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In studying the Hartree-Fock method for solving systems of interacting particles, I have often found that the method is referred to as a mean-field approach. Wikipedia's page for instance says that the mean-field approximation is implied. I don't see why this is the case though.

In the HF method, one considers a single Slater determinant and minimizes the energy with respect to the orbitals that compose the determinant. The equations that follow this procedure are the HF equations and while it's true that they resemble the Schrodinger equations of independent electrons in a mean-field, they also have an exchange term, which makes those equations different from the Hartree equations, which actually do look like a mean-field approximation. In any case, I don't see where a mean-field approximation is implied.

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2 Answers 2

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Hartree-Fock reduces the multi-particle problem to a one-particle one, representing its interaction with the rest of the world as an interaction between the particle and some average (mean) field. This is perhaps less obvious when treating a two-particle system, where both particles appear on an equal footing, but it becomes rather obvious when one is treating a system of N-particles.

Slater determinant is a manifestation of that: it gives an exact wave function for $N$ non-interacting particles, but once the interaction is on, the factorization into single-particle orbitals becomes an approximation.

Remarks: The exchange term may appear unusual in comparison to the usual form of the Coulomb interaction, but it remains a one-particle interaction with the rest. When applying more formal mean field frameworks, e.g., via the Habbard-Stratonovich transformation, both terms appear naturally, on equal footing.

Example: This is not to give a summary of all the mean-field approaches, but just to give an idea why it is natural to call HF mean field. Let us consider a second-quantized electron Hamiltonian with Coulomb interaction: $$ \mathcal{H} = \sum_{k,\sigma}\epsilon_{k,\sigma}c_{k,\sigma}^\dagger c_{k,\sigma} + \sum_{k,k',\sigma,\sigma'}V_{k,\sigma;k',\sigma'}c_{k,\sigma}^\dagger c_{k',\sigma'} + \\ \frac{1}{2}\sum_{k,k', k'', k''', \sigma, \sigma'}U_{k, k', k''', k''}c_{k,\sigma}^\dagger c_{k',\sigma'}^\dagger c_{k'',\sigma'} c_{k''',\sigma} $$ Here the second term is an interaction with an external field, which potentially can be spin-dependent (hence the spin indices). The last term is the usual Coulomb interaction, which we will factorize in a HF manner as $$ \frac{1}{2}\sum_{k,k', k'', k''', \sigma, \sigma'}U_{k, k', k''', k''}c_{k,\sigma}^\dagger c_{k',\sigma'}^\dagger c_{k'',\sigma'} c_{k''',\sigma} \approx\\ \frac{1}{4}\sum_{k,k', k'', k''', \sigma, \sigma'}U_{k, k', k''', k''}c_{k,\sigma}^\dagger c_{k''',\sigma}\langle c_{k',\sigma'}^\dagger c_{k'',\sigma'}\rangle - \frac{1}{4}\sum_{k,k', k'', k''', \sigma, \sigma'}U_{k, k', k''', k''}c_{k,\sigma}^\dagger c_{k'',\sigma'} \langle c_{k',\sigma'}^\dagger c_{k''',\sigma}\rangle $$ here the first term in the last expression is the Hartree term, whereas the last term is the Fock term - they differ just by the order of pairing operators that will be replaced by their mean, i.e. $$ c_{k',\sigma'}^\dagger c_{k'',\sigma'}\rightarrow \langle c_{k',\sigma'}^\dagger c_{k'',\sigma'}\rangle $$ We can further rewrite the Coulomb term as $$ \frac{1}{2}\sum_{k,k', k'', k''', \sigma, \sigma'}U_{k, k', k''', k''}c_{k,\sigma}^\dagger c_{k',\sigma'}^\dagger c_{k'',\sigma'} c_{k''',\sigma} \approx \sum_{k,k',\sigma,\sigma'}U^{eff}_{k,\sigma;k',\sigma'}c_{k,\sigma}^\dagger c_{k',\sigma'}, $$ which has the same form as the one-particle potential (external field term). The use of mean and field comes rather naturally in this context and has a clear meaning. Note also that in this formalism there is no really apparent difference between Hartree and Fock terms - they differ by the spin indices, which are necessarily the same in the Hartree term, but even these could be symmetrized, as is done in the classical reference on the field theory methods in condensed matter, although for more detailed derivations of this type and introduction to the second quantization I would recommend more pedestrian Fetter & Walecka. These books however may not call it mean field, which is a term used mainly by the people raised on path integrals, for which a good condensed matter reference is Negele & Orland.

