贝塔衰变 phase-space factor

All these are probably well known, I just take notes from these references below, and figure out some derivations.

References:

Christian Illiadis，《Nuclear Physics of Stars》，Wiley-VCH (1st Edition, 2007)

R. A. Herrera, C. W. Johnson, G. M. Fuller, PRC 105, 015801 (2022)

L. Tan, Y. X. Liu, L. J. Wang, et al., PLB 805, 135432 (2020)

1. Density of States of a free particle in a box

In the \(L \times L \times L\) box, a free particle is

\[\psi(x,y,z) = \frac{1}{\sqrt{V}} e^{i \vec{k} \cdot \vec{r}} = \frac{1}{\sqrt{V}} e^{i(k_x x + k_y y + k_z z)} \]

Periodic Boundary Condition:

\[\psi(x+L,y,z) = \psi(x,y,z),\\ \psi(x,y+L,z) = \psi(x,y,z),\\ \psi(x,y,z+L) = \psi(x,y,z). \]

Therefore

\[k_x L = 2n_x \pi ~~~\Rightarrow~~~ k_x = n_x\frac{2\pi}{L}. \]

Similar is \(k_y, k_z\), therefore in the \(k\) space, every volume \(\frac{(2\pi)^3}{L^3} = \frac{(2\pi)^3}{V}\) corresponds to one state.

Because \(p = \hbar k\), in the \(p\) space, every volume \(\frac{h^3}{V}\) corresponds to one state.

Therefore, considering \((p, p+dp)\), the density of state is

\[\frac{dn}{dE} = \frac{4\pi p^2 dp \frac{V}{h^3}}{ dE }. \]

2. \(\beta-\) decay

Starting from Fermi's Golden rule, the probability \(N(p)dp\) for electron or positron in \((p, p+dp)\) is emitted is,

\[d \lambda = N(p) dp = \frac{2\pi}{\hbar} \left| \int \Psi^*_f \hat{H}_{weak} \Psi_i \right|^2 \frac{dn}{dE_0}. \]

\(\Psi_f = \psi_f \psi_e \psi_\nu\)，but because in the nucleus the emitted electron and neutrino can be considered as plane wave with long wavelength relative to the nucleus radius, \(\psi_e, \psi_\nu\) are considered as constants \(\frac{1}{\sqrt{V}}\).

Then if the weak interaction is approximated as constant, \(|\langle \psi^*_f \hat{H}_{weak} \psi_i \rangle|^2\) somehow shrinks to Fermi or Gamow-Teller matrix elements,

\[| H_{fi}|^2 = \frac{1}{V^2}( G^2_V M^2_F + G^2_A M^2_{GT}). \]

The rest is density of states, using \(dn = 4\pi p^2 dp \frac{V}{h^3}\) in Eq. (4), we get

\[\frac{d n }{ d E } = \frac{ dn_e dn_\nu} { dE} = \frac{(4\pi)^2 V^2}{h^6} p_e^2 dp_e p_\nu^2 dp_\nu \frac{1}{dE_0}. \]

The $E_0 = Q_{\beta-} $ is energy release between parent and daughter atomic masses, i.e. counting the static energy of released electron together with the daughter nucleus,

\[Q_{\beta-} = [ m(^A_Z X_N) - m (^A_{Z+1}X'_{N-1})]c^2, \]

therefore $$E_0 = K_e + E_\nu$$, i.e. the kinetic energy of electron and total energy of neutrino.

Using \(p_\nu = \frac{E_\nu}{c} \Rightarrow dp_\nu = \frac{dE_0}{c}\), and \(p^2_\nu = \frac{(E_0 - K_e)^2}{c^2}\), and introducing \(F(Z', p_e)\) for correction due to Coulomb attraction/repulsion, finally one gets,

\[d\lambda = N(p_e)dp_e = \frac{1}{2\pi^3 \hbar^7 c^3}( G^2_V M^2_F + G^2_A M^2_{GT})F(Z', p_e) p^2 (E_0 - K_e)^2 d p_e. \]

So the total decay constant ( that means number of parent nuclei \(N_{parent} dt \lambda = -dN_{parent}\), so that \(N_{parent} = C e^{-\lambda t}, T_{1/2} = \frac{\ln 2}{\lambda}\)) is

\[\lambda = \frac{ \ln 2}{ T_{1/2}} = \frac{( G^2_V M^2_F + G^2_A M^2_{GT})}{2\pi^3 \hbar^7 c^3} \int^{p_{max}}_0F(Z', p_e) p^2 (E_0 - K_e)^2 d p_e. \]

