## Wavelet Analysis or MRA

Although the Fourier transform is a very powerful tool for data analysis, it has some limit due to lack of time information. From physics point of view, any time-data should live in time-frequency space. Since the Fourier transform has very narrow frequency resolution, according to  uncertainty principle, the time resolution will be very large, therefore, no time information can be given by Fourier transform.

Usually, such limitation would not be a problem. However, when analysis musics, long term performance of a device, or seismic survey, time information is very crucial.

To over come this difficulty, there a short-time Fourier transform (STFT) was developed. The idea is the applied a time-window (a piecewise uniform function, or Gaussian) on the data first, then FT. By applying the time-window on difference time of the data (or shifting the window), we can get the time information. However, since the frequency range of the time-window  always covers the low frequency, this means the high frequency  signal is hard to extract.

To improve the STFT, the time-window can be scaled (usually by 2). When the time window is shrink by factor of 2, the frequency range is expanded by factor of 2. If we can subtract the frequency ranges for the time-window and the shrink-time-window, the high frequency range is isolated.

To be more clear, let say the time-window function be

$\phi_{[0,1)}(t) = 1 , 0 \leq t < 1$

its FT is

$\hat{\phi}(\omega) = sinc(\pi \omega)$

Lets also define a dilation operator

$Df(t) = \sqrt{2} f(2t)$

the factor $\sqrt{2}$ is for normalization.

The FT of $D\phi(t)$ has smaller frequency range, like the following graph.

We can subtract the orange and blue curve to get the green curve. Then FT back the green curve to get the high-frequency time-window.

We can see that, we can repeat the dilation, or anti-dilation infinite time. Because of this, we can drop the FT basis $Exp(-2\pi i t \omega)$, only use the low-pass time-window to see the low-frequency behaviour of the data, and use the high-pass time-window to see the high-frequency behaviour of the data. Now, we stepped into the Multi-resolution analysis (MRA).

In MRA, the low-pass time-window is called scaling function $\phi(t)$, and the high-pass time-window is called wavelet $\psi(t)$.

Since the scaling function is craetd by dilation, it has the property

$\phi(t) = \sum_{k} g_{0}(k) \phi(2t-k)$

where $k$ is integer. This means the vector space span by ${\phi(t-k)}_{k}=V_0$ is a subspace of the dilated space $DV_0 =V_1$. The dilation can be go one forever, so that the whole frequency domain will be covered by $V_{\infty}$.

Also, the space span by the wavelet, ${\psi(t-k)}=W_0$, is also a subspace of $V_1$. Thus, we can imagine the structure of MRA is:

Therefore, any function $f(t)$ can also be expressed into the wavelet spaces. i.e.

$f(t) = \sum_{j,k} w_{j,k} 2^{j/2}\psi(2^j t - k)$

where $j, k$ are integers.

I know this introduction is very rough, but it gives a relatively smooth transition from FT to WT (wavelet transform), when compare to the available material on the web.

## Phase shift of elastics scattering

I found that the derivation of most “google result” is not clear enough. So here is my derivation. Before process, people may need to review the pervious post.

Most people start from,

$u_l = A( \hat{h}_l^- - S_l \hat{h}_l^+ )$

that annoying me because it is somehow not “natural”. Why there is a “minus” sign? Why the $\hat{h}_l^-$ is the first term? For my self, a more natural way is,

$u_l = a \hat{h}_l^+ + b \hat{h}_l^-$

but that is not so natural at all, because in numerical calculation, for simplicity, there is no complex number, we only have,

$u_l = \alpha \hat{j}_l + \beta \hat{n}_l$

The first term is alway there as it is the free solution and bounded at $r = 0$. the second term is caused by the potential.

The goal is to find a solution take the form

$\displaystyle \psi = A \left( e^{i \vec{k} \cdot \vec{r}} + f(\theta) \frac{e^{ikr}}{r} \right)$

where the first term is free wave and the second term is scattered wave. The solution for elastics scattering is

$\displaystyle \psi = \sum C_l P_l (\cos\theta) \frac{u_l}{kr} \rightarrow \sum C_l P_l(\cos\theta) (\alpha \hat{j}_l + \beta \hat{n}_l)$

we used the substitution,

$\displaystyle R_l(r) = \frac{u_l(\rho)}{\rho}, \rho = kr$.

The radial function can be solved using Rungu-Kutta method on the equation,

$\displaystyle \frac{d^2}{d\rho^2} u_l = \frac{2 m_\mu}{\hbar^2} (V-E) u_l + \frac{l(l+1)}{\rho^2}$

and the solution of $u_l$ at far away is,

$u_l \rightarrow \alpha \hat{j}_l + \beta \hat{n}_l$.

the arrow means $r \rightarrow \infty$. So, the problem is how to rewrite the solution. In the way, we will see how the phase shift or the S-matrix was found.

