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In the book "Statistical Physics, Part I ($3^{{\rm rd}}$ edition)" by Landau and Lifshitz, at $\S59$ when he treats the diamagnetic part of the magnetisation of a degenerate electron gas for weak fields ($\mu_BH\ll\varepsilon_F$, where $\varepsilon_F$ is the Fermi energy), he says that the energy levels of the orbital motion are given by

$$ \varepsilon=\frac{p_z^2}{2m}+(2n+1)\mu_BH,\;n=0,1,2,\ldots $$

where $p_z$ is the component of the momentum in the direction of the field $H$ and $\mu_B=e\hbar/2m_ec$ is the Bohr magneton. That's ok!

But then he says that the number of states of the particle, at a given value of $n$, such that its momentum lies in the interval $p_z$ and $p_z+{\rm d}p_z$ is

$$ 2\frac{VeH}{(2\pi\hbar)^2c}{\rm d}p_z,$$

where the factor $2$ accounts for the two possible spin orientations.

I tried to reach this result, but I had no success. And that's strange, because the number of states depends on the field $H$.

Well, if you could help me understand this, I would be very grateful.

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