\frac{\pi}{2} + n\pi) = (-1)^{n+1}, we can simplify this to c_{n} = \frac{