Neutron cross section of antiferromagnetic spin waves

For the calculation of the neutron scattering cross section for antiferromagnetic spin waves, we again invoke the master equation \eqref{eq:magn_cross_master} and use the solutions for the antiferromagnetic spin waves above.

Firstly, we calculate the correlation functions. As for the ferromagnetic spin waves, the \(zz\) term is elastic and other terms involving \(z\) vanish. also the \(++\) and \(--\) terms vanish. Hence, we need only to calculate

\begin{align} \left\langle S_{\mathbf q}^- S_{\mathbf q}^+ \right\rangle &= \left\langle (u_{\mathbf q} \hat\alpha_{\mathbf q}^+ + v_{\mathbf q}\hat\beta_{\mathbf q}) (u_{\mathbf q}\hat\alpha_{\mathbf q} + v_{\mathbf q}\beta_{\mathbf q}^+) \right\rangle \label{dummy141100595}\\ &= u_{\mathbf q}^2 n_{\mathbf q} + v_{\mathbf q}^2 (n_{\rm B}(\hbar\omega_{\mathbf q}/k_{\rm B}T)+1) .\label{dummy1267455531} \end{align}

Transforming to real space and \(x-y\) coordinates, the final expression reads:

\begin{align}\label{dummy785616823} \left( \dfrac{d^2\sigma}{d\Omega dE_{\rm f}}\right)_{\rm magn.} &= \left(\gamma r_0 \right)^2 \dfrac{k_{\rm f}}{k_{\rm i}} \left[ \dfrac{g}{2} F(q)\right]^2 \exp(-2W)\left( 1+\hat{q}_z^2 \right) \dfrac{1}{4} \dfrac{(2\pi)^3}{v_0} \\ &\quad \times \displaystyle\sum_{ {\mathbf q},{\boldsymbol\tau} ,a} \biggr[ \biggr(n_{\rm B}\biggr( \dfrac{\hbar\omega_{a, \mathbf q}}{k_{\rm B}T} \biggr) +1\biggr) \delta\left(\hbar\omega_{\mathbf q'}-\hbar\omega \right) \delta\left( {\mathbf q'} - {\mathbf q} - {\boldsymbol\tau} \right) \nonumber\\ &\qquad\qquad\qquad + n_{\rm B}\biggr( \dfrac{\hbar\omega_{a, \mathbf q}}{k_{\rm B}T} \biggr) \delta\left(\hbar\omega_{\mathbf q'}+\hbar\omega \right) \delta\left( {\mathbf q'} + {\mathbf q} - {\boldsymbol\tau} \right) \biggr] \nonumber\\ &\quad \times \left[ u_{\mathbf q}^2 + v_{\mathbf q}^2 + 2 u_{\mathbf q} v_{\mathbf q} \cos({\boldsymbol\rho} \cdot {\boldsymbol\tau}) \right] , \nonumber \end{align}

where \({\boldsymbol\tau}\) is a reciprocal lattice vector of a sublattice, \({\boldsymbol\rho}\) is a vector joining the sublattices, and \(a\) is the sublattice index. The difference between this cross section and the one for ferromagnetic spin waves is the presence of the coherence factor, \(u_{\bf q}^2 + v_{\bf q}^2 + 2 u_{\bf q} v_{\bf q} \cos({\boldsymbol\rho} \cdot {\boldsymbol\tau})\). It should be noted that the term \(\cos({\boldsymbol\rho} \cdot {\boldsymbol\tau})\) assumes the value 1 for \({\boldsymbol\tau}\) being a nuclear peak and \(-1\) for \({\boldsymbol\tau}\) being an AFM peak.

We now proceed to calculate the cross section in a few simple cases.

*The simple nearest neighbour antiferromagnet

In some lattice types, like the bcc lattice, all nearest neighbours are ions from the opposite sublattice. Hence, for a nearest neighbour system, \(J_1({\bf q})=0\). We assume absence of anisotropy, \({\bf B}_{\rm A}=0\). Hence, the expressions simplify to

\begin{align} \Omega_{\mathbf q} &= 2S \sqrt{J({\mathbf 0})^2 - J({\mathbf q})^2} , \label{dummy235263662}\\ u_{\mathbf q}^2 + v_{\mathbf q}^2 &= \dfrac{(2S)^2 J({\mathbf 0})}{\Omega_{\mathbf q}} , \label{dummy983085806}\\ 2 u_{\mathbf q} v_{\mathbf q} &= \dfrac{-(2S)^2 J({\mathbf q})}{\Omega_{\mathbf q}} .\label{dummy80770084} \end{align}

For scattering close to an AFM peak (\(\cos({\boldsymbol\rho} \cdot {\boldsymbol\tau})=-1\)), the coherence factor can be written as

\begin{equation}\label{eq:coh_simple} \left[ u_{\mathbf q}^2 + v_{\mathbf q}^2 + 2 u_{\mathbf q} v_{\mathbf q} \cos({\boldsymbol\rho} \cdot {\boldsymbol\tau}) \right] = 2 S \sqrt{\dfrac{J({\mathbf 0})+J({\mathbf q})}{J({\mathbf 0})-J({\mathbf q})}} , \end{equation}

which diverges as \(1/\Omega_{\bf q}\) for low values of \(|{\bf q}|\). Close to a nuclear Bragg peak (\(\cos({\boldsymbol\rho} \cdot {\boldsymbol\tau})=1\)), the solution is the reciprocal of equation \eqref{eq:coh_simple}, having a value of zero for \(|{\bf q}|=0\). At the zone boundary, \(J({\bf q})=0\), and the coherence factor equals \(2S\).

*Antiferromagnetic nanoparticles in zero field

In nanoparticles, the spin wave spectrum is quantized. However, only the \({\bf q}=0\) mode has ever been observed. The energy of the lowest mode, the collective magnetic precessions, is thus completely determined by the anisotropy:

\begin{align} \Omega_{\mathbf 0} &= g \mu_{\rm B} B_{\rm A} ,\label{dummy728084531}\\ u_{\mathbf 0}^2 + v_{\mathbf 0}^2 &= \dfrac{(2S)^2 J({\mathbf 0})+g \mu_{\rm B} B_{\rm A}} {\Omega_{\mathbf 0}} ,\label{dummy2008801156}\\ 2 u_{\mathbf 0} v_{\mathbf 0} &= \dfrac{-(2S)^2 J({\mathbf 0})}{\Omega_{\mathbf 0}} .\label{dummy597972028} \end{align}

For \({\boldsymbol\tau}\) being an AFM Bragg peak, \(\cos({\boldsymbol\rho} \cdot {\boldsymbol\tau})=-1\), the coherence factor is approximately

\begin{equation}\label{eq:coh_nano} \left[ u_{\mathbf 0}^2 + v_{\mathbf 0}^2 + 2 u_{\mathbf 0} v_{\mathbf 0} \cos({\boldsymbol\rho} \cdot {\boldsymbol\tau}) \right] \approx 2 S \dfrac{2 J({\mathbf 0})}{g \mu_{\rm B} B_{\rm A}} , \end{equation}

which usually is a very large number. On the other hand, for \({\boldsymbol\tau}\) being a nuclear Bragg peak, \(\cos({\boldsymbol\rho} \cdot {\boldsymbol\tau})=1\), the coherence factor equals unity, much lower than for the AFM Bragg peak. This explains why collective magnetic excitations are not observed at the position of nuclear Bragg peaks.