Categories: Perturbation, Physics, Quantum mechanics.


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 VV, and impose periodic boundary conditions on it, such that the single-particle orbitals are (suppressing spin):

r|ψk=ψk(r)=1Vexp(ikr)k=2πV1/3(nx,ny,nz)\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 nx,ny,nzZn_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 H^0\hat{H}_0, where 2k2/(2m)\hbar^2 |\vb{k}|^2 / (2 m) is the kinetic energy of the orbital with wavevector k\vb{k}, and ss is the spin:

H^0=sk2k22mc^s,kc^s,k\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=0T = 0, the NN-electron ground state of this Hamiltonian is known as the Fermi sea or Fermi sphere FS\Ket{\mathrm{FS}}, and is constructed by filling up the single-electron states starting from the lowest energy:

FS=sj=1N/2c^s,kj0\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=0T = 0, all the electrons stay in their assigned state. The energy and wavenumber k|\vb{k}| of the highest filled orbital are called the Fermi energy ϵF\epsilon_F and Fermi wavenumber kFk_F, and obey the expected kinetic energy relation:

ϵF=22mkF2\begin{aligned} \boxed{ \epsilon_F = \frac{\hbar^2}{2 m} k_F^2 } \end{aligned}

The Fermi sea can be visualized in k\vb{k}-space as a sphere with radius kFk_F. Because k\vb{k} is discrete, the sphere’s surface is not smooth, but in the limit VV \to \infty it becomes perfect.

Now, we would like a relation between the system’s parameters, e.g. NN and VV, and the resulting values of ϵF\epsilon_F or kFk_F. The total population NN must be given by:

N=skFSc^s,kc^s,kFS=sV(2π)3FSc^s,kc^s,kFSdk\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 k\vb{k} into an integral with a constant factor, by using that each orbital exclusively occupies a volume (2π)3/V(2 \pi)^3 / V in k\vb{k}-space.

At zero temperature, this inner product can only be 00 or 11, depending on whether k\vb{k} is outside or inside the Fermi sphere. We can therefore rewrite using a Heaviside step function:

N=sV(2π)3Θ(kFk)dk=2V(2π)3Θ(kFk)dk\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 ss by a factor 22. In order to evaluate this 3D integral, we go to spherical coordinates (k,θ,φ)(|\vb{k}|, \theta, \varphi):

N=V4π302π0π0Θ(kFk)k2sin(θ)dkdθdφ=V4π34π0kFk2dk=Vπ2[k33]0kF=V3π2kF3\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/Vn = N/V, we thus arrive at the following relation:

kF3=3π2n\begin{aligned} \boxed{ k_F^3 = 3 \pi^2 n } \end{aligned}

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

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

E(0)=FSH^0FS=sk2k22mFSc^s,kc^s,kFS\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 k\vb{k} into an integral, and recognize the spin’s irrelevance:

E(0)=sV(2π)32k22mFSc^kc^kFSdk=2V8π3mk2Θ(kFk)dk\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)E^{(0)} is proportional to kF5k_F^5:

E(0)=2V8π3m02π0π0(k2Θ(kFk))k2sin(θ)dkdθdφ=2V8π3m4π0kFk4dk=2V2π2m[k55]0kF=2V10π2mkF5\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)/NE^{(0)} / N, which we find to be as follows, using that kF3=3π2nk_F^3 = 3 \pi^2 n:

E(0)N=3210mkF2=35ϵFn2/3\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 rsr_s, defined as the radius of a sphere containing a single electron, measured in Bohr radii a04πε02/(e2m)a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m):

4π3(a0rs)3=1n=3π2kF3    rs=(34πa03n)1/3=(9π4)1/31a0kF\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:

E(0)N=3210m4πε0e24πε0e2a02kF2a02=3e240πε0(9π4)2/31a0rs22.21rs2  Ry\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 n1/3n^{-1/3}, so the interaction energy EintE_\mathrm{int} should scale as n1/3n^{1/3}. We already know that the kinetic energy Ekin=E(0)E_\mathrm{kin} = E^{(0)} scales as n2/3n^{2/3}, meaning perturbation theory should be reasonable if 1Eint/Ekinn1/31 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}, so in the limit of high density nn \to \infty.

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

W^=12Vs1s2k1k2q0e2ε0q2c^s1,k1+qc^s2,k2qc^s2,k2c^s1,k1\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)E^{(1)} to the ground state (i.e. Fermi sea) energy is then given by:

E(1)=FSW^FS=e22ε0Vs1s2k1k2q01q2FSc^s1,k1+qc^s2,k2qc^s2,k2c^s1,k1FS\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 c^\hat{c}^\dagger are for the same orbitals as the two annihilation operators c^\hat{c}. Since q0\vb{q} \neq 0, this means that s1=s2s_1 = s_2, and that momentum is conserved: k2=k1 ⁣+ ⁣q\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}. And of course both k1\vb{k}_1 and k1 ⁣+ ⁣q\vb{k}_1 \!+\! \vb{q} must be inside the Fermi sphere, to avoid annihilating an empty orbital. Let s=s1s = s_1 and k=k1\vb{k} = \vb{k}_1:

