Past exam of the condensed matter physics major option course of the University of Oxford

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The cover identifies the Part C major option, Trinity Term 2019, 17 June 2019, A11046W1; the bundle's 2018 filename does not determine the examination year. All six questions are treated. Figure 2(A) has missing image tiles, including part of its lower-left plot and some path labels. The readable portions and the standard silicon valley model support the conclusions below; the missing image content is not reconstructed. Numerical values read from plotted curves are estimates. Write $e>0$ for the elementary electric charge.

TODO: Recover the missing tiles of original Figure 2(A), affecting questions 2(a) and 2(e). The solutions distinguish the surviving figure and standard silicon-band interpretation from unavailable path labels and precise missing valence-sheet contours.

Infrared spectroscopy measures absorption as the incident photon frequency is swept through molecular normal mode frequencies. In the electric-dipole approximation the interaction is $-\mathbf{p}\cdot \mathbf{E}$. Expanding the molecular electric dipole moment in a normal coordinate $Q$ givesThe fundamental one-quantum vibrational transition dipole moment is proportional to the derivative multiplying $Q$. Thusis the criterion for vibrational infrared activity; the derivative must also have a component along the incident polarization of an electromagnetic wave. A permanent molecular electric dipole moment is not necessary. Absorption occurs near $\hslash \omega =E_{v+1}-E_v$, with broadened lines reflecting finite lifetimes and the molecular environment.

Use the indicated out-of-plane displacements and the equivalence of the four oxygen sites and of the two carbon sites in the isolated planar oxalate ion. In $V_1$, the two oxygen atoms on one side move oppositely to those on the other side, and the two carbon atoms also move oppositely. Their dipole changes cancel. The mode is even under molecular inversion symmetry, which independently forbids an electric-dipole transition.

In $V_2$, all four oxygen atoms move together out of the plane while both carbon atoms move in the opposite direction. The positive and negative charge distributions acquire a relative out-of-plane displacement, producing a nonzero derivative of the molecular electric dipole moment.

In $V_3$, two oxygen atoms move out and two move in, in a checkerboard arrangement, while the carbon atoms are stationary. The equivalent oxygen contributions cancel. This mode illustrates that being odd under inversion symmetry is necessary but not sufficient for vibrational infrared activity: its full symmetry does not transform like any component of an electric dipole moment.These conclusions concern the isolated-molecule displacements drawn in the source; a different crystal environment can alter selection rules through additional symmetry breaking.

There are six atoms in the nonlinear oxalate ion, hence $3\cdot 6=18$ positional degrees of freedom. Subtract three rigid translations and three rigid rotations to obtain $3N-6=12$ vibrational normal modes. Planarity does not remove out-of-plane motions. Since three modes have been displayed,

The primitive unit cell contains two carbon atoms and four edge-shared oxygen atoms, each contributing one half: $n=2+4/2=4$ atoms. Equivalently, one molecular repeat contributes its two carbon atoms but only half of each of its four shared oxygen atoms. There are therefore $3n=12$ phonon branches.

At the Brillouin zone centre, three rigid translations give three acoustic phonon branches; the remaining $3n-3$ are optical phonon branches:A free layer has two in-plane acoustic modes and an out-of-plane flexural acoustic mode. A two-dimensional lattice does not imply that its atoms can move only in two directions. Artificially suppressing all out-of-plane motion would instead define a different model with two acoustic and six optical branches.

The primitive translations connect the two centred-rectangular sublattices. In the orientations marked in the source, ${\mathbf{a}}_p$ points right and down, whereas ${\mathbf{b}}_p$ points right and up. ThusTheir parallelogram has half the area of the rectangular cell, confirming that the latter contains two primitive unit cells.

Let the reciprocal lattice vector be $\mathbf{G}=h_r{\mathbf{a}}_r^{\ast }+k_r{\mathbf{b}}_r^{\ast }=h_p{\mathbf{a}}_p^{\ast }+k_p{\mathbf{b}}_p^{\ast }$, with each dual pair obeying $\mathbf{a}\cdot {\mathbf{a}}^{\ast }=\mathbf{b}\cdot {\mathbf{b}}^{\ast }=2\pi$ and zero cross products. Its coefficients are obtained by dotting with the corresponding direct-space primitive lattice translation vectors:HenceFor actual integer-indexed reciprocal lattice vectors, this also explains the centred-rectangular condition that $h_r$ and $k_r$ have the same parity. General phonon wave vectors need not have integer coefficients.

Comparing the two coefficients in the reciprocal-coordinate transformation givesThe inverse transformation is $h_r=h_p+k_p$, $k_r=k_p-h_p$. This provides an independent check on the signs associated with the two primitive arrows in the source.

