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ucph>Peter at 11:24, 18 September 2019

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We will now proceed to calculate the inelastic scattering cross section for
phonons in a crystal. As a first attempt, and to build intuition, we use a classical interpretation of equation (\ref{eq:sum_lambda_f})
for general inelastic nuclear scattering equation.

== Inelastic cross section of atoms in a lattice ==
As a start, we make the assumption that the nuclei vibrate around their equilibrium positions,
eq. (\ref{eq:vibration}). The cross section then becomes
\begin{align}\label{eq:vibration}
\frac{d^2\sigma}{d\Omega dE_{\rm f}} &=
\frac{k_{\rm f}}{k_{\rm i}} \sum_{j,j'}
\frac{b_j b_{j'}}{2 \pi \hbar} \exp(i {\bf q} \cdot (-{\bf r}_j+{\bf r}_{j'})) \\
&\quad \times \int_{-\infty}^{\infty}
\big\langle \exp(-i {\bf q} \cdot {\bf u}_j(0))
\exp(i {\bf q} \cdot {\bf u}_{j'}(t)) \big\rangle
\exp(-i\omega t) dt \nonumber .
%} \, .
\end{align}

We next make the assumption that the atoms are placed in a crystal, {\i.e.} with translational symmetry (as for crystal diffraction).
Then, we replace the sum over \(j\) and \(j'\) with \(N\) times the sum over \(j\), and \(-{\bf r}_j+{\bf r}_{j'}\) by \({\bf r}_j\). For reasons
of simplicity, we assume a Bravais lattice
\begin{align}
\left.\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right|_{\rm Bravais} &=
\frac{k_{\rm f}}{k_{\rm i}}
\frac{N b^2}{2 \pi \hbar} \sum_{j} \exp(i {\bf q} \cdot {\bf r}_j) \\
&\quad \times \int_{-\infty}^{\infty}
\big\langle \exp(-i {\bf q} \cdot {\bf u}_0(0))
\exp(i {\bf q} \cdot {\bf u}_{j}(t)) \big\rangle
\exp(-i\omega t) dt \nonumber .
%} \, .
\end{align}
We now utilize that the displacement from equilibrium, \(u_j(t)\), is small, so that we can Taylor expand the exponential functions.
We also utilize that \(\langle u_j(t) \rangle = 0\), so that only even powers of \(u_j(t)\) contribute, reaching:
\begin{equation}
\big\langle \exp(-i {\bf q} \cdot {\bf u}_0(0))
\exp(i {\bf q} \cdot {\bf u}_{j}(t)) \big\rangle \approx \langle 1 - ({\bf q} \cdot u_0(0))^2/2 - ({\bf q} \cdot u_j(t))^2/2
+ ({\bf q} \cdot u_0(0))({\bf q} \cdot u_j(t)) \rangle .
\end{equation}
The factor 1 and the square terms above each have a constant time average and therefore leads to diffraction. In fact, they becomes the Debye-Waller factor, already discussed under diffraction.
%ToDo: develop this better
We here concentrate on the mixed term leading to phonon cross section:
\begin{align} \label{eq:phonon_cross_classical_3}
\left.\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right|_{\rm Bravais, 1~ph.} &=
\frac{k_{\rm f}}{k_{\rm i}}
\frac{N b^2}{2 \pi \hbar} \sum_{j,j'} \exp(i {\bf q} \cdot {\bf r}_j) \\
&\quad \times \int_{-\infty}^{\infty}
\big\langle ({\bf q} \cdot {\bf u}_0(0))
({\bf q} \cdot {\bf u}_{j}(t)) \big\rangle
\exp(-i\omega t) dt \nonumber .
%} \, .
\end{align}
To calculate the cross section, we need to know the vibration function, \({\bf u}_j(t)\). However, since this is a linear problem, it is
sufficient to calculate for one vibration (or phonon) frequency, and then add the results in the end. We thus assume the vibrations
to be described by (in one dimension for simplicity)
\begin{equation} \label{eq:vibration_phase}
{\bf u}_j(t) = A_q \left( \exp(i q j a - i \omega_q t + i \phi) + {\rm complex\, conjugate}\right) ,
\end{equation}
where \(\phi\) is a phase factor.
It is important here to remember that \({\bf u}_j(t)\) is always a real number, even though it is convenient to describe it
with complex exponentials.

