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Categories: Perturbation, Physics, Quantum mechanics.

Jellium

Jellium, also called the uniform or homogeneous electron gas, is a theoretical material where all electrons are free, and the ions’ positive charge is smeared into a uniform background “jelly”. This simple model lets us study electron interactions easily.

Without interactions

Let us start by neglecting electron-electron interactions. This is clearly a dubious assumption, but we will stick with it for now. For an infinitely large sample of jellium, the single-electron states are simply plane waves. We consider an arbitrary cube of volume $V$, and impose periodic boundary conditions on it, such that the single-particle orbitals are (suppressing spin):

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

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 to write the non-interacting Hamiltonian $\hat{H}_0$, where $\hbar^2 |\vb{k}|^2 / (2 m)$ is the kinetic energy of the orbital with wavevector $\vb{k}$, and $s$ is the spin:

$$\begin{aligned} \hat{H}_0 = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}} \end{aligned}$$

Assuming that the temperature $T = 0$, the $N$-electron ground state of this Hamiltonian is known as the Fermi sea or Fermi sphere $\Ket{\mathrm{FS}}$, and is constructed by filling up the single-electron states starting from the lowest energy:

$$\begin{aligned} \Ket{\mathrm{FS}} = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0} \end{aligned}$$

Because $T = 0$, all the electrons stay in their assigned state. The energy and wavenumber $|\vb{k}|$ of the highest filled orbital are called the Fermi energy $\epsilon_F$ and Fermi wavenumber $k_F$, and obey the expected kinetic energy relation:

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

The Fermi sea 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 $V \to \infty$ it becomes perfect.

Now, we would like a relation between the system’s parameters, e.g. $N$ and $V$, and the resulting values of $\epsilon_F$ or $k_F$. The total population $N$ must be given by:

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

Where we have turned the sum over $\vb{k}$ into an integral with a constant factor, by using that each orbital exclusively occupies a volume $(2 \pi)^3 / V$ in $\vb{k}$-space.

At zero temperature, this inner product can only be $0$ or $1$, depending on whether $\vb{k}$ is outside or inside the Fermi sphere. We can therefore rewrite using a Heaviside step function:

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

Where we realized that spin does not matter, and replaced the sum over $s$ by a factor $2$. In order to evaluate this 3D integral, we go to spherical coordinates $(|\vb{k}|, \theta, \varphi)$:

$$\begin{aligned} N &= \frac{V}{4 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi} \\ &= \frac{V}{4 \pi^3} 4 \pi \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} = \frac{V}{\pi^2} \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_0^{k_F} = \frac{V}{3 \pi^2} k_F^3 \end{aligned}$$

Using that the electron density $n = N/V$, we thus arrive at the following relation:

$$\begin{aligned} \boxed{ k_F^3 = 3 \pi^2 n } \end{aligned}$$

This result also justifies our assumption that $T = 0$: we can accurately calculate the density $n$ for many conducting materials, and this relation then gives $k_F$ and $\epsilon_F$. It turns out that $\epsilon_F$ is usually very large compared to the thermal energy $k_B T$ at reasonable temperatures, so we can conclude that thermal fluctuations are negligible.

Now, $\epsilon_F$ is the highest single-electron energy, but about the total $N$-particle energy $E^{(0)}$?

$$\begin{aligned} E^{(0)} = \matrixel{\mathrm{FS}}{\hat{H}_0}{\mathrm{FS}} = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \end{aligned}$$

Once again, we turn the sum over $\vb{k}$ into an integral, and recognize the spin’s irrelevance:

$$\begin{aligned} E^{(0)} &= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}} \\ &= \frac{\hbar^2 V}{8 \pi^3 m} \int_{-\infty}^\infty |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \dd{\vb{k}} \end{aligned}$$

In spherical coordinates, we evaluate the integral and find that $E^{(0)}$ is proportional to $k_F^5$:

$$\begin{aligned} E^{(0)} &= \frac{\hbar^2 V}{8 \pi^3 m} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Big( |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \Big) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi} \\ &= \frac{\hbar^2 V}{8 \pi^3 m} 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|} = \frac{\hbar^2 V}{2 \pi^2 m} \bigg[ \frac{|\vb{k}|^5}{5} \bigg]_0^{k_F} = \frac{\hbar^2 V}{10 \pi^2 m} k_F^5 \end{aligned}$$

