View source for Neutron cross section from ferromagnetic spin waves

For calculating the neutron scattering cross section for ferromagnetic spin waves, we begin with the master equation \eqref{eq:magn_cross_master} and use the description of spin waves above.

We describe the spin-spin correlation functions in terms of the "coordinates" \(+\), \(-\) and \(z\). The sum of the \(z\)-component of the spins, \(S^z=\sum_{j}s_{j}^z\), commute with \(\hat{H}\), making this sum a constant of motion. Hence, the number of \(+\) and \(-\) operators must be the same for the operator to have an expectation value. It is then easy to see that spin correlation functions of the types \(\langle s^+(0) s^+(t) \rangle\) and \(\langle s^+(0) s^z(t)\rangle\) vanish, leaving only three non-zero terms: \(\langle s^z(0) s^z(t)\rangle\), \(\langle s^+(0) s^-(t) \rangle\), and \(\langle s^-(0) s^+(t) \rangle\). Since \(S^z\) is a constant of motion, we have \(s_{j}^z(t)=s_{j}^z(0)\), and hence the \(zz\) term gives rise to elastic scattering. 

The spin wave cross section comes from the term

\begin{align}\label{dummy132877240}
\langle s_{j}^-(0) s_{j'}^+(t) \rangle 
  &= \dfrac{1}{N} \displaystyle\sum_{\mathbf q'} \exp(i{\mathbf q'}\cdot ({\mathbf r}_{j'}-{\mathbf r}_j))
    \langle S_{\mathbf q'}^-(0) S_{\mathbf q'}^+(t) \rangle \\
  &= \dfrac{2\langle S^z \rangle}{N} \displaystyle\sum_{\mathbf q'} \exp(i{\mathbf q'}\cdot ({\mathbf r}_{j'}-{\mathbf r}_{j}))
    \exp(-i \omega_{\mathbf q'}t) n_{\rm B}\biggr( \frac{\hbar \omega_{\mathbf q'}}{k_{\rm B}T} \biggr) .\nonumber
\end{align}

We note the value of the expectation value \(\langle S_{\bf q'}^- S_{\bf q'}^+ \rangle = 2 \langle S^z \rangle n_{\rm B}(\hbar\omega_{\bf q'}/k_{\rm B}T)\), where the thermal population factor for Bosons is given by [[Scattering from lattice vibrations#label-eq:nB|this equation]] on the [[Scattering from lattice vibrations]] page<!--(\ref{eq:nB})-->.

Since \([S_{\bf q'}^+,S_{\bf q'}^-]=2 S^z\), we immediately have

\begin{align}\label{dummy1302615768}
\langle s_{j}^+(0) s_{j'}^-(t) \rangle 
  &= \dfrac{1}{N} \displaystyle\sum_{\mathbf q'} \exp(-i{\mathbf q'}\cdot ({\mathbf r}_{j'}-{\mathbf r}_j))
    \langle S_{\mathbf q'}^-(0) S_{\mathbf q'}^+(t) + 2 S^z \rangle \\
  &= \dfrac{2 \langle S^z \rangle}{N} \displaystyle\sum_{\mathbf q'} \exp(-i{\mathbf q'}\cdot ({\mathbf r}_{j'}-{\mathbf r}_j))
    \exp(i \omega_{\mathbf q'}t) \biggr( n_{\rm B}\biggr( \frac{\hbar\omega_{\mathbf q'}}{k_{\rm B}T}\biggr)+1\biggr) .  \nonumber
\end{align}

To transform back to the \(x\) and \(y\) operators, we note that the \(x\) and \(y\) axes are equivalent, whence \(\langle s_{j}^x(0)s_{j'}^x(t)\rangle = \langle s_{j}^y(0)s_{j'}^y(t)\rangle\). Moreover, the sum of the two mixed terms \(\langle s_{j}^x(0)s_{j'}^y(t) + s_{j}^y(0)s_{j'}^x(t)\rangle\) is seen to be zero. We calculate directly

\begin{equation} \label{dummy1512931648}
\langle s_{j}^x(0)s_{j'}^x(t)\rangle = 
  \dfrac{1}{4}\langle s_{j}^+(0)s_{j'}^-(t) + s_{j}^-(0) s_{j'}^+(t) \rangle .  
\end{equation}

This allows us to calculate the coordinate sum entering the cross section expression:

\begin{align}\label{dummy1389285453}
\displaystyle\sum_{\alpha\beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right)
\langle s_{j}^\alpha(0)s_{j'}^\beta(t)\rangle &=
(2-\hat{q}_x^2-\hat{q}_y^2) \langle s_{j}^x(0)s_{j'}^x(t)\rangle \\
&= \dfrac{1}{4} (1+\hat{q}_z^2) \langle s_{j}^+(0)s_{j'}^-(t) 
  + s_{j}^-(0)s_{j'}^+(t) \rangle  .\nonumber
\end{align}

Inserting into equation \eqref{eq:magn_cross_master}, the expression for the inelastic neutron scattering cross section for a ferromagnet reads: 

\begin{align}\label{dummy2135302037}
\left( \dfrac{d^2\sigma}{d\Omega dE_{\rm f}}\right)_{\text{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{\langle S^z \rangle}{2} \\
 &\quad\times 
    \dfrac{(2\pi)^3}{v_0} \displaystyle\sum_{ {\mathbf q},{\boldsymbol\tau}} 
    \biggr[ \biggr(n_{\text{B}}\biggr( \frac{\hbar\omega_{\mathbf q}}{k_{\text{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\quad  + n_{\text{B}}\biggr( \frac{\hbar\omega_{\mathbf q'}}{k_{\text{B}}T} \biggr) \delta\left(\hbar\omega_{\mathbf q'}+\hbar\omega \right) 
       \delta\left( {\mathbf q'} + {\mathbf q} - {\boldsymbol\tau} \right) \biggr], \nonumber
\end{align} 

where \({\boldsymbol\tau}\) is a reciprocal lattice vector and \(v_0\) is the volume of the unit cell. We can see that the cross section depends upon energy and temperature only through the thermal population factor, \(n_{\rm B}\), and the ordered moment, \(\langle S^z \rangle\). At low temperatures, compared to the spin wave energy, the spin wave spectral weight is constant, apart from the slow variation with \(q\) of the magnetic structure factor, \(F(q)\), and the Debye-Waller factor, \(\exp(-2W)\). Notice that the spin wave intensity is highest when \({\bf q}\) is in the direction of the ordered moment. This can be understood by the spin wave being a purely transverse excitation.