Formalism I - Resonant Magnetic Xray Scattering Intensity

The resonant magnetic scattering intensity is calculated as outlined for example in [25] for the magnetic electric dipole scattering according to the squared absolute value of the scattering amplitude given for the ion $n$ in the lattice as

    $\displaystyle f_{nE1}^{RXMS}=$ (22)
    $\displaystyle =\left (
\sigma \rightarrow \sigma & \pi \right...
...ft ( \begin{array}{cc}
1 & 0 \\
0 & \cos2\Theta
\end{array} \right )
    $\displaystyle \times
\left ( \begin{array}{cc}
0 & z_1 \cos \Theta + z_3 \sin \...
...sin \Theta - z_1 \cos \Theta & -z_2 \sin 2\Theta
\end{array} \right ) + F^{(2)}$  
    $\displaystyle \times
\left ( \begin{array}{cc}
z_2^2 & -z_2(z_1 \sin \Theta - z...
...cos \Theta) & -\cos^2 \Theta (z_1^2 \tan^2 \Theta + z_3^2)
\end{array} \right )$  

Here, dropping the site index $n$, $\Theta$ is the Bragg scattering angle and $z_1, z_2$ and $z_3$ aree components of the magnetization vector with respect to the coordinate system $[\mathbf u_1,\mathbf u_2,\mathbf u_3]$. The scattering plane, defined by the direction of the incident and final wave vectors $\mathbf k$ and $\mathbf k'$, contains $\mathbf u_1$ lying perpendicular and in the sense of $\mathbf k$ and $\mathbf u_3$ parallel to the scattering vector $\mathbf Q=\mathbf k-\mathbf k'$. The scattering channels for $F^{(1)}$ and $F^(2)$ terms are considered separately and no interference is considered.

To determine the azimuthal dependence the $z_i$ are expanded in terms of the components $\mu_a,\mu_b$ and $\mu_c$ of the magnetic moment along the orthogonal crystallographic unit cell vectors $\mathbf a$,$\mathbf b$ and $\mathbf c$ (for non orthogonal crystal lattices currently magnetic xray scattering intensities cannot be calculated):

$\displaystyle z_1$ $\textstyle =$ $\displaystyle \mu_a \sin \alpha_1 \cos(\Psi+\delta_1)+\mu_b \sin \alpha_2 \cos(\Psi +\delta_2)$  
    $\displaystyle +\mu_c \sin \alpha_3 \cos (\Psi +\delta_3),$  
$\displaystyle z_2$ $\textstyle =$ $\displaystyle \mu_a \sin \alpha_1 \sin (\Psi +\delta_1)+\mu_b \sin \alpha_2 \sin (\Psi+\delta_2)$  
    $\displaystyle +\mu_c \sin \alpha_3 \sin (\Psi +\delta_3)$  
$\displaystyle z_3$ $\textstyle =$ $\displaystyle \mu_a \cos \alpha_1 + \mu_b \cos \alpha_2 + \mu_c \cos \alpha_3$ (23)

with angles $\alpha_i=\angle(\mathbf a_i \cdot \mathbf u_3)_{\Psi=0}$ and $\delta_i=\angle(\mathbf a_i^{\perp}\cdot %
\mathbf u_1)_{\Psi=0}$, where $\mathbf a_{1,2,3}=\mathbf a,\mathbf b,\mathbf c$ and $\mathbf a_i^{\perp}$ is the projection of $\mathbf a_i$ onto the plane perpendicular to $\mathbf Q$. In the chosen experimental geometry $\Psi=0$ when $\mathbf a$ points to the x-ray source.

Martin Rotter 2017-01-10