The free solution is the spherical wave,

$\displaystyle e^{i \vec{k} \cdot \vec{r}} = \sum_l (2l+1) i^l P_l(\cos\theta) j_l(kr)$

The spherical Bessel function $j_l(kr)$ cna be express as Heankel function

$h_l^{\pm} = n_l \pm i j_l \rightarrow e^{\pm i (kr - l \frac{\pi}{2})}$

The $+$ sign is outgoing wave.

$\displaystyle u_l \rightarrow (\alpha \hat{j}_l + \beta \hat{n}_l)$

$\displaystyle = \frac{\alpha}{2i} (\hat{h}_l^{+} - \hat{h}_l^{-}) + \frac{\beta}{2}(\hat{h}_l^{+} + \hat{h}_l^{-})$

$\displaystyle = \frac{\alpha + i \beta}{2i} (\hat{h}_l^{+} - \hat{h}_l^{-}) + \beta \hat{h}_l^{+}$

$\displaystyle = (\alpha - i \beta ) \left( \frac{\hat{h}_l^+ - \hat{h}_l^-}{2i} + \frac{\beta}{\alpha - i \beta} \hat{h}_l^+\right)$

$\displaystyle = (\alpha - i \beta ) \left( \hat{j}_l + \frac{\beta}{\alpha - i \beta} \hat{h}_l^+\right)$

Since the $u_l$ should be normalized, we can se $\alpha = \cos \delta$ and $\beta = \sin\delta$.

$\displaystyle \frac{\beta}{\alpha - i \beta } = \sin(\delta) e^{i\delta}$

We put $u_l$ back

$\displaystyle \psi \rightarrow \sum_l C_l P_l (cos\theta)(\alpha - i \beta ) \left( \hat{j}_l + \sin(\delta) e^{i\delta} \hat{h}_l^+\right)$

By setting

$\displaystyle C_l = A i^l \frac{2l+1}{\alpha - i \beta}$,

we have the first term is the free wave function. In the second term, $\hat{h}_l^+ \rightarrow e^{i(kr - l \frac{\pi}{2}}) / kr$. Notice that

$e^{i l \frac{\pi}{2}} = i^{-l}$

That cancel the $i^l$ term in $C_l$. And we have

$\displaystyle f(\theta) = \sum (2l+1) P_l (\cos\theta) \frac{\sin(\delta) e^{i\delta}}{k}$

some people will write the $u_l$ as $\hat{h}_l^{\pm}$ and the S-matrix,

$\displaystyle u_l = \frac{\alpha + i \beta} {2i} \hat{h}_l^+ - \frac{\alpha - i \beta}{2i} \hat{h}_l^-$

$\displaystyle = -\frac{\alpha - i \beta}{2i} \left( \hat{h}_l^- - \frac{\alpha + i \beta}{\alpha - i \beta} \hat{h}_l^+ \right)$

$\displaystyle = A' (\hat{h}_l^- - S_l \hat{h}_l^+)$

where

$\displaystyle S_l =\frac{\alpha + i \beta}{\alpha - i \beta} = e^{2i\delta}$.

Remember that this is the S-matrix for elastics scattering.

The time-independent Schrödinger equation is

$(-\frac{\hbar^2}{2m}\nabla^2 + V ) \Psi = E \Psi$

Using the Laplacian in spherical coordinate. and Set $\Psi = R Y$

$\nabla^2 R Y - \frac{2m}{\hbar^2}(V-E) R Y = 0$

$\nabla^2 = \frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d}{dr}) - \frac{1}{r^2} L^2$

The angular part,

$L^2 Y = l(l+1) Y$

$\frac{d}{dr}(r^2\frac{dR}{dr}) - l(l+1)R - \frac{2mr^2}{\hbar^2}(V-E) R = 0$

To simplify the first term,

$R = \frac{u}{r}$

$\frac{d}{dr}(r^2 \frac{dR}{dr})= r \frac{d^2u}{dr^2}$

A more easy form of the radial function is,

$\frac{d^2u}{dr^2} + \frac{l(l+1)}{r^2} u - \frac{2m}{\hbar^2} (V-E) u = 0$

The effective potential $U$

$U = V + \frac{\hbar^2}{m} \frac{l(l+1)}{r^2}$

$\frac{d^2u}{dr^2} + \frac{2m}{\hbar^2} (E - U) u = 0$

We can use Rungu-Kutta method to numerically solve the equation.

