Past exam of the physics A course of the University of Cambridge / 2011 / Paper 1 / Section C / C10

For this calculation define the net wavevector gained by the neutron as $\mathbf{Q}={\mathbf{k}}_f-{\mathbf{k}}_i$, the opposite of a common sample-transfer convention. Since $\mathbf{Q}\perp {\mathbf{k}}_i$,The perpendicular neutron recoil and phonon absorption condition therefore requires $\omega (Q)=\hslash Q^2/(2m_n)$. In the supplied periodic dispersion, interpret its $q$ as this transfer $Q$ in the extended-zone representation of a phonon dispersion. Put $x=Qa/2$ and $C=m_{n}av/\hslash \simeq 15.84$. Nonzero intersections obeyThere is no root in $0<x\le \pi /2$: $\sin x\ge 2x/\pi$ there and $C>{\pi }^2/4$. On $(\pi /2,\pi )$ the ratio $x^2/\sin x$ increases strictly from ${\pi }^2/4$ to infinity, so exactly one root occurs. Nor can there be a higher root: $x\ge \sqrt{C}$ gives $x^2\ge C\ge C|\sin x|$, with strict inequality at the possible boundary. For $\pi \le x\le \sqrt{C}$, use $|\sin x|\le x-\pi$; $x^2-C(x-\pi )>0$ throughout this short interval, since its minimum there is $C(1+\pi -\sqrt{C})>0$. Thus the only nonzero magnitude satisfiesSolving the exact dispersion, which is permitted instead of the linear approximation, givesA straight chord between the zone edge and the next zone centre gives $\omega \simeq 4v/a-2vQ/\pi$. Its intersection solves $(\hslash /2m_n)Q^2+(2v/\pi )Q-4v/a=0$, yielding $Q\simeq 1.006\times 10^{10}{\mathrm{m}}^{-1}$ and $f\simeq 0.508\mathrm{THz}$ as a less accurate estimate.

The reduced-zone phonon wavevector is $q_{\mathrm{red}}=Q-2\pi /a=-1.873\times 10^9{\mathrm{m}}^{-1}$ along the chosen positive transfer direction. The difference is a reciprocal-lattice vector, not an additional independent phonon outside the first zone. At low temperature the absorption probability contains the Bose-Einstein distribution factor $n_B(\omega )=[e^{\hslash \omega /k_{B}T}-1{]}^{-1}$, which tends to zero. Here $\hslash \omega /k_B\simeq 27.6\mathrm{K}$, so this absorption channel is strongly suppressed well below that scale. elastic scattering and kinematically allowed spontaneous phonon creation need not disappear.

For elastic scattering, $E_f=E_i$ and ${\mathbf{k}}_f-{\mathbf{k}}_i=\mathbf{G}$, where $\mathbf{G}$ is a reciprocal lattice vector. In the ideal static crystal, coherent scattering obeys this lattice selection rule; the crystal as a whole can take the mechanical recoil with negligible energy. crystal momentum is defined modulo $\hslash \mathbf{G}$, not as unconstrained free-particle momentum.

For annihilation of a phonon $(\mathbf{q},\omega )$,For phonon creation,The signs follow whether the neutron gains or loses the phonon's energy and crystal momentum. These are one-phonon coherent inelastic neutron scattering conditions, with $\mathbf{G}$ accounting for lattice periodicity.