Scattering of neutrons from magnetic ions

We now develop the formalism for magnetic neutron scattering. This is performed in a general way that gives us an expression for the elastic and inelastic scattering simultaneously.

*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} = - \boldsymbol\mu \cdot {\bf B} = - \gamma \mu_{\rm N} \hat{\boldsymbol\sigma} \cdot {\bf B} , \end{equation}

where \(\hat{\boldsymbol\sigma}\) represents the three Pauli matrices for the neutron.

The external field that scatters the neutron comes from the individial electrons, that combine to ionic moments as described above.

Since the magnetic moment of an electronic spin, \({\bf s}_j\) at position \({\bf r}_j\) is given by

\begin{equation} \boldsymbol\mu_{\rm e} = - g \mu_{\rm B} {\bf s}_j \end{equation}

and the field from a dipole can be described as

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

equation \eqref{eq:nuclear_Zeeman} becomes

\begin{equation} H_{{\rm Z},j} = \frac{\mu_0}{4\pi} g \mu_{\rm B} \gamma \mu_{\rm N} \hat{\boldsymbol\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, \(\hat{V}\), in the master equation for neutron scattering from the Basics of neutron scattering page. As when developing the inelastic nuclear cross section, the Scattering from lattice vibrations page, we must perform a thermal average over the initial states of the sample, \(|\lambda_{\rm i} \rangle\), and sum over the final states, \(|\lambda_{\rm f}\rangle\), which are consistent with the observed momentum transfer, \({\bf q}\) and energy transfer, \(\hbar \omega\).

By simple substitution, the resulting equation for the scattering cross section becomes

\begin{align} \label{eq:cross_spinonly} \left. \frac{d^2 \sigma}{d\Omega dE_{\rm f}} \right|_{\sigma_{\rm i} \rightarrow \sigma_{\rm f} } &= \frac{k_{\rm i}}{k_{\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 \sum_{\lambda_{\rm i},\lambda_{\rm f}} p_{\lambda_{\rm i}} \\ &\quad\times \biggr| \biggr\langle {\bf k}_{\rm f} \lambda_{\rm f} \sigma_{\rm f} \biggr| \sum_j \hat{\boldsymbol\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} \lambda_{\rm i} \sigma_{\rm i} \biggr\rangle \biggr|^2 \nonumber \\ &\quad\times \delta\left( \hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}} \right) . \nonumber \end{align}

*The magnetic matrix element

We now turn to the calculation of the complicated matrix element in \eqref{eq:cross_spinonly}. We utilize the mathematical identity

\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} \lambda_{\rm f} \sigma_{\rm f} \biggr| \sum_j \hat{\boldsymbol\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} \lambda_{\rm i} \sigma_{\rm i} \biggr\rangle \nonumber \\ &\quad= \frac{1}{2\pi^2} \biggr\langle \lambda_{\rm f} \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{\boldsymbol\sigma} \cdot (\hat{\bf q}' \times ({\bf s}_j \times \hat{\bf q}')) \biggr| \lambda_{\rm i} \sigma_{\rm i} \biggr\rangle \nonumber \\ &\quad= 4\pi \biggr\langle \lambda_{\rm f} \sigma_{\rm f} \biggr| \sum_j \exp(i {\bf q} \cdot {\bf r}_j) \hat{\boldsymbol\sigma} \cdot (\hat{\bf q} \times ({\bf s}_j \times \hat{\bf q})) \biggr| \lambda_{\rm i} \sigma_{\rm i} \biggr\rangle . \label{eq:magn_matrix} \end{align}

In the last step we used that the integration over \(d^3{\bf r}\) gives \((2\pi)^3 \delta({\bf q}+{\bf q}')\). 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 \eqref{eq:magn_matrix} and \eqref{eq:spinperp} reveal that the spin component parallel to \({\bf q}\) is invisible to neutrons. This is a completely general result and is essential to all magnetic neutron scattering.

