Past exam of the condensed matter physics course of the University of Oxford / 2016 (A13485W1, B6)

Source paper: www.whatdotheyknow.com/request/physics_past_exam_papers_2015_16/response/973997/attach/3/Cowell%20Attachment.zip?cookie_passthrough=1

Take $(x_j,y_j,z_j)$ as fractional coordinates in the chosen unit cell. In the kinematic approximation the crystal basis structure factor isEquivalently, in Cartesian coordinates the phase is ${\mathbf{G}}_{hkl}\cdot {\mathbf{r}}_j$. Reversing every phase sign complex-conjugates the nonabsorbing amplitude and leaves its intensity unchanged. The atomic scattering factor is the Fourier transform of the atom's electron density, in electron scattering units:At zero scattering angle the normal form factor counts electrons: $f_j(0)=Z_j$ for a neutral atom, modified by the ionic charge for an ion. At increasing $Q$, different parts of the electron cloud scatter with different phases, reducing the coherent amplitude. Typical nonresonant atomic curves fall from their forward value toward small values at large angle. A localized cloud falls more slowly in reciprocal space. Anomalous scattering near an absorption edge adds energy-dependent complex terms and is excluded from this simple sketch.

For a fixed reflection and fixed instrumental and geometric corrections,The full amplitude also contains the lattice sum. At its reciprocal-lattice peak, $N_c$ coherently illuminated identical cells give a peak amplitude $N_{c}F$; peak height and integrated peak intensity have different size dependences because the peak narrows with crystal size. Powder multiplicity, polarization, absorption, thermal motion and measurement geometry also affect observed intensities. The basis factor determines relative interference, not every experimental prefactor.

The two sites give phases one and $\exp [i\pi (h+k+l)]=(-1{)}^{h+k+l}$. ThusFor real nonresonant form factors, even $h+k+l$ gives the square of the sum and odd $h+k+l$ the square of the difference. The factor $K_{hkl}$ represents the remaining experimental corrections, not another basis-interference term. Since the two ions differ, a half-body-diagonal translation does not preserve the decorated crystal. Its Bravais lattice is therefore simple cubic, with a two-ion basis; CsCl is not a monatomic body-centered cubic Bravais lattice.

Replacing both sites by identical Fe atoms changes the amplitude toNow the body-centering translation does preserve the structure. It is a body-centered cubic lattice with one Fe atom per primitive cell, or two per conventional cubic cell.

For 111 the parity is odd: CsCl generally has a nonzero amplitude $f_{\mathrm{C}{\mathrm{s}}^{+}}-f_{\mathrm{C}{\mathrm{l}}^{-}}$, whereas the identical-Fe 111 reflection is systematically absent. A numerical absolute CsCl intensity requires the angle-dependent form factors and experimental normalization; no such values are supplied. Forward electron counts, 54 for Cs+ and 18 for Cl-, explain why their amplitudes are substantially unequal, but are not substituted for the actual finite-angle factors without an additional approximation.

The isothermal volume compressibility is $\kappa =-(1/V)(\partial V/\partial p{)}_T$. For isotropic cubic compression $V\propto a^3$ and $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$, soDifferentiate Bragg's law $2d_{hkl}\sin \theta =\lambda$ at fixed incident wavelength and fixed reflection:This derivative uses radians. It remains locally valid with a pressure-dependent compressibility, provided the cubic isotropic phase is retained. For a constant stipulated $\kappa$, integration givesThus $a_0(1-\kappa p/3)$ is only the first-order approximation, not the exact consequence of a constant logarithmic compressibility. The finite-pressure calculation also assumes that the reflection still exists, the crystal remains in the same structure and $\sin {\theta }_0{e}^{\kappa p/3}\le 1$. These limitations matter for a large extrapolation.