Apply the same change of reciprocal lattice coordinates to the phonon wave vectors:Although $(1,0{)}_r$ is an integer coordinate of the rectangular reciprocal grid, it is not a reciprocal lattice vector of the actual primitive crystal. Thus $P_1$ is not equivalent to the Brillouin zone centre.

For ordinary one-phonon infrared optical absorption, both an energy and a momentum selection rule must hold. The optical photon has negligible crystal momentum on the scale of the Brillouin zone, so crystal momentum conservation selects $\mathbf{q}\simeq 0$ modulo a genuine reciprocal lattice vector. In addition, the mode must change the macroscopic electric polarization, with a component along the incident polarization of an electromagnetic wave.

In terms of atomic displacements ${\mathbf{u}}_j$ and Born effective charge tensors $Z_j^{\ast }$, the relevant derivative is proportional to ${\sum }_j{Z}_j^{\ast }{\mathbf{u}}_j$ per primitive unit cell. Thus an optical phonon at $\mathbf{q}=0$ is not automatically infrared active: its dipole contributions may cancel. Disorder-assisted and multiphonon processes relax these simple first-order selection rules.

$P_1$ has primitive coordinates $(1/2,1/2)$ and therefore fails the first-order optical momentum condition; its local $V_1$ pattern also has zero dipole derivative. $P_2$ is a zone-centre mode with the dipole-active $V_2$ pattern, so its molecular dipole changes add coherently.The allowed absorption for $P_2$ uses an polarization of an electromagnetic wave with an out-of-plane component. At normal incidence on an ideal free-standing layer the electric field is in the plane, so that specific measurement geometry would suppress it despite the mode's intrinsic vibrational infrared activity.

The lower curve is a valence band and the upper curve is a conduction band, separated by an indirect band gap. The readable upper curve has a minimum just inside the X boundary and slightly below the plotted chemical potential. It is therefore partially occupied, while the visible valence states lie below the chemical potential and are filled at low temperature. The sample is electron-doped, with electron pockets around the six symmetry-related conduction minima of silicon.

The full high-symmetry path and part of the lower-left dispersion are absent from the held image. Source gap: those missing path labels and curve segments cannot be read or reconstructed from this copy. The identification of the conventional six silicon valleys and the valence maximum at $\Gamma$ uses the usual silicon band model, rather than purported readings of missing pixels.

Two methods for calculating such an electronic band structure are density functional theory, solving a self-consistent periodic electronic problem, and a tight-binding model, diagonalizing a hopping Hamiltonian built from localized atomic orbitals. Neither is a measurement of the plotted dispersion relation.

For a nondegenerate band $E(\mathbf{k})$, define the reciprocal effective mass tensor byIt is the curvature tensor of the energy band. Near an extremum, $E\simeq E_0+{\hslash }^2\delta {\mathbf{k}}^T{M}^{-1}\delta \mathbf{k}/2$, so positive principal curvatures give positive electron effective masses. Together with the group velocity $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$ and $\hslash \dot{\mathbf{k}}=\mathbf{F}$, it yields ${\dot{v}}_i=(M^{-1}{)}_{ij}{F}_j$. Negative electron curvature at a valence band maximum is conventionally recast as a positive band effective mass for missing electrons.

In a parabolic conduction valley, $\mathbf{v}=\hslash M^{-1}\mathbf{k}$. The Lorentz force on an electron gives $M\dot{\mathbf{v}}=-e\mathbf{v}\times \mathbf{B}=e\mathbf{B}\times \mathbf{v}$. In the principal axes of the stated effective-mass tensor,Call the matrix on the right $A$. Its characteristic polynomial isOne mode has zero frequency, corresponding to constant drift along the field; the other two have eigenvalues $\pm i\omega$. ThusMore generally the ellipsoidal cyclotron mass is $m_{\mathrm{cr}}^2=\det M/({\hat{\mathbf{B}}}^{T}M\hat{\mathbf{B}})$. The tensor axes must be rotated for the differently oriented valleys.

The photon energy fixes the resonant cyclotron frequency by $\hslash \omega =575\mu \mathrm{eV}$. ConsequentlyRead the two absorption maxima as approximately $B_L=0.95$--$0.97\mathrm{T}$ and $B_H=2.10\mathrm{T}$. For $\mathbf{B}\parallel [001]$, the two valleys whose long axis is along $z$ have $m_{\mathrm{cr}}=m_1$, whereas the other four have $m_{\mathrm{cr}}=\sqrt{m_1{m}_2}$. HenceFor the representative reading $B_L=0.96\mathrm{T}$, the values are $m_1=0.193m_e$ and $m_2=0.925m_e$. The graph does not warrant additional precision. The larger longitudinal effective mass means flatter dispersion and a longer constant-energy ellipsoid along the valley axis.

