Probability of being in ground & 1st excited state of widened infinite well

[Vighnahara]

Homework Statement

A particle is initially in the ground state of a one-dimensional infinite square well extending from $$x=0$$ to $$x = L/2$$. Its wave function, correctly normalized, is given by $$\psi (x) = \dfrac{2}{\sqrt{L}} \sin{(\dfrac{2 \pi x}{L})}}$$ for $$0 \leq x \leq L/2$$

Suddenly, the right hand wall of the well is moved to $$x = L$$

(a) Find the probability that the particle is in the ground state of the widened well.
(b) Find the probability that the particle is in the second state of the widened well.

Homework Equations

$$E_n = \dfrac{n^2 \pi^2 \hbar^2}{2 m L}$$ for a particle in an infinite well from $$x=0$$ to $$x=L$$.

The Attempt at a Solution

I'm assuming there's something like $$\Sigma_{n=0}^\inifinity p(E_n) = 1$$, i.e. that the particle has to be in SOME energy state. And I'm looking for $$p(E_1)$$ and $$p(E_2)$$. I'm also assuming that the particle is initially at the ground state energy for the $$x=0..L/2$$ box, but I'm not sure how to put this together into the probability of a different energy state of the new system. Also, for $$x=0..L$$, I have $$E_1 = \dfrac{\pi^2 \hbar^2}{2 m L}$$ and $$E_2 = \dfrac{2 \pi^2 \hbar^2}{m L}$$ so the particle from the $$x=0..L/2$$ box is part-way between these two energy states and would have to resolve into one of them. It's closer to $$E_1$$ than $$E_2$$ so intuitively, I would expect $$p(E_1)>p(E_2)$$.

 

[fzero]

The energy eigenvalues are less important than the actual wavefunctions for the new well. Determine the new energy eigenfunctions and write the original state in terms of them.

 

[Vighnahara]

So I got a hold of my prof who advised to do $$<\phi | \phi_1>$$... and that "worked" in that it gave me an answer... now my new question is: if I do this for $$\phi_1, \phi_2, \phi_3, ...$$, I get answers that sum to >1. Should I be normalizing these somehow? Or are some energy transitions "forbidden" ?