Following the previous post, I tested my understanding on the Hartree method. Now, I move to the Hartree-Fock method. The “mean field” energy of the Hartree-Fock is

I also use the same method, in which, the trial wave function is replaced every iteration and the integration is calculated when needed.

Since the total wave function must be anti-symmetry under permutation, therefore one state can be occupied by only one particle. Thus, if we use the ground state in the mean field, the “meaningful” wave functions are the other states.

It is interesting that the mean field energy is zero when , the consequence is no mean field for the same state. Suppose the mean field energy is constructed using the ground state, and we only use 3 states, the direct term is

The exchange term is

Due to the symmetry of the mutual interaction. We can see that some off-diagonal terms are cancelled. For example,

These two integrals are the same. In fact,

$latex \langle \alpha \nu | \beta \nu \rangle – \langle \alpha \nu | \nu \beta \rangle = $

whenever

We can see, when interchange , the direct term and the exchange term are identical, and then the mean field energy is zero. Also, when the mean field energy is also zero.

Due to the zero mean field, the off-diagonal terms of the Hamiltonian with or are zero. Then, the eigen energy is the same as the diagonal term and the eigen vector is un-change.

Back to the case, the direct matrix at the 1st trial is,

The exchange matrix is

Thus, the Fock matrix is

Therefore, the eigen states are the basis, unchanged

Only the eigen energies are changed, as , ,

The total wave function for 2 particles at state 1 and state-\mu is

I found that the “mean field” function is not as trivial as in the Hartree case, because of the exchange term. In principle, the mean field for particle-i at state- is,

However, the direct term is multiply with , but the exchange term is multiply with , which are two different functions. i.e., the “mean field” is affecting two functions, or the “mean field” is shared by two states.

Although the mean field for single state can be defined using exchange operator symbolically, but I don’t know how to really implement into a calculation. Thus, I don’t know how to cross-check the result.