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+---
+title: "Jellium"
+firstLetter: "J"
+publishDate: 2021-11-23
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-10-15T20:31:12+02:00
+draft: false
+markup: pandoc
+---
+
+# 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):
+
+$$\begin{aligned}
+    \braket{\vb{r}}{\psi_{\vb{k}}}
+    = \psi_{\vb{k}}(\vb{r})
+    = \frac{1}{\sqrt{V}} \exp\!(i \vb{k} \cdot \vb{r})
+    \qquad \quad
+    \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z)
+\end{aligned}$$
+
+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](/know/concept/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](/know/concept/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](/know/concept/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](/know/concept/time-independent-perturbation-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:
+
+$$\begin{aligned}
+    E^{(1)}
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|}
+    \\
+    %&= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    %\bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^3}{3} \frac{|\vb{q}|}{2 k_F} - \frac{4 k_F^2 |\vb{q}|^3}{48 |\vb{q}|^2} \bigg)
+    %\dd{|\vb{q}|}
+    %\\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F}
+    \\
+    %&= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg( \frac{k_F^3}{3} 2 k_F + \frac{16 k_F^4}{192} - \frac{k_F^2 4 k_F^2}{8} \bigg)
+    %\\
+    &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4
+    = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4
+    = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F
+\end{aligned}$$
+
+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.