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Consider a free fermion gas in a (cubic) volume $V$. The "density of states" function is

$$g(p)=\frac{V}{2 \pi^2 \hbar ^3} p^2$$ This is an example of a more general type of "gas in a box" model .

I have two main questions on $g(p)$:

  • How is $g(p)$ actually related to the degeneracy of states? (here the same function is derived in the paragraph "Thomas–Fermi approximation for the degeneracy of states"). If I choose $p$ (and hence the energy level) does $g(p)$ give me the degeneracy of that energy level?
  • The function is referred to density of "states", but could it be interpreted equivalently as density of "modes" (referring to the wavefunction solution to Shroedinger equation in the 3D box of volume $V$)?
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    $\begingroup$ To my knowledge, yes and yes. $\endgroup$ – probably_someone Nov 22 '17 at 0:53
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If I choose $p$ (and hence the energy level) does $g(p)$ give me the degeneracy of that energy level?

Yes, the density of states is indeed a quantity that measures the number of (degenerate) states per energy-unit. Degenerate states are states that have the same energy value. This condition happens because there exists some observable(s), in this particular case $\vec p$, that commutes with the Hamiltonian.

The function is referred to density of "states", but could it be interpreted equivalently as density of "modes" (referring to the wavefunction solution to Shroedinger equation in the 3D box of volume $V$)?

Yes, because "modes" basically represent different $\vec p$ (i.e. different momentum states) with the same energy.

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