Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

PHY452H1S Basic Statistical Mechanics. Lecture 16: Fermi gas. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]


Peeter’s lecture notes from class. May not be entirely coherent.

Fermi gas


Continuing a discussion of [1] section 8.1 content.

We found

\begin{aligned}n_{\mathbf{k}} = \frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}}\end{aligned} \hspace{\stretch{1}}(1.2.1)

With no spin

\begin{aligned}\int n_\mathbf{k} \times \frac{d^3 k}{(2\pi)^3} = \rho\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fig 1.1: Occupancy at low temperature limit


Fig 1.2: Volume integral over momentum up to Fermi energy limit


\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\end{aligned} \hspace{\stretch{1}}(1.2.3)


\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.4)

\begin{aligned}\sum_\mathbf{k} n_\mathbf{k} = N\end{aligned} \hspace{\stretch{1}}(1.2.5)

\begin{aligned}\mathbf{k} = \frac{2\pi}{L}(n_x, n_y, n_z)\end{aligned} \hspace{\stretch{1}}(1.2.6)

This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intended. We see that both achieve the same result}, where

\begin{aligned}\Psi(x + L) = \Psi(x)\end{aligned} \hspace{\stretch{1}}(1.2.7)

Moving on

\begin{aligned}\sum_{k_x} n(\mathbf{k}) = \sum_{p_x} \Delta p_x n(\mathbf{k})\end{aligned} \hspace{\stretch{1}}(1.2.8)


\begin{aligned}\Delta k_x = \frac{2 \pi}{L} \Delta p_x\end{aligned} \hspace{\stretch{1}}(1.2.9)

this gives

\begin{aligned}\sum_{k_x} n(\mathbf{k}) = \sum_{n_x} \frac{L}{2\pi} \Delta k_x \rightarrow \frac{L}{2\pi} \int d k_x\end{aligned} \hspace{\stretch{1}}(1.2.10)

Over all dimensions

\begin{aligned}\sum_{\mathbf{k}} n_\mathbf{k} = \left( \frac{L}{2\pi} \right)^3 \left( \int d^3 \mathbf{k} \right)n(\mathbf{k})=N\end{aligned} \hspace{\stretch{1}}(1.2.11)

so that

\begin{aligned}\rho = \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)


\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.13)

Example: Spin considerations


\begin{aligned}\sum_{\mathbf{k}, m_s} = N\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}\sum_{\mathbf{k}, m_s} \frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}} = (2 S + 1)\left( \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} n(\mathbf{k}) \right)L^3\end{aligned} \hspace{\stretch{1}}(1.2.15)

This gives us

\begin{aligned}k_{\mathrm{F}} = \left( \frac{ 6 \pi^2 \rho }{2 S + 1} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.16)

and again

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\end{aligned} \hspace{\stretch{1}}(1.2.17)


High Temperatures

Now we want to look at the at higher temperature range, where the occupancy may look like fig. 1.3

Fig 1.3: Occupancy at higher temperatures


\begin{aligned}\mu(T = 0) = \epsilon_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.2.18)

\begin{aligned}\mu(T \rightarrow \infty) \rightarrow - \infty\end{aligned} \hspace{\stretch{1}}(1.2.19)

so that for large T we have

\begin{aligned}\frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}} \rightarrow e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.2.20)

\begin{aligned}\rho &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} e^{\beta \mu} e^{-\beta \epsilon_k} \\ &= e^{\beta \mu} \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} e^{-\beta \epsilon_k} \\ &= e^{\beta \mu} \int dk \frac{4 \pi k^2}{(2 \pi)^3} e^{-\beta \hbar^2 k^2/2m}.\end{aligned} \hspace{\stretch{1}}(1.2.21)

Mathematica (or integration by parts) tells us that

\begin{aligned}\frac{1}{{(2 \pi)^3}} \int 4 \pi^2 k^2 dk e^{-a k^2} = \frac{1}{{(4 \pi a )^{3/2}}},\end{aligned} \hspace{\stretch{1}}(1.2.22)

so we have

\begin{aligned}\rho &= e^{\beta \mu} \left( \frac{2m}{ 4 \pi \beta \hbar^2} \right)^{3/2} \\ &= e^{\beta \mu} \left( \frac{2 m k_{\mathrm{B}} T 4 \pi^2 }{ 4 \pi h^2} \right)^{3/2} \\ &= e^{\beta \mu} \left( \frac{2 m k_{\mathrm{B}} T \pi }{ h^2} \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.23)

Introducing \lambda for the thermal de Broglie wavelength, \lambda^3 \sim T^{-3/2}

\begin{aligned}\lambda \equiv \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}},\end{aligned} \hspace{\stretch{1}}(1.2.24)

we have

\begin{aligned}\rho = e^{\beta \mu} \frac{1}{{\lambda^3}}.\end{aligned} \hspace{\stretch{1}}(1.2.25)

Does it make any sense to have density as a function of temperature? An inappropriately extended to low temperatures plot of the density is found in fig. 1.4 for a few arbitrarily chosen numerical values of the chemical potential \mu, where we see that it drops to zero with temperature. I suppose that makes sense if we are not holding volume constant.