The 110 plane spacing of the conventional cubic cell isFor the first-order Bragg condition,The quoted experimental deflection is twice the angle between the beam and the reflecting planes; reporting only $22.298^{\circ }$ would report the Bragg angle instead of the requested scattering angle. The 110 reflection is allowed for the identical-Fe body-centered lattice because $1+1+0$ is even. Wavelength and spacing have been expressed in the same units, so their ratio is dimensionless. Higher Bragg orders are conventionally represented by the corresponding higher-index reflection, rather than inserted as an unannounced extra order here.

Use the integrated constant-compressibility model from c, with $\kappa p=0.0059\times 100=0.59$:ThereforeCompression reduces the spacing and moves the same allowed reflection to larger deflection, as the positive pressure derivative predicts. The input compressibility corresponds to a constant bulk modulus $1/\kappa \simeq 169.5$ GPa in this ideal model.

This is the stipulated unchanged cubic-phase, constant-compressibility extrapolation, rather than a reliable prediction for real Fe at 100 GPa. Real Fe exhibits a pressure-induced body-centered-cubic to hexagonal-close-packed transformation, and compressibility can change with pressure.^[ref] A real high-pressure diffraction interpretation would require its actual phase and equation of state. The source gives neither, so its numerical exercise is answered within its explicitly stated idealization.

In the free electron model, conduction electrons occupy independent plane-wave states with energy ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ in a uniform mean background. Crystal-periodic band structure and explicit electron–electron correlations are omitted. Electrons still obey the Pauli exclusion principle and Fermi–Dirac statistics; free does not mean distinguishable classical particles. A scattering time can be added phenomenologically to describe transport.

The Fermi energy is the highest occupied single-particle energy at zero temperature, measured here from the bottom of the parabolic band. The Fermi temperature is the corresponding scale $T_F=E_F/k_B$, not the actual thermal temperature of the sample. Metallic electron densities are large, so filling all distinct states up to the required momentum gives Fermi energies of several electronvolts, equivalent to tens of thousands of kelvin or more. At ordinary temperatures, only a narrow region around the Fermi surface has appreciably changing occupations.

A room-temperature metal and its lattice are nevertheless at room temperature, not at $T_F$. High occupied kinetic energies are present even in the zero-temperature ground state and cannot all be liberated as heat without violating the allowed occupancy structure. A metal below skin temperature often feels cold because it transfers heat rapidly out of the skin; its thermal conductivity and thermal effusivity matter, not the value of its Fermi temperature. At the same actual temperature, a poorly conducting material can therefore feel warmer while neither material is thermodynamically hotter.

Use periodic boundaries in a box of side $L$ and include two spin states for each wavevector. One orbital state occupies $(2\pi /L{)}^d$ in wavevector space. Let $v_d$ be the volume of the unit ball in dimension $d$, namely $2,\pi ,4\pi /3$. Filling the ball of radius $k_F$ givesSince $E_F={\hslash }^2{k}_F^2/(2m)$,The density $n$ means number per length, area or volume as appropriate; it is not assigned three-dimensional units in every case.

For the same spin-summed states, the cumulative number per $d$-dimensional volume below energy $E$ is $n(E)=2v_d(2mE/{\hslash }^2{)}^{d/2}/(2\pi {)}^d$. Differentiating yields $g(E)=(d/2)n(E)/E$. Evaluating at the filling energy,In one, two and three dimensions the density of states varies respectively as $E^{-1/2}$, a constant and $E^{1/2}$. The normalization counts both signs of momentum and both spin states exactly once. A single-spin density of states or extra valley degeneracy would require a different counting convention, not an extra unexplained factor in these constants.

A conventional face-centered cubic cell contains four atoms. With three conduction electrons per atom,Using the bare electron mass,The Drude transport relation is $\rho =m_e/(n{e}^2\tau )$, givingFor a degenerate electron gas, transport samples electrons near the Fermi surface, so use $v_F=\hslash k_F/m_e=2.025\times 10^6$ m/s:This is a transport relaxation time and associated transport mean free path inferred from the specified resistivity. It need not equal a single-particle quantum lifetime when scattering is strongly angle dependent. The free-electron valence count also neglects the detailed band structure of aluminium, as the stipulated model requires. A classical thermal speed is not the appropriate velocity scale for this strongly degenerate metal.

