A missing factor of 2 in the standard Hartree-Fock mean field?

Question

Let's start from a very simple argument: If $A$ and $B$ are some operators, then I can write their product as
$$AB = (A-\langle A\rangle)(B - \langle B \rangle) + \langle A \rangle B + A \langle B \rangle + \langle A \rangle \langle B \rangle$$ In a mean-field approximation, I would then go on to neglect the first product on the right hand side under the assumption that fluctuations around the mean are small and thus terms quadratic in fluctuations can be neglected.

So far, so good. But then I look at the typical electron-electron interaction term: $$H_{int} = \frac{1}{2}\sum_{k,p,q,\alpha,\beta} V(q) c_{k+q,\alpha}^\dagger c_{p-q,\beta}^\dagger c_{p\beta} c_{k\alpha}$$ To form the operators $A$ and $B$, I have to pair a creation and destruction operator, but there are two different ways to do so.

For the sake of brevity, let's call the operators $c_1^\dagger c_2^\dagger c_3 c_4$. Then in the mean field approach one gets four different terms, one for each way to pick one creation and one destruction operator and everage over them: $$c_1^\dagger c_2^\dagger c_3 c_4 \approx -\langle c_1^\dagger c_3 \rangle c_2^\dagger c_4 -\langle c_2^\dagger c_4 \rangle c_1^\dagger c_3 +\langle c_1^\dagger c_4 \rangle c_2^\dagger c_3 +\langle c_2^\dagger c_3 \rangle c_1^\dagger c_4$$

Naively, I'd have expected a factor of 2 here, because here I used two ways of forming operators $A$ and $B$ (in the sense from above) to obtain my mean-field approximation.

Answer 1

I think I worked it out myself. What I found is that trying to arrive at the Hartree-Fock Hamiltonian via this mean-field procedure isn't really that appropriate.

Instead, one says: Given a two-particle operator $\hat O = c_1^\dagger c_2^\dagger c_3 c_4$, how can I find a single-particle operator $\hat O^{MF}$ that is a good approximation?

If we have a complete set of single-particle basis states that include states $1$ to $4$ of the two-particle operator, we would then look at the basis elements of $\hat O$ and $\hat O^{MF}$ and try to find the best possible agreement.

Now, the two-particle operator can do three different things: 1. Leave a state as is. This is the case if either $1 = 3$ and $2 = 4$ or $1 = 4$ and $2 = 3$. 2. Move one particle from a previously occupied state into a previously unoccupied state, e.g. if $1 = 3$ but $2 \not = 4$ or $1 = 4$ and $2 \not= 3$. 3. Move two particles into new states, i.e. if $1$ and $2$ are different from $3$ and $4$.

With our single particle operator, we can only hope to somehow reproduce the behavior of case 1 and 2, but not case 3.

So, given states of the type $|\psi_0\rangle$ that can be written as $\prod_i c_{n_i}^\dagger |0\rangle$, we look at all matrix elements of $\hat O$ between $|\psi_0\rangle$ and states of the type $c_i^\dagger c_j |\psi_0\rangle$, $i \not= j$, that arise from $|\psi_0\rangle$ by exchanging two particles.

It is a bit tedious to work these matrix elements out in detail, but one can then indeed confirm that by chosing $$\hat O^{MF} = \langle c_1^\dagger c_4\rangle c_2^\dagger c_3 + \langle c_2^\dagger c_3\rangle c_1^\dagger c_4 - \langle c_1^\dagger c_3\rangle c_2^\dagger c_4 - \langle c_2^\dagger c_4\rangle c_1^\dagger c_3$$ the matrix elements for the two-particle operator $\hat O$ and the mean-field operator $\hat O^{MF}$ will turn out to be the same (for the states of the type introduced above).