Past exam of the condensed matter physics major option course of the University of Oxford / 2018 / 1 / c

For neutron energy loss, $E_f=30-20=10\mathrm{meV}$. Write $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$. The neutron dispersion $E={\hslash }^2{k}^2/(2m_n)$, or $E[\mathrm{meV}]=2.0721k[{\mathring{\mathrm{A}}}^{-1}{]}^2$, givesA triangle of these wavevectors exists ifWith $c=4.50\mathring{\mathrm{A}}$, the requested momentum transfers have $Q=(2\pi /c)|l|=1.3963|l|{\mathring{\mathrm{A}}}^{-1}$. ConsequentlyFor example $|l|=1$ is too small even for forward scattering at these unequal neutron energies, and $|l|=5$ exceeds the backscattering maximum. This identifies accessible reciprocal lattice points, not guaranteed nonzero intensity: a one-phonon structure factor contains the atomic scattering lengths, eigenvectors and factors $\mathbf{Q}\cdot {\mathbf{e}}_j$. A transverse eigenvector perpendicular to $\mathbf{Q}$ may be invisible, and basis interference can cause extinctions. The supplied geometry and cell constants do not specify that full intensity selection rule.

For $l=2$, $Q=2.7925{\mathring{\mathrm{A}}}^{-1}$. The scattering triangle givesTo state signed sample angles without an implicit axis convention, take the incident beam along $+x$, positive angular rotation counterclockwise, and at $\omega =0$ choose $+a$ along $+x$ and $+c$ along $+y$. ThenThe positive-$l$ condition is the vector equation ${\mathbf{k}}_i-{\mathbf{k}}_f=Q\hat{\mathbf{c}}$. PutThe two detector sides then give, modulo full rotations,These pairs obey the stated same-sense sign convention and transfer momentum along $+c$. The lower-right panel below illustrates the first choice. If the initial $+c$ direction is chosen on the opposite side of $+a$, a commonly used positive-angle pair is $(55.20^{\circ },46.51^{\circ })$. Merely taking absolute values without fixing that crystal-axis convention can exchange $l=+2$ and $l=-2$. The measurable scattering angle and the vector-triangle construction are invariant.