For one spin-degenerate parabolic two-dimensional band, $k_F=\sqrt{2\pi n}=1.3729\times 10^8$ m${}^{-1}$. The free-electron model with $m=m_e$ givesThe given $10^3\Omega$ is a sheet resistivity, not a bulk resistivity in ohm-metres. The two-dimensional Drude sheet conductance is $n{e}^2\tau /m_e$, soThus $v_F=\hslash k_F/m_e=1.589\times 10^4$ m/s andThe sample temperature is not given. This is specifically the Fermi-surface mean free path, appropriate in the degenerate regime $T\ll 8.3$ K for the bare-mass model. At higher temperatures the characteristic thermal speed should replace $v_F$ for a thermal transport distance. If a real two-dimensional band has mass $m^{\ast }$, the stated results scale as $E_F,T_F\propto 1/m^{\ast }$, $\tau \propto m^{\ast }$ and $v_F\propto 1/m^{\ast }$; the inferred ${\ell }_F=\hslash k_F/(n{e}^2{\rho }_{\square })$ is mass independent under these same Drude assumptions. No semiconductor effective mass or valley degeneracy is silently introduced.

Write the two allowed moment projections as $\pm \mu$, where $\mu =g{\mu }_B/2$ for spin one-half. Their Zeeman energies in the prescribed flux density are $\mp \mu B$. The single-particle partition function and the mean moment along the field areFor noninteracting moments, magnetization is moment per volume:For electron-like spin-only moments $g\simeq 2$, these become $M_s=N{\mu }_B/V$ and $\eta ={\mu }_B/k_B$.

Differentiate with respect to the specified $B$ to obtainThis $B$ derivative is not a dimensionless SI volume susceptibility. In the weak-feedback approximation $B\simeq {\mu }_{0}H$, the usual dimensionless differential susceptibility is ${\chi }_{\mathrm{diff}}\simeq {\mu }_0{\chi }_B$. The small-field limit gives Curie's law ${\chi }_0={\mu }_{0}N{\mu }^2/(V{k}_{B}T)$. At finite bias a secant response $M/H\simeq {\mu }_{0}M/B={\chi }_0\tanh (x)/x$ differs from the differential slope ${\chi }_0{\mathrm{sech}}^{2}x$. For an internal $B={\mu }_0(H+M)$ retained self-consistently, the exact local conversion is ${\chi }_H={\mu }_0{\chi }_B/(1-{\mu }_0{\chi }_B)$; demagnetizing geometry can further affect an applied-field measurement. Stating the field convention avoids a hidden factor of ${\mu }_0$ or confusion between a slope and a ratio.

A $3d^9$ shell has one hole in the otherwise filled $d$ shell. Hund's rules give $S=1/2$, $L=2$, and, for a more-than-half-filled shell, $J=L+S=5/2$: the free-ion term is ${}^2{D}_{5/2}$. Its Landé factor isDistinguish the Curie effective moment from a saturated projection:The question then explicitly asks us to use orbital quenching in the crystal. Ligand fields split and mix orbital states, suppressing their freely orientable angular momentum. The spin-only approximation has $g\simeq 2$, $S=1/2$, so ${\mu }_{\mathrm{eff}}=\sqrt{3}{\mu }_B$ but the two field-axis projections are $\pm {\mu }_B$. The latter, not $\sqrt{3}{\mu }_B$, belong in the tanh argument and saturation magnetization.

