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Dipole matrix elements: $\hat{O} = {\mathbf r}$

For $k_a \neq k_b$ we have for each component $j = 1 \dots d$ of the dipole (position) operator,

$\displaystyle \left\langle a\right\vert r_j\left\vert b\right\rangle _{\mathcal{D}}$ $\textstyle =$ $\displaystyle \frac{1}{\varepsilon ^2} \oint_\Gamma \!\! d{\mathbf s} \,
...r_j (a^* n_m \partial_m b - a^*\!{\leftrightarrow}b)
\rule{0in}{0.25in} \right.$  
    $\displaystyle \hspace{1in} \left. \rule{0in}{0.25in}
- \; 2 ( a^* n_m \partial_m \partial_j b + a^*\!{\leftrightarrow}b) \right\} ,$ (H.22)

where as before $\varepsilon \equiv k_a^2 - k_b^2$. This follows from row 3 of the matrix ${\mathcal{T}}^{-1}$. Summation over $m$ is implied. A mixture of vector and Einstein notations has been avoided. In the Dirichlet BC case this becomes
\left\langle a\right\vert r_j\left\vert b\right\rangle _{\m...
...l_n a^* \partial_n b , \hspace{0.5in}
\mbox{(Dirichlet BCs)}.
\end{displaymath} (H.23)

Note that the integral is proportional to the matrix element $(\partial {\mathcal{H}} / \partial x)_{ab}$ of the billiard deformation corresponding to translation in the $j$ direction.

No form for $k_a = k_b$ has been found. The corresponding unit vector $e_\alpha = \delta_{\alpha 3}$ does not lie in the row space of ${\mathcal{T}}^{-1}$, indicating that other boundary derivatives are needed as input.

Alex Barnett 2001-10-03