Is Hartree-Fock a standard mean field approximation?

I have read many times that Hartree-Fock is a mean field approximation, but I struggle to reconcile it with a standard mean field approach. In the simplest form of mean field approximation, we utilize the equality $$AB = (A-\langle A\rangle)(B - \langle B \rangle) + \langle A \rangle B + A \langle B \rangle - \langle A \rangle \langle B \rangle.\tag{1}$$ We usually assume the first term on the right hand side is negligible and we drop the last term because it is a constant. In a Hartree-Fock approximation we write (see these lecture notes or this question) $$c_1^\dagger c_2^\dagger c_3 c_4 \approx -\langle c_1^\dagger c_3 \rangle c_2^\dagger c_4 -\langle c_2^\dagger c_4 \rangle c_1^\dagger c_3 +\langle c_1^\dagger c_4 \rangle c_2^\dagger c_3 +\langle c_2^\dagger c_3 \rangle c_1^\dagger c_4,\tag{2}$$ while expecting from the mean field theory \begin{align} c_1^\dagger c_2^\dagger c_3 c_4 =-\frac{1}{2}(c_1^\dagger c_3) (c_2^\dagger c_4)+\frac{1}{2}(c_1^\dagger c_4) (c_2^\dagger c_3)\\\approx \frac{1}{2}\left[-\langle c_1^\dagger c_3 \rangle c_2^\dagger c_4 -\langle c_2^\dagger c_4 \rangle c_1^\dagger c_3 +\langle c_1^\dagger c_4 \rangle c_2^\dagger c_3 +\langle c_2^\dagger c_3 \rangle c_1^\dagger c_4\right].\tag{3} \end{align} In other words, application of the mean field theory according to the equation (1) gives an extra factor of $$1/2$$ as compared to the effective Hamiltonian of the Hartree-Fock approximation. Am I missing anything, or Hartree-Fock is indeed a non-standard mean-field approximation and cannot be obtained from Eq. (1)?

Edit. Essentially, I would like to arrive at Hartree-Fock approximation, when starting from either Eq. (1) or Hubbard-Stratonovich transformation; or, alternatively, confirm that this is not possible. I am well aware about the standard way of deriving Hartree-Fock approximation by minimizing the energy of the wave function of a non-interacting fermionic system.

• where do you get (3) from? Oct 11, 2021 at 14:16
• Equation (3) is what I would expect to get if I were to derive something similar to (2) when starting from equation (1) (which is what I want). Oct 11, 2021 at 14:44
• Why should you expect factors 1/2?? Oct 11, 2021 at 15:27
• Equation (3) is essentially my unsuccessful attempt to derive Eq. (2) using Eq. (1). I am taking $c_1^\dagger c_2^\dagger c_3 c_4$ and trying to reshuffle it such way, that after one applies Eq. (1), the result would look like the Hartree-Fock approximation (Eq. (2)). Breaking $c_1^\dagger c_2^\dagger c_3 c_4$ into two terms as in the first line of Eq. (3) was the only thing I could come up with. Oct 11, 2021 at 15:40
• There is no way to get factors 1/2 from (1). Work out ABC by applying (1) to BC in place of B, and then to BC in place of AB. Then do something similar for ABCD. Then use that single c/a operators have zero vacuum expectation. Oct 11, 2021 at 16:21

The spirit and the practice of the mean field theory
The spirit of the MFT is replacing an interacting (multi-particle) Hamiltonian by a non-interacting (single-particle) one, for particles moving in an effective field, thus making the problem solvable. (So Hartree-Fock nicely filles the bill.) In practice there are several "standard" ways to do this, specifically:

• Weiss mean field theory, as described in the OP, which is usually applied in the equations of motion
• Habbard-Stratonovich transformation in path integral (I recommend Negele&Orland's book for coincise presentation, although it is deep in the text)
• Partial resummation of diagrams in Feynman-Dyson expansion

My personal impression was always that the trend to call Hartree-Fock a mean field approximation comes from the path integral people, since it requires a leap of imagination to obtain it by Weiss method. On the other hand, in Feynman-Dyson expansion the Hartree and Fock terms appear naturally as a result of applying Wick's theorem, but no one (to my knwoledge) is calling it mean field in that context.

There are four operators in the expression of interest, so let us do mean field honestly, as we would do it for four operators: $$c_1^\dagger c_2^\dagger c_3 c_4 = (\langle c_1^\dagger\rangle + \delta c_1^\dagger) (\langle c_2^\dagger\rangle + \delta c_2^\dagger) (\langle c_3\rangle + \delta c_3)(\langle c_4\rangle +\delta c_4)$$ This is obviously a problematic expression, since the averages are zero when we are dealing with a particle-number-conserving Hamiltonian, as is often the case - but formally we can write it this way. If we now keep only the terms up to the second order in $$\delta c^\dagger,\delta c$$, we have $$c_1^\dagger c_2^\dagger c_3 c_4 = \langle c_1^\dagger\rangle \langle c_2^\dagger\rangle c_3 c_4 + \langle c_1^\dagger\rangle c_2^\dagger \langle c_3\rangle c_4 + \langle c_1^\dagger\rangle c_2^\dagger c_3 \langle c_4\rangle + c_1^\dagger \langle c_2^\dagger\rangle \langle c_3\rangle c_4 + c_1^\dagger \langle c_2^\dagger\rangle c_3 \langle c_4\rangle + c_1^\dagger c_2^\dagger \langle c_3\rangle \langle c_4 \rangle$$ We throw the first and the last terms, since they are not single-particle ones, and unite the averages under the same bracket, as if they were uncorrelated, arriving at $$c_1^\dagger c_2^\dagger c_3 c_4 = -\langle c_1^\dagger c_3\rangle c_2^\dagger c_4 + \langle c_1^\dagger c_4\rangle c_2^\dagger c_3 + \langle c_2^\dagger c_3\rangle c_1^\dagger c_4 - \langle c_2^\dagger c_4\rangle c_1^\dagger c_3,$$ which is the desired result.