For $\mathbf{B}\parallel [111]$, its squared projection onto the long axis of every valley is $1/3$. All six valleys therefore have the same cyclotron effective mass:At the fixed photon energy,Thus the two-peak spectrum is replaced by one peak near $1.4\mathrm{T}$. The source does not determine the linewidth or absolute oscillator strength for the new direction; the sketch below shows only a normalized illustrative line at the computed position.

For the electron-doped sample, sketch six closed ellipsoids centred on the $\Gamma$--X axes, just inside the six square faces of the face-centred-cubic first Brillouin zone, with long axes pointing radially toward X. In the parabolic approximation,The radial-to-transverse semiaxis ratio is $\sqrt{m_2/m_1}\simeq 2.2$. The readable conduction minimum near X and its crossing of $\mu$ support these electron Fermi surface pockets. The two $[001]$ cyclotron resonance fields arise from the two distinct orientations of the six ellipsoids, rather than two unrelated carrier species. The left panel of the original sketch includes axes, $\Gamma$, and the three positive-axis X points; the negative-axis X points are symmetry-equivalent.

The surviving valence maximum lies roughly $1.2\mathrm{eV}$ below the original $\mu$. Lowering $\mu$ by $1.5\mathrm{eV}$ therefore places it in the upper valence band in the conventional silicon band interpretation. The electron pockets disappear and one sketches a hole pocket centred at $\Gamma$, as in the middle panel. Detailed valence-band warping and the number of spin-orbit-resolved sheets cannot be extracted from the damaged figure, and the single drawn surface is schematic rather than a claim that all such sheets coincide.

A discriminating physical property is the Hall coefficient: electron-dominated transport gives a negative value, whereas hole-dominated transport gives a positive value in the ordinary single-dominant-carrier regime. One should not infer a numerical resistivity change without the carrier densities and mobilities.

Use time dependence $e^{-i\omega t}$ and assume the dopants supply $N$ mobile electrons per unit volume. The Drude model equation isThe free-carrier electric polarization is ${\mathbf{P}}_c=-Ne\mathbf{x}$, so, after adding the bound-electron background,ThereforeThe symbol ${\omega }_p$ here is the background-screened plasma frequency; the unscreened carrier frequency is $\sqrt{{ϵ}_{\infty }}{\omega }_p$. For positive $\gamma$, the imaginary part of ${ϵ}_r$ is positive in this time convention, consistent with absorption. Incomplete dopant ionization would replace the nominal doping density by the actual mobile carrier density.

For a nonmagnetic sample bordering air, the normal-incidence Fresnel equations giveAt short infrared wavelengths the carrier response becomes small and $R$ approaches the background value $R_{\infty }$. Extrapolating the short-wavelength trend gives about $0.29$, henceThe leftmost point at $4\mu \mathrm{m}$ is about $0.27$ and has not quite reached that asymptote. As a consistency check, using the minimum below and $R(4\mu \mathrm{m})\simeq 0.27$ gives ${ϵ}_{\infty }\simeq 11$ without treating that point as the limiting value.

The minimum is near ${\lambda }_{\min }\simeq 13\mu \mathrm{m}$. For weak damping, zero optical reflectance means refractive-index matching, ${ϵ}_r=1$, rather than ${ϵ}_r=0$. Thususing $N=5.4\times 10^{24}{\mathrm{m}}^{-3}$. Given graphical uncertainty, about $0.08m_e$ is an appropriate estimate. This is a carrier effective mass from a reflectivity minimum, not an independent determination of the bare electron mass.

Just beyond the minimum, $\mathrm{Re}{ϵ}_r$ passes through zero near ${\lambda }_p={\lambda }_{\min }/\sqrt{1-1/{ϵ}_{\infty }}\simeq 13.6\mu \mathrm{m}$. At longer wavelengths it becomes negative and reflection rises sharply toward the conducting regime. Finite damping rounds the edge, prevents a strictly exact zero minimum, and keeps the plotted long-wavelength reflectivity below unity.

The broad feature has the scale expected for a carrier plasmon, a longitudinal collective oscillation of the mobile electron density. Using the parameters just estimated and $N=10^{24}{\mathrm{m}}^{-3}$ givesConversely a bare screened plasma oscillation at $340{\mathrm{cm}}^{-1}$ would require $N\simeq 1.16\times 10^{18}{\mathrm{cm}}^{-3}$ for the same parameters, consistent with the source's approximate density. Its broadness is also consistent with carrier damping, unlike the much sharper lattice line near $270{\mathrm{cm}}^{-1}$.

