# Low Temperature Expansion of Chemical Potential

1. Jun 16, 2008

### QuasiParticle

I'm trying to derive a low temperature series expansion for the chemical potential of a weakly interacting Fermi gas. The starting point is, of course, the Fermi-Dirac distribution function (p is the particle momentum):

$$f(p) = \frac{1}{e^{\beta(\epsilon(p) - \mu)}+1} ,$$

where, in the Hartree-Fock approximation, we have

$$\epsilon(p) = \frac{p^2}{2m} + n V(0) - \frac{1}{(2\pi \hbar)^3} \int d^3p' V(\textbf{p} - \textbf{p}' ) f(p').$$

Here, $$m$$ is the effective mass, $$n$$ is the particle density, $$V(0)$$ is the interaction potential $$V(q)$$ at zero momentum transfer. The potential may be assumed to depend only on the momentum transfer $$V(\textbf{p} - \textbf{p}' ) = V(| \textbf{p} - \textbf{p}' | ) = V(q)$$. The F-D distribution $$f(p')$$ in the exchange term may be approximated with the non-interacting one. The chemical potential is determined by the condition (spin-1/2):

$$n = \frac{2}{(2\pi \hbar)^3} \int d^3p f(p) = \frac{1}{\pi^2 \hbar^3} \int_0^\infty p^2 f(p) dp$$

Now, the right-hand side should somehow be expanded as a series in $$( k_B T/ \mu)^2$$, which can then be inverted to give $$\mu$$ as a function of $$T$$. It seems that the Sommerfeld method used for a non-interacting system is not easy to use in this case. I know the result should be the following:

$$\mu (T) = \mu_F (T) + n V(0) - \frac{1}{2} n \left[ F + G \frac{\pi^2}{12} \left( \frac{T}{T_F} \right)^2 \right] ,$$

where
$$F = \frac{3}{2 k_F^3} \int_0^{2 k_F} k^2 \left( 1 - \frac{k}{2 k_F} \right) V(k) dk .$$
and
$$G = 3 \left( V(2 k_F) - \frac{1}{4} \int_0^{2 k_F} \frac{k^3}{k_F^4} V(k) dk \right) .$$

The potential is now written as a function of the Fermi wave vector ($$p = \hbar k$$). $$\mu_F (T)$$ is the chemical potential of a non-interacting Fermi gas. The zero temperature limit, i.e. $$F$$, is rather simple to derive.

Has anyone come across this problem or know any good references? I would really appreciate any assistance.

2. Dec 29, 2008

### QuasiParticle

I guess I could update this thread a little bit. I was able to derive the requested expansion somewhat after posting the above message. My approach was, however, slightly different. The result also had an extra term and reads:

$$G = 3 \left( V(2 k_F) + \frac{1}{4} \int_0^{2 k_F} \left( \frac{k}{k_F^2} - \frac{3}{2} \frac{k^3}{k_F^4} \right) V(k) dk \right) .$$

The reason for this small discrepancy is unclear. By comparing the two results to numerical calculations using the exact equations, I find that my $$G$$ is a better approximation.