Quantum mechanics of magnetic diffraction

We now develop the quantum mechanical formalism for elastic magnetic neutron scattering. The formalism for inelastic magnetic scattering is developed in Scattering theory for magnetic dynamics.

The magnetic interaction

The interaction responsible for magnetic neutron scattering is the nuclear Zeeman term for a neutron in an external magnetic field:

\begin{equation} \label{eq:nuclear_Zeeman} H_{\rm Z} = - {\bf \mu} \cdot {\bf B} = - \gamma \mu_{\rm N} \hat{\bf \sigma} \cdot {\bf B} , \end{equation}

where \(\hat{\bf \sigma}\) represents the spin of the neutron and is composed by the Pauli matrices.

The external field that scatters the neutron comes from the individial electrons on the atoms.

The magnetic moment of one electronic spin, \({\bf s}_j\), is given by (\ref{eq:emoment})

The field from a dipole placed at the Origin can be described as

\begin{equation} B = \frac{\mu_0}{(4\pi)} \nabla \times ({\bf \mu} \times {\bf r}/r^3). \end{equation}

We now position the spin at position \({\bf r}_j\), and transform (\ref{eq:nuclear_Zeeman}) into

\begin{equation} H_{{\rm Z},j} = \frac{\mu_0}{4\pi} g \mu_{\rm B} \gamma \mu_{\rm N} \hat{\bf \sigma} \cdot \nabla\times\left(\frac{{\bf s}_j\times({\bf r}-{\bf r}_j)}{|{\bf r}-{\bf r}_j|^3}\right) . \end{equation}

The neutron interaction with the magnetic ions is given as the total nuclear Zeeman interaction, summed over all magnetic sites, \(j\). This we use as the scattering potential,

\begin{equation} \label{eq:magnetic_potential} \hat{V} = \sum_j H_{{\rm Z},j} , \end{equation}

that we insert in the master equation for magnetic diffraction (\ref{eq:master_scatt}).

\begin{equation} \frac{d\sigma}{d\Omega} = \frac{k_{\rm f}}{k_{\rm i}} \left(\frac{m_{\rm n}}{2\pi \hbar^2}\right)^2 \big| \big\langle \psi_{\rm i} \big|\hat{V}\big| \psi_{\rm f}\big\rangle \big|^2 . \end{equation}

By substitution, the resulting equation for the magnetic diffraction cross section becomes

\begin{align} \label{eq:diffract_spinonly} \left. \frac{d \sigma}{d\Omega} \right|_{\sigma_{\rm i} \rightarrow \sigma_{\rm f} } &= \left( \frac{\mu_0}{4\pi} \right)^2 \left( \frac{m_{\rm N}}{2\pi\hbar^2} \right)^2 \left( g \mu_{\rm B} \gamma \mu_{\rm N} \right)^2 \\ &\quad\times \biggr| \biggr\langle {\bf k}_{\rm f} \sigma_{\rm f} \biggr| \sum_j \hat{\bf \sigma} \cdot \nabla\times\left(\frac{{\bf s}_j \times({\bf r}-{\bf r}_j)}{|{\bf r}-{\bf r}_j|^3}\right) \biggr| {\bf k}_{\rm i} \sigma_{\rm i} \biggr\rangle \biggr|^2 . \nonumber \end{align}

We here note that we have explicitly added the spin value to the initial and final state of the neutron.

The magnetic matrix element

We now turn to the calculation of the complicated matrix element in (\ref{eq:diffract_spinonly}).

We utilize a mathematical identity \cite{squires}, \begin{equation} \nabla \times \left( \frac{{\bf s} \times {\bf r}}{r^3} \right) = \frac{1}{2\pi^2} \int \hat{\bf q}' \times ({\bf s} \times \hat{\bf q}') \exp(i {\bf q}' \cdot {\bf r}) d^3{\bf q}' \end{equation} to reach \begin{align} &\biggr\langle {\bf k}_{\rm f} \sigma_{\rm f} \biggr| \sum_j \hat{\bf \sigma} \cdot \nabla\times\left(\frac{{\bf s}_j \times({\bf r}-{\bf r}_j)}{|{\bf r}-{\bf r}_j|^3}\right) \biggr| {\bf k}_{\rm i} \sigma_{\rm i} \biggr\rangle \nonumber \\ &\quad= \frac{1}{2\pi^2} \biggr\langle {\bf k}_{\rm f} \sigma_{\rm f} \biggr| \sum_j \int d^3{\bf q'} \hat{\bf \sigma} \cdot (\hat{\bf q}' \times ({\bf s}_j \times \hat{\bf q}')) \exp(i {\bf q}' \cdot ({\bf r}-{\bf r}_j)) \biggr| {\bf k}_{\rm i} \sigma_{\rm i} \biggr\rangle \nonumber\\ &\quad= \frac{1}{2\pi^2} \biggr\langle \sigma_{\rm f} \biggr| \sum_j \int d^3{\bf r} d^3{\bf q'} \exp(i {\bf q}\cdot {\bf r}) \nonumber\\ &\quad\quad\times \exp(i {\bf q}' \cdot ({\bf r}-{\bf r}_j)) \hat{\bf \sigma} \cdot (\hat{\bf q}' \times ({\bf s}_j \times \hat{\bf q}')) \biggr| \sigma_{\rm i} \biggr\rangle \nonumber \\ &\quad= 4\pi \biggr\langle \sigma_{\rm f} \biggr| \sum_j \exp(i {\bf q} \cdot {\bf r}_j) \hat{\bf \sigma} \cdot (\hat{\bf q} \times ({\bf s}_j \times \hat{\bf q})) \biggr| \sigma_{\rm i} \biggr\rangle . \label{eq:magn_matrix} \end{align}