The sulfate and water environment supplies mainly closed-shell, weakly magnetic surroundings, stabilizes localized Cu2+ sites and separates their moments so that exchange can be weak on the temperature scale considered. Paramagnetism here is an approximation of weakly interacting local spins, not a claim that dilution guarantees no magnetic ordering at any temperature. Elemental copper has a filled $3d$ band and itinerant conduction electrons rather than one localized Cu2+ spin per atom. Pauli spin response and orbital/core contributions then replace a local-spin Curie law; ordinary pure copper has a weak net diamagnetic response.^[ref]

There is one Cu2+ per formula unit, hence $n_s=N_A/V_m=5.5148\times 10^{27}$ m${}^{-3}$. The spin-only weak-field SI volume susceptibility isThese neglect exchange, anisotropic $g$ factors and background diamagnetism. At high field and low temperature, saturation requires the nonlinear response in a; a field-independent Curie value then describes only the zero-field slope.

For $B=0.05$ T, use the two spin-only moment states. Their probabilities and normalized magnetization areAt 300 K, $x=1.11952\times 10^{-4}$; at 2 K, $x=0.0167928$. ThereforeThe phrase “fraction aligned” is ambiguous if it is used to mean net alignment. Half the moments are already in the parallel state at zero field, but the net magnetization is then zero because the antiparallel half cancels them. The table reports both the literal parallel-state population and the imbalance that produces magnetization. At this weak field even the 2 K sample is far from saturation, so the Curie approximation is accurate within this independent-spin model.

For $B=5$ T, the same spin-only two-state probabilities apply, but the low-temperature argument is no longer small. At 300 K, $x=0.0111952$; at 2 K, $x=1.67928$. HenceThe 2 K sample is about $93.3\%$ of its saturation magnetization, with about $96.6\%$ of the moments in the aligned state. The two percentages describe different observables. A linear extrapolation would predict $M/M_s\simeq x=1.679$, exceeding saturation and therefore contradicting the bounded two-state population.

For completeness, at this 2 K, 5 T bias the dimensionless secant susceptibility is approximately ${\chi }_0\tanh x/x=0.01199$, while the differential susceptibility is ${\chi }_0{\mathrm{sech}}^{2}x=0.002805$ under the same prescribed-field approximation. Neither equals the zero-field Curie slope $0.02159$. Thus saturation reduces the incremental response even while the magnetization itself is large.

Inelastic neutron scattering measures both momentum and energy exchanged with the crystal. Comparing incident and outgoing neutron wavevectors gives $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$, while their energy difference gives a created or annihilated phonon energy $\hslash \omega$. Crystal momentum conservation identifies the reduced phonon wavevector modulo a reciprocal-lattice vector. Varying scattering geometry maps dispersion curves and polarization-dependent intensities across a substantial part of the Brillouin zone.

Inelastic X-ray scattering provides a complementary energy- and momentum-resolved probe, particularly when small samples or high-pressure environments favor photon beams. Raman and infrared spectroscopy measure optical-frequency resonances subject to different symmetry selection rules. Photon momentum at these frequencies is usually small on the Brillouin-zone scale, so these methods mostly sample near the zone center rather than an arbitrary phonon wavevector. Brillouin scattering and ultrasonic measurements probe low-frequency acoustic motion and infer sound velocities or elastic constants.

Heat-capacity measurements constrain the thermally weighted vibrational density of states and its low-frequency behavior, but do not by themselves reconstruct a unique wavevector-resolved dispersion. Elastic diffraction and temperature-dependent displacement factors also constrain mean-square motion, while energy-resolved inelastic measurements distinguish individual excitations directly. A monatomic scalar chain has only the acoustic branch derived below; it should not be assigned an optical branch merely because Raman spectroscopy exists for more complicated crystals.

The right spring changes length by $u_{n+1}-u_n$ and pulls atom $n$ to the right with force $C(u_{n+1}-u_n)$. The left spring contributes $-C(u_n-u_{n-1})$. Adding these restoring forces and applying Newton's second law givesUniform displacement makes both spring extensions vanish, as translational invariance requires. Search for a normal mode $u_n=u_0{e}^{i(kna-\omega t)}$. Substitution yieldsOne representative first Brillouin zone is $-\pi /a\le k\le \pi /a$. This is one positive-frequency acoustic branch: near zero wavevector, $\omega \simeq a\sqrt{C/m}|k|$, so the sound speed is $v_s=a\sqrt{C/m}$. Opposite signs of $k$ label opposite propagation directions, not a second polarization branch.

