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

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.

For the electronic and lattice heat-capacity crossover in a monovalent metal, to obtain a numerical estimate from $E_F$ alone, assume monovalent potassium is a three-dimensional free electron gas with a parabolic band. Including both spins, the molar density of states is $g(E_F)=3N_A/(2E_F)$. Using the supplied approximate coefficient $3$, rather than silently replacing it by ${\pi }^2/3$, givesEquating it to the stated low-temperature Debye heat capacity yields the nonzero crossoverWe used $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$ and $E_F=2\mathrm{eV}$. This is self-consistently far below ${\theta }_D=100\mathrm{K}$ and the Fermi temperature. Using the exact Sommerfeld coefficient instead gives $0.954\mathrm{K}$; that is a change of approximation, not a discrepancy in the algebra. Without a band/DOS assumption, the supplied Fermi energy does not by itself determine $g(E_F)$.

For independent conduction-electron states labelled by band, wavevector and spin, combine those labels into $j$. ThenHere $g$ counts the same spin and band states as the sum, and $\mu (T)$ is fixed by the number constraint $N=\int g{f}_{\mathrm{FD}}dϵ$. The energy zero is a fixed band reference. Use the total DOS for total internal energy or the molar DOS for molar internal energy; these conventions must not be mixed.

In the harmonic approximation, sum over wavevector $\mathbf{q}$ and phonon branch $s$:The occupation factor is the Bose-Einstein distribution with zero phonon chemical potential; $\frac{1}{2}\hslash \omega$ is the zero-point energy. A temperature-independent zero-point term does not contribute to the heat capacity in this fixed-frequency approximation.

Only states within an energy shell of width roughly $k_{B}T$ around $E_F$ appreciably change occupation at low temperature. There are roughly $g(E_F)k_{B}T$ such states and each excitation costs energy of order $k_{B}T$. Thus $\Delta U\propto g(E_F)(k_{B}T{)}^2$, giving $C_{el}\propto g(E_F)k_B^{2}T$. This is the thermal-shell estimate of electronic heat capacity; filled states far below $E_F$ are blocked by the Pauli exclusion principle. A smooth DOS and fixed particle number give the more precise Sommerfeld result $C_{el}=({\pi }^2/3)g(E_F)k_B^{2}T$ to leading order.

For three-dimensional acoustic modes, $\omega \propto q$ and the mode count in a spherical shell is proportional to $q^{2}dq$, giving phonon DOS proportional to ${\omega }^2$. Thermally occupied modes have $\hslash \omega$ of order $k_{B}T$, so $U_{la}-U_0\propto \int {\omega }^3{n}_B(\omega )d\omega \propto T^4$. Differentiation gives $C_{la}\propto T^3$. At $T\gg {\theta }_D$, all $3N$ harmonic modes are classically populated; each has mean energy $k_{B}T$ by equipartition. Thus the molar heat capacity tends to $3R$, the Dulong-Petit law. Anharmonic expansion and electronic contributions are not included in that lattice limit.