I'm studying how to calculate the density of states in the final configuration in order to apply Fermi golden rule. For free EM field the following expression is the starting point: $$d^3n=\frac V {(2\pi\hbar)^3}d^3P$$ while for non-zero mass particle this expression is the starting point: $$d^3n=\frac V {(2\pi\hbar)^3}d^3P(2S+1)$$ where $S$ is the spin number and $2S+1$ are the different spin states. Can you explain me in which cases and why it is necessary to write the $2S+1$ factor in the formula? To me it looks strange that in one case we don't write it.
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The quick answer is that the second expression includes spin degeneracy, i.e. the number of states which have the same energy. As you probably know, for a given principal quantum number $s$, which determines the magnitude of the spin vector, there will be $2s+1$ possible projections of this spin which give the same energy (usually on the $z$ axis). Normally, in quantum statistics, you want to assume there is spin degeneracy and is sometimes written as a function $g$ so it's simpler to write.
Sometimes it is easier to think in statistics on whether your system has distinguishable or indistinguishable particles, if you have to apply Pauli's exclusion principle, etc.
