# Hartree-Fock correction to $e$-$e$ interaction

The corrections to the energy per electron in a jellium model (uniform distribution of positive ion charge approximation to the regulated long range order ionic array) is given by (in units of Ry) $$\frac{E}{N} = \frac{2.21}{(r_s/a_o)^2} − \frac{0.916}{(r_s/a_o)} +0.0622\ln(r_s/a_o)−0.096+ O(r_s /a_o ).$$ ($r_s$ is the radius of a sphere that holds the same volume as one conduction electron and $a_o$ is the Bohr radius, but I don't think is so important to my questions). See e.g. the derivation and this same result in Ashcroft and Mermin P.336.

1. My question is why does the last three terms contain contributions to the kinetic and potential energy? It seems to me that it should only be potential energy since the source of the extra terms comes from the perturbation which is precisely due to the e-e interactions, but I have been told otherwise.

2. My other question is this whole procedure was implemented to better account for the interaction between the electrons in a metal. This expansion above is only valid for $r_s /a_o \ll 1$ and for most metals this value is from 2 to 6. So what is the usefulness of this result then if it does not apply to the very thing it sought out to describe?

Even if it's not really appropriate to derivate quantitative results about electrons properties in solids, Jellium model still have some interesting qualitative features. As you correctly pointed, Jellium still a mean-field theory and so fails about dealing with strongly correlated systems.

Lets remind a few about jellium model. The starting point is the hamiltonian : $$\mathcal{H}=\mathcal{H}_{e^-}+\mathcal{H}_{i^+}+\mathcal{H}_{e^-/e^-}+\mathcal{H}_{e^-/i^+}$$ where $\mathcal{H}_{e^-}$ is the free electrons kinetic energy term, $\mathcal{H}_{i^+}$ is the positive electrostatic ionic background, $\mathcal{H}_{e^-/e^-}$ the electron-electron repulsion term and $\mathcal{H}_{e^-/i^+}$ the interaction between the electrons and ionic background.

By introducing a scale $\gamma$ at which Coulomb interactions between electrons are consistant, which consists basically in taking : $$\mathcal{H}_{e^-/e^-}=\lim_{\gamma\rightarrow 0}\frac{e^2}{2}\sum_{i\neq j}\frac{e^{-\gamma|\textbf{x}_i-\textbf{x}_j|}}{|\textbf{x}_i-\textbf{x}_j|}\equiv\lim_{\gamma\rightarrow 0}\frac{1}{2\Omega}\sum_{\textbf{k},\textbf{p},\textbf{q}}\frac{4\pi e^2}{q^2+\gamma^2}a^\dagger_{\textbf{k}-\textbf{q}}a^\dagger_{\textbf{p}-\textbf{q}}a_{\textbf{p}}a_{\textbf{k}}$$

Such procedure allows us to get rid of divergences introduced by the $1/r$ decreasing Coulomb interaction. And interestingly, it can be shown that the $\mathcal{H}_{i^+}$ background energy is compensated by the $\textbf{q}=0$ term of $\mathcal{H}_{e^-/e^-}$ at the thermodynamic limit ($\Omega\rightarrow+\infty$, where $\Omega$ is the volume of the system). Then, one can end up with : $$\mathcal{H}=\mathcal{H}_{e-}+\mathcal{V}=\sum_{\textbf{k}}\frac{\hbar^2\textbf{k}^2}{2m}a^\dagger_{\textbf{k}}a_{\textbf{k}}+\frac{e^2}{2\Omega}\sum_{\textbf{k},\textbf{p},\textbf{q}\neq 0}\frac{4\pi}{q^2}a^\dagger_{\textbf{k}-\textbf{q}}a^\dagger_{\textbf{p}-\textbf{q}}a_{\textbf{p}}a_{\textbf{k}}$$ Regarding your first question, it is now clear the $\mathcal{V}$ term is the result of the "mixing" (said mutual counterbalance) of various contributions of the initial hamiltonian, namely : $$\mathcal{V}=\lim_{\Omega\rightarrow+\infty}\lim_{\gamma\rightarrow 0}\,\mathcal{H}_{i^+}+\mathcal{H}_{e^-/i^+}+\frac{1}{2\Omega}\sum_{\textbf{k},\textbf{p},\textbf{q}}\frac{4\pi e^2}{\gamma^2+q^2}a^\dagger_{\textbf{k}}a^\dagger_{\textbf{p}}a_{\textbf{p}}a_{\textbf{k}}$$

To obtain the $E/N$ OP's expression, one has just to compute $\frac{\langle F|\mathcal{H}|F\rangle}{N}$ by using a perturbation theory, where $|F\rangle$ is the Fermi sea : $$E=E^{(0)}+E^{(1)}+E^{(2)}+...$$ Such perturbative expansion is valid only if Coulomb interaction is negligible compared to kinetic energy. Typical kinetic energy scales like : $$E_{\text{kin}}\sim E_F\sim n^{2/3}\sim 1/r_s^2\quad\text{where}\;r_s=1/n^3\;\text{with}\;n\;\text{is the density}$$ $r_s$ can be interpreted as the typical distance between two electrons. Typical Coulomb energy scales like : $$E_{\text{Coulomb}}\sim\frac{e^2}{r_s}\sim\frac{1}{r_s}$$ Then, it should verify : $$\frac{E_{\text{Coulomb}}}{E_{\text{kin}}}\sim\frac{1/r_s}{1/r_s^2}\sim r_s<<1$$

Regarding your second question, let just plot $E/N$ at first order of perturbation :

As you can see, it exhibits a negative minimum around $\frac{r_s}{a_0}=4.83$. This negative energy minimum actually refers to metallic bonding (attractive interaction), where electrons are somewhat "gluing" ions between them.

• Many thanks for your answer. I have not studied the Hartree-approximation using your method of annihilation operators so I could not follow your mathematics exactly. (I am following the method in Ashcroft and Mermin (AAM)). I could not understand the part where you say the second term of $\mathcal H$ is the result of mixing of the subsequent terms. Could you maybe describe this part in words or at the level of the notation of AAM? I think the method of AAM is the same you used, just not in the operator formulism. Thanks – CAF Nov 29 '14 at 22:42
• In addition to dolan's answer, I recommend reading first chapter of Fetter and Walecka's classic many-body book. This book provides the best introduction to second quantization method, with thorough and detailed derivations. By doing so, you can gain acquaintance with second quantization method. Finally, the Jellium model is treated as an example of applying second quantization, at the end of the chapter. – Hadi Vafaei Nov 30 '14 at 13:22
• Actually, I don't really remember how they are doing in AAM. But still, the only thing you need to understand about my notation is just that it's a Fourier analysis of $\mathcal{H}$ and that $\textbf{q}$ refers to the momentum transfert between two electrons interacting through Coulomb interaction. And then, you can see that the $\mathcal{V}$ term is the result of various cancellation of background energy. – dolun Nov 30 '14 at 13:40
• Thanks for the edits. But how is $\mathcal V$ explicitly got terms in kinetic energy? As far as I can see, there are only contributions from the positive ionic background, electron-ion interaction and electron-electron interaction which all contribute to the potential energy just? – CAF Nov 30 '14 at 14:05
• Yes you're right and actually I don't really understand why you are speaking about kinetic terms in $\mathcal{V}$ since there is none... Kinetic energy is something well separated of $\mathcal{V}$ in jellium model since you are doing a perturbative treatment on $\mathcal{V}$. Such perturbative expansion would be totally messed up if $\mathcal{H}_{e^-}$ and $\mathcal{V}$ where coupled. – dolun Nov 30 '14 at 14:17