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diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md index 5c50f80..5cd8483 100644 --- a/source/know/concept/jellium/index.md +++ b/source/know/concept/jellium/index.md @@ -12,32 +12,24 @@ layout: "concept" **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. +This is a version of the [Fermi gas](/know/concept/fermi-gas/) model, +which we extend by including electron-electron interactions using +[time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/). -## 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): +## 0th order -$$\begin{aligned} - \Inprod{\vb{r}}{\psi_{\vb{k}}} - = \psi_{\vb{k}}(\vb{r}) - = \frac{1}{\sqrt{V}} \exp(i \vb{k} \cdot \vb{r}) - \qquad \quad - \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z) -\end{aligned}$$ +Let us start with the 0th order of the perturbation expansion. +Without interactions or potentials, this is simply a Fermi gas, +so the single-electron wavefunctions are just plane waves. +For mathematical convenience, we consider these waves +in a cube of volume $$V$$ with periodic boundaries, +leading to a discrete spectrum of allowed wavevectors $$\vb{k}$$, +which becomes continuous for $$V \to \infty$$. -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 $$\hat{H}_0$$, +The unperturbed many-particle Hamiltonian $$\hat{H}_0$$ is given below in the +[second quantization](/know/concept/second-quantization/), where $$\hbar^2 |\vb{k}|^2 / (2 m)$$ is the kinetic energy of the orbital with wavevector $$\vb{k}$$, and $$s$$ is the spin: @@ -46,10 +38,9 @@ $$\begin{aligned} = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}} \end{aligned}$$ -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 +Which, at absolute zero $$T = 0$$, has an $$N$$-electron ground state +known as the *Fermi sphere* $$\Ket{\mathrm{FS}}$$ +that is constructed by filling up the single-electron states starting from the lowest energy: $$\begin{aligned} @@ -57,63 +48,9 @@ $$\begin{aligned} = \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$$, -we thus arrive at the following relation: +From our analysis of the Fermi gas, we have an important result +for the wavenumber $$k_F = |\vb{k}_{N/2}|$$ of the highest filled orbital, +as a function of the particle density $$n = N / V$$: $$\begin{aligned} \boxed{ @@ -122,103 +59,137 @@ $$\begin{aligned} } \end{aligned}$$ -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 $$E^{(0)}$$? +Now, let us calculate the total $$N$$-particle ground state +energy $$E^{(0)}$$ of the unperturbed system: $$\begin{aligned} E^{(0)} = \matrixel{\mathrm{FS}}{\hat{H}_0}{\mathrm{FS}} - = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} + = 2 \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \end{aligned}$$ -Once again, we turn the sum over $$\vb{k}$$ into an integral, -and recognize the spin's irrelevance: +Where we have recognized the spin's irrelevance, +by replacing the sum over $$s$$ with a factor $$2$$. +Next, we turn the sum over the allowed $$\vb{k}$$-values into an integral, +which is a [common trick](/know/concept/discrete-spectrum-summation/) +enabled by our periodic boundary conditions, yielding: $$\begin{aligned} E^{(0)} - &= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m} + &= \frac{2 V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}} - \\ +\end{aligned}$$ + +The matrix element +$$\matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}$$ +is either $$0$$ or $$1$$, depending on whether $$\vb{k}$$ +is outside the Fermi sphere, or, equivalently, +whether $$|\vb{k}|$$ is above or below $$k_F$$. +We can write this fact by introducing a +[Heaviside step function](/know/concept/heaviside-step-function/) $$\Theta(k)$$: + +$$\begin{aligned} + E^{(0)} &= \frac{\hbar^2 V}{8 \pi^3 m} \int_{-\infty}^\infty |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \dd{\vb{k}} \end{aligned}$$ -In spherical coordinates, -we evaluate the integral and find that $$E^{(0)}$$ is proportional to $$k_F^5$$: +We evaluate this in +[spherical coordinates](/know/concept/spherical-coordinates/) +and find that $$E^{(0)}$$ is proportional to $$k_F^5$$: $$\begin{aligned} E^{(0)} &= \frac{\hbar^2 V}{8 \pi^3 m} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Big( |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \Big) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi} \\ - &= \frac{\hbar^2 V}{8 \pi^3 m} 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|} - = \frac{\hbar^2 V}{2 \pi^2 m} \bigg[ \frac{|\vb{k}|^5}{5} \bigg]_0^{k_F} - = \frac{\hbar^2 V}{10 \pi^2 m} k_F^5 + &= \frac{\hbar^2 V}{8 \pi^3 m} \: 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|} + \\ + &= \frac{\hbar^2 V}{10 \pi^2 m} k_F^5 \end{aligned}$$ 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 $$k_F^3 = 3 \pi^2 n$$: +which we find to be as follows, +using that $$k_F^3 = 3 \pi^2 N / V$$: $$\begin{aligned} \boxed{ \frac{E^{(0)}}{N} = \frac{3 \hbar^2}{10 m} k_F^2 - = \frac{3}{5} \epsilon_F } - \:\sim\: n^{2/3} + \:\:\propto\: n^{2/3} \end{aligned}$$ -Traditionally, this is expressed using a dimensionless parameter $$r_s$$, +Traditionally, this is rewritten using the **Wigner-Seitz radius** $$r_s$$, defined as the radius of a sphere containing a single electron, -measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$$: +measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e_0^2 m)$$: $$\begin{aligned} \frac{4 \pi}{3} (a_0 r_s)^3 - = \frac{1}{n} - = \frac{3 \pi^2}{k_F^3} - \quad \implies \quad + \equiv \frac{1}{n} + \qquad \implies \qquad r_s - = \Big( \frac{3}{4 \pi a_0^3 n} \Big)^{1/3} - = \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 k_F} + = \bigg( \frac{3}{4 \pi a_0^3 n} \bigg)^{1/3} \end{aligned}$$ -Such that the ground state energy can be rewritten in Rydberg units of energy like so: +Note that this is dimensionless due to our choice of $$a_0$$ as a unit. +In the Fermi gas, we have: + +$$\begin{aligned} + r_s + = \bigg( \frac{9 \pi}{4} \bigg)^{1/3} \frac{1}{a_0 k_F} + \qquad \implies \qquad + k_F + = \bigg( \frac{9 \pi}{4} \bigg)^{1/3} \frac{1}{a_0 r_s} +\end{aligned}$$ + +By inserting