For long time, i know the Hartree-Fock method is a way to find the mean-field, and the method is a mean-field theory. But how exactly the Hartree-Fock connects to the mean field, I have no idea. And occasionally, I mentioned the Hartree-Fock in its simplest form. Here, I will makes the connection crystal clear.

Hartree = self-consistence field

Fock = anti-symmetry wave function

Mean-field approximation, or in chemistry, self-consistence field approximation.

The full Hamiltonian is

$\displaystyle H = H_1 + H_2$

$\displaystyle H_1 = \sum_{i}^{N} \left( -\frac{1}{2}\nabla^2_i - \frac{Z}{r_i} \right) = \sum_{i} H_i$

$\displaystyle H_2 =\sum_{i

Here, $H_1$ is one-body operator, and $H_2$ is two-body operator. I used Coulomb potential in the one-body operator, but it can be generalized as $H_i$, and the mutual interaction can also be generalized as $G_{ij}$.

The idea of the mean-field approximation is that, what if, we can find a one-body potential $V(r)$, the mean-field, such that

$\displaystyle H = H_0 + H_R$

$\displaystyle H_0 = \sum_{i}^N \left( -\frac{1}{2}\nabla^2_i + V(r_i) \right)$

$\displaystyle H_R = \sum_{i

Here, $H_0$ is the mean-field Hamiltonian, which represent most of the effective interaction to a particle, such that $H_R$ is the residual interaction, which is very small, and can be later treated as perturbation.

So, the problem is, how to find this mean field $V(r)$?

Back to the form $H = H_1 + H_2$, we first construct a trial wave function,

$\displaystyle \Phi = \frac{1}{\sqrt{N!}} \sum_{P} (-1)^P P \Phi_H = \sqrt{N!} A \Phi_H$

$\displaystyle \Phi_H = \phi_\alpha(1)\phi_\beta(2) ... \phi_{\nu}(N)$

$\displaystyle A = \frac{1}{N!} \sum_{P}(-1)^P P = \frac{1}{N!} (1 - P_{ij} + P_{ijk} + .. )$

Here, $P$ is the permutation operator, it can be 1-body exchange, 2-body exchange, and so on, but we will see that, only 1-body exchange (which is no change at all) and 2-body exchange are needed. $\Phi_H$ is the Hartree wave function, it is a simple product of wave function of difference particles of difference states, or the diagonal product of the Slater determinant. In $\phi_\lambda(i)$, $\latex \lambda$ represents the state and $i$ is the “id” of the particle. Notice that

$\langle \phi_\mu | \phi_\nu \rangle = \delta_{\mu \nu}$

$A$ is the anti-symmetrization operator, it commute with Hamiltonian and a kind of projector operator,

$[A,H] = 0, A^2 = A.$

Now, we evaluate the energy using this trial wave function.

$E_{\Phi} = \langle \Phi | H_1 | \Phi \rangle + \langle \Phi | H_2 | \Phi \rangle$

$\displaystyle \langle \Phi | H_1 | \Phi \rangle = N! \langle \Phi_H | A H_1 A | \Phi_H \rangle = N! \langle \Phi_H | H_1 A | \Phi_H \rangle$

$\displaystyle N! \langle \Phi | H_1 | \Phi \rangle = \sum_{i} \langle \Phi_H| H_i (1 - P_{ij} + P_{ijk} + .. ) | \phi_\alpha(1)\phi_\beta(2) ... \phi_{\nu}(N) \rangle$

Since the one-body operator $H_i$ only acts on the $i$ particle, any exchange will make the operator do nothing on the $j$ particle, then the orthogonality of the wave function makes the integration zero. Lets demonstrate on 2 particles case.

$\displaystyle \langle \phi_a(1) \phi_b(2) | H_1 | \phi_a(2) \phi_b(1) \rangle = \langle \phi_a(1)| H_1 | \phi_b(1) \rangle \langle \phi_b(2) | \phi_a(2) \rangle = 0$

Notice that this $H_1$ is the one-body operator act on particle 1.