The initial condition of $u$ has to be 0. (home work)

I used excel to calculate a scattered state of L = 0 of energy 30 MeV. The potential is a Wood-Saxon of depth 50 MeV, radius 3.5 fm, diffusiveness 0.8 fm.

Another example if bound state of L = 0. I have to search for the energy, so that the wavefunction is flat at large distance. The outermost eigen energy is -7.27 MeV. From the radial function, we know it is a 2s orbit.

## A reminder for the use of Dirac notation and Translation

The momentum operator $\hat{P}$ acts on a position ket or bar is tricky.

$\left

$B A \hat{P}\left|x\right> = B i\frac{d}{dx} ( A \left|x\right>)$

I added $A$ and $B$ to emphasize the tricky point, where $A$ or $B$ can be an operator, a ket, a bar, or a variable. We can see, after the first quantization, the position bar or ket is always waiting for something and the derivative is differentiate the whole thing, not only the bar or ket. Miss use will result incorrect result. for example, the non-commute property of the position and the momentum operator,

$\left= -i\frac{d}{dx} (\left =-i\frac{d}{dx} (\left$

$=-i x \frac{d}{dx} (\left - i = x \left - i = \left$

If you use it wrongly, you may have something like

$\left = -i\frac{d\left$

or

$\left = -i\frac{d}{dx} (\left) = -i\frac{d}{dx}(x) = -i$

The translation operator.

In many text book, the translation operator is stated as

$D(\epsilon) = 1 - i \hat{P} \epsilon$,

where $\hat{P}$ is momentum operator or the generator of displacement.

*********************

Actually, there is a reason, notice that the translation operator is unitary operator,

$D(\epsilon)D^{\dagger}(\epsilon) = 1$

then , since any unitary operator can be written as

$D(\epsilon) = \exp(-i \epsilon \hat{Q}) \sim 1 - i \epsilon \hat{Q} +....$

where $\hat{Q}$ is a Hermitian operator.

But what is $\hat{Q}$ ? since $\hat{Q}$ is Hermitian, it has eigen ket and eigenvalue,

$\hat{Q} \left|q\right> = \left|q\right> q$.

The, we can check

$\left = \left = \left =\exp(-i \epsilon q) \left$

also

$\left = \left = \exp(-i \epsilon q) \left$

we set a wave function $\psi_q(x) = \left$

$\psi_q(x-\epsilon) = \exp(-i\epsilon q ) \psi_q(x)$

Then, we can have a derivative of the wave function

$\frac{d}{dx}\phi_q(x) =\lim_{\epsilon\to 0} \frac{\exp(i\epsilon q ) -1 }{\epsilon} \psi_q(x) = i q \psi_q(x)$

the solution is $\psi_q(x) = \exp(iqx)$, which is a plane wave with momentum q. Therefore, the operator $\hat{Q}$ is the momentum operator, and it is really a generator a translation on the wave function $\psi_q(x) \rightarrow \psi_q(x+\epsilon)$. To find the form of the momentum operator acting on position bar,

$\left = q \exp(-i q x) = -i \frac{d}{dx}( \exp(-iqx)) = -i\frac{d}{dx} (\left)$

$\left

*********************************

another mathematical way to do is, using Taylor series,

$D(\epsilon) \left|x\right> = \left|x+\epsilon\right> = \left|x\right> + \epsilon \frac{d}{dx}\left|x\right> + ...$

compare with the exponential, than

$\hat{Q}\left|x\right> = i\frac{d}{dx} ( \left|x\right>)$.

$\left = -i \frac{d}{dx}(\left) =\left q$.

The time propagation and angle rotation can also do in the same way. The rotation around z-axis is trivial, but a general rotation is not so easy in detail, but the idea can be generalized that the rotation generator is $\vec{\hat{J}}\cdot \vec{n}$.

Why the general translation operator is in exponential form? because it is a unitary operator. Why it is unitary? because it is symmetric that translate back and forth result no change. In general, if we have a quantity that can be “translate” and the translation is symmetry, than, the generator must take the same exponential form.

It is interesting that, the translation generator $\hat{Q}$ is a differential of the position $\alpha$, here $\alpha$ is a position of a general coordinate.  When I started QM, I always feel insecure that change of position like this way, because classically, the change of a position means it moves in time, and there is no time in the $\hat{P}$. However, look at the rotation generator. The same derivation (between the *****) is applicable on the rotation about z-axis that replace the $x \rightarrow \phi$, azimuth angle. Then, at the end, we will recognize the generator $\hat{Q} = -i\frac{d}{d\phi} = L_z$, which is an angular momentum operator on the z-axis, and classically in the sense that the $L_z = \frac{d}{d\phi}$ is known in EM in spherical coordinate (the wave equation). The key point is that the rotation on z-axis takes the form $\frac{d}{d\phi}$!

