(i) Briefly describe the microcanonical ensemble.

(ii) For quantum mechanical systems the energy levels are discrete. Explain why we can write the probability distribution in this case as

$$p\left(\left\{n_{i}\right\}\right)= \begin{cases}\text { const }>0 & \text { for } E \leqslant E\left(\left\{n_{i}\right\}\right)<E+\Delta E \\ 0 & \text { otherwise }\end{cases}$$

What assumption do we make for the energy interval $\Delta E$ ?

Consider $N$ independent linear harmonic oscillators of equal frequency $\omega$. Their total energy is given by

$$E\left(\left\{n_{i}\right\}\right)=\sum_{i=1}^{N} \hbar \omega\left(n_{i}+\frac{1}{2}\right)=M \hbar \omega+\frac{N}{2} \hbar \omega \quad \text { with } \quad M=\sum_{i=1}^{N} n_{i}$$

Here $n_{i}=0,1,2, \ldots$ is the excitation number of oscillator $i$.

(iii) Show that, for fixed $N$ and $M$, the number $g_{N}(M)$ of possibilities to distribute the $M$ excitations over $N$ oscillators (i.e. the number of different choices $\left\{n_{i}\right\}$ consistent with $M$ ) is given by

$$g_{N}(M)=\frac{(M+N-1) !}{M !(N-1) !}$$

[Hint: You may wish to consider the set of $N$ oscillators plus $M-1$ "additional" excitations and what it means to choose $M$ objects from this set.]

(iv) Using the probability distribution of part (ii), calculate the probability distribution $p\left(E_{1}\right)$ for the "first" oscillator as a function of its energy $E_{1}=n_{1} \hbar \omega+\frac{1}{2} \hbar \omega$.

(v) If $\Delta E=\hbar \omega \ll E$ then exactly one value of $M$ will correspond to a total energy inside the interval $(E, E+\Delta E)$. In this case, show that

$$p\left(E_{1}\right) \approx \frac{g_{N-1}\left(M-n_{1}\right)}{g_{N}(M)} .$$

Approximate this result in the limit $N \gg 1, M \gg n_{1}$.