## Correlation energy

The correlation is from the off-diagonal terms of the residual interaction, which is the TBME of the shell model. Consider following Hamiltonian of these nuclei

$H(Z,N-1) = H(Z,N-2) + h_n + R_n$

$H(Z,N) = H(Z,N-2) + h_{n_1} + h_{n_2} + R_{n_1} + R_{n_2} + R_{nn}$

the correlation energy between the two neutrons is the term $R_{nn}$. Be aware that this is a residual interaction, not the nucleon-nucleon interaction $V_{ij}$.

$V_{ij}= \begin{pmatrix} V_{11} & V_{12} & V_{1C} \\ V_{21} & V_{22} & V_{2C} \\ V_{C1} & V_{C2} & V_{CC} \end{pmatrix}$

Thus,

$R_{n} = \begin{pmatrix} V_{11} - U_1 & V_{1C} \\ V_{C1} & V_{CC}-U_C \end{pmatrix}$,

where $U_i$ is mean field.

The separation energies is proportional to the terms

$S_n(Z,N-1) \sim h_n + V_n$

$S_{2n}(Z,N)\sim h_{n_1} + h_{n_2} + V_{n_1} + V_{n_2} + V_{nn}$

Thus, the neutron-neutron correlation energy is

$\Delta_{pn}(N,Z) = 2*S_n(Z,N-1) - S_{2n}(N,Z)$

For 18O, $S_n(^{17}O) = 4.1431$ MeV, and $S_{2n}(^{18}O) = 12.1885$ MeV, thus, $\Delta_{2n}(^{18}O) = 3.9023$ MeV.

In the shell model calculation, the single particle energy of the 1d5/2 and 2s1/2 neutron are -4.143 MeV and -3.27 MeV respectively. The residual interaction is

$V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

The eigenenergy of the ground state of 18O is -10.539 on top of 16O.

The non-correlated binding energy of 18O is  2*-4.143 = -8.286 MeV.

Therefore, the theoretical correlated energy is 2.258 MeV.

## Nuclear correlation & Spectroscopic factor

In the fundamental, correlation between two objects is

$P(x,y) \neq P(x) P(y)$

where $P$ is some kind of function. To apply this concept on nuclear physics, lets take a sample from 18O. 18O can be treated as 16O + n + n. In the independent particle model (IPM), the wave function can be expressed as

$\left|^{18}O\right>= \left|^{16}O\right>\left|\phi_a\right>\left|\phi_b\right>= \left|^{16}O\right>\left|2n\right>$

where the wave function of the two neutrons is expressed as a direct product of two IPM eigen wave functions, that they are un-correlated. Note that the anti-symmetry should be taken in to account, but neglected for simplicity.

We knew that IPM is not complete, the residual interaction has to be accounted. According to B.A. Brown, Lecture Notes in Nuclear Structure Physics [2011], Chapter 22, the 1s1/2 state have to be considered. Since the ground state spins of 18O and 16O are 0, thus, the wavefunction of the two neutrons has to be spin 0, so that only both are in 1d5/2 or 2s1/2 orbit. Thus, the two neutrons wave function is

$\left| 2n \right> = \alpha \left|\phi_1\right>\left|\phi_1\right>+\beta \left|\phi_2\right>\left|\phi_2\right>$

when either $\alpha$ or $\beta$ not equal 0, thus, the two neutrons are correlation. In fact, the $\alpha = 0.87$ and $\beta = 0.49$.

The spectroscopic factor of the sd-shell neutron is the coefficient of $\alpha$ times a isospin-coupling factor.

From the above example, the correlation is caused by the off-diagonal part of the residual interaction. To be more specific, lets take 18O as an example. The total Hamiltonian is

$H_{18} = H_{16} + h_1 + h_2 + V$

where $h_1 = h_2 = h$ is the mean field or single particle Hamiltonian

$h\left|\phi_i\right>= \epsilon_i\left|\phi_i\right>$

since $H_{16}$ is diagonal and not excited (if it excited, then it is called core polarization in shell model calculation, because the model space did not included 16O.), i.e. $H_{16} = \epsilon_{16} I$, we can neglect it in the diagonalization of the $h_1+h_2 + V$ and add back at the end. In the 1d5/2 and 2s1/2 model space, in order to form spin 0, there is only 2 basis,

$\left|\psi_1\right> = \left|\phi_1\right>\left|\phi_1\right>$ and

$\left|\psi_2\right> = \left|\phi_2\right>\left|\phi_2\right>$

express the Hamiltonian in these basis,

$V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

Because of the diagonalization, the two states $\left|\psi_1\right>$ and$\left|\psi_2\right>$ are mixed, than the two neutrons are correlated. This is called configuration mixing.

According to B.A. Brown,

the configuration mixing on the above is long-ranged correlation (LRC). It is near the Fermi surface and the energy is up to 10 MeV.

The short-ranged correlation (SRC) is caused by the nuclear hard core that scattered a nucleon to highly single particle orbit up to 100 MeV.

The LRC is included in the two-body-matrix element. The SRC is included implicitly through re-normalization of the model space.

