## Physics behind Woods-Saxon energy levels

In the last post, I hope I explained how to find the Woods-Saxon energy levels for a given parameters. I just searched the best fit Woods-Saxon parameters to best fit the neutron single-particle energy for  209Pb. This is a double magic nucleus, and the there is no large fragmentation for the neutron excitation energy, thus, the outermost neutron in this nucleus can be considered as a good single-particle.

The best fit parameters are The rms difference is just 78 keV!! For each level, the difference is not more than 50 keV!

So, how can we interpret this fitting result? It means, the energy levels can be well explained by WS mean field.

Recall that, the mean field actually included a lot things, it is the effective single particle potential that a nucleon is feeling. When we pushing the rms value to be minium. We are actually finding the best mean field. And the difference is due to the residual interaction.

But what contained in the mean field and residual interaction? The total Hamiltonian is $\displaystyle H = H_{N} + H_{NN} + H_{NNN} + ....$

the mean field approach is to add an artificial potential $V$ $\displaystyle H = (H_{N} + V) + (H_{NN} + H_{NNN} +... - V) = H_{m} + H_{R}$

such that $latex H_{R}$ is minimum.

And by fitting energy levels using Woods-Saxon energy levels, we are basically doing the same thing! And the mean field is explicitly containing 2-body force, 3N-force and so on. So, can we say “because the Woods-Saxon can explain the energy level very well, the tensor force or other force is insignificant” ? the answer is no. Because the Woods-Saxon potential explicitly contains tensor force and other force.

## 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.

## 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 $U$