I want to show that
$ \frac{1}{k^2} \int \, (\nabla \times \vec M)\cdot (\nabla \times \vec M)^* d^3 r= \int \vec M \cdot \vec M^*d^3 r$.
$\vec M$ is defined as
$\vec M_{jm} = i\frac{m}{\sin \theta} e^{i m\phi} P_j^m (\cos \theta) z_j(k r) \hat e_{\theta} - \,e^{i m \phi} \frac{d P_j^m (\cos \theta)}{d \theta} z_j (k r) \hat e_{\phi} $,
where j is an integer number $>0$, $m$ is between $-j$ and $j$, $P_j^m$ are the Legendre Polynomials, $z_j (kr)$ is the spherical Bessel function and the unit vectors $\hat e_i$ are those of spherical coordinates. The divergence of $\vec M$ is equal to zero.
Has anyone an idea how to show this? I couldn't find a formula like this in Bronstein or other formula tables and I couldn't figure it out by myself.
If $\mathbf M$ is a constant vector field then $\nabla \times {\bf M} = {\mathbf 0}$, in which case the left integral is $0$. If $\mathbf M$ is nonzero, the right integral is positive (assuming the domain of integration has positive measure anyway), and so the equation cannot hold for general $\mathbf M$.