In this posting, I introduce electromagnetic multipole moments. The static electric and magnetic multipole are important observables of nuclear structure. One can test a nuclear model by comparing computed multipole moments and measured ones.
The $2^\lambda$-pole moment of a nucleus in a certain state is defined as
\begin{eqnarray} {\cal M}(\sigma \lambda) = \langle \xi \, J \, M = J | {\cal M}_{\sigma \lambda 0} | \xi \, J \, M = J \rangle ,\end{eqnarray}
where $\xi$ other quantum numbers, $J$ is the angular momentum, $M$ its $z$ projection, and $\mu=0$ component of the multipole tensor operator ${\pmb \cal M}(\sigma \lambda)$ (or ${\cal M}_{\sigma \lambda \mu}$). By the Wigner-Eckart theorem, we can rewrite it as follows:
\begin{eqnarray} {\cal M}(\sigma \lambda) = \begin{pmatrix} J \, \lambda \, J \\ -J \, 0 \, J \end{pmatrix} (\xi \, J || {\pmb \cal M}_{\sigma \lambda} || \xi \, J ). \end{eqnarray}
For the dipole $(\lambda =1)$ and quadrupole $(\lambda =2)$, we can evalaute the Wigner $3j$ symbols for thoes cases, which result in
\begin{eqnarray} && {\cal M}(\sigma 1) = \sqrt{\frac{J}{(J+1)(2J +1)}} ( \xi \, J || {\pmb M}_{\sigma 1} || \xi J ) \\[12pt] && {\cal M}(\sigma 2) = \sqrt{\frac{J(2J -1)}{(J+1)(2J+1) (2J+3)}} ( \xi \, J || {\pmb M}_{\sigma 2} || \xi \, J ). \end{eqnarray}
Using these results, we can define the conventional magnetic dipole moment $\mu$ and electric quadrupole moment $Q$ as follows:
\begin{eqnarray} &&\frac{\mu}{c} \equiv \zeta \sqrt{\frac{4\pi}{3}} {\cal M}(M1) = \zeta \sqrt{\frac{4\pi}{3}} \sqrt{\frac{J}{(J+1)(2J +1)}} ( \xi \, J || {\pmb M}_1 || \xi \, J) \\[12pt] && eQ \equiv \zeta \sqrt{\frac{16 \pi }{5}} {\cal M}(E2) = \zeta\sqrt{\frac{16 \pi }{5}} \sqrt{\frac{J (2J -1)}{(J+1)(2J + 1)(2J +3)}} ( \xi \, J || {\pmb Q}_{2} || \xi \, J ). \end{eqnarray}
From the single-particle matrix element, we can obtain the following reuslts:
\begin{eqnarray} Q &=&\zeta\sqrt{\frac{16 \pi }{5}} \sqrt{\frac{j (2j -1)}{(j+1)(2j + 1)(2j +3)}} ( \xi \, j || {\pmb Q}_{2} || \xi \, j ) \\[12pt] &=& \zeta^2 \sqrt{\frac{16\pi}{5}} \sqrt{\frac{j(2j-1)}{(j+1)(2j +1)(2j+3)}} \left[ \frac{(-1)^{j + 2 - \frac{1}{2}}}{\sqrt{4\pi}} \frac{1 + (-1)^{2l+2}}{2} \sqrt{5} (2j+1) \right] \begin{pmatrix} j & j & 2 \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{pmatrix} {\cal R}_{nl nl}^{(2)} \\[12pt] &=& \sqrt{\frac{4}{5}} \sqrt{\frac{j(2j-1)}{(j+1)(2j +1)(2j+3)}} \left[ (-1)^{J - \frac{1}{2}} \sqrt{5} (2j+1) \right] \\[12pt] &\times& \left( (-1)^{J-\frac{1}{2}} \frac{3 \cdot \frac{1}{4} - j(j+1)}{\sqrt{j(j+1)(2j-1)(2j+1)(2j+3)}} \right) {\cal R}_{nl nl}^{(2)} \\[12pt] &=& \sqrt{4} \sqrt{\frac{1}{(j+1)(2j+3)}} \left( \frac{3 \cdot \frac{1}{4} - j(j+1)}{\sqrt{(j+1)(2j+3)}} \right) {\cal R}_{nl nl}^{(2)} = \left( \frac{3 - 4 j(j+1)}{2(j+1)(2j+3)} \right) {\cal R}_{nl nl}^{(2)}, \end{eqnarray}
and for the magnetic moment,
\begin{eqnarray} \mu &=&\zeta \sqrt{\frac{4\pi}{3}} \sqrt{\frac{j}{(j+1)(2j +1)}} ( \xi \, j || {\pmb M}_1 || \xi \, j) \\[12pt] &=& \zeta^2 \mu_N \sqrt{\frac{4\pi}{3}} \sqrt{\frac{j}{(j+1)(2j +1)}} \left[ \frac{(-1)^{j + 1 - \frac{1}{2}}}{\sqrt{4\pi}} \frac{1 - (-1)^{2l+1}}{2} \sqrt{3} (2j+1) \right] \begin{pmatrix} j & j & 1 \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{pmatrix} \\[12pt] &\times& (1 - \kappa) \left[g_l \left( 1 + \frac{\kappa}{2} \right) - \frac{1}{2}g_s \right] {\cal R}_{nl nl}^{(0)}. \end{eqnarray}
Since ${\cal R}_{nl nl}^{(0)} = 1$ and $\kappa = (-1)^{l+j+ \frac{1}{2}} (j + \frac{1}{2}) + (-1)^{l+j+\frac{1}{2}} ( j + \frac{1}{2} ) = 2 (-1)^{l+j+\frac{1}{2}}(j+\frac{1}{2}),$
\begin{eqnarray} \mu &=& \mu_N \sqrt{\frac{j}{(j+1)(2j +1)}} \left[ (-1)^{j + \frac{1}{2}} (2j+1) \right] \left( \frac{(-1)^{j-\frac{1}{2}}\frac{1}{2}}{\sqrt{j(j+1)(2j+1)}} \right) \\[12pt] &\times& (1 - 2 (-1)^{l+j+\frac{1}{2}}(j+\frac{1}{2})) \left[g_l \left( 1 + \frac{ 2 (-1)^{l+j+\frac{1}{2}}(j+\frac{1}{2})}{2} \right) - \frac{1}{2}g_s \right] \\[12pt] &=& \mu_N (-1)^{2j} \frac{(1 - (-1)^{l+j+\frac{1}{2}}(2 j+ 1))}{4(j+1)} \left[g_l \left( 2 + (-1)^{l+j+\frac{1}{2}} (2j+1) \right) - g_s \right] \\[12pt] &=& \mu_N \frac{(1 - (-1)^{l+j+\frac{1}{2}}(2 j+ 1))}{4(j+1)} \left[ g_s - g_l \left( 2 + (-1)^{l+j+\frac{1}{2}} (2j+1) \right) \right]. \end{eqnarray}