There is still some things need to clarify, like the time translation operator is unitary, but the time-reversal operator is anti-unitary. Also, the detail of the general rotation operator $\hat{D}_n(\epsilon) \left|\vec{r}\right> = \left|\vec{r'}\right>$. we know that the general rotation operator is a matrix form, How?

## Gamma Transition

The gamma decay brings a excited nucleus to a lower energy state by emitting a photon. Photon carry angular momentum with intrinsic spin of 1. Its parity is positive. Thus, we have three constrain from conversation laws immediately.

$E_f = E_i + \hbar \omega$

$J_f^{\pi_f} = J_i^{\pi_i} + L$

where $E$ is the energy of the nucleus, $\hbar \omega$ and $L$ are the photon energy and angular momentum respectively. We also have to consider the parities of electric and magnetic transition are different.

To calculate the transition rate, we can start from a classical equation of power emission from an antenna, since the photon energy is quantized, the transition rate [number of photon emitted per time] is the power divided by a photon energy.

$T(qL) = \frac{2(2L+1)}{\epsilon_0 L [(2L+1)!!]^2 \hbar} (\frac{\omega}{c})^{2L+1} B(qL)$

where $qL$ is the electromagnetic multipole with angular momentum $L$, and $B(qL)$ the the reduced transition probability, it is equal to the square of the magnitude of the transition matrix element $M_{fi}(qL)$.

In the electric transition, the multipole is

$qL = e r^L$

we assume the transition is conduced by a single nucleon and the rest of the nucleus is unaffected. The transition matrix element than can be written as

$M_{fi}(qL) = \left$

The single particle wave function can be written as

$\left|j m\right>= R_{nl}(r) [Y_l \times \chi_{1/2}]_{jm}$

The matrix elements becomes,

$M_{fi}(qL) = e \int_{0}^{\infty} R_{n_f l_f}^* r^L R_{n_i l_i} r^2 dr \times \left$

To evaluate the radial integral, we make another assumption that the nucleus is a sphere of uniform density with radius $R=r_0 A^{1/3}$,

$R_{nl}(r) = \frac{\sqrt{3}}{R^{3/2}}$, for $r, so that $\int_{0}^{R} |R_{nl}(r)|^2 r^2 dr = 1$

$\left=\frac{3}{R}\int_{0}^{R} r^{L+2} dr = \frac{3}{L+3} r_0^L A^{L/3}$

The reduced transitional probability

$B_{sp}(qL)=\sum \limits_{M m_f} |\left|^2$

$= e^{2} \left< r^{L} \right> ^{2} \sum \limits_{M m_f} \left$

the angular part could be assumed as $1/4\pi$ as the total solid angle is $4\pi$. Thus, with these three assumptions, we have the Weisskopf single particle estimation for the L-pole reduced electric transition probability

$B_W(EL) = \frac{1}{4\pi}(\frac{3}{L+3})^2 r_0^{2L} A^{2L/3} [e^2 fm^{2L}]$

For the magnetic transition, we have to take into account of the spin and orbit angular momentum. The single particle reduced transition probability

$B_{sp}(ML) = \sum \limits_{M m_f} |\left|^2$

the result,

$B_{sp}(ML)=L(2L+1) \left< r^{(L-1)} \right> ^2$

$\sum \limits_{M m_f} ((g_s - \frac{2g_l}{L+1}) \left< [ Y_{L-1} \times \vec{s} ]_{LM} \right> \left< [ Y_{L-1} \times \vec{j} ]_{LM} \right> )^2$

The term

$L(2L+1)(g_s \frac{2g_l}{L+1})^2 \sim 10$

and the rest of the angular part assumed to be $1/4\pi$ again, then

$B_W(ML) = \frac{10}{\pi}(\frac{3}{L+3})^2 r_0^{(2L-2)} A^{(2L-2)/3} \mu_N^2 fm^{2L-2}$

and notice that $\mu_N = e\hbar / (2m_p)$.

Some results can be deduced form the calculation

$B_{sp}(ML)/B_{sp}(EL) \sim 0.3 A^{-2/3}$

$B_{sp}(EL)/B_{sp}(E(L-1)) \sim \frac{1}{7} \times 10^7 A^{-2/3} E_\gamma^{-2}$

$T(E1) = 1.0 \times 10^{14} A^{2/3} E_\gamma^3$

$T(E2) = 7.3 \times 10^{7} A^{4/3} E_\gamma^5$

$T(E3) = 34 A^{2} E_\gamma^7$

$T(M1) = 3.1 \times 10^{13} A^{0} E_\gamma^3$

$T(M2) = 2.2 \times 10^{7} A^{2/3} E_\gamma^5$

$T(M3) = 10 A^{4/3} E_\gamma^7$

The deviation from the single particle limit indicates a strong collective state.