This identifies a plasmon-related feature, not a uniquely determined undamped pure electronic mode. Gallium arsenide is polar, and its longitudinal optical phonon couples to the carrier plasmon. A more complete dielectric model has longitudinal eigenfrequenciesThe measured Raman scattering maximum can also shift with damping and the scattering matrix elements. Experimental Raman scattering from these coupled modes and carrier-collision broadening is established independently.^[ref] The supplied approximate density and one broad spectrum do not determine a unique coupled-mode fit. Treating $340{\mathrm{cm}}^{-1}$ as an exact pure-plasmon eigenfrequency would overstate the available information.

A possible assignment is a zone-centre transverse optical phonon of gallium arsenide: the two atomic sublattices oscillate oppositely, producing a narrow lattice Raman scattering line near $270{\mathrm{cm}}^{-1}$. Its energy is $hc\tilde{\nu }\simeq 33.5\mathrm{meV}$, a reasonable optical-phonon scale. Unlike the broad carrier feature, its frequency is set mainly by interatomic force constants and ionic masses.

Whether this transverse optical phonon is allowed in an ideal first-order Raman scattering geometry depends on the surface orientation and incident and scattered polarizations; disorder or imperfect geometry can relax a forbidden selection rule. Those experimental details are unspecified, so this is a possible explanation rather than an exclusive assignment.

For a free transition-metal ion, degeneracy among different orbital states can support an orbital contribution to its magnetic moment. In a crystal, the electric fields and bonding of the surrounding ligands split those states. If the orbital ground state becomes nondegenerate and is separated from excited orbital states, its wavefunction can usually be chosen real in zero field. Since the orbital angular momentum operator is imaginary in a real-orbital basis, its expectation value then vanishes.

This suppression is orbital quenching. The remaining leading magnetic moment is the spin contribution, approximately $g{\mu }_B\mathbf{S}$ with $g\simeq 2$. Spin-orbit coupling admixes excited orbital states and produces residual anisotropy and shifts in the g-factor; quenching does not mean that every orbital matrix element or every higher-order orbital response vanishes. A degenerate ground orbital multiplet need not be quenched until a further distortion removes its degeneracy.

For the regular octahedron, repulsion from the six oxygen ligands raises the two $e_g$ orbitals, $d_{z^2}$ and $d_{x^2-y^2}$, above the three $t_{2g}$ orbitals, $d_{xy},d_{xz},d_{yz}$. The latter point between the metal-ligand axes and experience less direct repulsion. The crystal field splitting is therefore a lower triplet and an upper doublet.

For the $3d^2$ V${}^{3+}$ ion, Hund's rules put the two electrons into different $t_{2g}$ orbitals with parallel spins, giving $S=1$. There are three choices for the unoccupied $t_{2g}$ orbital, so the orbital ground state is triply degenerate. Its spin degeneracy is separate. This orbital degeneracy makes a symmetry-lowering Jahn-Teller effect possible.

Take the elongation axis as $z$. The axial oxygen ligands move farther away while the equatorial ones move closer, at fixed mean bond length. Orbitals containing $z$ are stabilized relative to their in-plane partners. Thus $d_{xz},d_{yz}$ move below $d_{xy}$ within $t_{2g}$, while $d_{z^2}$ moves below $d_{x^2-y^2}$ within $e_g$. The schematic level ordering appears in the figure below; the octahedral gap remains open for a small distortion.

The two parallel-spin electrons occupy $d_{xz}$ and $d_{yz}$, one each. Filling the entire lower doublet gives a unique orbital configuration, so its orbital degeneracy is one and the spin is still $S=1$. This is the orbitally quenched configuration relevant to the later spin Hamiltonian.

The energetic preference can be made explicit in a traceless linear splitting model. For an elongation coordinate $Q>0$, write the $t_{2g}$ levels relative to their barycentre as $(-aQ,-aQ,2aQ)$ with $a>0$. Their two-electron ground energy decreases by $2aQ$. For an equal-magnitude compression, the levels are $(-2aQ,aQ,aQ)$ and the two-electron energy decreases only by $aQ$. With the same quadratic elastic cost, minimizing $-2aQ+k{Q}^2/2$ gives four times the stabilization of minimizing $-aQ+k{Q}^2/2$. Thus the small-distortion Jahn-Teller effect favors elongation for this high-spin $d^2$ filling. The argument assumes the linear coupling and harmonic elastic model, rather than claiming that symmetry alone fixes all real-material distortions.

