## Absolute polarization measurement by elastic scattering

The magnitude of proton polarization can be measured by NMR technique with a reference. Because the NMR gives the free-induction decay signal, which is a voltage or current. For Boltzmann polarization using strong magnetic field and low temperature, the polarization can be calculated. However, when a reference point is not available, the absolute magnitude of proton polarization can be measured using proton-proton elastic scattering. The principle is the nuclear spin-orbital coupling. That creates left-right asymmetry on the scattering cross section.

Because of spin-orbital interaction:

$V_{ls}(r) = f(r) \vec{l} \cdot \vec{s} ,$

where $f(r)$ is the distance function, $\vec{l}$ is the relative angular momentum, $\vec{s}$ is the spin of the incident proton. In the following picture, the spin of the incident proton can be either out of the plane ($\uparrow$ ) or into the plan ($\downarrow$). When the proton coming above, the angular momentum is into the plane ($\downarrow$). The 4 possible sign of the spin-orbital interaction is shown. We can see, when the spin is up, the spin-orbital force repulses the proton above and attracts the proton below. That creates an asymmetry in the scattering cross section.

The cross section is distorted and characterized using analysing power $A_y$. Analyzing power is proportional to the difference between left-right cross-section. By symmetry (parity, time-reversal) consideration, $A_y = 1 + P sin(2\theta)$ (why?), in center of mass frame. In past post, the transformation between difference Lorentz frame. The angle in the $A_y$ has to be expressed in lab angle. The cross section and $A_y$ can be obtained from http://gwdac.phys.gwu.edu/ .

In scattering experiment, the number of proton (yield) is counted in left and right detectors. The yield should be difference when either proton is polarized. The yield is

$Y(\theta, \phi) = L \epsilon \sigma_0 (1 + cos(\phi)A_y(\theta) P) ,$

where $L$ is the luminosity, $\epsilon$ is the detector efficiency, $\sigma_0$ is the integrated cross-section of un-polarized beam and target of the detector, $P$ is the polarization of either the target or beam. When both target and the beam are polarized, the cross section is

$\sigma = \sigma_0 (1 + (P + P_T)A_y + P P_T C_yy),$

where $C_yy$ is spin-spin correlation due to spin-spin interaction of nuclear force.

Using the left-right yield difference, the absolute polarization of the target or the beam can be found using,

$\displaystyle A_y P = \frac{Y_L - Y_R}{Y_L + Y_R} ,$

where $Y_L = Y(\phi =0)$ and $Y_R = Y(\phi=\pi)$.

## 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.

## Meson Theory on Strong Nuclear Force

The theory is very difficult, I only state the result.

There are 4 main mesons for strong nuclear force.

type mass [MeV] range symmetry force type
$\pi$  135  long  pseudo scalar  $-\sigma_{1}\cdot\sigma_{2},-S_{12}$
$\sigma$  400 ~ 2000  medium  scalar $-1, -L\cdot S$
$\rho$  775  short  vector  $-2\sigma_1\cdot\sigma_2, +S_{12}$
$\omega$  783  short vector  $+1, -3L\cdot S$

## Projection theorem

The simplest way to say is:

a operator can be projected on another one, for example, The orbital angular momentum cab be projected on the total angular momentum.

$L = L\cdot J \frac {J}{j(j+1)}$

a simple application is on the Zeeman effect on spin-orbital coupling. the Hamiltonian is:

$H_B = - \mu \cdot B = - ( \mu_l L + \mu_s S ) \cdot B$

by the Wigner-Eckart theorem:

$L = L\cdot J \frac {J}{j(j+1)}$

$S = S\cdot J \frac {J}{j(j+1)}$

then the Hamiltonian becomes:

$H_B = - \frac{1}{j(j+1)} ( \mu_l (L \cdot J) + \mu_s (S \cdot J) ) J\cdot B$

and introduce the Bohr Magneton and g-factor:

$H_B = - g \mu_B J \cdot B$

$g = - \frac{1}{j(j+1)} ( g_l (L \cdot J) + g_s (S \cdot J) )$

## magic number

we knew that for some atoms are more stable that others. like He, Ne, Ar, etc, which are belonged to noble gas. the reason for they are non-reactive is, there outer most electron shell is filled out.

similar things happened in nuclei. in the shell model of nuclei, protons and neutrons just like the electrons in atom. if the outer most shell of proton or neutron is filled out, the nuclei is very stable. and we called this number of proton or neutron be MAGIC NUMBER.

the first magic number is 2. nuclei with 2 protons is more stable then others. However, if only have 2 protons, with out neutron, it is very unstable because of coulomb force. and 2 neutron also unstable, if without a proton. the only stable 2 nucleons state is deuteron. If there are 2 protons and 2 neutrons, we called this double magic number, and this nuclei, which is He is very very stable.

the list of magic number is 2, 8, 20, 28, 50, 82, 126 in theory prediction.

however, when the nuclei become heavier and heavier, the stability of nuclei in the magic number lost. to understand this. we have to know that the magic number is come from the large spin-orbital coupling term in the Hamiltonian of the nuclei. and recent research suggest that, the spin-orbital coupling may change by the number of nucleons.