$0 \rightarrow 0$, forbidden

$1^+ \rightarrow 0^+$, M1

$2^+ \rightarrow 0^+$,  E2

## Shell model calculation and the USD, USDA, and USDB interaction

Form the mean field calculation, the single particle energies are obtained. However, the residual interaction is still there that the actual state could be affected. Because the residual interaction produces the off-diagonal terms in the total Hamiltonian, and that mixed the single particle state.

The Shell Model calculation can calculate the nuclear structure from another approach. It started from a assumed nuclear Hamiltonian, with a basis of wavefunctions. The Hamiltonian is diagonalized with the basis, then the eigenstates are the solution of the wavefunctions and the nuclear structure, both ground state and excited states. The basis is usually the spherical harmonic with some radial function. Or it could be, in principle, can take from the result of mean field calculation. Thus, the Shell Model calculation attacks the problem directly with only assumption of the nuclear interaction.

However, the dimension of the basis of the shell model calculation could be very huge. In principle, it should be infinitely because of the completeness of vector space. Fro practical purpose, the dimension or the number of the basis has to be reduced, usually take a major shell. for example the p-shell, s-d shell, p-f shell. However, even thought the model space is limited, the number of basis is also huge. “for $^{28}$Si the 12-particle state with M=0 for the sum of the $j_z$ quantum numbers and $T_z=0$ for the sum of the %Latex t_z\$ quantum numbers has dimension 93,710 in the m-scheme” [B. A. Brown and B. H. Wildenthal, Ann. Rev. Nucl. Part. Sci. 38 (1998) 29-66]. Beside the huge dimensions and the difficult for diagonalizing the Hamiltonian, the truncation of the model space also affect the interaction.

We can imagine that the effective interaction is different from the actual nuclear interaction, because some energy levels cannot be reached, for example, the short range hard core could produce very high energy excitation. Therefore, the results of the calculation in the truncated model space must be “re-normalized”.

There are 4 problems in the shell model calculation:

• the model space
• the effective interaction
• the diagonalization
• the renormalization of the result

The shell model can also calculate the excited state with $1\hbar \omega$ (1 major shell). This requires combination of the interactions between 2 major shell.

For usage, say in the code OXBASH, user major concern is the choice of the interaction and model space. The shell model are able to calculate

• The binding energy
• The excitation energies
• The nucleons separation energies
• The configuration of each state
• The magnetic dipole matrix elements
• The Gamow-Teller (GT) transition
• The spectroscopic factor
• …… and more.

The W interaction (or the USD) for the s-d shell was introduced by B.H. Wildenthal around 1990s. It is an parametric effective interaction deduced from fitting experimental energy levels for some s-d shell nuclei. Before it, there are some theoretical interactions that require “no parameter”, for example the G-matrix interaction is the in-medium nucleon-nucleon interaction.

The problem for the USD interaction is the interpretation, because it is a black-box that it can reproduce most of the experimental result better than theoretical interactions, but no one know why and how. One possible way is translate the two-body matrix elements (TBME) back to the central, spin-orbit, tensor force. It found that the central and spin-orbit force are similar with the theoretical interactions, but the tensor force could be different. Also, there could be three-body force that implicitly included in the USD interaction.

In 2006, B.A. Brown and W.A. Richter improved the USD interaction with the new data from the past 20 years [B.A. Brown, PRC 74, 034315(2006)]. The new USD interaction is called USDA and USDB. The difference between USDA and USDB is the fitting (something like that, I am not so sure), but basically, USDA and USDB only different by very little. Since the USDB has better fitting, we will focus on the USDB interaction.

The single particle energy for the USDB is

• $1d_{3/2} = 2.117$
• $2s_{1/2} = -3.2079$
• $1d_{5/2} = -3.9257$

in comparison, the single particle energies of the neutron of 17O of $2s_{1/2} = -3.27$ and $1d_{5/2} = -4.14$.

Can to USD interaction predicts the new magic number N=16?

Yes, in a report by O. Sorlin and M.-G. Porquet (Nuclear magic numbers: new features far from stability) They shows the effective single particle energy of oxygen and carbon using the monopole matrix elements of the USDB interaction. The new magic number N=16 can be observed.

## 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 $U$