They denote a dimensionless quantity Fermi integral (said to be tabulated in literature),

\[f( Z', E^{max}_e ) = \frac{1}{m^5_e c^7} \int^{p_{max}}_0 F(Z', p) p^2_e (E^{max}_e - E_e)^2 dp_e, \]

which is only dependent on \(E^{max}_e = K^{max}_e + m_e c^2\) and daughter nucleus' proton number \(Z'\).

Therefore

\[\lambda = \frac{ \ln 2}{ T_{1/2}} = \frac{m^5_e c^4}{2\pi^3 \hbar^7} ( G^2_V M^2_F + G^2_A M^2_{GT}) f(Z', E^{max}_e). \]

Then the \(ft\)-value is

\[f(Z', E^{max}_e)T_{1/2} = \frac{ 2\pi^3 \hbar^7}{m^5_e c^4} \frac{\ln 2}{G^2_V M^2_F + G^2_A M^2_{GT}}. \]

To summarize, this is for a given final nuclear state, and \(f(Z', E^{max}_e)\) accounts for all the stories of electron-neutrino energy distributions.

3. electron capture from the continuum

Again, start from Fermi's Golden rule,

\[d \lambda = N(p) dp = \frac{2\pi}{\hbar} \left| \int \Psi^*_f \hat{H}_{weak} \Psi_i \right|^2 \frac{dn}{dQ}. \]

In the relevant environment, electrons are stripped off, so here the \(Q\) value is between naked nuclei, as I understand,

\[W + Q = E_{\nu}, \]

here \(W\) is the total energy of that free electron,

\[W^2 = m^2_e c^4 + p^2_e c^2, \]

\(E_{\nu}\) is the total energy of the neutrino,

\[E_{\nu} = p_\nu c. \]

The initial and final waves are

\[\Psi_i = \psi_i \psi_e, ~~~~ \Psi_f = \psi_f \psi_\nu. \]

Again, treating electron and neutrino as plane waves (this may be verified with their energy and De Broglie wavelength),

\[| H_{fi}|^2 \equiv \left| \int \Psi^*_f \hat{H}_{weak} \Psi_i \right|^2 = \frac{1}{V^2}( G^2_V M^2_F + G^2_A M^2_{GT}). \]

Then the density of state for electron and neutrino should be accounted,

\[(\frac{4\pi V}{h^3})^2 p^2_e dp_e p^2_\nu dp_\nu. \]

\(W^2 = m^2_e c^4 + p^2_e c^2 ~~~\Rightarrow WdW = p_e dp_e c^2\), and $p_e = \sqrt{W^2 -1} /c $ if \(m_e c^2\) is taken as unit energy.

And \(p_\nu c = E_\nu = W + Q\), so with given \(W\), \(dp_\nu c = dQ\), also \(p^2_\nu = \frac{E^2_\nu}{c^2} = \frac{ (W+Q)^2 }{c^2}\). So we get density of state,

\[(\frac{4\pi V}{h^3})^2 p^2_e dp_e p^2_\nu dp_\nu = (\frac{4\pi V}{h^3})^2 \frac{1}{c^6} W \sqrt{W^2-1} (W + Q)^2 dW. \]

Consider the Coulomb attraction in \(F(Z, W)\), and also the Fermi distribution of those free electrons, we get

\[\lambda = \frac{m^5_e c^{10}}{\hbar} ( G^2_V M^2_F + G^2_A M^2_{GT}) (\frac{4\pi }{h^3})^2 \frac{1}{c^6} \int^\infty_{W_{min}}W \sqrt{W^2-1} (W + Q)^2 F(Z, W) S_e(W) dW \\ = \frac{m^5_e c^{5}}{2\pi^3\hbar^7} ( G^2_V M^2_F + G^2_A M^2_{GT}) \int^\infty_{W_{min}}W \sqrt{W^2-1} (W + Q)^2 F(Z, W) S_e(W) dW, \]

where

\[S_e(W) = \frac{1}{ e^{ \frac{W-\mu_e}{kT} } +1 }. \]