Derivation of Eq. (34) in Magnetic tensor operator

In this posting, I derive Eq. (34) from Eq. (29).

We start from Eq. (29): 

$$ M_{\lambda \mu} = \int {\pmb \mu} ({\pmb r}) \cdot \nabla \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r.  $$

Here, we can use the following relation:

$$ ({\pmb \nabla} \times ( {\pmb \nabla} \times {\pmb r} ) ) r^\lambda Y_{\lambda \mu}(\theta, \phi) = (\lambda +1) {\pmb \nabla} \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right), \label{1}\tag{1} $$

(Please refer to the following link: https://djlab.tistory.com/19 for proof.) Then, the given equation becomes:

$$ M_{\lambda \mu} = \frac{1}{\lambda + 1}\int \left[ {\pmb \mu} ({\pmb r}) \cdot \left[{\pmb \nabla} \times \left( {\pmb \nabla} \times {\pmb r} \right) \right] \right] \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r. \label{2}\tag{2} $$

Using the following relation:

\begin{eqnarray} {\pmb \mu} ({\pmb r}) \cdot \left[ {\pmb \nabla} \times ( \nabla \times {\pmb r} ) \right] &=& \mu_i \left[ \epsilon_{ijk} \partial_j \epsilon_{klm} \partial_l r_m \right] =\mu_i \left[ \epsilon_{ijk} \epsilon_{klm} \partial_j \partial_l r_m \right] = \mu_i \left[ \epsilon_{ijk} \epsilon_{klm} \partial_j \left( \delta_{lm} + r_m\partial_l \right) \right]\\[12pt] &=& \mu_i \left[ \epsilon_{ijk} \epsilon_{klm} \partial_j \left( r_m \partial_l \right) \right] = \mu_i \left[ - \epsilon_{ijk} \partial_j \epsilon_{kml} r_m \partial_l \right] = - {\pmb \mu} ({\pmb r}) \cdot \left[ \nabla \times \left( {\pmb r} \times \nabla \right) \right], \label{3}\tag{3} \end{eqnarray}

One can obtain:

$$ M_{\lambda \mu} = - \frac{1}{\lambda + 1}\int \left[ {\pmb \mu} ({\pmb r}) \cdot \left[ \nabla \times \left( {\pmb r} \times \nabla \right) \right] \right] \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r. \label{4}\tag{4} $$

Let ${\pmb A} = {\pmb r} \times {\pmb \nabla} $. Then,

\begin{eqnarray} - M_{\lambda \mu} &=& \frac{1}{\lambda + 1}\int \left[ {\pmb \mu} ({\pmb r}) \cdot \left[ {\pmb \nabla} \times {\pmb A} \right] \right] \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r. \label{5}\tag{5} \end{eqnarray}

By the identity of $({\pmb A} \times {\pmb \nabla} ) \cdot {\pmb B} = {\pmb A} \cdot \left( {\pmb \nabla} \times {\pmb B} \right)$,

\begin{eqnarray} M_{\lambda \mu} &=& - \frac{1}{\lambda + 1}\int \left[ \left[ {\pmb \nabla} \times {\pmb \mu} ({\pmb r}) \right] \cdot {\pmb A} \right] \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r \\[12pt] &=& - \frac{1}{\lambda + 1}\int \left[ \left[ {\pmb \nabla} \times {\pmb \mu} ({\pmb r}) \right] \cdot \left[ {\pmb \nabla} \times {\pmb r} \right] \right] \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r . \label{6}\tag{6} \end{eqnarray}

The given scalar product is written as follows:

\begin{eqnarray} \left[ {\pmb \nabla} \times {\pmb \mu} ({\pmb r}) \right] \cdot \left[ {\pmb \nabla} \times {\pmb r} \right] &=& \left( \epsilon_{ijk} \partial_j \mu_k \right) \left( \epsilon_{ilm} \partial_l r_m \right) = \epsilon_{ijk} \epsilon_{ilm} \left( \partial_j \mu_k \right) \left( \partial_l r_m \right) = \left( \partial_l r_m \right) \left( \partial_l \mu_m - \partial_m \mu_l \right) \\[12pt] &=& \partial_l \left[r_m \left( \partial_l \mu_m - \partial_m \mu_l \right) \right] = \partial_l \left[r_m \partial_l \mu_m - r_m \partial_m \mu_l \right] . \label{7}\tag{7} \end{eqnarray}

Here the term in the square bracket is same as ${\pmb r} \times \left({\pmb \nabla} \times {\pmb \mu}(\pmb r) \right)$, because

\begin{eqnarray} \left[ {\pmb r} \times \left({\pmb \nabla} \times {\pmb \mu}(\pmb r) \right) \right]_i &=& \epsilon_{ijk} r_j \epsilon_{klm} \partial_l \mu_m = \epsilon_{kij} \epsilon_{klm}r_j \partial_l \mu_m \\[12pt] &=& \left( \delta_{il} \delta_{jm} -\delta_{im} \delta_{jl} \right) r_j \partial_l \mu_m = r_j \partial_i \mu_j - r_j \partial_j \mu_i. \label{8}\tag{8} \end{eqnarray}

Therefore,

\begin{eqnarray} \left[ {\pmb \nabla} \times {\pmb \mu} ({\pmb r}) \right] \cdot \left[ {\pmb \nabla} \times {\pmb r} \right] &=& {\pmb \nabla} \cdot \left( {\pmb r} \times \left({\pmb \nabla} \times {\pmb \mu}(\pmb r) \right) \right). \label{9}\tag{9} \end{eqnarray}

Substituting the above result into Eq. (\ref{4}),

\begin{eqnarray} M_{\lambda \mu} =- \frac{1}{\lambda + 1}\int \left[ {\pmb \nabla} \cdot \left( {\pmb r} \times \left({\pmb \nabla} \times {\pmb \mu}(\pmb r) \right) \right) \right] \left( r^\lambda Y_{\lambda \mu} (\theta, \phi) \right) d^3 r . \label{10}\tag{10} \end{eqnarray}

Then, integrating Eq. (\ref{8}) by parts,

\begin{eqnarray} M_{\lambda \mu} = \frac{1}{\lambda + 1}\int \left[ {\pmb r} \times \left({\pmb \nabla} \times {\pmb \mu}(\pmb r) \right) \right] \left( \nabla \left[ r^\lambda Y_{\lambda \mu} (\theta, \phi) \right] \right) d^3 r . \label{11}\tag{11} \end{eqnarray}

Since ${\pmb j}({\pmb r}) = c {\pmb \nabla} \times {\pmb \mu} ({\pmb r})$, we can obtain the following result:

\begin{eqnarray} M_{\lambda \mu} = \frac{1}{c(\lambda + 1)}\int \left[ {\pmb r} \times {\pmb j}(\pmb r) \right] \left( \nabla \left[ r^\lambda Y_{\lambda \mu} (\theta, \phi) \right] \right) d^3 r . \end{eqnarray}

This is the result of Eq.(34) in https://djlab.tistory.com/20.

 

Reference:

[1] The Nuclear Many-Body Problem (Ring, P. and Schuck, P. 2004), Eq. (B.21) & Eq. (B.22).