There is a correlation due to tensor force. Since the tensor force is also short-ranged, sometimes it is not clear what SRC is referring from the context. And the tensor force is responsible for the isoscalar pairing.

The discrepancy of the experimental spectroscopic factor and the shall model calculation is mainly caused by the SRC.

## Fermi and Gamow-Teller Transition

The beta decay is caused by the weak interaction. The weak interaction is very short range, because the mediate particles, the $W^{\pm}$ and $Z^0$ bosons are 80 GeV and 91 GeV respectively. The effective range is like $10^{-3}$ fm. So, the interaction can assumed to be a delta function and only the coupling constant matter. The Fermi coupling constant is

$1.17 \times 10^{-11} (\hbar c)^2~ \mathrm{MeV^2}$

The fundamental process of beta decay is the decay of quark.

$\displaystyle u \xrightarrow{W^+} d + e^+ + \nu_e$

Since a pion is made from up and down quark, the decay of pion into position and electron neutrino is also due to weak interaction.

The Hamilton of the beta decay is

$\displaystyle H_w(\beta^{\pm})=G_V \tau_{\mp} + G_A \sigma \tau_{\mp}$

where $G_V$ is the vector coupling constant, the term is called Fermi transition. The $\tau_{\pm}$ is the isospin ladder operator. The beta+ decay changes the isospin from +1/2 (neutron) to -1/2 (proton). The $G_A$ is the axial coupling constant, the term is called Gamow-Teller transition. $\sigma$ is spin operator. Because of this operator, the Gamow-Teller transition did not preserve parity.

The $G_A$ is different from $G_V$, which is caused by the effect of strong interaction. The Goldberger-Trieman relation

$\displaystyle g_A = \frac{G_A}{G_V} = \frac{f_\pi g_{\pi N}}{M_N c^2} = -1.3$

where $f_\pi \sim 93~\textrm{MeV}$ is the pion decay constant. $g_{\pi N} \sim 14 \times 4\pi$ is the coupling constant between pion and nucleon.  This, we can see the effect of the strong interaction, in which pion is the meson for strong nuclear force.

The transition probability can be estimated by Fermi-Golden rule

$\displaystyle W(p_e)=\frac{2\pi}{\hbar}|\left< \psi_f|H|\psi_0\right> |^2 \rho(E_f)$

the final state wavefunction

$\displaystyle \left|\psi_f\right> = \frac{1}{\sqrt{V}} e^{ik_e r} \frac{1}{\sqrt{V}} e^{ik_{\nu}r} \left|j_f m_f\right>$

$\displaystyle e^{ikr} = \sum \limits_{L}\sqrt{4\pi (2L+1)} i^L j_L(kr) Y_{L0}(\theta)$

using long wavelength approximation, the spherical Bessel function can be approximated by the first term.

$\displaystyle j_L(kr) \sim \frac{(kr)^L}{(2L+1)!!}$

$\displaystyle \left| \psi_f\right>=\frac{1}{V}(1 + i \sqrt{\frac{4\pi}{3}} Y_{10} + ...) \left|j_f m_f\right>$

The first term 1, or L=0 is called allowed decay, so that the orbital angular momentum of the decayed nucleus unchanged. The higher order term, in which the weak interaction have longer range has very small probability and called L-th forbidden decay.

The density of state is

$\displaystyle \rho(E_f) = \frac{V}{2\pi^2 \hbar^7 c^3} F(Z,E_e)p_e^2 (E_0-E_e) ( (E_0-E_e)^2-(m_{\nu} c^2)^2)^2$

where the $F(Z, E_e)$ is the Fermi function.

The total transition probability is the integration with respect to the electron momentum.

$\displaystyle W = \int W(p_e) dp_e = \frac{m_e^5 c^4}{2 \pi^3 \hbar^7} f(Z,E_0) |M|^2$

where $f(Z,E_0)$ is the Fermi integral. The half-life

$\displaystyle T_{1/2} = \frac{\ln{2}}{W}$

To focus on the beta decay from the interference of the density of state, the ft-value is

$\displaystyle ft = f(Z,E_0) T_{1/2} =\frac{2\pi^3\hbar^7}{m_e^5 c^4} \frac{\ln{2}}{|M|^2}$

The ft-value could be difference by several order.

There is a super-allowed decay from $0^{+} \rightarrow 0^{0}$ with same isospin, which the GT does not involve. an example is

$\displaystyle ^{14}\mathrm{O} \rightarrow ^{14}\mathrm{N} + e^+ + \nu_e$

The ft-value is 3037.7s, the smallest of known.

Fermi Gamow-Teller
$\Delta S=0$ $\Delta S=1$
$J_f=J_i + L$ $J_f=J_i + L+1$
$T_f=T_i + 1$

transition L $\log_{10} ft_{1/2}$ $\Delta J$ $\Delta T$ $\Delta \pi$
Fermi GT
Super allowed 3.1 ~ 3.6 $0^+ \rightarrow 0^+$ not exist 0 no
allowed 0 2.9 ~ 10 0 (0), 1 0, 1 ; $T_i=0\rightarrow T_f=0$ forbidden no
1st forbidden 1 5 ~ 19 (0),1 0, 1, 2 0,1 yes
2nd forbidden 2 10 ~18 (1), 2 2, 3 no
3rd forbidden 3 17 ~ 22 (2), 3 3, 4 yes
4th forbidden 4 22 ~ 24 (3), 4 4, 5 no

The () means not possible if either initial or final state is zero. i.e $1^{-} \rightarrow 0^+$ is not possible for 1st forbidden.

