Energy Measurement in a Quantum Well

Consider the operator $\hat{H}$, the energy operator (Hamiltonian) for the infinite potential well with the eigenstates $\psi_n$, Eq.(3.6),

$$\psi_n(x) = \sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right ),\quad E_n= \frac{n^2 \hbar^2 \pi^2}{2mL^2},\quad n=1,2,3,...$$

If the state $\psi$ before the measurement is an eigenstate $\psi=\psi_n$, measurement of the energy yields the value $E_n$ with probability $\vert\langle \psi\vert\psi_n\rangle\vert^2$, that is with probability 1, and other values $E_m$, $m\ne n$, with probability $\vert\langle \psi\vert\psi_m\rangle\vert^2 =\vert\langle \psi_n\vert\psi_m\rangle\vert^2=0$. If the state $\psi$ before the measurement is (cp. the problems)

$$\psi(x) = \sqrt{{30}/{L}}/{L^2} x (L - x) = \sum_{n=1}^{\infty} c_n \psi_n, \quad c_n = 2 \sqrt{60} \frac{1-(-1)^n}{n^3\pi^3},$$

the probability to obtain the value $E_n$ when measuring the energy is

$$prob(E_n) = \vert c_n\vert^2= \left\{ \begin{array}{cc} 16\cdot 60/(n^6\pi^6) & \mbox{\rm $n$ odd} \\ 0 & \mbox{\rm $n$ even} \end{array}\right.$$

In particular, we easily check (cp. the problems) that

$$\sum_{n=0}^{\infty}prob(E_n)= [n=2k+1]=\sum_{k=1}^{\infty}\frac{16\cdot 60 }{(2k+1)^6\pi^6}=1$$

as it must be.