A quantum system has a complete set of orthonormalised energy eigenfunctions $\psi_{n}(x)$ with corresponding energy eigenvalues $E_{n}, n=1,2,3, \ldots$

(a) If the time-dependent wavefunction $\psi(x, t)$ is, at $t=0$,

$$\psi(x, 0)=\sum_{n=1}^{\infty} a_{n} \psi_{n}(x)$$

determine $\psi(x, t)$ for all $t>0$.

(b) A linear operator $\mathcal{S}$ acts on the energy eigenfunctions as follows:

$$\begin{aligned} &\mathcal{S} \psi_{1}=7 \psi_{1}+24 \psi_{2} \\ &\mathcal{S} \psi_{2}=24 \psi_{1}-7 \psi_{2} \\ &\mathcal{S} \psi_{n}=0, \quad n \geqslant 3 \end{aligned}$$

Find the eigenvalues of $\mathcal{S}$. At time $t=0, \mathcal{S}$ is measured and its lowest eigenvalue is found. At time $t>0, \mathcal{S}$ is measured again. Show that the probability for obtaining the lowest eigenvalue again is

$$\frac{1}{625}(337+288 \cos (\omega t))$$

where $\omega=\left(E_{1}-E_{2}\right) / \hbar$.