In general, it is more useful to consider the average kinetic energy per electron $E^{(0)} / N$, which we find to be as follows, using that $k_F^3 = 3 \pi^2 n$:

$$\begin{aligned} \boxed{ \frac{E^{(0)}}{N} = \frac{3 \hbar^2}{10 m} k_F^2 = \frac{3}{5} \epsilon_F } \:\sim\: n^{2/3} \end{aligned}$$

Traditionally, this is expressed using a dimensionless parameter $r_s$, defined as the radius of a sphere containing a single electron, measured in Bohr radii $a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$:

$$\begin{aligned} \frac{4 \pi}{3} (a_0 r_s)^3 = \frac{1}{n} = \frac{3 \pi^2}{k_F^3} \quad \implies \quad r_s = \Big( \frac{3}{4 \pi a_0^3 n} \Big)^{1/3} = \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 k_F} \end{aligned}$$

Such that the ground state energy can be rewritten in Rydberg units of energy like so:

$$\begin{aligned} \frac{E^{(0)}}{N} = \frac{3 \hbar^2}{10 m} \frac{4 \pi \varepsilon_0 e^2}{4 \pi \varepsilon_0 e^2} \frac{a_0^2 k_F^2}{a_0^2} = \frac{3 e^2}{40 \pi \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{2/3} \frac{1}{a_0 r_s^2} \approx \frac{2.21}{r_s^2} \; \mathrm{Ry} \end{aligned}$$

With interactions

To include Coulomb interactions, let us try time-independent pertubation theory. Clearly, this will give better results when the interaction is relatively weak, if ever.

The Coulomb potential is proportional to the inverse distance, and the average electron spacing is roughly $n^{-1/3}$, so the interaction energy $E_\mathrm{int}$ should scale as $n^{1/3}$. We already know that the kinetic energy $E_\mathrm{kin} = E^{(0)}$ scales as $n^{2/3}$, meaning perturbation theory should be reasonable if $1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$, so in the limit of high density $n \to \infty$.

The two-body Coulomb interaction operator $\hat{W}$ is as follows in second-quantized form:

$$\begin{aligned} \hat{W} = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e^2}{\varepsilon_0 |\vb{q}|^2} \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} \end{aligned}$$

The first-order correction $E^{(1)}$ to the ground state (i.e. Fermi sea) energy is then given by:

$$\begin{aligned} E^{(1)} = \matrixel{\mathrm{FS}}{\hat{W}}{\mathrm{FS}} = \frac{e^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} \matrixel{\mathrm{FS}}{ \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} }{\mathrm{FS}} \end{aligned}$$

This inner product can only be nonzero if the two creation operators $\hat{c}^\dagger$ are for the same orbitals as the two annihilation operators $\hat{c}$. Since $\vb{q} \neq 0$, this means that $s_1 = s_2$, and that momentum is conserved: $\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$. And of course both $\vb{k}_1$ and $\vb{k}_1 \!+\! \vb{q}$ must be inside the Fermi sphere, to avoid annihilating an empty orbital. Let $s = s_1$ and $\vb{k} = \vb{k}_1$:

$$\begin{aligned} E^{(1)} &= \frac{e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} \matrixel{\mathrm{FS}}{ \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}} \hat{c}_{s, \vb{k}} }{\mathrm{FS}} \\ &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} \matrixel{\mathrm{FS}}{ \big( \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}}\big) \big(\hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k}}\big) }{\mathrm{FS}} \\ &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|) \end{aligned}$$

Next, we convert the sum over $\vb{q}$ into an integral in spherical coordinates. Clearly, $\vb{q}$ is the “jump” made by an electron from one orbital to another, so the largest possible jump goes from a point on the Fermi surface to the opposite point, and thus has length $2 k_F$. This yields the integration limit, and therefore leads to:

$$\begin{aligned} E^{(1)} &= \frac{- e^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}} \int_0^{2 \pi} \!\!\int_0^\pi \!\!\int_0^\infty \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \frac{|\vb{q}|^2}{|\vb{q}|^2} \sin(\theta_q) \dd{|\vb{q}|} \dd{\theta_q} \dd{\varphi_q} \\ &= \frac{- e^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}} \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|} \end{aligned}$$

