## Hartree-Fock for 2-body

Long time ago, we talked about the mean-field calculation, an touched Hartree-Fock method. In that time, we explained excited-state approach. Now, we explain another approach by variation of the wave functions. This approach is inevitable in atomic physics, because the potential is fixed.

The Hamiltonian is

$H = H_1 + H_2 + V_{12}$

Since the spin component is anti-parallel, the space part of the total wave function is

$\Psi = \phi_1(r_1) \phi_2(r_2)$

The Schrodinger equation is

$H\Psi = E\Psi$

Integrate both side with $\phi_1(r_1)$

$\int dr_1 \phi_1(r_1) H \Psi = \int dr_1 \phi_1(r_1) E \Psi$

using the normalization and define $\left = \phi_1(r_1)$

$\left<1|H|1\right> \phi_2(r_2) = E \phi_2(r_2)$

similarly

$\left<2|H|2\right> \phi_1(r_1) = E \phi_1 (r_1)$

expand $H = H_1 + H_2 + V_{12}$

$(H_2 + \left<1|V_{12}|1\right>) \phi_2(r_2) = (E - \left<1|H_1|1\right>) \phi_2(r_2)$

$(H_1 + \left<2|V_{12}|2\right>) \phi_1(r_1) = (E - \left<2|H_2|2\right>) \phi_1(r_1)$

Now, we have 2 equations,  an initial guess of $\Psi = \phi_1(r_1) \phi_2(r_2)$,

The difficulty is that, the $\left<1|V_{12}|1\right>$ contains 2 variables.

## From Mean field calculation to Independent particle model

The independent particle model (IPM) plays a fundamental role in nuclear model. The total Hamiltonian of a nucleus is:

$H = \sum{\frac{P_i^2}{2m_i}} + \sum_{i,j}{V_{ij}}$

The mean field is constructed by Hartree-Fork method, so that, the total Hamiltonian,

$H = \sum{h_i} + \sum_{i,j}{V_{ij} - U_i}, h_i = \frac{P_i^2}{2m_i} + U_i$

The second term is called residual interaction. The residual should be as small as possible. Under the mean filed, each nucleon can be treated as independent particle model that experience by $h_i$, so that,

$h_i \left|\phi_i\right> = \left|\phi_i\right> \epsilon_i,$

where $\epsilon_i$ is the single particle energy, and $\left|\phi_i\right>$ is the single particle wave function.

Hartree-Fock method

Start with a trial single particle function $\phi_i$, construct a first trial total wavefuction $\Psi_0$

$\Psi_0=\frac{1}{\sqrt{A!}} \left| \begin{array}{ccc} \phi_1(r_1) & \phi_1(r_2) & ... \\ \phi_2(r_1) & \phi_2(r_2) & ... \\ ... & ... & \phi_A(r_A) \end{array} \right|$

where $A$ is the number of particle. Using variation method,

$\delta \left<\Psi_0|H|\Psi_o\right> = 0 \Rightarrow \left<\delta \Psi_0|H|\Psi_0\right>$

$H = \sum\limits_{i} h_i + \sum\limits_{i\neq j} V_{ij}$

we don’t need to variate the ket and bar, since they are related.

The variation can be made on

1. $\phi_i$ or
2. the particle-hole excitation.

For the particle-hole excitation,

$\left|\delta \Psi_0\right> = \sum \eta_{kt} \left|\Psi_{kt}\right>$

$\left|\Psi_{kt}\right>=\left|1,2,...,t-1, t+1,...,A,k\right>$

Then, the variation of the energy becomes,

$\sum \eta_{kt} \left< \Psi_{kt}|H| \Psi_0 \right> =0$

$\left<\Psi_{kt}|H|\Psi_0\right> = \left+ \sum\limits_{r} \left< kr | V_{ij}| rt \right>=0$

we used the one-body expression for the two-body interaction,

$\sum \limits_{ij} V_{ij} = \sum \limits_{\alpha \beta \gamma \delta} \left|\alpha \beta\right> V_{\alpha \beta \gamma \delta} \left<\gamma \delta\right|$

$V_{\alpha \beta \gamma \delta} = \left<\alpha \beta |V_ij |\gamma \delta\right>$

Thus, we take out $\left and $\left|t\right>$, we get the Hartree-Fock single particle Hamiltonian,

$h_{HF} = h + \sum \limits_{r} \left< r | V_{ij} | r \right>$

using this new single particle Hamiltonian, we have a better single particle wavefunction

$h_{HF} \phi_i^1 = \epsilon_i^1 \phi_1^1$

$U = \sum \limits_{i} h_{HF}$

with this new wavefunction, the process start again until convergence. After that, we will get a consistence mean field (in the sense that the wavefunction and the potential are consistence), the single particle energy and the total binding energy.

I am not sure how and why this process can minimized the mean field $U$