## Hartree-Fock and Mean-field

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.

## Exchange energy

In QM, when treating a system with identical particles, the exchange term raised because the system must be the same after particle exchanged.

I am reluctant to use the term “exchange interaction” or “exchange force”, “exchange potential”, because there is no interaction, no force carrier, no potential at all.

Suppose the Hamiltonian for single fermion is

$H\psi = \epsilon \psi$

The 2-particle system, the Schrodinger equation is

$H_T \Psi = H_1 + H_2 + H'= E \Psi$

where $H'$ is the interaction between the 2 particles.

Now, guessing the total wave function to be

$\Psi = a \psi_1(r_1) \psi_2(r_2) + b \psi_1(r_2)\psi_2(r_1) = a \Psi_n + b\Psi_e$

The first term is “normal”, the second term is “exchanged”. substitute to the equation

$a H_T \Psi_n + b H_T \Psi_e = a E \Psi_n + b E \Psi_e$

multiply both side with $\Psi_1$ or $\Psi_2$, and integrate $r_1, r_2$, we have

$\begin{pmatrix} J & K \\ K & J \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = E \begin{pmatrix} a \\ b \end{pmatrix}$

where

$J = \int \psi_1^*(r_1) \psi_2^*(r_2) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2$

$K = \int \psi_1^*(r_2) \psi_2^*(r_1) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2$

Here, we assumed $H'$ is symmetric against $r_1, r_2$. This is a fair assumption as $H'$ is the mutual interaction, and $H_1, H_2$ are independent.

The $J$ is the energy from the interaction of the particles.

The $K$ is the energy due to EXCHANGE of the particles.

As we can see, the energy $J = \epsilon_1 + \epsilon_2 + J'$.

The $H_T$ in $K$ can be reduced to $H'$ when the eigen wave function $\psi_1 , \psi_2$ are in difference orbits. When $\psi_1 = \psi_2$, $K = \epsilon_1 + \epsilon_2 + K'$.

Thus the individual energy can be subtracted, yield,

$\begin{pmatrix} J' & K' \\ K' & J' \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \Delta E \begin{pmatrix} a \\ b \end{pmatrix}$

The solution is

$\Delta E = J' + K', (a,b) = (1,1)$

This is the symmetric state. For fermion, the spin-part is anti-symmetric. This is the spin-singlet state.

$\Delta E = J' - K', (a,b) = (1,-1)$

This is the anti-symmetric state. For fermion, the spin-part is symmetric, or spin-triplet state.

It is worth to notice that, in the symmetric state, the energy is higher because the two particles can be as close as possible, that create large energy. In the opposite, in the anti-symmetric state, the two particle never contact each other, thus, the interaction energy reduced.

We can summarized.

The energy $J'$ is the shift of energy due to the mutual interaction.

But the system of identical particle subjects to the exchange symmetry. The exchanged wave function is also a state, that the total wave function has to include the exchanged wave function. This exchanged wave function creates the exchange term or exchange energy $K'$.

In the system of Fermion, the exchange energy is related to Pauli exclusion principle.