D.10 The curl is Hermitian

For later reference, it will be shown that the curl operator, $\nabla\times$ is Hermitian. In other words,

\begin{displaymath}
\int_{\rm all} \skew3\vec A^*\cdot\nabla\times \vec B {\rm...
...nabla\times\skew3\vec A^*\cdot \vec B {\rm d}^3{\skew0\vec r}
\end{displaymath}

The rules of engagement are as follows:

In index notation, the integral in the left hand side above reads:

\begin{displaymath}
\sum_i \int A_i^* (B_{{\overline{\overline{\imath}}},{\ove...
...h}}}-B_{{\overline{\imath}},{\overline{\overline{\imath}}}} )
\end{displaymath}

which is the same as

\begin{displaymath}
\sum_i \int [ (A_i^* B_{\overline{\overline{\imath}}})_{\o...
...A_{i,{\overline{\overline{\imath}}}}^* B_{\overline{\imath}}]
\end{displaymath}

as can be checked by differentiating out the first two terms. Now the third and fourth terms in the integral are $\nabla$ $\times$ $\skew3\vec A^*\cdot\vec{B}$, as you can see from moving all indices in the third term one unit forward in the cyclic sequence, and those in the fourth term one unit back. (Such a shift does not change the sum; the same terms are simply added in a different order.)

So, if the integral of the first two terms is zero, the fact that curl is Hermitian has been verified. Note that the terms can be integrated. Then, if the system is in a periodic box, the integral is indeed zero because the upper and lower limits of integration are equal. An infinite domain will need to be truncated at some large distance $R$ from the origin. Then shift indices and apply the divergence theorem to get

\begin{displaymath}
- \int_S (\skew3\vec A^* \times \vec B)\cdot {\hat\imath}_r {\,\rm d}S
\end{displaymath}

where $S$ is the surface of the sphere $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R$ and ${\hat\imath}_r$ the unit vector normal to the sphere surface. It follows that the integral is zero if $\skew3\vec A$ and $\vec{B}$ go to zero at infinity quickly enough. Or at least their cross product has to go to zero quickly enough.