## 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

## Electromagnetic multi-pole moment

Electromagnetic multipole comes from the charge and current distribution of the nucleons.

Magnetic multipole in nucleus has 2 origins, one is the spin of the nucleons, another is the relative orbital motion of the nucleons.  the magnetic charge or monopoles either not exist or very small. the next one is the magnetic dipole, which cause by the current loop of protons.

Electric multipole is solely by the proton charge.

From electromagnetism, we knew that the multipole has  different radial properties, from the potential of the fields:

$\displaystyle \Psi(r) = \frac{1}{4\pi\epsilon_0} \int\frac{\rho(r')}{|r-r'|}d^3r'$

$\displaystyle A(r) = \frac{\mu_0}{4\pi}\int\frac{J(r')}{|r-r'|}d^3r'$

and expand them into spherical harmonic by using:

$\displaystyle \frac{1}{|r-r'|} = 4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l} \frac{1}{2l+1}\frac{r_{<}^l}{r_>^{l+1}} Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)$

we have

$\displaystyle \Psi(r) = \frac{1}{\epsilon_0} \sum_{l,m}\frac{1}{2l+1}\int Y_{lm}^*(\theta',\phi') r'^l\rho(r')d^3r' \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$

$\displaystyle A(r)=\mu_0 \sum_{l,m}\frac{1}{2l+1}\int Y_{lm}^*(\theta',\phi') r'^l J(r') d^3r' \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$

we can see the integral give us the required multipole moment. the magnetic and electric are just different by the charge density and the current density. we summarize in this way :

$q_{lm} = \int Y^*_{lm}(\theta',\phi') r'^l \O(r') d^3 r'$

where O can be either charge or current density. The l determine the order of multipole. and the potential will be simplified :

$M(r)=\sum_{l,m}\frac{1}{2l+1} q_{lm} \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}$

were M can be either electric or magnetic potential, and i dropped the constant. since the field is given by 1st derivative, thus we have:

1. monopole has $1/r^2$ dependence
2. dipole has $1/r^3$
3. quadrapole has $1/r^4$
4. and so on

The above radial dependences are same for electric or magnetic. for easy name of the multipole, we use L-pole, which L can be 0 for monopole, 1 for dipole, 2 for quadrapole, etc.. and we use E0 for electric monopole, M0 for magnetic monopole.

Since the nucleus must preserver parity, and the parity for electric and magnetic moment are diffident.the different come from the charge density and current density has different parity. The parity for charge density is even, but for the current density is odd. and $1/r^2$ has even parity, $1/r^3$ has odd parity. therefore

• electric L-pole — $(-1)^{L}$
• magnetic L-pole — $(-1)^{L+1}$

for easy compare:

• E0, E2, E4… and M1,M3, M5 … are even
• E1,E3,E5…. and M0, M2, M4…. are odd

The expectation value for L-pole, we have to calculate :

$\int \psi^* Q_{lm} \psi dx$

where $Q_{lm}$ is multipole operator ( which is NOT $q_{lm}$), and its parity is follow the same rule. the parity of the wave function will be canceled out due to the square of itself. thus, only even parity are non-Zero. those are:

• E0, E2, E4…
• M1,M3, M5 …

that make sense, think about a proton orbits in a circular loop, which is the case for E1, in time-average, the dipole momentum should be zero.