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Differing wavenumbers $k_a \neq k_b$

The following is an application of Green's theorem. I start with the vector identity

\begin{displaymath}
\nabla \cdot (a^* \nabla b) \; = \;
a^* \nabla^2 b + \nabla a^* \cdot \nabla b .
\end{displaymath} (H.3)

Integrating over $\mathcal{D}$, applying Gauss' Theorem to the LHS and Helmholtz to the first term on the RHS gives
\begin{displaymath}
\oint_\Gamma \!\! d{\mathbf s} \,a^* \partial_n b \; = \; \...
...{\mathbf r} \,
( -k_b^2 a^* b + \nabla a^* \cdot \nabla b) .
\end{displaymath} (H.4)

Note that $\partial_n$ is an abbreviation for ${\mathbf n} \cdot \nabla$, the local normal derivative operator at a location ${\mathbf s}$ on $\Gamma $, and should not be confused with the cartesian derivatives $\partial_i$, $\partial_j$, etc which appear later. Subtraction of the same equation with the swap $a^*\!{\leftrightarrow}b$ gives the useful overlap formula
\begin{displaymath}
\left\langle a \left\vert b\right.\right\rangle _{\mathcal{...
...ma \!\! d{\mathbf s} \,(a^* \partial_n b - b \partial_n a^*) .
\end{displaymath} (H.5)

This formula is undefined for $k_a = k_b$, where the integral vanishes. Note that if $a$ and $b$ are chosen to obey the same (general mixed) BCs, the formula gives zero as one would expect for orthogonal eigenstates.



Alex Barnett 2001-10-03