*Matrix element for unpolarized neutrons

For the remainder of this page, 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 and that we can therefore sum over \(\sigma_{\rm f}\) and average over the initial spin state, \(\sigma_{\rm i}\).

We now calculate the spin part of the matrix element \eqref{eq:cross_spinonly} using \eqref{eq:magn_matrix}:

\begin{equation} \label{eq:magn_matrix_spin} \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| \hat{\boldsymbol\sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 . \end{equation}

Now, the dot product will contain terms of the type \(\sigma^x s_{\perp}^x\), where the first factor depends only on the neutron spin coordinate, \(\sigma\), and the second only on the sample coordinate, \(\lambda\). We further assume that the initial neutron state is not correlated with the initial state of the sample. Hence, we can factorize the two inner products:

\begin{align} \label{eq:magn_matrix_spin2} &\sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| \hat{\boldsymbol\sigma} \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 \\ &\quad= \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \biggr| \sum_\alpha \left\langle \sigma_{\rm f} \left| \sigma^\alpha \right| \sigma_{\rm i} \right\rangle \left\langle \lambda_{\rm f} \left| {\bf s}_{\perp}^\alpha \right| \lambda_{\rm i}\right\rangle \biggr|^2 \nonumber \\ &\quad= \sum_{\alpha, \beta, \sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\beta | \sigma_{\rm f} \rangle \langle \sigma_{\rm f} | \sigma^\alpha | \sigma_{\rm i} \rangle \langle \lambda_{\rm i} | {\bf s}_{\perp}^\beta | \lambda_{\rm f}\rangle \langle \lambda_{\rm f} | {\bf s}_{\perp}^\alpha | \lambda_{\rm i}\rangle \nonumber \\ &\quad= \sum_{\alpha, \beta, \sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\beta \sigma^\alpha | \sigma_{\rm i} \rangle \langle \lambda_{\rm i} | {\bf s}_{\perp}^\beta | \lambda_{\rm f}\rangle \langle \lambda_{\rm f} | {\bf s}_{\perp}^\alpha | \lambda_{\rm i}\rangle , \nonumber \end{align}

where we in the last step have used the completeness relation \(\sum_{\sigma_{\rm f}} |\sigma_{\rm f}\rangle \langle \sigma_{\rm f} | = 1\). For unpolarized neutrons, \(\alpha = \beta\) leads to

\begin{equation} \sum_{\sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\alpha \sigma^\beta | \sigma_{\rm i} \rangle = 1. \end{equation}

Likewise, if \(\alpha \neq \beta\), we have that \(\sum_{\sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i} | \sigma^\alpha \sigma^\beta | \sigma_{\rm i} \rangle = 0\). Using this to perform the sum over \(\sigma_{\rm i}\), we obtain

\begin{equation} \label{eq:magn_matrix_spin3} \sum_{\sigma_{\rm i}, \sigma_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| \boldsymbol\sigma \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 = \sum_\alpha \left\langle \lambda_{\rm i}| s_\perp^\alpha | \lambda_{\rm f}\right\rangle \left\langle \lambda_{\rm f}| s_\perp^\alpha | \lambda_{\rm i}\right\rangle . \end{equation}

When summed over the final states, \(|\lambda_{\rm f} \rangle\), we obtain

\begin{equation} \sum_{\sigma_{\rm i}, \sigma_{\rm f}, \lambda_{\rm f}} p_{\sigma_{\rm i}} \left| \left\langle \sigma_{\rm f} \lambda_{\rm f} \left| \boldsymbol\sigma \cdot {\bf s}_{\perp} \right| \sigma_{\rm i} \lambda_{\rm i}\right\rangle \right|^2 = \left\langle \lambda_{\rm i} \left| {\bf s}_{\perp} \cdot {\bf s}_{\perp} \right| \lambda_{\rm i}\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'} - ({\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 scattering

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}} = \gamma r_0 , \end{equation}

where \(r_0\) is the classical electron radius \(r_0=e^2\mu_0/(4\pi m_{\rm e})=2.818\) fm.

