Density of states of Fermi gas derivation

Question

I'm going over this book. While deriving the gensity of states for a gas of fermions the author makes the following argument:

Remember that we are treating the gas as having a set of states that can be occupied by various numbers of its particles. Those states are just the cells of the everyday position–momentum space that the gas can occupy: three spatial dimensions and three momentum dimensions. Counting those states then equates to counting the number of states available to one free particle moving in three spatial dimensions, given the constraint of some total energy $E$.

And then he differentiates the formula for number of states of 1 particle in 3d to obtain the density (and multiplies the density by $2s+1$ to account for spin).

I do not understand this. We have a gas of, let's say, $N$ fermions. Each of these particles has some particular coordinate in it's 6-D (or rather 7-D because 3-D position + 3-D momentum + 1D for spin). Then any combination of coordinates of each particle in their position-momentum space gives a valid state for the whole fermion gas. I don't see where the analogy with a single particle in 3D comes from.

You might find this useful: Why is the density of states required conceptually? Should it be seen as a mathematical trick related to Fourier series? — Roger V., Mar 06 '23 at 13:00

score 0 · Accepted Answer · answered Mar 06 '23 at 10:55

The problem is that we are not actually finding the gensity of microstates of a gas of fermions! We are simply counting the denisty of available microstates for a single fermion in its position-momentum (or, more precisely, position-momentum-spin) space.

As to why not the microstates of a gas as a whole. It is because we use this density of single-fermion states together with occupation number which gives the average number of particles occupying the same cell in position-momentum space available to one fermion.

Density of states of Fermi gas derivation

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