Fig 1.4: Density as a function of temperature


We can write

\begin{aligned}\boxed{e^{\beta \mu} = \left( \rho \lambda^3 \right)}\end{aligned} \hspace{\stretch{1}}(1.2.26)

\begin{aligned}\frac{\mu}{k_{\mathrm{B}} T} = \ln \left( \rho \lambda^3 \right)\sim -\frac{3}{2} \ln T\end{aligned} \hspace{\stretch{1}}(1.2.27)

or (taking \rho (and/or volume?) as a constant) we have for large temperatures

\begin{aligned}\mu \propto -T \ln T\end{aligned} \hspace{\stretch{1}}(1.2.28)

The chemical potential is plotted in fig. 1.5, whereas this - k_{\mathrm{B}} T \ln k_{\mathrm{B}} T function is plotted in fig. 1.6. The contributions to \mu from the k_{\mathrm{B}} T \ln (\rho h^3 (2 \pi m)^{-3/2}) term are dropped for the high temperature approximation.

Fig 1.5: Chemical potential over degenerate to classical range

Fig 1.6: High temp approximation of chemical potential, extended back to T = 0


\begin{aligned}P = - \frac{\partial {E}}{\partial {V}}\end{aligned} \hspace{\stretch{1}}(1.2.29)

For a classical ideal gas as in fig. 1.7 we have

Fig 1.7: Ideal gas pressure vs volume


\begin{aligned}P = \rho k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.2.30)

For a Fermi gas at T = 0 we have

\begin{aligned}E &= \sum_\mathbf{k} \epsilon_k n_k \\ &= \sum_\mathbf{k} \epsilon_k \Theta(\mu_0 - \epsilon_k) \\ &= \frac{V}{(2\pi)^3} \int_{\epsilon_k < \mu_0} \frac{\hbar^2 \mathbf{k}^2}{2 m} d^3 \mathbf{k} \\ &= \frac{V}{(2\pi)^3} \int_0^{k_{\mathrm{F}}} \frac{\hbar^2 \mathbf{k}^2}{2 m} d^3 \mathbf{k} \\ &= \frac{V}{(2\pi)^3} \frac{\hbar^2}{2 m} \int_0^{k_{\mathrm{F}}} k^2 4 \pi k^2 d k\propto k_{\mathrm{F}}^5\end{aligned} \hspace{\stretch{1}}(1.2.31)


\begin{aligned}E(T = 0) = V \times \frac{3}{5} \underbrace{\epsilon_{\mathrm{F}}}_{\sim k_{\mathrm{F}}^2}\underbrace{\rho}_{\sim k_{\mathrm{F}}^3}\end{aligned} \hspace{\stretch{1}}(1.2.32)


\begin{aligned}\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}E = \frac{3}{5} N \frac{\hbar^2}{2 m} \left( 6 \pi^2 \frac{N}{V} \right)^{2/3} = a V^{-2/3},\end{aligned} \hspace{\stretch{1}}(1.2.34)

so that

\begin{aligned}\frac{\partial {E}}{\partial {V}} = -\frac{2}{3} a V^{-5/3}.\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}P &= -\frac{\partial {E}}{\partial {V}}  \\ &= \frac{2}{3} \left( a V^{-2/3} \right)V^{-1} \\ &= \frac{2}{3} \frac{E}{V} \\ &= \frac{2}{3} \left( \frac{3}{5} \epsilon_{\mathrm{F}} \rho \right) \\ &= \frac{2}{5} \epsilon_{\mathrm{F}} \rho.\end{aligned} \hspace{\stretch{1}}(1.2.36)

We see that the pressure ends up deviating from the classical result at low temperatures, as sketched in fig. 1.8. This low temperature limit for the pressure 2 \epsilon_{\mathrm{F}} \rho/5 is called the degeneracy pressure.

Fig 1.8: Fermi degeneracy pressure



[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: