Wikiadmin: 1 revision imported

2020-02-18T22:15:13Z

1 revision imported

← Older revision	Revision as of 22:15, 18 February 2020
(No difference)

ucph>Tommy at 22:44, 28 August 2019

2019-08-28T22:44:10Z

New page

We now develop the formalism for inelastic magnetic neutron scattering.

===The magnetic master equation for inelastic scattering===

First, we lift the requirement from [[Quantum mechanics of magnetic diffraction]] that the state of the sample is unchanged in the scattering process.
We insert the scattering potential (\ref{eq:magnetic_potential})
in the master equation for inelastic neutron scattering (\ref{eq:master_scatt_inel}).

We note that we do not measure the final state of the sample.
Furthermore, we assume that the sample is initially in thermal equilibrium.
Therefore, we perform a thermal average over the initial states of the sample,
\(|\lambda_{\rm i} \rangle\), and sum over those final states,
\(|\lambda_{\rm f}\rangle\), which
are consistent with the observed momentum transfer, \({\bf q}\) and energy
transfer, \(\hbar \omega\).
This treatment is similar to the derivation of the inelastic nuclear cross section, [[Scattering theory for nuclear dynamics]].

By 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{\bf \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}

===Matrix element for unpolarized neutrons===

We start with the spin part of the matrix element (\ref{eq:cross_spinonly}), transformed
using (\ref{eq:magn_matrix}) to reach
\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{\bf \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, \(\lambda\).
The initial neutron state cannot be 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{\bf \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 in the form,
\(\sum_{\sigma_{\rm f}} |\sigma_{\rm f}\rangle \langle \sigma_{\rm f} | = 1\).
For unpolarized neutrons, \(\alpha = \beta\) leads to
\begin{equation}
\sum_{\sigma_{\rm i}} p_{\sigma_{\rm i}} \langle \sigma_{\rm i}
| \sigma^\alpha \sigma^\alpha | \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| {\bf \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}
Summing over the final states, \(|\lambda_{\rm f} \rangle\), we again use the completeness relation to 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| {\bf \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}

===The master equation for magnetic scattering===

Collecting the prefactors as in the previous section and all other equations from above,
we end up with {\em the master equation}
for the partial differential magnetic scattering cross section for unpolarized neutrons <ref name="marshall">W. Marshall and S.W. Lovesey. Theory of Thermal Neutron Scattering. Oxford, 1971</i> (Wiley, 2001)</ref>:

\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}

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

\begin{align} \label{eq:magnetic_master_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}

===Inelastic magnetic neutron scattering from a lattice===
As a final part of this derivation, we will now specialize the master equation
for magnetic scattering on a lattice, starting from (\ref{eq:magnetic_master_final}):

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 (\ref{eq:delta_hw}):
\begin{equation} \label{eq:delta_transform}
\delta(\hbar\omega+E_{\lambda_{\rm i}}-E_{\lambda_{\rm f}})
= \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} dt
\exp\left(-\frac{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}
&\biggr\langle \lambda_{\rm i} \biggr| \exp\biggr( \frac{i E_{\lambda_{\rm i}} t}{\hbar} \biggr)
\hat{A} \exp\biggr( -\frac{i E_{\lambda_{\rm f}} t}{\hbar} \biggr) \biggr| \lambda_{\rm f} \biggr\rangle \\
&\quad= \biggr\langle \lambda_{\rm i} \biggr| \exp\biggr( \frac{i \hat{H}_\lambda t}{\hbar} \biggr)
\hat{A} \exp\biggr( -\frac{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 completeness 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:

\begin{align}
\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 \frac{1}{2\pi\hbar} \sum_{j,j'} \int_{-\infty}^{\infty} dt
\exp(-i \omega t) \nonumber\\
&\quad\times \left\langle \exp(-i {\bf q} \cdot {\bf R}_{j}(0)) {\bf s}_{j}^\alpha(0)
\exp(i {\bf q} \cdot {\bf R}_{j'}(t)) {\bf 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 (\ref{eq:sum_lambda_f}).

The nuclear positions \({\bf R}(t)\) and the atomic spin \({\bf s}(t)\)
are both operators.
Each of these gives 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.~<ref name="marshall">W. Marshall and S.W. Lovesey. Theory of Thermal Neutron Scattering. Oxford, 1971</i> (Wiley, 2001)</ref>.

The two contributions that are elastic in the phonon channel
are reached in analogy with [[Scattering from lattice vibrations]].
We replace the general nuclear position operators \({\bf R}_j(t)\) with the
nuclear equilibrium positions in a lattice, \({\bf r}_j\)

and utilize the periodicity of the lattice.
Furthermore, we multiply with the Debye-Waller factor, \(\exp(-2W)\):

\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))
\left\langle {\bf s}_{j}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle \exp(-i \omega t) dt. \nonumber
\end{align}
As a final step, we utilize the translational symmetry of the lattice to see that the double sum of \(j\) and \(j'\) equals \(N\) times the sum over \(j-j'\).
The final result becomes:
\begin{align} \label{eq:magn_cross_master}
\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'})
\left\langle {\bf s}_{0}^\alpha(0) {\bf s}_{j'}^\beta(t) \right\rangle \exp(-i \omega t) dt. \nonumber
\end{align}

Scattering theory for magnetic dynamics - Revision history

Wikiadmin: 1 revision imported

ucph>Tommy at 22:44, 28 August 2019