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Categories: Physics, Quantum mechanics.

Fermi gas

A Fermi gas is a system of many fermions that do not interact directly, only indirectly through the Pauli exclusion principle, and hence obey Fermi-Dirac statistics.

There are several real-life systems for which this model is relevant, but most notably it serves as the foundation of the quantum-mechanical study of electrons (or electron holes) in materials. Obviously, electrons do interact strongly via the Coulomb force, but it is nevertheless a useful starting point to neglect that fact, and to then add the interactions later (see e.g. jellium).

Consider a collection of infinitely many non-interacting fermions. For mathematical convenience, we restrict ourselves to a cube with side $L$, and impose periodic boundary conditions. Then, at the end of our calculation, we should in theory take the limit $L \to \infty$ to recover the “true” system.

In the absence of any potentials, all the fermions’ wavefunctions are simply plane waves $\ket{\psi_\vb{k}}$ with wavevector $\vb{k}$. Due to the cube’s finite size and its periodic boundary conditions, those waves have a discrete spectrum of allowed wavevectors $\vb{k}$, meaning that each particle’s wavefunction $\ket{\psi_\vb{k}}$ is as follows in $\vb{r}$-space (modulo a constant phase):

$$\begin{aligned} \psi_{\vb{k}}(\vb{r}) = \frac{1}{\sqrt{L^3}} \exp(i \vb{k} \cdot \vb{r}) \qquad \qquad \vb{k} = \frac{2 \pi}{L} (n_x, n_y, n_z) \end{aligned}$$

Where $n_x, n_y, n_z \in \mathbb{Z}$. This is a discrete (but infinite) set of independent orbitals, so it is natural to use the second quantization’s operators $\hat{c}^\dagger$ and $\hat{c}$ in our analysis.

Let the temperature $T = 0$, then the $N$ fermions inside our cube fill the $N$ lowest-energy orbitals. The resulting $N$-particle ground state is known as the Fermi sea or Fermi sphere $\ket{\mathrm{FS}}$, and can be written as follows, where $S$ is the spin degeneracy, i.e. for each $\vb{k}$ there are $S$ orbitals with the same energy but different spin $s$ (for most relevant fermions $S = 2$):

$$\begin{aligned} \ket{\mathrm{FS}} = \prod_{s} \prod_{j = 1}^{N/S} \hat{c}_{s,\vb{k}_j}^\dagger \ket{0} \end{aligned}$$

The energy and wavenumber $|\vb{k}|$ of the highest filled orbital are called the Fermi energy $\varepsilon_F$ and Fermi wavenumber $k_F$, and obey the expected kinetic energy relation:

$$\begin{aligned} \boxed{ \varepsilon_F = \frac{\hbar^2}{2 m} k_F^2 } \end{aligned}$$

The Fermi sphere can be visualized in $\vb{k}$-space as a sphere with radius $k_F$. Because $\vb{k}$ is discrete, the sphere’s surface is not smooth, but in the limit $L \to \infty$ that “roughness” disappears.

Now, we would like a relation between the system’s parameters, e.g. $N$ and $L$, and the resulting values of $\varepsilon_F$ or $k_F$. The total number $N$ of fermions in our cube is given by:

$$\begin{aligned} N = \sum_{s} \sum_{\vb{k}} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} = \sum_{s} \frac{L^3}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}} \end{aligned}$$

Where the periodic boundary conditions have enabled us to convert the sum over $\vb{k}$ to an integral. For $T = 0$, the matrix element $\matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}$ is either $0$ or $1$, depending on whether $\vb{k}$ is outside or inside the Fermi sphere. We can write this using a Heaviside step function:

$$\begin{aligned} N = \sum_{s} \frac{L^3}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}} = \frac{S L^3}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}} \end{aligned}$$

Where we realized that spin does not matter, to replace the sum with a factor $S$. To evaluate this 3D integral, we transition to spherical coordinates $(|\vb{k}|, \theta, \varphi)$:

$$\begin{aligned} N &= \frac{S L^3}{8 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi} \\ &= \frac{S L^3}{8 \pi^3} 4 \pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \\ &= \frac{S L^3}{2 \pi^2} \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \\ &= \frac{S L^3}{6 \pi^2} k_F^3 \end{aligned}$$

Since the particle density $n = N / L^3$, we can rearrange this result to the following relation:

$$\begin{aligned} \boxed{ k_F^3 = \frac{6 \pi^2}{S} n } \qquad \end{aligned}$$

Consequently, the Fermi energy $\varepsilon_F$ and the corresponding orbital’s velocity $v_F = \hbar k_F / m$ can be expressed as a function of the density $n$:

$$\begin{aligned} \boxed{ \varepsilon_F = \frac{\hbar^2}{2 m} \bigg( \frac{6 \pi^2}{S} \bigg)^{2/3} n^{2/3} } \qquad \qquad \boxed{ v_F = \frac{\hbar}{m} \bigg( \frac{6 \pi^2}{S} \bigg)^{1/3} n^{1/3} } \end{aligned}$$

This is an important result, especially for electrons in metals. We know the electron density $n$ for many conductors, and then these relations tell us that $v_F \ll c$, and that the “Fermi temperature” $T_F = \varepsilon_F / k_B$ is very large (e.g. $T_F \approx 8 \cdot 10^4 \: \mathrm{K}$ for copper). This justifies our implicit assumptions that relativity and thermal fluctuations are negligible under normal circumstances.

We now have an expression for $\varepsilon_F$ as a function of $n$, which we can control by adding or removing fermions from the system. But it is also useful to isolate this relation for $n$ instead:

$$\begin{aligned} n &= \frac{S}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2} \end{aligned}$$

The total population $N = L^3 n$ can therefore be expressed as a function of $\varepsilon_F$:

$$\begin{aligned} N(\varepsilon_F) &= \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2} \end{aligned}$$

And from this we obtain a formula for the density of states $g$ of a 3D Fermi gas:

$$\begin{aligned} \boxed{ g(\varepsilon_F) = \dv{N}{\varepsilon_F} = \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{1/2} } \end{aligned}$$

Now, $\varepsilon_F$ is the highest energy of a single fermion, but what about the total $N$-particle energy $E$? This is easy to calculate using the density of states:

$$\begin{aligned} E &= \int_0^{\varepsilon_F} \varepsilon \: g(\varepsilon) \dd{\varepsilon} \\ &= \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \int_0^{\varepsilon_F} \varepsilon^{3/2} \dd{\varepsilon} \\ &= \frac{3}{2} \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \: \frac{2}{5} \varepsilon_F^{5/2} \end{aligned}$$

Here, we recognize $N(\varepsilon_F)$ from earlier, leading to the following expression for the total $E$:

$$\begin{aligned} \boxed{ E = \frac{3}{5} N \varepsilon_F } \end{aligned}$$

This model is a strong foundation for many more advanced calculations.

References

1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.

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