## 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.