this into the ground state energy +and using the definition of $$a_0$$, we can write: $$\begin{aligned} \frac{E^{(0)}}{N} - = \frac{3 \hbar^2}{10 m} \frac{4 \pi \varepsilon_0 e^2}{4 \pi \varepsilon_0 e^2} \frac{a_0^2 k_F^2}{a_0^2} - = \frac{3 e^2}{40 \pi \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{2/3} \frac{1}{a_0 r_s^2} - \approx \frac{2.21}{r_s^2} \; \mathrm{Ry} + &= \frac{3 \hbar^2}{10 m} \bigg( \frac{9 \pi}{4} \bigg)^{2/3} \frac{1}{a_0^2 r_s^2} + \\ + &= \frac{3}{5} \bigg( \frac{9 \pi}{4} \bigg)^{2/3} \bigg( \frac{e_0^2}{8 \pi \varepsilon_0 a_0} \bigg) \frac{1}{r_s^2} \end{aligned}$$ +Where the last parenthesized expression is +the Rydberg unit of energy $$\mathrm{Ry} \approx 13.6 \:\mathrm{eV}$$, so: + +$$\begin{aligned} + \boxed{ + \frac{E^{(0)}}{N} + \approx \frac{2.21}{r_s^2} \; \mathrm{Ry} + } +\end{aligned}$$ + +This result is found in a lot of literature. +The choice of Rydberg units is simply a tradition. + + -## With interactions +## 1st order -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. +In the next term of the perturbation expansion, +we start to include Coulomb interactions. +Clearly, this will give better results when the interaction is relatively weak, +but is that ever the case? 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 $$n \to \infty$$. +so the interaction energy $$E_\mathrm{int}$$ scales as $$n^{1/3}$$. +We also know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$ +is proportional to $$n^{2/3}$$, +meaning that it is reasonable to use perturbation theory +as long as $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \propto n^{-1/3}$$, +i.e. in the limit of high density $$n \to \infty$$. The two-body Coulomb interaction operator $$\hat{W}$$ is as follows in second-quantized form: $$\begin{aligned} \hat{W} - = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e^2}{\varepsilon_0 |\vb{q}|^2} + = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e_0^2}{\varepsilon_0 |\vb{q}|^2} \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} \end{aligned}$$ @@ -228,7 +199,7 @@ is then given by: $$\begin{aligned} E^{(1)} = \matrixel{\mathrm{FS}}{\hat{W}}{\mathrm{FS}} - = \frac{e^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} + = \frac{e_0^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2} \matrixel{\mathrm{FS}}{ \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1} }{\mathrm{FS}} @@ -246,17 +217,17 @@ Let $$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} + &= \frac{e_0^2}{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} + &= \frac{- e_0^2}{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} + &= \frac{- e_0^2}{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}$$ @@ -269,11 +240,11 @@ This yields the integration limit, and therefore leads to: $$\begin{aligned} E^{(1)} - &= \frac{- e^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}} + &= \frac{- e_0^2}{(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}} + &= \frac{- e_0^2}{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}$$ @@ -285,11 +256,11 @@ when we go to spherical coordinates $$(|\vb{k}|, \theta_k, \varphi_k)$$ for $$\v $$\begin{aligned} E^{(1)} - &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty + &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty \!\Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} \\ - &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F} + &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F} \!\Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} \end{aligned}$$ @@ -335,13 +306,13 @@ substituting $$\xi \equiv \cos(\theta_k)$$: $$\begin{aligned} E^{(1)} - &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}} + &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}} \!\!\int_{|\vb{q}|/(2 \cos{\theta_k})}^{k_F} |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|} \\ - &= \frac{e^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)} + &= \frac{e_0^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)} \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \frac{\sin(\theta_k)}{\sin(\theta_k)} \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} \\ - &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 + &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|} \end{aligned}$$ @@ -350,23 +321,23 @@ Evaluating these integrals: $$\begin{aligned} E^{(1)} - &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 + &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|} \\ - &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 + &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1 \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|} \\ - &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} + &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|} \\ - &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} + &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|} \\ - &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F} + &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F} \\ - &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4 - = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4 - = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F + &= \frac{- e_0^2 V}{16 \pi^4 \varepsilon_0} k_F^4 + = \frac{- e_0^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4 + = -\frac{3 e_0^2 N}{16 \pi^2 \varepsilon_0} k_F \end{aligned}$$ Per particle, the first-order energy correction $$E^{(1)}$$ @@ -375,7 +346,7 @@ is therefore found to be as follows: $$\begin{aligned} \boxed{ \frac{E^{(1)}}{N} - = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} k_F + = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} k_F } \end{aligned}$$ @@ -383,8 +354,8 @@ This can also be written using the parameter $$r_s$$ introduced above, leading t $$\begin{aligned} \frac{E^{(1)}}{N} - = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0} - = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s} + = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0} + = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s} \end{aligned}$$ Consequently, for sufficiently high densities $$n$$, |