Most Probable energy after infinite square well expands

[dark_matter_is_neat]

Homework Statement

Consider an infinite square well in one dimension extending from -a/2 to a/2. A particle of mass m is sitting in the ground state. At time t = 0 the size of the well doubles so it now goes from -a to a. If one then measures the energy, what is the most probable result and what is its probability.

Relevant Equations

E$\Psi(x)$ = $-\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi(x)}{\partial x^{2}}$
For wave function of the initial well:
$\Psi(\frac{a}{2}) = \Psi(-\frac{a}{2}) = 0$
For the wave function of the final square well:
$\Psi(a) = \Psi(-a) = 0$

Going through the schrodinger wave equation, $-\frac{2mE}{\hbar^{2}}\Psi(x) = \frac{\partial^{2} \Psi(x)}{\partial x^{2}}$, so $Psi(x) = C_{1}sin(\frac{\sqrt{2mE}}{\hbar}x) + C_{2}cos(\frac{\sqrt{2mE}}{\hbar}x)$.
Enforcing the boundary conditions:
$cos(\frac{\sqrt{2mE}}{\hbar} \frac{a}{2}) = 0$, so $\frac{\sqrt(2mE)}{\hbar} \frac{a}{2} = n \frac{\pi}{2}$ for odd n.
$sin(\frac{\sqrt{2mE}}{\hbar} \frac{a}{2}) = 0$, so $\frac{\sqrt(2mE)}{\hbar} \frac{a}{2} = n \pi$.
So this leads to:
$\Psi(x) = C_{2}cos(\frac{n \pi x}{a})$ for odd n
$\Psi(x) = C_{1}sin(\frac{n \pi x}{a})$ for even n
$E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2 m a^{2}}$
Enforcing normalization, $C_{1} = C_{2} = \sqrt{\frac{2}{a}}$.

The wave function for the expanded well is identical except you replace a with 2a.

My first thought for what would be the most probable energy, was the energy of the ground state of the initial square well, $\frac{\pi^{2} \hbar^{2}}{2 m a^{2}}$ which corresponds to the n = 2 state for the widened well , however when I calculate this probability I get:
P = $|\int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sqrt{2}}{a} cos(\frac{\pi x}{a}) sin(\frac{\pi x}{a}) dx|^{2}$ which is obviously 0.

I believe the actual most probable energy is the ground state energy of the new well, which has probability:
P = $|\int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sqrt{2}}{a} cos(\frac{\pi x }{a}) cos(\frac{\pi x}{2a}) dx|^{2}$ = $\frac{64}{9 \pi^{2}}$
Which I've checked against the probabilities for n>1 up to n = 3 and it had the highest probability.

What confuses me about this answer is if I were to look at a physically identical well that instead starts at 0 (so the first well would go from 0 to a and the widened well would go from 0 to 2a), I would get that the probability of E = $\frac{\pi^{2} \hbar^{2}}{2ma^{2}}$ to be non-zero and in fact be the most probable energy.

These to wells are identical, I just changed my coordinate system, but I end up getting two different results for the most probable energy. I find this really strange, shouldn't the system be translationally invariant? What is going on here? Did I just make a mistake in my calculations or is there something else I'm not seeing?

 

[dark_matter_is_neat]

Never mind I realize that the two wells I'm comparing are different. The well the expansion of a well going from 0 to a to 0 to 2a is different than a well expanding from -a/2 to a/2 to -a to a, since in the first situation the well is just expanding rightwards while in the second situation the well is expanding both leftwards and rightwards.

It's still interesting how the probabilities change depending on how you expand the well.

 

[Vanadium 50]

Think classically. It takes (or maybe recovers) energy to expand a well when there is a particle bouncing around inside it. There is no guarantee that this energy is the same when the particle is in different places when the expansion occurs.