Inelastic magnetic neutron scattering: Difference between revisions
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Latest revision as of 22:15, 18 February 2020
In this section we will specialize the master equation for magnetic scattering to describe spin waves. We start with repeating the magnetic scattering master equation from the Magnetic neutron scattering page:
\begin{align}\label{dummy1597252741} \dfrac{d^2\sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \dfrac{k_{\rm f}}{k_{\rm i}} \left[ \dfrac{g}{2} F(q)\right]^2 \displaystyle\sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{ {\mathbf q}}_\alpha\hat{ {\mathbf q}}_\beta\right) \\ &\quad \times \displaystyle\sum_{\lambda_{\rm i} \lambda_{\rm f}} p_{\lambda_{\rm i}} \displaystyle\sum_{j,j'} \left\langle \lambda_{\rm i}| \exp(-i {\mathbf q} \cdot {\mathbf r}_{j}) {\mathbf s}_{j}^\alpha | \lambda_{\rm f}\right\rangle \left\langle \lambda_{\rm f}\left| \exp(i {\mathbf q} \cdot {\mathbf r}_{j'}) {\mathbf s}_{j'}^\beta \right| \lambda_{\rm i}\right\rangle \nonumber\\ &\quad \times \delta\left( \hbar\omega + E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}\right) . \nonumber \end{align}
The next step closely follows the treatment of phonon scattering. First, we rewrite the energy-conserving \(\delta\)-function in terms of a time integral. To repeat the equation from the Scattering from lattice vibrations page:
\begin{equation}\label{eq:delta_transform} \delta(\hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}}) = \dfrac{1}{2\pi\hbar} \displaystyle\int_{-\infty}^{\infty} dt \exp\left(-\dfrac{i(\hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}})t}{\hbar}\right) . \end{equation}
Subsequently, we transform the stationary operators, here denoted \(\hat{A}\), to time-dependent Heisenberg operators, \(\hat{A}(t)\), through
\begin{align}\label{dummy1091740695} &\biggr\langle \lambda_{\rm i} \biggr| \exp\biggr( \dfrac{i E_{\lambda_{\rm i}} t}{\hbar} \biggr) \hat{A} \exp\biggr( -\dfrac{i E_{\lambda_{\rm f}} t}{\hbar} \biggr) \biggr| \lambda_{\rm f} \biggr\rangle \\ &\quad= \biggr\langle \lambda_{\rm i} \biggr| \exp\biggr( \dfrac{i \hat{H}_\lambda t}{\hbar} \biggr) \hat{A} \exp\biggr( -\dfrac{i \hat{H}_\lambda t}{\hbar} \biggr) \biggr| \lambda_{\rm f} \biggr\rangle \nonumber\\ &\quad= \langle \lambda_{\rm i} | \hat{A}(t) | \lambda_{\rm f} \rangle .\nonumber \end{align}
Making this transformation and performing the closure sum over the final states, (\(\sum_{\lambda_{\rm f}}|\lambda_{\rm f}\rangle\langle\lambda_{\rm f}| = 1\)), we reach the equation for the magnetic scattering cross section in the Heisenberg picture[1]:
\begin{align}\label{dummy72416411} \dfrac{d^2\sigma}{d\Omega dE_{\rm f}} &= \left(\gamma r_0 \right)^2 \dfrac{k_{\rm f}}{k_{\rm i}} \left[ \dfrac{g}{2} F(q)\right]^2 \displaystyle\sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{q}_\alpha\hat{q}_\beta\right) \\ &\quad\times \dfrac{1}{2\pi\hbar} \displaystyle\sum_{j,j'} \displaystyle\int_{-\infty}^{\infty} dt \exp(-i \omega t) \nonumber\\ &\quad\times \left\langle \exp(-i {\mathbf q} \cdot {\mathbf r}_{j}(0)) {\mathbf s}_{j}^\alpha(0) \exp(i {\mathbf q} \cdot {\mathbf r}_{j'}(t)) {\mathbf s}_{j'}^\beta(t) \right\rangle , \nonumber \end{align}
where \(\langle \hat{A} \rangle\) means the thermal average of the expectation value of an operator \(\hat{A}\), in the same way as the calculations leading to the phonon expression on the Scattering from lattice vibrations page.
The nuclear positions \({\bf r}(t)\) and the atomic spin \({\bf s}(t)\) are both operators. Each of these give rise to both an elastic and an inelastic contribution; in total the cross section becomes a sum of four terms. The one that is elastic in both channels is the magnetic diffraction signal, while the contribution that is elastic in the nuclear positions (the phonon channel) and inelastic in the spins (the magnetic channel) is the true magnetic inelastic signal. The two other contributions involve creation/annihilation of phonons via the magnetic interactions, and are not discussed further here; we refer to Ref. [2].
The two contributions that are elastic in the phonon channel are reached in analogy with the page Scattering from lattice vibrations. We replace the general nuclear position operators \({\bf r}_j\) with the nuclear equilibrium positions in a lattice and utilize the periodicity of the lattice. Furthermore, we multiply with the Debye-Waller factor, \(\exp(-2W)\)[3]:
\begin{align} \label{eq:magn_cross_master} \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) \displaystyle\sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{ {\mathbf q}}_\alpha\hat{ {\mathbf q}}_\beta\right) \\ &\quad\times N \dfrac{1}{2\pi\hbar} \displaystyle\int_{-\infty}^{\infty} dt \exp(-i \omega t) \displaystyle\sum_{j'} \exp(i {\mathbf q} \cdot ({\mathbf r}_j)) \left\langle {\mathbf s}_{0}^\alpha(0) {\mathbf s}_{j'}^\beta(t) \right\rangle . \nonumber \end{align}
This equation should be understood as the space and time Fourier transform of the spin-spin correlation function, and it is the starting point for most calculations of magnetic scattering cross sections.
- ↑ W. Marshall and S.W. Lovesey. Theory of Thermal Neutron Scattering (Oxford, 1971), equation (8.2)
- ↑ W. Marshall and S.W. Lovesey. Theory of Thermal Neutron Scattering (Oxford, 1971), equation (8.10)
- ↑ W. Marshall and S.W. Lovesey. Theory of Thermal Neutron Scattering (Oxford, 1971), equation (8.13) and (8.25)