6.2 Electromagnetic Transitions in One-Particle and One-Hole Nuclei - 6.2.1 Reduced Transition Probabilities

In this blog posting, for the simple case of one-particle and one-hole nucleus, I will introduce how to obtain reduced transition probabilities:

\begin{eqnarray} (\xi_f \, J_f || {\pmb {\cal M}}_{\sigma \lambda} || \xi_i \, J_i ) = \hat{\lambda}^{-1} \sum_{ab} (a || {\pmb {\cal M}}_{\sigma \lambda} || b) (\xi_f \, J_f || [c_a^\dagger \tilde{c}_b ]_\lambda || \xi_i \, J_i). \tag{1}\label{1}\end{eqnarray}

 

One-particle nuclei

For a one-particle nucleus, we can set the initial and final states as follows:

\begin{eqnarray} &&|\Psi_i \rangle = | n_i \, l_i\, j_i \,m_i \rangle = c_i^\dagger |{\rm CORE} \rangle \\[12pt] &&|\Psi_f \rangle = | n_f \, l_f \, j_f \, m_f \rangle = c_f^\dagger |{\rm CORE} \rangle. \label{2} \tag{2} \end{eqnarray}

Then, the one-body transition density $\langle \Psi_f | [c_a^\dagger \tilde{c}_b]_{\lambda \mu} | \Psi_i \rangle $ is given by

\begin{eqnarray} \langle \Psi_f | [c_a^\dagger \tilde{c}_b]_{\lambda \mu} | \Psi_i \rangle &=& \sum_{m_\alpha m_\beta} (j_a \, m_\alpha \, j_b \,m_\beta | \lambda \, \mu) \langle {\rm CORE} | c_f c_\alpha^\dagger \tilde{c}_\beta c_i^\dagger | {\rm CORE} \rangle \\[12pt] &=& \sum_{m_\alpha m_\beta} (j_a \, m_\alpha \, j_b \,m_\beta | \lambda \, \mu) (-1)^{j_b + m_\beta}\langle {\rm CORE} | c_f c_\alpha^\dagger c_{-\beta} c_i^\dagger | {\rm CORE} \rangle \\[12pt] &=& \delta_{af} \delta_{bi} (-1)^{j_i - m_i} (j_f \, m_f\, j_i \, -m_i | \lambda \, \mu ), \label{3} \tag{3} \end{eqnarray} 

and the left-hand side can be rewritten by the Wigner-Eckart theorem as follows:

\begin{eqnarray} \langle \Psi_f | [c_a^\dagger \tilde{c}_b]_{\lambda \mu} | \Psi_i \rangle &=& \hat{j}_f^{-1} (j_i \, m_i \, \lambda \, \mu | j_f \, m_f) \langle \Psi_f || [c_a^\dagger \tilde{c}_b]_{\lambda \mu} || \Psi_i \rangle \\[12pt] &=& \hat{j}_f^{-1} (-1)^{j_i - m_i} \frac{\hat{j}_f}{\hat{\lambda}}(j_f \, m_f \, j_i \, -m_i | \lambda \, \mu ) \langle \Psi_f || [c_a^\dagger \tilde{c}_b]_{\lambda \mu} || \Psi_i \rangle \\[12pt] &=& (-1)^{j_i - m_i} \hat{\lambda}^{-1} (j_f \, m_f \, j_i \, -m_i | \lambda \, \mu ) \langle \Psi_f || [c_a^\dagger \tilde{c}_b]_{\lambda \mu} || \Psi_i \rangle. \label{4} \tag{4} \end{eqnarray} 

Therefore, we can obtain the following reduced matrix element for the one-body transition density:

\begin{eqnarray} \langle \Psi_f || [c_a^\dagger \tilde{c}_b]_{\lambda \mu} || \Psi_i \rangle = \hat{\lambda}\delta_{af} \delta_{bi} \label{5}\tag{5} \end{eqnarray} 

We can substitute this result into Eq. (\ref{1}), which leads to

\begin{eqnarray} (\xi_f \, J_f || {\pmb {\cal M}}_{\sigma \lambda} || \xi_i \, J_i ) = \hat{\lambda}^{-1} \sum_{ab} (a || {\pmb {\cal M}}_{\sigma \lambda} || b) \hat{\lambda} \delta_{af} \delta_{bi} = (f || {\pmb {\cal M}}_{\sigma \lambda} || i) \label{6}\tag{6}. \end{eqnarray}

Finally, we then obtain the reduced transition probability for a one-particle nucleus as follows:

\begin{eqnarray} B(\sigma \lambda; \Psi_i \rightarrow \Psi_f) = \frac{1}{2J_i + 1} | (f || {\pmb {\cal M}}_{\sigma \lambda} || i) |^2 \label{7}\tag{7} \end{eqnarray}

 

One-hole nuclei

For one-hole nuclei, we can write the initial and final states as follows:

\begin{eqnarray} &&|\Phi_i \rangle = | (n_i \, l_i\, j_i \,m_i)^{-1} \rangle = h_i^\dagger |{\rm HF} \rangle \\[12pt] &&|\Phi_f \rangle = | (n_f \, l_f \, j_f \, m_f)^{-1} \rangle = h_f^\dagger |{\rm HF} \rangle. \label{8} \tag{8} \end{eqnarray}

