# Microcanonical ensemble confusion

I am a bit confused about the difference between macrostate and microstate in the microcanonical ensemble. So I have read that for the microcanonical ensemble, the probabilities of each microstate are equal $$p = 1/\Omega$$ where $\Omega$ is the number of microstates. For a given number of particles $N$ the number of microstates of particles distributed over discreet energy levels is given by

$$\Omega = \frac{N!}{\prod_i n_i}$$

where $n_i$ is the number of particles in the $i$th energy level. Maximising $\Omega$ given the constraint that the particle number is constant $\sum_i n_i = N$ and the energy is constant $\sum_i \varepsilon_i n_i = E$ gives

$$p_i \propto e^{-\varepsilon_i/kT}$$

This gives a probability for a particle to be in the $i$th energy level. I thought if we were in the microcanonical ensemble all probabilities are equal?

Thanks

• What is equal is the probability of each microstate, that is, a given distribution of particles with total energi E. Each microstate will have particles with different energies with some distribution. Then, after averaging across all posible microstates, you get the probability $p_i$ – user126422 Apr 21 '17 at 1:32
• Firstly , you lost the factorial in the formula for $\Omega$: $n_i!$. Secondly , $\ p_i$ is not the probability of a microstate of the microcanonical ensemble – Aleksey Druggist Jun 27 '17 at 9:33

Here is how to derive the microcanonical and the canonical distributions. In all cases $$\Omega(\{n_i\}) = \frac{N!}{\prod_i n_i!}$$
All $$n_i$$ in all distributions of the ensemble have the same energy: $$E_i=E$$ for all $$i$$. The problem is to find the distribution that maximizes $$\Omega$$ under the sole constraint $$\sum_i n_i = N$$ The solution is $$\frac{n_i^*}{N} = \frac{1}{\Omega} \doteq p_i$$ is the multiplicity of distirbution $$\{n_i\}$$.
We don't know the energy if microstate $$i$$ but we know that the average energy per microstate is the same in all distributions of the ensemble: $$E_\text{tot}/N = \bar E$$ for all distributions $$\{n_i\}$$. The problem now is to find $$\{n_i\}$$ that maximizes $$\Omega$$ under the following two conditions: $$\sum_i n_i = N,\quad \sum_i n_i E_i = N \bar E$$ The solution now is $$\frac{n_i^*}{N} = \frac{e^{-\beta E_i}}{Q} \doteq p_i,$$