The magnetic scattering length

Unpaired electrons may introduce a magnetic moment localized on (some of) the atoms in the material. The details of this is elaborated in Elastic magnetic scattering, but let us for now assume that each atom has an effective spin, \({\bf s}_j\), that is zero for non-magnetic atoms. The magnetic dipole moment generated by the spin is \begin{equation} \label{eq:emoment} {\bf m}_j = g \mu_{\rm B} {\bf s}_j , \end{equation} where \(g=2.0023\) is the electronic magnetogyric ratio (the g-factor) and \(\mu_{\rm B} = e\hbar/(2m_{\rm e}) = 9.2741 \cdot 10^{-24}\)~J/T is known as the Bohr magneton.

As we discussed earlier, the neutron also has a magnetic moment, given by \begin{equation} {\bf \mu} = \gamma \mu_{\rm N} {\bf \sigma} , \end{equation}

where \(\gamma= - 1.91304\) is the neutron magnetogyric ratio and the nuclear magneton is given by \(\mu_{\rm N} = e \hbar / (2 m_{\rm p}) = 5.05078 \cdot 10^{-27}\)~J/T. \({\bf \sigma}\) represents the direction of the nuclear spin and has a length of unity. (For readers with a knowledge of quantum mechanics, \({\bf \sigma}\) is given by the three Pauli matrices.)

Each magnetic moment in the material gives rise to a magnetic dipole field, and during the scattering process, the neutron is affected by the forces that arises from variations of this field. These forces give rise to scattering in a rather similar way as the scattering of neutrons from nuclei that we discussed earlier. The detailed calculation of this scattering is lengthy and involves plenty of electromagnetism and quantum mechanics. For this reason, we present it elsewhere; in Quantum mechanics of magnetic diffraction.

The final result of the magnetic scattering, however, can be stated in three simple points:

• The scattering involves only \({\bf s}_{j,\perp} \), which is the component of \({\bf s}_j\) perpendicular to the scattering vector, \({\bf q}\).
• The magnetic scattering length decays for larger values of \(|{\bf q}|\) due to the smearing`` of the atomic spins across their respective orbitals.
• The magnetic scattering length depends on the relative orientation of the nuclear spin, \(\bf \sigma\), and the atomic spin, \({\bf s}_j\).

In brief, the magnetic scattering length for the individual atom, \(j\), can be written as

\begin{equation} b_{{\rm m},j} = \gamma r_0 {\bf \sigma} \cdot {\bf s}_{j,\perp} F_{\rm m}({\bf q}), \end{equation}

where \(r_0=e^2\mu_0/(4\pi m_{\rm e})=2.8179\)~fm is the classical electron radius.

The magnetic form factor, corresponding to point 2 above, is denoted by the symbol \(F_{\rm m}({\bf q})\) and is equivalent to the atomic form factor in X-ray scattering. It is given by

\begin{equation} F({\bf q}) = \int \exp(i {\bf q} \cdot {\bf r}) \rho_s({\bf r}) d^3{\bf r} \, , \end{equation}

where \(\rho_s({\bf r})\) is the normalised spin density on the unfilled orbitals. For small values of \(q\), below around 2~\AA , the magnetic form factor is close to unity, \(F(| {\bf q}| )=1\), and it falls off smoothly to zero for large scattering vectors.

In the following, we assume that the magnetic form factor is identical for all magnetic ions in the material under investigation, even though this may be too simple an approach, in particular for materials containing more than one magnetic element.

Experimental considerations

The magnitude of the magnetic scattering length is similar to the nuclear scattering lengths, since \(\gamma r_0 = 5.39\)~fm and \(s F(q) g/2\) is of the order unity. Hence, the magnetic and nuclear scattering cross sections have similar magnitudes. For \(S > 1\), e.g. in some transition metal salts, like Fe-oxides, or in most rare-earth compounds, the magnetic scattering can even completely dominate the diffraction patterns.