Problem:Derivation of the cartesian formulation of the perpendicular spin component
Expand the steps in equation this equation from the Magnetic neutron scattering page:
\begin{equation} {\bf s}_{j \perp} \cdot {\bf s}_{j' \perp} = {\bf s}_j \cdot {\bf s}_{j'} - ({\bf s}_j \cdot \hat{\bf q})({\bf s}_{j'} \cdot \hat{\bf q}) = \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) s_j^\alpha s_{j'}^\beta. \end{equation}
Use the definition of the scalar product: \({\bf a} \cdot {\bf b} = \sum_\alpha a^\alpha b^\alpha\)
Use that \(\sum_{\alpha \beta} \delta_{\alpha \beta} a^\alpha b^\beta = \sum_{\alpha} a^\alpha b^\alpha\)
The middle term in the expression is given directly from the left term using the definition above. From there, we expand the dot products: \begin{align}\nonumber &{\bf s}_j \cdot {\bf s}_{j'} - ({\bf s}_j \cdot \hat{\bf q})({\bf s}_{j'} \cdot \hat{\bf q})\\\nonumber &= \sum_{\alpha} s_j^\alpha s_{j'}^\alpha - \sum_{\alpha}\hat{q}_\alpha s_j^\alpha \times \sum_{\beta}\hat{q}_\beta s_{j'}^\beta\\\nonumber &= \sum_{\alpha \beta} \delta_{\alpha \beta} s_j^\alpha s_{j'}^\beta - \sum_{\alpha}\hat{q}_\alpha s_j^\alpha \times \sum_{\beta}\hat{q}_\beta s_{j'}^\beta\\\nonumber &= \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) s_j^\alpha s_{j'}^\beta \end{align}