To perform the last step we used that \(\int \exp(i ({\bf q}+{\bf q'})\cdot {\bf r}) d^3 {\bf r} = (2\pi)^3 \delta({\bf q}+{\bf q}')\). The equation contains a term

\begin{equation} \label{eq:spinperp} \hat{\bf q} \times ({\bf s}_j \times \hat{\bf q}) \equiv {\bf s}_{j,\perp}, \end{equation}

which is simply the component of the spin on site \(j\) perpendicular to the scattering vector. Equation (\ref{eq:magn_matrix}) therefore proves that the spin component parallel to \({\bf q}\) is invisible to neutrons, as mentioned in The magnetic scattering length.

Matrix element for unpolarized neutrons

For the remainder of this notes, we assume that the neutrons are unpolarized, \(p_\uparrow = p_\downarrow = 1/2\). We also assume that we do not observe the final spin state, \(\sigma_{\rm f}\), of the neutron. To obtain the cross section for unpolarised neutrons, we therefore sum over \(\sigma_{\rm f}\) and average over the initial spin state, \(\sigma_{\rm i}\).

We save this derivation for later and just state the result here:

\begin{equation} \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \left| {\bf \sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \right\rangle \right|^2 = \left\langle \left| {\bf s}_{\perp} \cdot {\bf s}_{\perp} \right| \right\rangle. \end{equation}

When this expression is used in the calculation for the cross section, we will encounter terms of the general type \({\bf s}_{j \perp} \cdot {\bf s}_{j' \perp}\). We here utilize that the perpendicular projection is defined as \({\bf s}_{j \perp} \equiv {\bf s}_j - ({\bf s}_j \cdot \hat{\bf q}) \hat{\bf q}\), where \(\hat{\bf q}\) is a unit vector in the direction of \({\bf q}\), to reach

\begin{equation} \label{eq:cartesian_perp} {\bf s}_{j \perp} \cdot {\bf s}_{j' \perp} = {\bf s}_j \cdot {\bf s}_{j'} - 2({\bf s}_j \cdot \hat{\bf q})({\bf s}_{j'} \cdot \hat{\bf q}) + ({\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}

where the indices \(\alpha\) and \(\beta\) run over the Cartesian coordinates (\(x\), \(y\), and \(z\)), and \(\hat{q}^{\alpha}\) and \(s_j^{\alpha}\) etc. are now scalar variables.

The master equation for magnetic diffraction

We now collect the prefactors from the calculations above, assuming that the proton and neutron masses are identical: \begin{equation} \frac{m_{\rm N}}{2\pi\hbar^2} g \mu_{\rm B} \gamma \mu_{\rm N} \mu_0 = \gamma \frac{\mu_0}{4\pi}\frac{e^2}{m_{\rm e}}\frac{g}{2} = \gamma r_0 \frac{g}{2}, \end{equation}

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

Collecting all equations, we end up with the master equation for the magnetic differential scattering cross section for unpolarized neutrons:

\begin{equation} \label{eq:magnetic_master_diffract} \frac{d \sigma}{d\Omega } = \left(\gamma r_0 \right)^2 \left(\frac{g}{2}\right)^2 \frac{k_{\rm f}}{k_{\rm i}} \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \left\langle Q_\alpha \right\rangle \left\langle Q_\beta \right\rangle , \end{equation}

where we have defined \({\bf Q}\) as the Fourier transform of the spins \({\bf s}_j\)

positioned at \({\bf r}_j\), with respect to the scattering vector, \({\bf q}\):

\begin{equation} {\bf Q}({\bf q}) = \sum_j \exp(i {\bf q} \cdot {\bf r}_j) {\bf s}_j . \end{equation}

The magnetic form factor

We assume the electrons causing the magnetism to be located in orbitals around particular ions as discussed in Elastic magnetic scattering. The electron coordinates are therefore replaced by the nuclear positions, \({\bf r}_j\), plus a small deviation from this, \({\bf r}\), representing the extension of the particular electron orbital. We thus make the substitution

\begin{eqnarray} {\bf Q}({\bf q}) &=& \sum_{j}\int \exp(i {\bf q} \cdot ({\bf r}_j+{\bf r})) {\bf s}_j d^3{\bf r} \nonumber \\ &=& \sum_{j} \exp(i {\bf q} \cdot {\bf r}_{j}) {\bf s}_{j} F({\bf q}) \nonumber \\ &=& {\bf M}({\bf q}) F({\bf q}), \end{eqnarray} where \({\bf M}({\bf q})\) and \(F({\bf q})\) are defined as in Correlation between nuclear and magnetic scattering.

Orbital contributions

When taking contributions from orbital magnetism into account, e.g. from rare-earth ions, the term \(g{\bf s}\) is replaced by \(g_{\rm L}{\bf J}\), where \(g_{\rm L}\) is the Land\'e factor:

\begin{equation} g_{\rm L} = 1 + \frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)} , \end{equation}

which is a number between 1 and 2, and \({\bf J}\) is the total angular momentum However, in the expressions to follow we keep the notation \({\bf s}\) for simplicity.

The derivation of the contribution from orbital moment is lengthy and adds nothing to the general understanding of magnetic neutron scattering, so we simply omit it here. Details of this derivation are found in Theory of Thermal Neutron Scattering.