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  • $\begingroup$ The Hartree term looks exactly like the average field generated by the charge distribution of other particles, but the exchange term mixes the orbitals together. Because of the exchange term, we don't actually have an eigenvalue equation. Is it called mean-field anyway because we have single-particle Schrodinger equations? If that's the case it seems the name isn't that fitting. $\endgroup$ Commented Jun 23, 2021 at 13:02
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    $\begingroup$ @KarimChahine Yes, this is the case. The name is much more appropriate when looked from the point of view of general many-body phsyics approaches or the theory of critical phenomena, where HF arises as a result of typical mean-field approximation schemes. $\endgroup$
    – Roger V.
    Commented Jun 23, 2021 at 13:40
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    $\begingroup$ @KarimChahine it is no different from the Hartree term. It does not represent any bare interaction, like Coulomb potential - is this what bothers you? $\endgroup$
    – Roger V.
    Commented Jun 24, 2021 at 5:41
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    $\begingroup$ @KarimChahine I added an example... I hope I didn't amke too many errors, since I did it on the top of my head. $\endgroup$
    – Roger V.
    Commented Jun 24, 2021 at 7:57
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    $\begingroup$ @RogerVadim I have a related question: (mattermodeling.stackexchange.com/questions/9509/…) which I think it can be answered by your great answer here. The first three paragraphs of your answer clicked for me especially the Remarks section. Would you be interested to have a look at my question please? I would be more than glad to have your answer there and accept it shall you decide to answer my question. Thank you in advance. $\endgroup$
    – Sha
    Commented Nov 17, 2022 at 17:39
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In HF we assume that we can separate the wavefunction into a product ansatz $$ \Psi(r_1, r_2, ...)\Psi^*(r_1, r_2, ...)=|\Psi(r_1, r_2, ...)|^2=\rho(r_1, r_2, ...) $$

$$ \rho(r_1, r_2, ...)\approx\rho_1(r_1)\rho_2(r_2)\cdots $$

$$ \rho_1(r_1)=\psi_1(r_1)\psi^*_1(r_1) $$

This allows us to define single particle potentials that are summed to form the mean field potential $V_H$, $$ \begin{aligned} V_H(r_i) &=\frac{-e}{4\pi\varepsilon_0}\sum_{j=1(\neq i)}^N \int dr_j \frac{\rho_j(r_j) }{|r_j-r_i|}\\[1.5em] V_H(r_i) &=\frac{-e}{4\pi\varepsilon_0}\sum_{j=1(\neq i)}^N \int dr_j \frac{\psi_j(r_j)\psi^*_j(r_j) }{|r_j-r_i|} \end{aligned} $$

The potential energy of an particle is then calculated by interaction with this mean field potential of the remaining particles, $$ \begin{aligned} E_{pot,i} &= q_e\int_V dr_i \psi_i(r_i)\psi^*_i(r_i)\frac{e^2}{4\pi\varepsilon_0}\sum_{j=1(\neq i)}^N \int dr_j \frac{\psi_j(r_j)\psi^*_j(r_j) }{|r_j-r_i|}\\ &=q_e\int_V dr_i \psi_i(r_i)\psi^*_i(r_i)V_{H}(r_i) \end{aligned} $$

I would consider this the reason why HF is called a mean field method.

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    $\begingroup$ Isn't that the Hartree method? In Hartree-Fock, the ansatz is not a simple product, which doesn't have the correct symmetry, but a Slater determinant. This way the potential energy includes also an exchange term, which doesn't appear in the Hartree (without Fock) method $\endgroup$ Commented Jun 23, 2021 at 12:57
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    $\begingroup$ @KarimChahine You still have this potential inside HF even if you have an additional exchange term due to the determinant Ansatz. This additional exchange term does not change the fact that you are still using a mean field potential for the Coulomb interaction. It does introduce some correlation but the correlation due to the determinant Ansatz is purely "spin-based" and does not restore the exact correlation due to the Coulomb interaction. $\endgroup$
    – Hans Wurst
    Commented Jun 23, 2021 at 14:38
  • $\begingroup$ So you would say HF is a mean-field method because the Coulomb interaction is treated as a mean of all other orbitals? $\endgroup$ Commented Jun 23, 2021 at 16:11
  • $\begingroup$ @KarimChahine Yes, as long as only a single determinant is used. $\endgroup$
    – Hans Wurst
    Commented Jun 23, 2021 at 16:21

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