For those states, the one-body transition density is given as

\begin{eqnarray} \langle \Phi_f | [c_a^\dagger \tilde{c}_b]_{\lambda \mu} | \Phi_i \rangle &=& \sum_{m_\alpha m_\beta} (j_a \, m_\alpha \, j_b \,m_\beta | \lambda \, \mu) \langle {\rm HF} | h_f c_\alpha^\dagger \tilde{c}_\beta h_i^\dagger | {\rm HF} \rangle \\[12pt] &=&\sum_{m_\alpha m_\beta} (j_a \, m_\alpha \, j_b \,m_\beta | \lambda \, \mu) \langle {\rm HF} | h_f (-\tilde{h}_\alpha) h^\dagger_\beta h_i^\dagger | {\rm HF} \rangle \\[12pt] &=& - \sum_{m_\alpha m_\beta} (-1)^{j_a + m_\alpha}(j_a \, m_\alpha \, j_b \,m_\beta | \lambda \, \mu) \langle {\rm HF} | h_f h_{-\alpha} h^\dagger_\beta h_i^\dagger | {\rm HF} \rangle \\[12pt] &=& - \sum_{m_\alpha m_\beta} (-1)^{j_a + m_\alpha}(j_a \, m_\alpha \, j_b \,m_\beta | \lambda \, \mu) \left(\delta_{-\alpha \beta} \delta_{f i} - \delta_{f \beta} \delta_{-\alpha i} \right) \\[12pt] &=& - \delta_{f i} \sum_{m_\alpha} (-1)^{j_a + m_\alpha}(j_a \, m_\alpha \, j_a \, -m_\alpha | \lambda \, \mu) \\[12pt] &+& \delta_{ai} \delta_{bf} (-1)^{j_i - m_i} (j_i \, -m_i \, j_f \, m_f | \lambda \, \mu). \label{9} \tag{9} \end{eqnarray}

The summation in the first term can be evaluated by the property of the Clebsch-Gordan coefficient

\begin{eqnarray} (j \, m \, j \, -m | 0 \, 0 ) = (-1)^{j-m} \hat{j}^{-1} &\Rightarrow & (-1)^{j_a - m_\alpha} = \hat{j}_a (j_a \, m_\alpha \, j_a \, -m_\alpha | 0 \, 0 ) \\[12pt] &\Rightarrow & (-1)^{j_a + m_\alpha} = (-1)^{j_a - m_\alpha} (-1)^{+2m_\alpha} \\[12pt] &=& - \hat{j}_a (j_a \, m_\alpha \, j_a \, -m_\alpha | 0 \, 0 ), \label{10} \tag{10} \end{eqnarray}

and therefore,

\begin{eqnarray} \langle \Phi_f | [c_a^\dagger \tilde{c}_b]_{\lambda \mu} | \Phi_i \rangle &=& - \delta_{f i} \sum_{m_\alpha} \left( - \hat{j}_a (j_a \, m_\alpha \, j_a \, -m_\alpha | 0 \, 0 ) \right) (j_a \, m_\alpha \, j_a \, -m_\alpha | \lambda \, \mu) \\[12pt] &+& \delta_{ai} \delta_{bf} (-1)^{j_i - m_i} (j_i \, -m_i \, j_f \, m_f | \lambda \, \mu) \\[12pt] &=& \hat{j}_a \delta_{f i} \delta_{\lambda 0} \delta_{\mu 0} + \delta_{ai} \delta_{bf} (-1)^{j_i - m_i} (j_i \, -m_i \, j_f \, m_f | \lambda \, \mu). \\[12pt] \label{11} \tag{11} \end{eqnarray}

Since $\lambda \ge 1$ for electromagnetic operators, the first term vanishes.

\begin{eqnarray} \langle \Phi_f | [c_a^\dagger \tilde{c}_b]_{\lambda \mu} | \Phi_i \rangle &=& \delta_{ai} \delta_{bf} (-1)^{j_i - m_i} (j_i \, -m_i \, j_f \, m_f | \lambda \, \mu) \\[12pt] &=& \delta_{ai} \delta_{bf} (-1)^{j_i - m_i} (-1)^{j_i + j_f + \lambda}(j_f \, m_f \, j_i \, -m_i | \lambda \, \mu). \label{12} \tag{12} \end{eqnarray}

Using Eq. (\ref{4}), we can obtain the reduced matrix elements of the one-body transition density as follows:

\begin{eqnarray} \langle \Phi_f || [c_a^\dagger \tilde{c}_b]_{\lambda \mu} || \Phi_i \rangle &=& \delta_{ai} \delta_{bf} (-1)^{j_i + j_f + \lambda} \hat{\lambda}. \label{13} \tag{13} \end{eqnarray}

As a result, the reduced matrix element of electromagnetic operators for one-hole nuclei is given as

\begin{eqnarray} ( \Phi_f || {\pmb {\cal M}}_{\sigma \lambda} || \Phi_i) = (-1)^{j_i + j_f + \lambda} (f || {\pmb {\cal M}}_{\sigma \lambda} || i ), \label{14} \tag{14} \end{eqnarray}

which results in the reduced transition probability as follows:

\begin{eqnarray} B(\sigma \lambda; \Phi_i \rightarrow \Phi_f) = \frac{1}{2J_i + 1} | (f || {\pmb {\cal M}}_{\sigma \lambda} || i) |^2. \label{15} \tag{15} \end{eqnarray}

Note that this result is the same as that of Eq. (\ref{7}).