Where we have used that the direction of $\vb{q}$, i.e. $(\theta_q,\varphi_q)$, is irrelevant, as long as we define $\theta_k$ as the angle between $\vb{q}$ and $\vb{k} \!+\! \vb{q}$ when we go to spherical coordinates $(|\vb{k}|, \theta_k, \varphi_k)$ for $\vb{k}$:

$$\begin{aligned} E^{(1)} &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty \!\Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} \\ &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F} \!\Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} \end{aligned}$$

Unfortunately, this last step function is less easy to translate into integration limits. In effect, we are trying to calculate the intersection volume of two spheres, both with radius $k_F$, one centered on the origin (for $\vb{k}$), and the other centered on $\vb{q}$ (for $\vb{k} \!+\! \vb{q}$). Imagine a triangle with side lengths $|\vb{k}|$, $|\vb{q}|$ and $|\vb{k} \!+\! \vb{q}|^2$, where $\theta_k$ is the angle between $|\vb{k}|$ and $|\vb{k} \!+\! \vb{q}|$. The law of cosines then gives the following relation:

$$\begin{aligned} |\vb{k}|^2 = |\vb{q}|^2 + |\vb{k} \!+\! \vb{q}|^2 - 2 |\vb{q}| |\vb{k} \!+\! \vb{q}| \cos(\theta_k) \end{aligned}$$

We already know that $|\vb{k}| < k_F$ and $0 < |\vb{q}| < 2 k_F$, so by isolating for $\cos(\theta_k)$, we can obtain bounds on $\theta_k$ and $|\vb{k}|$. Let $|\vb{k}| \to k_F$ in both cases, then:

$$\begin{aligned} \cos(\theta_k) = \frac{|\vb{k} \!+\! \vb{q}|^2 + |\vb{q}|^2 - |\vb{k}|^2}{2 |\vb{k} \!+\! \vb{q}| |\vb{q}|} &\:\:\underset{|\vb{q}| \to 0}{>}\:\:\: \frac{k_F^2 + |\vb{q}|^2 - k_F^2}{2 k_F |\vb{q}|} = \frac{|\vb{q}|}{2 k_F} \\ &\underset{|\vb{q}| \to 2 k_F}{<}\:\: \frac{k_F^2 + 4 k_F ^2 - k_F^2}{2 k_F 2 k_F} = 1 \end{aligned}$$

Meaning that $0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$. To get a lower limit for $|\vb{k}|$, we “cheat” by artificially demanding that $\vb{k}$ does not cross the halfway point between the spheres, with the result that $|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$. Then, thanks to symmetry (both spheres have the same radius), we just multiply the integral by $2$, for $\vb{k}$ on the other side of the halfway point.

Armed with these integration limits, we return to calculating $E^{(1)}$, substituting $\xi \equiv \cos(\theta_k)$:

$$\begin{aligned} E^{(1)} &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}} \!\!\int_{|\vb{q}|/(2 \cos{\theta_k})}^{k_F} |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} \\ &= \frac{e^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)} \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \frac{\sin(\theta_k)}{\sin(\theta_k)} \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} \\ &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} \end{aligned}$$

Where we have used that $\varphi_k$ does not appear in the integrand. Evaluating these integrals:

$$\begin{aligned} E^{(1)} &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|} \\ &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|} \\ &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|} \\ &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|} \\ &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F} \\ &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4 = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4 = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F \end{aligned}$$

Per particle, the first-order energy correction $E^{(1)}$ is therefore found to be as follows:

$$\begin{aligned} \boxed{ \frac{E^{(1)}}{N} = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} k_F } \end{aligned}$$

This can also be written using the parameter $r_s$ introduced above, leading to:

$$\begin{aligned} \frac{E^{(1)}}{N} = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0} = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s} \end{aligned}$$

Consequently, for sufficiently high densities $n$, the total energy $E$ per particle is given by:

$$\begin{aligned} \boxed{ \frac{E}{N} \approx \bigg( \frac{2.21}{r_s^2} - \frac{0.92}{r_s} \bigg) \; \mathrm{Ry} } \end{aligned}$$

Unfortunately, this is as far as we can go. In theory, the second-order energy correction $E^{(2)}$ is as shown below, but it turns out that it (and all higher orders) diverge:

$$\begin{aligned} E^{(2)} = \sum_{\Psi_n \neq \mathrm{FS}} \frac{\big| \matrixel{\mathrm{FS}}{\hat{W}}{\Psi_n} \big|^2}{E^{(0)} - E_n} \end{aligned}$$

The only cure for this is to go to infinite order, where all the infinities add up to a finite result.

References

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

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