Consider a two-state quantum system with a Hamiltonian \begin{equation} \hat{H}\doteq \begin{pmatrix} E_1&0\\ 0&E_2 \end{pmatrix} \end{equation} Another physical observable \(M\) is described by the operator \begin{equation} \hat{M}\doteq \begin{pmatrix} 0&c\\ c&0 \end{pmatrix} \end{equation} where \(c\) is real and positive. Let the initial state of the system be \(\left|{\psi(0)}\right\rangle =\left|{m_1}\right\rangle \), where \(\left|{m_1}\right\rangle \) is the eigenstate corresponding to the larger of the two possible eigenvalues of \(\hat{M}\). What is the expectation value of \(M\) as a function of time? What is the frequency of oscillation of the expectation value of \(M\)?

1. Find the entropy of a set of \(N\) oscillators of frequency \(\omega\) as a function of the total quantum number \(n\). Use the multiplicity function: \begin{equation} g(N,n) = \frac{(N+n-1)!}{n!(N-1)!} \end{equation} and assume that \(N\gg 1\). This means you can make the Sitrling approximation that \(\log N! \approx N\log N - N\). It also means that \(N-1 \approx N\).

2. Let \(U\) denote the total energy \(n\hbar\omega\) of the oscillators. Express the entropy as \(S(U,N)\). Show that the total energy at temperature \(T\) is \begin{equation} U = \frac{N\hbar\omega}{e^{\frac{\hbar\omega}{kT}}-1} \end{equation} This is the Planck result found the hard way. We will get to the easy way soon, and you will never again need to work with a multiplicity function like this.