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L2L translation formula

From (B.11) we obtain
 
$\displaystyle R_{n,m}(\overrightarrow{\mbox{\boldmath$ x $ }_0\mbox{\boldmath$ ...
...{n-n',m-m'}(\overrightarrow{\mbox{\boldmath$ x $ }_0\mbox{\boldmath$ x $ }_1}).$     (E.4)

Substituting (E.4) into the right-hand side of (3.26) we have
 
$\displaystyle { \sum_{n=0}^{\infty}\sum_{m=-n}^{n}
\frac{\partial R_{n,m}(\over...
... }_0\mbox{\boldmath$ x $ }})}{\partial n_x}
L_{n,m}(\mbox{\boldmath$ x $ }_0) }$
  = $\displaystyle \sum_{n=0}^{\infty}\sum_{m=-n}^{n}
\frac{\partial}{\partial n_x}
...
...th$ x $ }_0\mbox{\boldmath$ x $ }_1})
\right)
L_{n,m}(\mbox{\boldmath$ x $ }_0)$  
  = $\displaystyle \sum_{n=0}^{\infty}\sum_{m=-n}^{n}
\sum_{n'=0}^{n}\sum_{m'=-n'}^{...
...\boldmath$ x $ }_0\mbox{\boldmath$ x $ }_1})
L_{n,m}(\mbox{\boldmath$ x $ }_0).$ (E.5)

Exchanging an order of summation in (E.5) ( $\displaystyle \sum_{n}\sum_{n'} \rightarrow
\sum_{n'}\sum_{n}$, See Fig.B) we obtain

\begin{eqnarray*}\lefteqn{ \sum_{n=0}^{\infty}\sum_{m=-n}^{n}
\frac{\partial R_...
...math$ x $ }})}{\partial n_x}
L_{n,m}(\mbox{\boldmath$ x $ }_1),
\end{eqnarray*}


where $L_{n,m}(\mbox{\boldmath$\space x $ }_1)$ is the coefficient of the local expansion centred at $\mbox{\boldmath$\space x $ }_1$ defined as

\begin{eqnarray*}L_{n,m}(\mbox{\boldmath$ x $ }_1)=\sum_{n'=n}^{\infty}\sum_{m'=...
...mbox{\boldmath$ x $ }_1})
L_{n',m'}(\mbox{\boldmath$ x $ }_0).
\end{eqnarray*}




Ken-ichi Yoshida
2001-07-28