6.1.3 Single-Particle Matrix Elements of the Multipole Operators

In this posting, I derive the single-particle matrix elements of the multipole operators.

 

1. Single particle matrix elements for electric tensor operator ($\sigma = E$)

The electric tensor operator is given as

\begin{eqnarray} Q_{\lambda \mu} = \zeta^{\rm E \lambda} \sum_{j=1}^A e(j) r^\lambda_j Y_{\lambda \mu } (\Omega_j), \end{eqnarray}

so we can simply rewrite it as follows:

\begin{eqnarray} Q_{\lambda \mu} = \sum_{j=1}^A q^{\lambda \mu} ({\bf r}_j) \end{eqnarray}

where $ q^{\lambda \mu} ({{\bf r}_i}) \equiv e (j) r^\lambda_j Y_{\lambda \mu} (\Omega_j) $. Then, by the completeness relation, 

\begin{eqnarray} q^{\lambda \mu} ({\bf r}_j) = \sum_{\alpha \beta} | \alpha \rangle \langle \alpha | q^{\lambda \mu} ( {\bf r}_i ) | \beta \rangle \langle |\beta |  \equiv \sum_{\alpha \beta} q^{\lambda \mu}_{\alpha \beta} |\alpha \rangle \langle \beta | .\end{eqnarray}

Then, for the many-body state, one can obtain

\begin{eqnarray} Q_{\lambda \mu} | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_A \rangle &=& \sum_{j=1}^A q^{\lambda \mu} ( {\bf r}_j ) | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_j, \cdots, \alpha_A \rangle \\[12pt] &=& \sum_{j=1}^A \sum_{\alpha \beta} q^{\lambda \mu}_{\alpha \beta} |\alpha \rangle \langle \beta | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_j, \cdots, \alpha_A \rangle \\[12pt] &=& \sum_{j=1}^A \sum_{\alpha \beta} q^{\lambda \mu}_{\alpha \beta} |\alpha \rangle (-1)^{j-1} \langle \beta | \alpha_j \rangle | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_{j-1}, \alpha_{j+1}, \cdots, \alpha_A \rangle \\[12pt] &=& \sum_{j=1}^A \sum_{\alpha \beta} q^{\lambda \mu}_{\alpha \beta} \delta_{\beta, \alpha_j} c_\alpha^\dagger | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_{j-1}, 0 , \alpha_{j+1}, \cdots, \alpha_A \rangle \\[12pt] &=& \sum_{j=1}^A \sum_{\alpha \beta} q^{\lambda \mu}_{\alpha \beta} \delta_{\beta, \alpha_j} c_\alpha^\dagger c_{\alpha_j} | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_{j-1}, \alpha_j , \alpha_{j+1}, \cdots, \alpha_A \rangle \\[12pt] &=& \sum_{\alpha \beta} q^{\lambda \mu}_{\alpha \beta} c_\alpha^\dagger c_{\beta} | \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_{j-1}, \alpha_j , \alpha_{j+1}, \cdots, \alpha_A \rangle \end{eqnarray}

Therefore, we can write

\begin{eqnarray} Q_{\lambda \mu} = \sum_{\alpha \beta} \langle \alpha | q^{\lambda \mu} | \beta \rangle c_\alpha^\dagger c_{\beta} .\end{eqnarray}

Then, the expectation value of the electric tensor operator is given as

\begin{eqnarray} \langle n_f \, l_f \, (s_f = \frac{1}{2}) \, j_f, m_f | Q_{\lambda \mu} | n_i \, l_i \, (s_i = \frac{1}{2}) \, j_i, m_i \rangle &=& \sum_{\alpha \beta} \langle \alpha | q^{\lambda \mu}_{\alpha \beta} | \beta \rangle \\[12pt] &\times& \langle n_f \, l_f \, (s_f = \frac{1}{2}) \, j_f, m_f | c_\alpha^\dagger c_{\beta} | n_i \, l_i \, (s_i = \frac{1}{2}) \, j_i, m_i \rangle, \end{eqnarray}

By the Wigner-Eckart theorem, the given equation can be rewritten as follows:

\begin{eqnarray} ( n_f \, l_f \, \frac{1}{2} \, j_f || {\pmb Q}_{\lambda} || n_i \, l_i \, \frac{1}{2} \, j_i ) = \lambda^{-1} \sum_{ab} ( a || {\pmb q}^{\lambda}_{\alpha \beta} || b ) ( n_f \, l_f \, \frac{1}{2} \, j_f || [c_\alpha^\dagger \tilde{c}_{\beta}]_\lambda || n_i \, l_i \, \frac{1}{2} \, j_i ), \end{eqnarray}

