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## Large volume Fermi gas density

Posted by peeterjoot on September 5, 2013

Here’s part of a problem from our final exam. I’d intended to redo the whole exam over the summer, but focused my summer study on world events instead. Perhaps I’ll end up eventually doing this, but for now I’ll just post this first part.

## Question: Large volume Fermi gas density (2013 final exam pr 1)

Write down the expression for the grand canonical partition function $Z_{\mathrm{G}}$ of an ideal three-dimensional Fermi gas with atoms having mass $m$ at a temperature $T$ and a chemical potential $\mu$ (or equivalently a fugacity $z = e^{\beta \mu}$). Consider the high temperature “classical limit” of this ideal gas, where $z \ll 1$ and one gets an effective Boltzmann distribution, and obtain the equation for the density of the particles

\begin{aligned}n = \frac{1}{{V \beta}} \frac{\partial {\ln Z_{\mathrm{G}}}}{\partial {\mu}}\end{aligned} \hspace{\stretch{1}}(1.1)

by converting momentum sums into integrals. Invert this relationship to find the chemical potential $\mu$ as a function of the density $n$.

Hint: In the limit of a large volume $V$:

\begin{aligned}\sum_\mathbf{k} \rightarrow V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}.\end{aligned} \hspace{\stretch{1}}(1.2)

Since it was specified incorrectly in the original problem, let’s start off by verifing the expression for the number of particles (and hence the number density)

\begin{aligned}\frac{1}{{\beta}} \frac{\partial {}}{\partial {\mu}} \ln Z_{\mathrm{G}} &= \frac{1}{{\beta}} \frac{\partial {}}{\partial {\mu}} \ln \sum_N z^N e^{-\beta E_N} \\ &= \frac{1}{{\beta}} \frac{1}{{\Omega}}\frac{\partial {}}{\partial {\mu}} \sum_N z^N e^{-\beta E_N} \\ &= \frac{1}{{\beta}} \frac{1}{{\Omega}}\sum_N e^{-\beta E_N} \frac{\partial {}}{\partial {\mu}} e^{\mu \beta N} \\ &= \frac{1}{{\Omega}}\sum_N N z^N e^{-\beta E_N} \\ &= \left\langle{{N}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.3)

Moving on to the problem, we’ve seen that the Fermion grand canonical partition function can be written

\begin{aligned}Z_{\mathrm{G}} = \prod_\epsilon \left( { 1 + z e^{-\beta \epsilon}} \right),\end{aligned} \hspace{\stretch{1}}(1.4)

so that our density is

\begin{aligned}n &= \frac{N}{V} \\ &= \frac{1}{{ V \beta }}\frac{\partial {}}{\partial {\mu}} \ln \prod_\epsilon \left( { 1 + z e^{-\beta \epsilon}} \right) \\ &= \frac{1}{{ V \beta }}\sum_\epsilon \frac{\partial {}}{\partial {\mu}} \ln \left( { 1 + z e^{-\beta \epsilon}} \right) \\ &= \frac{1}{{ V \beta }}\sum_\epsilon \frac{\partial {z}}{\partial {\mu}}\frac{e^{-\beta \epsilon}}{1 + z e^{-\beta \epsilon}} \\ &= \frac{1}{{ V }}\sum_\epsilon z\frac{e^{-\beta \epsilon}}{1 + z e^{-\beta \epsilon}}.\end{aligned} \hspace{\stretch{1}}(1.5)

In the high temperature classical limit, where $z \ll 1$ we have

\begin{aligned}n \\ &\approx \frac{1}{ V } \sum_\epsilon z e^{-\beta \epsilon} \\ &\approx z \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}e^{-\beta \epsilon} \\ &= \frac{2 z}{(2 \pi)^2} \int_0^\infty k^2 dk e^{-\beta \frac{\hbar^2 k^2}{2m} } \\ &= \frac{z}{2 \pi^2} \left( { \frac{\beta \hbar^2}{2m} } \right)^{-3/2} \int_0^\infty x^2 e^{-x^2} dx \\ &= \frac{z}{2 \pi^2} \left( { \frac{2m}{\beta \hbar^2} } \right)^{3/2} \frac{\sqrt{\pi}}{4} \\ &= \frac{z}{8 \pi^{3/2}} \frac{\left( {8 \pi^2 m k_{\mathrm{B}} T } \right)^{3/2}}{h^3} \\ &= z \frac{\left( { 2 \pi m k_{\mathrm{B}} T } \right)^{3/2}}{h^3}.\end{aligned} \hspace{\stretch{1}}(1.6)

This is

\begin{aligned}\boxed{n = \frac{z}{\lambda^3},}\end{aligned} \hspace{\stretch{1}}(1.7)

where

\begin{aligned}\lambda = \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}}.\end{aligned} \hspace{\stretch{1}}(1.8)

Inverting for $\mu$ we have

\begin{aligned}\mu = \frac{1}{\beta} \ln z\end{aligned} \hspace{\stretch{1}}(1.10)

or

\begin{aligned}\boxed{\mu = k_{\mathrm{B}} T \ln \left( { n \lambda^3 } \right).}\end{aligned} \hspace{\stretch{1}}(1.10)