## 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.

## a GEANT4 Simulation

The GEANT4 program structure was borrow from the example B5. I found that the most confusing part is the Action. Before that, let me start with the main().

GEANT4 is a set of library and toolkit, so, to use it, basically, you add a alot GEANT4 header files on your c++ programs. And, every c++ program started with main(). I write the tree diagram for my simplified exampleB5,

main()
+- DetectorConstruction.hh
+- Construct()
+- HodoscopeSD.hh     // SD = sensitive detector
+- HodoscopeHit.hh //information for a hit
+- ProcessHits()   //save the hit information
+- ConstructSDandField() //define which solid is SD
+- ConstructMaterials()  //define material
+- PhysicsList.hh  // use FTFP_BERT, a default physics for high energy physics
+- ActionInitialization.hh
+- PrimaryGeneratorAction.hh // define the particle gun
+- RunAction.hh // define what to do when a Run start, like define root tree
+- Analysis.h  // call for g4root.h
+- EventAction.hh //fill the tree and show some information during an event

A GEANT4 program contains 3 parts: Detector Construction, Physics, and Action. The detector construction is very straight forward. GEANT4 manual is very good place to start. The physics is a kind of mystery for me. The Action part can be complicated, because there could be many things to do, like getting the 2ndary beam, the particle type, the reaction channel, energy deposit, etc…

Anyway, I managed to do a simple scattering simulation, 6Li(2mm) + 22Ne(60A MeV) scattering in vacuum. A 100 events were drawn. The detector is a 2 layers plastic hodoscope, 1 mm for dE detector, 5 mm for E detector. I generated 1 million events. The default color code is Blue for positive charge, Green for neutral, Red for negative charge. So, the green rays could be gamma or neutron. The red rays could be positron, because it passed through the dE detector.

The histogram for the dE-TOF is

We can see a tiny spot on (3.15,140), this is the elastics scattered beam, which is 20Ne. We can see 11 loci, started from Na on the top, to H at the very bottom.

The histogram of dE-E

For Mass identification, I follow the paper T. Shimoda et al., Nucl. Instrum. Methods 165, 261 (1979).

I counted the 20Na from 0.1 billion beam, the cross section is 2.15 barn.

## [Pol. p Target] resume experiment

after a long summer break due to lab maintain.

we check the system to see weather it gives same results before.

the Hall probe reading is weird, so we calibrate it with water NMR signal.

and we redo the 30% laser duty, found that it is small, much smaller then expected.

===============================

after discussion with my professor on my PhD topic.

i like to study the spin by nuclear scattering experiment. the polarized spin target is a good spin detector.

one possibility is on the EPR paradox and the Bell’s inequality. my professor gave me a PhD thesis on proton-neutron spin experiment on EPR.

another possibility is on the localized special relativity and quantum entanglement. since these two are strongly related by spin. my professor gave me a book on spin statistic about that.

another unclear way is through the study of spin group, Lorentz group and Mobius group. by some transformation, a 3D rotation can transform into a 2X2 matrix and then reveal that spin can have classical picture with the help of complex number. that is a suggestion that L, the orbital momentum, and S, the spin, may be the same thing. moreover, the mathematical structure of L and S are the same for s=1. can we find a counterpart of l=1/2???

## β-delayed proton emission branches in 43Cr

β-delayed proton emission means, a positron emitted firstly, then proton emit. so, a proton converted to a neutron and a positron, then  a proton emit. as a result, the mass number reduced by 1 from lost 2 proton and gain 1 neutron. therefore, this process should be only happened in neutron-deficient nuclei.

Chromium-43 has atomic number 24, the neutron number is 19.

it was reported that a first time observation of β-delayed 3 proton decay in 45Fe. and the same decay was discovered in 43Cr in this paper. both process was recorded by an Imaging Time Projection Chamber.

the feature of this chamber is, it can capture the photos for decay process. in the paper, they shows a clear picture of 3-protons decay. and also, the measurement is very accurate. they can found 12 events of 3 protons decay among the total 12524 events of proton decay, which is about 1044 to 1. base on this precision, they deduced the relative branching ratio to be 91.8% for 1-proton case, 8.1% for 2 protons case, and 0.096% of 3 protons case.

by the chamber, they also recorded non-proton decay events, which may come from non-decay or β-deacy. the energy of β-decay is too small to detected.

However, they use the Maximum likelihood method to deduce the decay probability of β-decay. they found that it was 12%. since 2 β-decay has not been observed or either not possible, they deduced the absolute branching ratio.

a discussion on the extreme small branching ration of 3-protons decay was presented at the end of paper.