Correlation between nuclear and magnetic scattering

To describe the total scattering from a magnetic material, we must treat both the nuclear and the magnetic contributions to the scattering length. We therefore expand (\ref{eq:diffraction}) to reach. \begin{equation} \label{eq:diffraction_nandm} \left.\frac{d\sigma}{d\Omega}\right|_{\rm coh} = \biggr|\sum_j \left(b_j + b_{{\rm m},j}\right) \exp(i {\bf q} \cdot {\bf r}_j ) \biggr|^2 \, . \end{equation} Expanding the square, we obtain a double sum \begin{equation} \left.\frac{d\sigma}{d\Omega}\right|_{\rm coh} = \sum_{j,j'} (b_j + b_{{\rm m},j})(b_{j'} + b_{{\rm m},j'}) \exp(i {\bf q} \cdot ({\bf r}_j-{\bf r}_{j'})) \, . \end{equation} We now rearrange this to obtain four double sums: \begin{eqnarray} \label{eq:foursums} \left.\frac{d\sigma}{d\Omega}\right|_{\rm coh} &=& \sum_{j,j'} (b_j \exp(i {\bf q} \cdot {\bf r}_j)) (b_{j'}\exp(- i {\bf q} \cdot {\bf r}_{j'}))\\ \nonumber &+& \sum_{j,j'} (b_{{\rm m},j} \exp(i {\bf q} \cdot {\bf r}_j)) ( b_{j'}\exp(- i {\bf q} \cdot {\bf r}_{j'})) \\ \nonumber &+& \sum_{j,j'} (b_j \exp(i {\bf q}\cdot {\bf r}_j)) (b_{{\rm m},j'}\exp(- i {\bf q} \cdot {\bf r}_{j'})) \\ \nonumber &+& \sum_{j,j'} (b_{{\rm m},j} \exp(i {\bf q} \cdot {\bf r}_j)) ( b_{{\rm m},j'}\exp(- i {\bf q} \cdot {\bf r}_{j'})) \, . \end{eqnarray}

Let us for simplicity redefine the constituents of (\ref{eq:foursums}) as

\begin{eqnarray} N({\bf q}) &=& \sum_j b_j \exp(i {\bf q} \cdot {\bf r}_j) \nonumber \\ {\bf M}({\bf q}) &=& \gamma r_0 F_{\rm m}({\bf q}) \sum_j {\bf S}_{j,\perp} \exp(i {\bf q} \cdot {\bf r}_j) . \end{eqnarray}

Here, the symbol \(N({\bf q})\) is the Fourier transform of the nuclear scattering length, and \({\bf M}({\bf q})\) represents the Fourier transform of the magnetic scattering length except the dot product with the nuclear spin \({\bf \sigma}\).

With these definition, the cross section (\ref{eq:foursums}) becomes

\begin{equation} \label{eq:blumeminusone} \left.\frac{d\sigma}{d\Omega}\right|_{\rm coh} = |N({\bf q})|^2 + {\bf M}({\bf q}) \cdot {\bf M}^*({\bf q}) + 2 \Re \left\{ {\bf \sigma} \cdot {\bf M}({\bf q}) N^*({\bf q})\right\} . \end{equation}

To be able to continue this derivation, we have to choose an axis of quantization of the neutron spin. For convenience, let us take the \(z\)-direction. We define the neutron polarization along this direction as

\begin{equation} P_z = \langle \sigma_z \rangle = \frac{n^\uparrow- n^\downarrow}{n^\uparrow + n^\downarrow} , \end{equation}

where \(n^\uparrow\) represents the number of neutrons in the beam that has its spin up`` and similar for spin down``. (In quantum mechanics, the spin of a spin-\(1/2\) particle along any given direction can take only two values). The maximal value of \(P_z\) is unity. The beam can also be polarized in other directions, leading to the polarization vector \({\bf P}\), which has a maximal length on unity.

Most neutron scattering is performed with unpolarzed neutron beams. This means that for any direction, there will be as many spin up`` as spin down`` neutrons, meaning that \(\langle \sigma_z \rangle = 0\). This also holds for \(\langle \sigma_x \rangle\) and \(\langle \sigma_y \rangle\). When we now take the average cross section over the polarization direction, the term \({\bf \sigma} \cdot {\bf M}({\bf q}) N^*({\bf q})\) will be averaged to \(\langle {\bf \sigma} \rangle \cdot {\bf M}({\bf q}) N^*({\bf q})\), which vanishes, leaving only the \(|N({\bf q})|^2\) and \(|{\bf M}({\bf q})|^2\) terms non-zero.

We hereby see that for diffraction with unpolarized neutrons, the nuclear-magnetic interference term vanish, and therefore the scattering cross section is simply a sum of the nuclear and the magnetic term:

\begin{equation} \label{eq:nuclearmagnetic} \left.\frac{d\sigma}{d\Omega}\right|_{\rm coh} = |N({\bf q})|^2 + {\bf M}({\bf q}) \cdot {\bf M}^*({\bf q}) = \left.\frac{d\sigma}{d\Omega}\right|_{\rm nuc, coh} +\left.\frac{d\sigma}{d\Omega}\right|_{\rm magn, coh} \, . \end{equation}

It should already here be said that the expression (\ref{eq:blumeminusone}) is not entirely correct. In the (rare, but possible) case that both the magnetic structure has a chiral order, e.g. a spiral structure, and the neutron beam is polarized, an additional term will appear in the cross section : \begin{equation} {\bf P} \cdot \Im \left\{ {\bf M}({\bf q}) \times {\bf M}^*({\bf q}) \right\}. \end{equation}

This term will be elaborated in a future chapter on polarized neutron scattering.

Experimental considerations

Also magnetic scattering can be incoherent. In a magnetic system at sufficiently high temperatures, the spin directions are random, also known as a paramagnet. Such an unordered phase gives rise to uncorrelated directions of \({\bf b}_{{\rm m},j}\), and therefore incoherent scattering. This scattering does not correlate with the nuclear incoherent scattering, as will be shown in section \ref{sect:paramagnet}. Hence, also the incoherent cross section can be written as a sum of a nuclear and a magnetic contribution.