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ucph>Tommy at 19:27, 1 September 2019

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In the same way as for the inelastic nuclear scattering described above, the inelastic magnetic scattering
requires a quantum mechanical treatment. This is detailed in [[Scattering theory for magnetic dynamics]].
These quantum mechanical calculations lead directly to the observable scattering cross section,
which covers both elastic and inelastic magnetic scattering from magnetic moments on a lattice:
\begin{align}
\left(\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right)_{\rm magn.} &=
\left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2
\exp(-2W)
\sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{\bf q}_\alpha\hat{\bf q}_\beta\right)
\\
&\quad\times
\frac{1}{2\pi\hbar} \int_{-\infty}^{\infty}
\sum_{j,j'} \exp(i {\bf q} \cdot ({\bf r}_{j'}-{\bf r}_j)) %+\Deltabold_i))
\left\langle {\bf s}_{j}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle \exp(-i \omega t) dt. \nonumber
\end{align}
Due to the translational symmetry of the lattice, the index \(j\) can be chosen to be the Origin (\(j \rightarrow 0\)). We can then relabel \( ({\bf r}_{j'}-{\bf r}_j) \rightarrow {\bf r}_{j'}\), and thereby all terms in the sum over \(j\) becomes equal, giving a factor \(N\).
The final result becomes:
\begin{align} \label{eq:inel_final_mag}
\left(\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right)_{\rm magn.} &=
\left(\gamma r_0 \right)^2 \frac{k_{\rm f}}{k_{\rm i}} \left[ \frac{g}{2} F(q)\right]^2
\exp(-2W)
\sum_{\alpha \beta} \left( \delta_{\alpha\beta}-\hat{\bf q}_\alpha\hat{\bf q}_\beta\right)
\\
&\quad\times
\frac{N}{2\pi\hbar} \int_{-\infty}^{\infty}
\sum_{j'} \exp(i {\bf q} \cdot {\bf r}_j) %+\Deltabold_i))
\left\langle {\bf s}_{0}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle \exp(-i \omega t) dt. \nonumber
\end{align}
In reality, we here describe contribution that is elastic in the nuclear positions
(the phonon channel) and elastic or inelastic in the spins (the magnetic channel).
To obtain this, we have multiplied with the Debye-Waller factor, \(\exp(-2W)\).
In these note, we will not describe magnetic scattering that is inelastic in the phonon channel;
this is discussed to some extend in Squires \cite{squires}.

The expression for the cross section should be understood as the space and time Fourier transform
of the spin-spin correlation function, \(\left\langle {\bf s}_{0}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle\),
and it is the starting point for most calculations of
magnetic scattering cross sections.
We note that since the spin operators do not appear as argument of the exponential function, there is no need
to perform the series expansion of the complex exponential as needed for the lattice vibrations.

Scattering from magnetic dynamics - Revision history

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ucph>Tommy at 19:27, 1 September 2019