E(1)=e22ε0Vskq01q2FSc^s,k+qc^s,kc^s,k+qc^s,kFS=e22ε0Vskq01q2FS(c^s,k+qc^s,k+q)(c^s,kc^s,k)FS=e22ε0Vskq01q2Θ(kFk)Θ(kFk ⁣+ ⁣q)\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 q\vb{q} into an integral in spherical coordinates. Clearly, q\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 2kF2 k_F. This yields the integration limit, and therefore leads to:

E(1)=e2(2π)3ε0k02π ⁣ ⁣0π ⁣ ⁣0Θ(kF ⁣ ⁣k)Θ(kF ⁣ ⁣k ⁣+ ⁣q)q2q2sin(θq)dqdθqdφq=e22π2ε0k02kFΘ(kF ⁣ ⁣k)Θ(kF ⁣ ⁣k ⁣+ ⁣q)dq\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 q\vb{q}, i.e. (θq,φq)(\theta_q,\varphi_q), is irrelevant, as long as we define θk\theta_k as the angle between q\vb{q} and k ⁣+ ⁣q\vb{k} \!+\! \vb{q} when we go to spherical coordinates (k,θk,φk)(|\vb{k}|, \theta_k, \varphi_k) for k\vb{k}:

E(1)=e2V16π5ε002kF ⁣ ⁣ ⁣ ⁣02π ⁣ ⁣ ⁣0π ⁣ ⁣ ⁣0 ⁣Θ(kF ⁣ ⁣k)Θ(kF ⁣ ⁣k ⁣+ ⁣q)k2sin(θk)dkdθkdφkdq=e2V16π5ε002kF ⁣ ⁣ ⁣ ⁣02π ⁣ ⁣ ⁣0π ⁣ ⁣ ⁣0kF ⁣Θ(kF ⁣ ⁣k ⁣+ ⁣q)k2sin(θk)dkdθkdφkdq\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 kFk_F, one centered on the origin (for k\vb{k}), and the other centered on q\vb{q} (for k ⁣+ ⁣q\vb{k} \!+\! \vb{q}). Imagine a triangle with side lengths k|\vb{k}|, q|\vb{q}| and k ⁣+ ⁣q2|\vb{k} \!+\! \vb{q}|^2, where θk\theta_k is the angle between k|\vb{k}| and k ⁣+ ⁣q|\vb{k} \!+\! \vb{q}|. The law of cosines then gives the following relation:

k2=q2+k ⁣+ ⁣q22qk ⁣+ ⁣qcos(θk)\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 k<kF|\vb{k}| < k_F and 0<q<2kF0 < |\vb{q}| < 2 k_F, so by isolating for cos(θk)\cos(\theta_k), we can obtain bounds on θk\theta_k and k|\vb{k}|. Let kkF|\vb{k}| \to k_F in both cases, then:

cos(θk)=k ⁣+ ⁣q2+q2k22k ⁣+ ⁣qq>q0kF2+q2kF22kFq=q2kF<q2kFkF2+4kF2kF22kF2kF=1\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<θk<arccosq/(2kF)0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}. To get a lower limit for k|\vb{k}|, we “cheat” by artificially demanding that k\vb{k} does not cross the halfway point between the spheres, with the result that kcos(θk)>q/2|\vb{k}| \cos(\theta_k) > |\vb{q}|/2. Then, thanks to symmetry (both spheres have the same radius), we just multiply the integral by 22, for k\vb{k} on the other side of the halfway point.

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

E(1)=e2V16π5ε0202kF ⁣ ⁣ ⁣02π ⁣ ⁣0arccosq/(2kF) ⁣ ⁣q/(2cosθk)kFk2sin(θk)dkdθkdφkdq=e2V8π5ε02π02kF ⁣ ⁣ ⁣1q/(2kF) ⁣ ⁣q/(2ξ)kFk2sin(θk)sin(θk)dkdξdq=e2V4π4ε002kF ⁣ ⁣ ⁣q/(2kF)1 ⁣ ⁣q/(2ξ)kFk2dkdξdq\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 φk\varphi_k does not appear in the integrand. Evaluating these integrals:

E(1)=e2V4π4ε002kF ⁣ ⁣ ⁣q/(2kF)1[k33]q/(2ξ)kFdξdq=e2V4π4ε002kF ⁣ ⁣ ⁣q/(2kF)1(kF33q324ξ3)dξdq=e2V4π4ε002kF[kF33x+q348ξ2]q/(2kF)1dq=e2V4π4ε002kF(kF33+q348kF2q4)dq=e2V4π4ε0[kF3q3+q4192kF2q28]02kF=e2V16π4ε0kF4=e2N16π4ε0nkF4=3e2N16π2ε0kF\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)E^{(1)} is therefore found to be as follows:

E(1)N=3e216π2ε0kF\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 rsr_s introduced above, leading to:

E(1)N=3e216π2ε0a0kFa0=3e216π2ε0(9π4)1/31a0rs\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 nn, the total energy EE per particle is given by:

EN(2.21rs20.92rs)  Ry\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)E^{(2)} is as shown below, but it turns out that it (and all higher orders) diverge:

E(2)=ΨnFSFSW^Ψn2E(0)En\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.


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