# Microcanonical quantum statistics

I am solving exercise 7.1 of Sethna's book on statistical mechanics. It says:

A system has two single-particle quantum states, with energies (measured in degrees Kelvin) $E_0 /k_B = − 10$ and $E_2 /k_B = 10$. Experiments are performed by adding three non-interacting particles to these two states, either identical spin 1/2 fermions, identical spinless bosons or spinless identical particles obeying Maxwell- Boltzmann statistics.

The system is first held at constant energy. Which of the following curves represents the entropy of fermions, bosons and classical particles?

I started with bosons. At a certain energy, the bosons will occupy that energy level. Therefore we have only one possible configuration, i.e. $\Omega(E_1) = \Omega(E_2) = 1$. Then $S = k_B\log\Omega = 0$. Then B is the curve for bosons.

Now, since fermions must have different quantum states, and in our case both energy and spin determine the quantum states, I can write the state of the three particles as the occupation numbers $|n,m\rangle$, where $n,m=0,1,2$ with $n+m=3$, meaning that there are $n$ fermions in the energy level $E_1$ and $m$ fermions in the energy level $E_2$. Then it must be clear that from Pauli's exclusion principle, it is not possible that 3 fermions share the same energy level. Then the possible configurations are $|2,1\rangle,|1,2\rangle$, no matter the energy. Then $\Omega(E_1) = \Omega(E_2) = 2$. Then D would be the curve.

However, I am not sure about this. What does it mean to have the system held at constant energy?

I also do not understand how to compute $\Omega$ in the case of classical particles. I appreciate if you could give me any hints.