We now further utilize that we can express the perpendicular spin component as

\begin{equation} s^\alpha_{\perp} = \sum_\beta (\delta_{\alpha,\beta}-\hat{q}_\alpha \hat{q}_\beta) s^\alpha , \end{equation}

where the indices \(\alpha\) and \(\beta\) run over the Cartesian coordinates (\(x\), \(y\), and \(z\)).

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

\begin{align} \frac{d^2 \sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \nonumber \\ &\times \sum_{\lambda_{\rm i} \lambda_{\rm f}} p_{\lambda_{\rm i}} \left\langle \lambda_{\rm i}|Q_\alpha|\lambda_{\rm f}\right\rangle \left\langle \lambda_{\rm f}|Q_\beta|\lambda_{\rm i}\right\rangle \delta\left( \hbar\omega + E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}\right) , \end{align}

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} = \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 the section on Magnetic ions. 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{equation} {\bf Q} = \sum_{j}\int \exp(i {\bf q} \cdot ({\bf r}_j+{\bf r})) {\bf s}_j d^3{\bf r} = \sum_{j} \exp(i {\bf q} \cdot {\bf r}_{j}) {\bf S}_{j} F({\bf q}) . \end{equation}

Here, the magnetic form factor is given by

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

where \(s({\bf r})\) is the normalised spin density. For small values of \(q\), the magnetic form factor is close to unity, \(F(0)=1\), 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.

Orbital contributions

When taking contributions from orbital magnetism into account, e.g. from rare-earth ions, the spin operator \({\bf s}\) is replaced by \(g_{\rm L}{\bf J}/2\), 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 (in the equations 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 Marshall.

The final magnetic cross section

Performing all replacements above, the modified cross section for magnetic neutron scattering reads:

\begin{align} \label{eq:magnetic_master_almost_final} \frac{d^2\sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2 \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \\ &\quad \times \sum_{\lambda_{\rm i} \lambda_{\rm f}} p_{\lambda_{\rm i}} \sum_{j,j'} \big\langle \lambda_{\rm i}| \exp(-i {\bf q} \cdot {\bf r}_{j}) {\bf s}_{j}^\alpha | \lambda_{\rm f}\big\rangle \big\langle \lambda_{\rm f}\big| \exp(i {\bf q} \cdot {\bf r}_{j'}) {\bf s}_{j'}^\beta \big| \lambda_{\rm i}\big\rangle \nonumber \\ &\quad \times \delta\left( \hbar\omega + E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}\right) . \nonumber \end{align}

In \eqref{eq:magnetic_master_almost_final}, denotes the nuclear positions, which are not fixed. The quantum mechanical way of handling this is by treating ${\bf r}_j$ are operators, give rise to phonon scattering, as described earlier. We will here only consider magnetic scattering that does not involves phonons, i.e. is elastic in the phonon channel. Hence, we can interpret \({\bf r}_j\) as the lattice positions, while multiplying the magnetic scattering cross section with the Debye-Waller factor $\exp(-2W)$.

\begin{align} \label{eq:magnetic_master_final} \left.\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right|_{\rm magn} &= \exp(-2W) \left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2 \sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \\ &\quad \times \sum_{\lambda_{\rm i} \lambda_{\rm f}} p_{\lambda_{\rm i}} \sum_{j,j'} \big\langle \lambda_{\rm i}| \exp(-i {\bf q} \cdot {\bf r}_{j}) {\bf s}_{j}^\alpha | \lambda_{\rm f}\big\rangle \big\langle \lambda_{\rm f}\big| \exp(i {\bf q} \cdot {\bf r}_{j'}) {\bf s}_{j'}^\beta \big| \lambda_{\rm i}\big\rangle \nonumber \\ &\quad \times \delta\left( \hbar\omega + E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}\right) . \nonumber \end{align}