# Number of microstates in Einstein solid if the oscillators are distinguishable

In the Einstein solid model ($$N$$ oscillators of same frequency $$\omega$$) the quantum harmonic oscillators are considered distinguishable and the microstate is characterized by the set of occupation numbers of all oscillators $$k_i$$.

If the total energy is fixed (microcanonical enesemble) and therefore $$\sum k_i=n$$ (total number of energy quanta) has also a fixed value, the number of possible microstates is : $$\Omega=\frac{(N+n-1)!}{n! (N-1)!}\tag{1}$$

As shown here this is derived thinking about the number of possible arrangements of $$n$$ pebbles and $$N-1$$ partitions, which is $$(N+n-1)!$$, and the denominator $$n! (N-1)!$$ is there because

if (e.g.) partition #3 and partition #5 trade places, no one would notice. The same argument goes for quanta. To obtain the number of possible distinguishable arrangements one has to divide the total number of arrangements by the number of indistinguishable arrangements.

So how can the oscillators be distinguishable and, at the same time, the partitions (that represent the oscillators) be traded with no change in the state? In other words, since there is a $$(N-1)!$$ in the denominator it looks like in $$(1)$$ the oscillators are not considered distinguishable, but how can that be?

The formula for $\Omega$ corresponds to indistinguishable arrangments, exactly as you said. For example in Statistical Physics, Franz Mandl (1984) (section 2.2) gives the same result for dipoles in a lattice, which explicitly explains why they are not distinguishable.
Speaking specifically for our problem, Einstein solid is model which you consider the vibrations of atoms as indepedent quantum harmonic oscilations with some constant frequency $\omega_E$. Therefore the energy level of every oscillator is given by $E_i=\hbar\omega(\frac{1}{2}+n_i)$, where n the number of quanta (phonons) corresponds to each atom vibration. Also we consider that the number of phonons are fixed. The final part to understand, there is no point in labelling the differents oscillations. By labelling, I mean that you can't consider as different one atom from another. If you "exchange" the modes between two atoms, you don't have a different result. So you cannot distinguish them. Therefore you follow the whole analysis you have given initially.