OK,

$\displaystyle \langle \Phi | H_1 | \Phi \rangle = \sum_{i} \langle \phi_\mu(i) | H_i | \phi_\mu(i) \rangle = \sum_\mu e_\mu$

$\displaystyle \langle \Phi | H_2 | \Phi \rangle =\frac{1}{2}\sum_{ij} \langle \Phi_H | H_2 (1 - P_ij) | \Phi_H \rangle$

$\displaystyle = \frac{1}{2} \sum_{ij} \begin{matrix} \langle \phi_\mu(i) \phi_\nu(j)|G_{ij}|\phi_\mu(i) \phi_\nu(j) \rangle \\ - \langle \phi_\mu(i) \phi_\nu(j)|G_{ij}|\phi_\nu(i) \phi_\mu(j) \rangle \end{matrix} = \sum_{\mu<\nu} \left( \langle \mu\nu | \mu\nu \rangle - \langle \mu\nu | \nu\mu \rangle \right)$

The $\langle \mu\nu | \mu\nu \rangle$ is direct term, and the $\langle \mu\nu | \nu\mu \rangle$ is the exchange term. This is a simplified notation, the first position is for the $i$-th particle, and the second position is always for the $j$-th particle.

Thus, the total energy is

$\displaystyle E_{\Phi} = \sum_{\mu} e_{\mu} + \sum_{\mu<\nu} \left(\langle \mu\nu | \mu\nu \rangle - \langle \mu\nu | \nu\mu \rangle\right)$

We can factor out the ket $|\phi_\mu(i) \rangle$ in the above equation and get the Fock-operator, with a notation for the exchange term operator

$\displaystyle F_i = H_i + \sum_{i

$J_{ij} |\phi_\mu(i) \rangle = \langle \phi_\nu(j)|G_{ij} | \phi_\nu(j) \rangle |\phi_\mu(i) \rangle$

$K_{ij} |\phi_\mu(i) \rangle = \langle \phi_\nu(j)|G_{ij} | \phi_\mu(j) \rangle |\phi_\nu(i) \rangle$

Becareful on the $\mu , \nu$! I know the notation is messy, I know….

The Fock-operator is an effective one-body operator.

First, we put the trial basis wave function $\phi_\lambda(i)$ can get the Fock-matrix $\langle \phi_\nu(i) | F_i | \phi_\mu(i) \rangle$, then diagonalize it, get the new eigen-states. Use this set of new eigen-states to calculate agian and again until converged!

So, where is the mean-field? Lets expand the Fock-operator into the Hartree-Fock equation.

$F_i |\phi_\mu(i) \rangle = \epsilon_\mu |\phi_\mu(i) \rangle$

$\displaystyle \left( -\frac{1}{2} \nabla^2_i - \frac{Z}{r_i} + \frac{1}{2}\sum_{j} \left( J_{ij} - K_{ij} \right) \right) \phi_\mu(i) = \left( -\frac{1}{2} \nabla^2_i + V(r_i) \right) \phi_\mu(i) = E_\mu \phi_\mu(i)$

$\displaystyle V(r_i) = - \frac{Z}{r_i} + \sum_{i

This is the mean-field!

Using the trial basis, we evaluate the Fock-matrix, that is equivalent to evaluate the mean-field.

In this post, the Hartree-fock for 2-body ground state is discussed. Unfortunately, that method is not the same in here. I would said, that method is only Hartree but not Fock. Since the method in that post can find a consistence field, but the ground state spatial is not anti-symmetric.

Since the spin-state is factored out, the spatial wave function of the case is identical. Thus, the exchange term is gone. The Slater determinant is

$\displaystyle \Phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi(1)\alpha & \phi(1) \beta \\ \phi(2) \alpha & \phi(2) \beta \end{pmatrix} = \phi(1) \phi(2) \frac{1}{\sqrt{2}}(\alpha(1) \beta(2) - \beta(1) \alpha(2))$

Here, $\alpha, \beta$ are spin-state.

In general, the 2 particle system, the energy is

$\langle \Phi | H_1 |\phi \rangle = \langle \phi_a(1) | H_1|\phi_a(1) \rangle + \langle \phi_b(2) | H_1|\phi_b(2) \rangle$

$\langle \Phi | H_2 |\phi \rangle = \langle \phi_a(1) \phi_b(2) | G_{12}|\phi_a(1) \phi_b(2) \rangle - \langle \phi_a(1) \phi_b(2) | G_{12}|\phi_b(1) \phi_a(2) \rangle$

When $a = b$, the direct term and exchange term cancelled. Thus, the “mean-field” is simply the Coulomb potential. Therefore, the method in that post is kind of getting around.

Some people may found that the Fock operator is

$\displaystyle F_i = H_i + \frac{1}{2} \sum_{j} \left(2 J_{ij} - K_{ij} \right)$

In this way, the direct term for $i = j$ will not be cancelled.  In the case of 2 particle, the mean field is the Coulomb potential plus the average mutual interaction from the other particle.