quantum mechanics - Calculating the probability of a given energy

Given a normalised wavefunction say

$$\psi(x) = A\sin(n\pi x),$$ (where $A$ is a normalisation constant) I can calculate the probability of finding the particle being between a position $x$ and $x + dx$ as $$\int_{x}^{x+dx} \psi(x)^*\psi(x) dx.$$ Likewise I can calculate the expected value of the energy $$\langle E \rangle = \int_{-\infty}^\infty \psi(x)^*\hat{H} \psi(x) dx.$$ However, how can I calculate the probability of a given value of the energy say $$E = \frac{\hbar^2 \pi^2}{2}$$ as my wavefunction $\psi(x)$ is defined in terms of position not energy?

Answer

$\newcommand{\d}{\;\mathrm{d}}$

Elaborating on the answer of zeldredge, I want to say why the following expression works:

$$P(E_0) = \left| \int_{-\infty}^{\infty} \ \Phi^* \psi \d x \right|^2 \tag{1}$$

Notice that the eigenfunctions of the Hamiltonian spans the space. That is you can write any function $\psi$ as a linear combination of eigenfunctions of the Hamiltonian ie.

$$\psi=a_0\Phi_0+a_1\Phi_1+a_2\Phi_2 \dots = \sum_i a_i \Phi_i \tag{2}$$

with $a_i \in \mathbb C$ and $\sum_i \left|a_i \right|^2=1$. Notice that as in all superposition of states the probability of finding particle in a particular state is given by $\left| a_i \right|^2$. Assuming that you have chosen your states cleverly such that they are orthonormal ie they satisfy the following property,

$$\int \Phi_i^*\Phi_j \d x = \delta_{ij}$$

you can see immediately why Eqn. (1) works.

$$\int \Phi_i^* \psi \d x = \int \Phi_i^* \sum_j a_j \Phi_j \d x = \sum_j a_j\int \Phi_i^* \Phi_j \d x = \sum_j \delta_{ij} a_j =a_i$$

Therefore the norm squared of this expression just gives the probability to find the state in the energy eigenfunction $\Phi_i$.

Notice I assumed that the energy levels are discrete to write the equation (2). Furthermore I dropped $x$'s in $\psi(x)$ and $\Phi(x)$ because they were cluttering the equations.