Inserting (\ref{eq:vibration_phase}) into (\ref{eq:phonon_cross_classical_3}), we reach terms of the form
\begin{equation}
\big\langle (\exp(i \phi) + \exp(-i \phi))
(\exp(i q j a) \exp(-i \omega_q t) \exp(i \phi) + \exp(-i q j a) \exp(i \omega_q t) \exp(-i \phi)) \big\rangle .
\end{equation}
The thermal average is in effect equivalent to averaging over the phase factor, as well as replacing \(A_q\) with \(\langle A_q \rangle\).
The above equation then simplifies to
\begin{equation}
(\exp(i q j a) \exp(-i \omega_q t) + \exp(-i q j a) \exp(i \omega_q t) ) .
\end{equation}
Inserting into (\ref{eq:phonon_cross_classical_3}), we reach
\begin{align} \label{eq:phonon_cross_classical_4}
\left.\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right|_{\rm Bravais, 1~ph.} &=
\frac{k_{\rm f}}{k_{\rm i}}
\frac{N b^2 \langle A_q^2 \rangle ({\bf q} \cdot \hat{e}_p)^2}{2 \pi \hbar} \sum_{j} \exp(i {\bf q} \cdot {\bf r}_j) \\
&\quad \times \int_{-\infty}^{\infty}
(\exp(i q j a) \exp(-i \omega_q t) + \exp(-i q j a) \exp(i \omega_q t) )
\exp(-i\omega t) dt \nonumber ,
%} \, .
\end{align}
for which the Fourier transforms over space and time are performed to give
\begin{align} \label{eq:phonon_cross_classical_5}
\left.\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right|_{\rm Bravais, 1~ph.} &=
\frac{k_{\rm f}}{k_{\rm i}}
N \frac{(2 \pi)^3}{V_0}\langle A_q^2 \rangle b^2 ({\bf q} \cdot \hat{e})^2 \sum_{bf \tau} \\
&\quad \times \left[
\delta(\omega-\omega_{q}) \delta({\bf q}-{\bf q}'+{bf \tau}) + \delta(\omega+\omega_{q}) \delta({\bf q}+{\bf q}'+{bf \tau})
\right] \nonumber ,
%} \, .
\end{align}
where \(\hat{e}\) is a unit vector in the direction of the vibration.

The full quantum mechanical description, given in section \ref{sect:phonon_quantic}, essentially agrees with the classical calculation above.
Here, the mean squared vibration amplitude, \(\langle A_q^2 \rangle\), is found to be related to the thermal Bose factor, \(n_q\). In addition, quantum effects produce a difference between the prefactors of the two sets of delta functions. The final cross section reads

\begin{align} \label{eq:cross_one_phonon}
\left.\frac{d^2\sigma}{d\Omega dE_{\rm f}}\right|_{\rm Bravais, 1~ph.} &=
\frac{k_{\rm f}}{k_{\rm i}}
\frac{b^2(2\pi)^3}{2 M V_0}
\exp(-2W) \sum_{q,p,\tau} \frac{({\bf q}\cdot {\bf e}_{q,p})^2}{\omega_{q,p}} \\
&\quad\times
\left[ (n_{q,p}+1) \delta(\omega-\omega_{q,p}) \delta({\bf q}-{\bf q}'+{bf \tau}) \right. \nonumber \\
&\qquad \left. + n_{q,p} \delta(\omega+\omega_{q,p}) \delta({\bf q}+{\bf q}'+{bf \tau}) \right]
, \nonumber
\end{align}
where a polarization index, \(p\), has been added to the phonon modes.

== Understanding the one-phonon cross section ==
We now like to discuss the interpretation of the phonon cross section (\ref{eq:cross_one_phonon}),
starting with the last two lines.
In the first term in the square brackets,
the neutron looses energy, while the lattice gains the same amount, \(\hbar\omega_{q,p}\).
Hence, this corresponds to the creation of one phonon.
Likewise, in the second term the neutron gains energy, while the lattice loses,
representing annihilation of one phonon.
The population factors, \(n_{q,p}\) and \(n_{q,p}+1\) give the physically meaningful
result that at zero temperature (\(n_{q,p}=0\)),
phonon annihilation has zero cross section (no phonons are present),
while phonon creation is still allowed.

The term \(({\bf q}\cdot {\bf e}_{q,p})^2\) in (\ref{eq:cross_one_phonon})
implies that the neutron only senses vibrations {\em parallel} to
the total scattering vector.
This intuitively makes sense, since a displacement of the nucleus perpendicular
to {\bf q} would not change the phase of the scattering.

==== Experimental considerations ====
The dependence of the scattering cross section on
\(({\bf q}\cdot {\bf e}_{q,p})^2\) has strong implications on the experimental strategy.
If, {\em e.g.}, in a cubic system you like to study longitudinal phonons,
you first like to find the highest reasonable value of \(q\), for example \((0\, 0\, 6)\).
Next, you like to investigate the scattering along \((0\, 0\, 6+l)\) to pick up the longitudinal vibrations.
In contrast, for transverse phonons, you should look along another direction, for example \((h\, 0\, 6)\), since
these phonons have polarization transverse to the \((h\, 0\, 0)\) direction, not to \({\bf q}\).

If one wants to study all vibration directions in a crystal of low symmetry,
it may be necessary to investigate the phonon spectrum in
different Brillouin zones to change the angle between the
scattering vector, \({\bf q}\), and the phonon wave vector, \({\bf q}'\).

The scattering cross section for phonons - Revision history

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ucph>Peter at 11:24, 18 September 2019