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M2X translation formula

Using (4.2), we can transform the right-hand side of (3.52)

\begin{eqnarray*}\lefteqn{ \sum_{n=0}^{\infty} \sum_{m=-n}^{n}
\left( \frac{\pa...
...(\overrightarrow{O\mbox{\boldmath$ x $ }})_2 \sin \alpha_q(p))},
\end{eqnarray*}


where we have used the result obtained in J.1 and X1j(p,q;O) and X2(p,q;O) are the coefficients of the exponential expansion at O, defined as

\begin{eqnarray*}X^1_j(p,q;O)=\frac{\omega_p}{M(p)d}\sum_{m=-\infty}^{\infty}(-i...
...p)} \sum_{n=\vert m\vert}^{\infty} (\lambda_p/d)^n M^2_{n,m}(O),
\end{eqnarray*}


and $\cal F,G$ are operators defined as

\begin{eqnarray*}{\cal F}_{kij}(\overrightarrow{O\mbox{\boldmath$ x $ }}) &=& \f...
...2\mu}\frac{\partial}{\partial x_k}\frac{\partial}{\partial x_i}.
\end{eqnarray*}




Ken-ichi Yoshida
2001-07-28