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Final Is the number of spin states necessary in the density of states function?

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final Final density of states

I'm studying how to calculate the density of states in the final configuration in order to apply Fermi golden rule. For free emEM field the following expression is the starting point: $$d^3n=\frac V {(2\pi\hbar)^3}d^3P$$ while for non zero-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. Thank you

final density of states

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. Thank you

Final density of states

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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SimoBartz
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