Energy and probability for the particle in a box

The question statement: we have a particle of mass m in a box of length a. The wavefunction is given as

$$\psi(x)=x\sin \left(\frac{\pi x}{a}\right)$$

It asks to normalize the wave function and find the expectation value of x, which I did no problem. However, it then asks for the expectation value of the energy. I know that this value is given by

$$\frac{\langle p^2\rangle}{2m}$$

But it then asks for the energy in as a fraction of $E_1$, the ground state energy of the particle in this box. How do I find the ground state energy for this wave function? There is no "n" value, so how can I find $E_n$ for $n=1$ (I am sure it is not a typo; I checked). I assume we are already at the n=1 level, so we just take

$$E_1=\frac{\hbar^2 \pi^2}{2mL^2}$$

Is this the correct procedure?

As well, it also asks two more questions that I am confused with: find the probability that the energy will be $E_1$ if we take a measurement of the energy, and obtain an expression for the probability that if a measurement of the energy is made, the result will be $E_n$. For the latter, do I just integrate over all n's for the wavefunction squared, where

$$\psi(x)_n=A x\sin \left(\frac{\pi n x}{a}\right)$$

Where A is our normalization constant? For the former, do I simply take $n=1$ and integrate?

Since you didn't mention anything about the Hamiltonian, I assume it's just a particle in a box. The wavefunction vanishes at the boundaries.

You are obviously familiar with the eigenfunctions of the Hamiltonian! Is $\psi(x)$ such an eigenfunction? If not, can it be expressed in terms of eigenfunctions? How?

For the last part, think how probabilities relate to the coefficients in the expansion wrt. a certain basis.

[Update:] According to your comment you only know the energies, eg. the eigenvalues of the Hamiltonian $E_n = \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}n^2$ And you know - at least you say so yourself - how to calculate the expectation-value of the enery. Divide this number by $E_1$ then! The wavefunction $\psi(x)$ does not carry the quantum number $n$, because it is not an eigenstate. It's rather a superposition of different eigenstates (different $n$).

[/Update]

• @user29716 I've updated my answer. As to the last task: i'm not sure how to do this w/o knowing the eigenfunctions, because it amounts to taking overlaps $\langle n\vert \psi \rangle$ Commented Sep 16, 2013 at 16:21