Then, we can evaluate the reduced matrix elements for the electric tensor operator as follows:

\begin{eqnarray} ( a || {\pmb q}^{\lambda}_{\alpha \beta} || b ) &=& \zeta^{\rm E \lambda} ( n_a \, l_a \, \frac{1}{2} \, j_a || e r^\lambda Y_{\lambda \mu } || n_b \, l_b \, \frac{1}{2} \, j_b ) = \zeta^{\rm E \lambda} e \langle a | r^\lambda | b\rangle ( a|| Y_{\lambda \mu } || b ) \\[12pt] &=& \zeta^{\rm E \lambda} e {\cal R}_{ab}^{(\lambda)} \left( \frac{1}{\sqrt{4\pi}} (-1)^{j_b - \frac{1}{2} + \lambda} \frac{1 + (-1)^{l_a + l_b + \lambda}}{2} \hat{j}_a \hat{j}_b \hat{\lambda} \right) \begin{pmatrix} j_b & j_a & \lambda \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{pmatrix}, \end{eqnarray}

where ${\cal R}_{ab}^{(\lambda)}$ is

\begin{eqnarray} {\cal R}_{ab}^{(\lambda)} = \int_0^\infty g_{n_a l_a} (r) r^\lambda g_{n_b l_b} (r) r^2 dr.\end{eqnarray}

Here I used the relation Eq. (2.57) in Ref. [1] for the reduced matrix elements of spherical harmonics $Y_{\lambda \mu}$. 

 

2. Single particle matrix elements for magnetic tensor operator ($\sigma = M$)

Similarly, for the magnetic tensor operator, we can write it as follows:

\begin{eqnarray}  M_{\lambda \mu} = \sum_{\alpha \beta} m^{\lambda \mu}_{\alpha \beta} c_\alpha^\dagger c_\beta = \sum_{\alpha \beta} \langle \alpha | \frac{\mu_N}{\hbar c} \zeta^{\rm M \lambda} \left[ \frac{2}{\lambda +1} g_l {\pmb l} + g_s {\pmb s} \right]  \cdot \nabla \left[ r^\lambda Y_{\lambda \mu} \right]| \beta \rangle c_\alpha^\dagger c_\beta. \end{eqnarray}

Thus, we can write the expectation value of the operator as follows:

\begin{eqnarray} ( n_f \, l_f \, \frac{1}{2} \, j_f || {\pmb M}_{\lambda} || n_i \, l_i \, \frac{1}{2} \, j_i ) = \lambda^{-1} \sum_{ab} ( a || {\pmb m}^{\lambda}_{\alpha \beta} || b ) ( n_f \, l_f \, \frac{1}{2} \, j_f || [c_\alpha^\dagger \tilde{c}_{\beta}]_\lambda || n_i \, l_i \, \frac{1}{2} \, j_i ), \end{eqnarray}

Then, the reduced matrix elements for the electric tensor operator are given as [1]

\begin{eqnarray} ( a || {\pmb m}^{\lambda}_{\alpha \beta} || b ) &=& \zeta_{ab}^{\rm M \lambda} \frac{\mu_N/c}{\sqrt{4\pi}} (-1)^{jb + \lambda - \frac{1}{2}} \frac{1 - (-1)^{l_a + l_b + \lambda }}{2} \hat{\lambda} \hat{j}_a \hat{j}_b \begin{pmatrix} j_a & j_b & \lambda \\ \frac{1}{2} & -\frac{1}{2} & 0\end{pmatrix} \\[12pt] &\times& (\lambda - \kappa) \left[ g_l \left( 1 + \frac{\kappa}{\lambda +1} \right) - \frac{1}{2} g_s \right] {\cal R}^{(\lambda -1)}_{ab}, \end{eqnarray}

where $\kappa \equiv (-1)^{l_a + j_a + \frac{1}{2}} (j_a + \frac{1}{2} ) + (-1)^{l_b + j_b + \frac{1}{2}} (j_b + \frac{1}{2} )$.

 

The charges and $g$ factors are respectively given as

$$ e(p) = 3, \ \ \ e(n) = 0, \ \ \  g_l(p) = 1, \ \ \  g_l(n) =0 , \ \ \ g_s(p) = g_p , \ \ \ g_s(n) = g_n.$$

However, these values are valid for particles in free space. In nuclei, to treat the particle-hole excitations, effective charges and $g$ factors can be used. We discuss that later. 

 

Symmetry properties

The given matrix elements for the electromagnetic operators have the following symmetry properties:

\begin{eqnarray} (b || {\pmb q}_\lambda || a)_{\rm CS} &= & (-1)^{j_a + j_b +1} (a || {\pmb q}_\lambda || b )_{\rm CS} \\[12pt] (b || {\pmb m}_\lambda || a)_{\rm CS} &= & (-1)^{j_a + j_b +1} (a || {\pmb m}_\lambda || b )_{\rm CS} \\[12pt] (b || {\pmb q}_\lambda || a)_{\rm BR} &= & (-1)^{j_a + j_b + \lambda +1} (a || {\pmb m}_\lambda || b )_{\rm BR} \\[12pt] (b || {\pmb m}_\lambda || a)_{\rm BR} &= & (-1)^{j_a + j_b +\lambda} (a || {\pmb m}_\lambda || b )_{\rm BR} , \end{eqnarray}

where 'CS" and 'BR' denote the Condon-Shortly (CS) and Biedenharn-Rose (BR) phases, respectively.

 

 

Reference

[1] From Nucleons to Nucleus: Concepts of Microscopic Nuclear Theory, J. Suhonen, Springer, 2007