The model assumes small displacements, harmonic nearest-neighbor forces, one scalar displacement per atom and uniform equilibrium spacing. It excludes damping, anharmonic expansion and transverse polarizations. In a three-dimensional monatomic crystal there would normally be three acoustic polarizations; that counting cannot be imported into this explicitly one-dimensional scalar model.

The supplied sound speed determines the spring-to-mass ratio without needing either parameter separately:ConsequentlyThe curve is even in $k$, rises linearly from the origin with one-sided slopes $\pm 4.0$ km/s, and flattens at both zone edges. Its maximum is ${\omega }_m=4.0\times 10^{13}$ rad/s, or ordinary frequency ${\nu }_m={\omega }_m/(2\pi )=6.366$ THz. Keeping angular frequency distinct from cycles per second avoids a factor-of-$2\pi$ error. Repeating the curve outside the first zone is a representation of equivalent crystal momenta, not the discovery of additional modes.

Differentiate the positive-frequency branch on either side of the origin:It tends to $+4.0$ km/s as $k\to 0^{+}$ and to $-4.0$ km/s as $k\to 0^{-}$, and vanishes at either zone boundary. A positive-$k$ wavepacket therefore transports energy to the right, while a negative-$k$ one transports it to the left. At the zone edge, neighboring atoms oscillate out of phase and the dispersion is flat, giving zero group velocity even though the frequency is nonzero.

The graph has separate one-sided limits at zero because $\omega (k)$ has a cusp there. Assigning one differentiable slope at exactly $k=0$ would erase the distinction between the two propagation directions; a precisely uniform translation has no preferred propagation direction. Plotting only $0\le k\le \pi /a$ is also legitimate if the missing negative branch is stated. Group velocity and phase velocity $\omega /k$ are different except in the long-wavelength linear limit.

For a long periodic chain of length $\mathcal{L}=Na$, neighboring allowed signed wavevectors differ by $2\pi /\mathcal{L}$. Counting both signs once gives a state density per length $dk/\pi$ on the positive half-zone. Therefore the density per angular-frequency interval and per length isIt is zero outside this band, starts at $g(0)=1/(\pi v_s)=7.958\times 10^{-5}$ s/m and has an integrable inverse-square-root Van Hove singularity as $\omega \to {\omega }_m^{-}$. The state-count normalization isThus the scalar chain has one vibrational mode per atom, with no extra factor of two for the already counted propagation directions. If the horizontal variable were ordinary frequency $\nu$, the density would be $g_{\nu }(\nu )=2\pi g_{\omega }(2\pi \nu )$. A finite chain has discrete lines instead of an actual infinite peak; the smooth divergent curve is the large-chain continuum description.

At fixed chain length, each harmonic mode contributes a Bose–Einstein thermal occupation and a temperature-independent zero-point energy. The latter contributes no heat capacity. With $x=\hslash \omega /(k_{B}T)$,At low temperature only the approximately linear part of the acoustic dispersion is excited, so $g(\omega )\simeq 1/(\pi v_s)$. Using ${\int }_0^{\infty }{x}^2{e}^x/(e^x-1{)}^{2}dx={\pi }^2/3$ givesIt is linear in temperature, rather than the three-dimensional Debye $T^3$ law. At high temperature every scalar oscillator tends to $k_B$, henceThe cutoff temperature is ${\Theta }_m=\hslash {\omega }_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; ${\Theta }_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(N{k}_B)\simeq (2\pi /3)T/{\Theta }_m$. The low-temperature continuum limit also requires temperature above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode energy a finite chain does not retain a continuum linear law down to arbitrary temperature.