For axial compression along $z$, the ordering is reversed: $d_{xy}$ lies below the $d_{xz},d_{yz}$ doublet, and $d_{x^2-y^2}$ lies below $d_{z^2}$ within $e_g$. One electron occupies $d_{xy}$ and the other occupies either $d_{xz}$ or $d_{yz}$, with parallel spins in the weak-distortion regime. The orbital degeneracy is therefore two, with $S=1$.

The retained orbital degeneracy allows further symmetry lowering and explains why this compressed arrangement does not provide the fully quenched equilibrium of the elongated case in the specified simple model.

The term $-\alpha ϵ$ is the leading electronic energy gain from splitting the initially degenerate orbitals and occupying the lower levels. Its negative sign expresses the lowering of the occupied-electron energy for the chosen elongation branch of the Jahn-Teller effect. The positive coefficient $\alpha$ is a linear electron-lattice coupling in this reduced model.

The term $\beta {ϵ}^2/2$ is the harmonic elastic cost of moving the ligands away from the undistorted reference structure. Mechanical stability requires $\beta >0$. The competition between linear gain and quadratic cost produces a finite equilibrium distortion rather than indefinite elongation.

Complete the square:The minimum is stable because $d^{2}E/d{ϵ}^2=\beta >0$. HenceHere the Jahn-Teller stabilization energy is the positive magnitude of the energy reduction.

The low-temperature bond-length difference is ${ϵ}_0=2.05-1.98=0.07\text{A˚}$. With the question's thermal criterion,Use $\alpha =\beta {ϵ}_0$ and $E_{\mathrm{JT}}=\beta {ϵ}_0^2/2$ to obtainIn SI units these are about $7.89\times 10^{-10}\mathrm{N}$ and $113\mathrm{N}{\mathrm{m}}^{-1}$, respectively. They are estimates within the stipulated energy criterion; a cooperative transition also involves interactions and entropy, so its temperature is not in general exactly an isolated-ion stabilization energy divided by $k_B$.

Above the transition, $ϵ=0$ and both site types have the same g-factor $g$. For the given spin-only Zeeman Hamiltonian, adjacent $S=1$ levels have the same energy spacing, so fixed-frequency electron paramagnetic resonance gives a single resonance field $B_0=h\nu /(g{\mu }_B)$, ignoring unresolved additional interactions.

Below the transition, half the elongation axes lie along $x$ and half along $y$. A field along $x$ is parallel to the long bond on one sublattice and perpendicular on the other. The line splits intoThe separation grows on cooling as the distortion grows. Equal site populations give comparable integrated weights when the microwave coupling to the two sites is equivalent; populations alone do not fix all polarization-dependent intensities.

A suitable staggered structural order parameter isFor an $x$-elongated A site and a $y$-elongated B site, $\Psi =ϵ$, with the opposite sign for the domain interchanging the pattern. The electron paramagnetic resonance data determine its magnitude fromFor small splitting this is equivalently proportional to $B_{\perp }-B_{\parallel }$, but the reciprocal-field formula follows directly from the stated g-factors. It does not fix the sign of the structural domain. In a continuous transition $|\Psi |$ tends continuously to zero at $T_{\mathrm{JT}}$ and grows below it, eventually saturating. A mean-field sketch has $|\Psi |\propto (T_{\mathrm{JT}}-T{)}^{1/2}$ close to the transition; continuity alone does not determine that exponent. The last panel of the original orbital figure shows this illustrative behavior.

Choose the four basis-spin signs as $(+,-,-,+)$ in the order of positions listed in the source; reversing all signs changes no diffraction intensity. Apart from the common magnetic form factor and perpendicular-moment factor, the magnetic structure factor isThe first factor is nonzero exactly when $h$ and $k$ have opposite parity, and the second exactly when $l$ is odd. Thus the magnetic Bragg peaks requireWhen both conditions hold, $|F_M|=4|m|$ before geometric and magnetic-form-factor weights. These are magnetic selection rules for the antiferromagnetic component, not conditions forbidding all nuclear diffraction at other indices.

Measure several allowed magnetic reflections from an oriented single crystal by magnetic neutron diffraction, preferably with polarization analysis. A neutron couples to the component of the ordered magnetic moment perpendicular to the scattering vector $\mathbf{Q}$. For a collinear arrangement, the geometric intensity factor isCompare reflections whose $\mathbf{Q}$ point in different crystal directions, correcting for the magnetic form factor and known experimental geometry. Their relative intensities distinguish candidate spin directions; a moment parallel to $\mathbf{Q}$ gives no magnetic signal from that component. Polarized neutrons help separate nuclear and magnetic contributions. Observing only peak positions establishes periodicity, not the spin direction.