What is the probability to measure a particle's energy in it's ground state?

Question

The following wave equation describes a particle in a 1D box that is indfinitely deep (where a is the length of the box)

$\psi (x) = Nx(a-x) e^{i \pi x / a}$ where $ 0 \leq x \leq a$ and

$\psi (x) = 0 $ otherwise

I have already computed the normalization constant N. Now I want to compute the probability to measure the particle's energy in it's ground state. So far, I have:

$p_1 = |c_1|^2 = | < \phi_1 | \psi > |^2 = \bigg( \int_0^a \sqrt{ \frac{2}{a}} \sin(\frac{\pi x}{a}) N x (a - x) e^{i \pi x / a} \bigg)^2$

However, I cant find a way to solve this integral, but I assume there should be some kind of trick that I could use?

Furthermore, how can I determine whether $\psi$ is an eigenstate of the time-independant Schrödinger equation?

Answer 1

You can test whether or not $\psi(x)$ is an eigenstate by checking if $H\psi(x)$ is a multiple of $\psi(x)$. If yes, the multiplicative factor is the eigenvalue; if no, $\psi(x)$ is not an eigenstate.

To compute $\int_0^a dx \sin\left(\frac{\pi x}{a}\right) x(a-x) e^{i\pi x/a}$ expand the exponential $e^{i\pi x/a}=\cos(\pi x/a)+i\sin(\pi x/a)$, proceed to evaluate \begin{align} \langle \phi_1\vert \psi\rangle& = N \sqrt{\frac{2}{a}}\int_0^a dx\sin\left(\frac{\pi x}{a}\right) x(a-x) \cos\left(\frac{\pi x}{a}\right) \\ &\quad + i N \sqrt{\frac{2}{a}}\int_0^a dx\sin\left(\frac{\pi x}{a}\right) x(a-x) \sin\left(\frac{\pi x}{a}\right) \end{align} and then continue from there.