## 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.

## Woods-Saxon Shape

this is a collective model of the nuclei density vs radius. it has another name Fermi-shape.

$\rho(r) = \frac { \rho_0} {1+ Exp( (r - R_0)/a) }$

where $\rho_0$ is central density, or density at r = 0. $R_o$ is the radius of half density and $a$ is the diffuseness. when a is large, the “tail” of the shape will be longer.

the radius is measured in unit of fm $1 fm = 10^{-15} m$.

$R_0$ can be vary from 1 fm to 7 or 8 fm.  for 16O, it is about 2 fm. and for 208Pb, it is about 6fm .

$a$ is more or less the same for different nuclei.

the density of nuclei can be mass density or charge density, the Woods-Saxon also gives a good approximation.

for mass density, a $10^{17} kg m^{-3}$. for comparison, water density is 1 kg per meter cube.

and charge density is about $0.15 e fm^{-3}$.