Past exam of the general relativity and cosmology and condensed matter physics paper of Part B of the physics course of the University of Oxford / 2011 / Section VI

For the mutually orthogonal direct primitive lattice translation vectors ${\mathbf{a}}_i$, define the reciprocal lattice basis byThus $\mathbf{G}={\sum }_i{h}_i{\mathbf{b}}_i$, with integer $h_i$, satisfies $\exp (i\mathbf{G}\cdot \mathbf{R})=1$ for every lattice vector $\mathbf{R}$.

Write $\mathbf{r}={\sum }_i{s}_i{\mathbf{a}}_i$. The electron density is periodic with period one in each $s_i$, so its three-dimensional Fourier series has integer harmonics:If $\Omega$ is a primitive unit cell of volume ${\Omega }_c$, Fourier orthogonality givesConversely, translation by ${\mathbf{a}}_i$ multiplies a general plane wave by $e^{i\mathbf{q}\cdot {\mathbf{a}}_i}$. Periodicity permits only $\mathbf{q}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$, precisely the reciprocal lattice vectors. For a physical bounded periodic electron density, the expansion exists in the usual square-integrable sense.

Use the scattering vector $\mathbf{q}={\mathbf{k}}_{\mathrm{in}}-{\mathbf{k}}_{\mathrm{out}}$; reversing its sign conjugates the following amplitudes without changing the intensities. In the kinematic approximation, the atomic scattering factor of atom $s$ isAtoms at $\mathbf{R}+{\mathbf{d}}_s$ therefore give the scattering amplitudeHere $A_e$ is the single-electron amplitude; $F$ is the unit cell structure factor, and $S_L$ is the geometrical lattice sum. If the crystal has $N_i$ repeats along ${\mathbf{a}}_i$, the latter factor is a product of geometric series, whose magnitudes containThe sharp constructive peaks occur when every $\mathbf{q}\cdot {\mathbf{a}}_i=2\pi h_i$. Thus a permitted reflection has wavevector change equal to a reciprocal lattice vector, with $F(\mathbf{G})\ne 0$ and the additional elastic condition $|{\mathbf{k}}_{\mathrm{out}}|=|{\mathbf{k}}_{\mathrm{in}}|$. A finite crystal broadens these peaks; it does not impose an exact delta-function selection on every scattered photon.

In the conventional cubic unit cell, the four face-centered cubic lattice positions contributeThe two-atom atomic basis of a crystal gives the zinc blende structure factorMixed-parity Miller indices have $S_{\mathrm{FCC}}=0$. With all indices even or all odd, $S_{\mathrm{FCC}}=4$. For real atomic scattering factors,Hence all same-parity reflections are generally permitted in GaAs, although the last class is weak when the two atomic scattering factors nearly coincide. These are the structural selection rules; a particular experiment must also satisfy the elastic Bragg scattering condition.

In the electron-count estimate $f_{\mathrm{Ga}}\simeq 31$, $f_{\mathrm{As}}\simeq 33$, the structure factors are $F_{111}=4(31-33i)$ and $F_{200}=4(31-33)$. Thus a single-directed-reflection estimate isA powder X-ray diffraction peak collects all equivalent reflections. The powder diffraction multiplicities are eight for $\{111\}$ and six for $\{200\}$, so the multiplicity-and-electron-count estimate isAt $(200)$ the two sublattice amplitudes are opposite in phase and nearly cancel; at $(111)$ their relative phase is a quarter cycle, so this cancellation does not occur. In silicon, the two atomic scattering factors are identical and the diamond cubic structure has an exactly forbidden $(200)$ reflection, while $(111)$ remains allowed in the kinematic approximation.

The numerical ratio neglects the angular dependence of the atomic scattering factors, the Lorentz-polarization factor and thermal-motion corrections. No wavelength or measurement geometry is supplied to refine those factors. This is therefore a stated structural estimate, rather than a unique precision ratio of measured peak areas.

Let $a$ be a stable equilibrium separation: $V^{\prime }(a)=0$, $K=V^{\prime \prime }(a)>0$. With bond extension $\delta =x-a$, the Taylor expansion isThe harmonic approximation applies when the typical thermal and zero-point bond fluctuations satisfy $|\delta |\ll a$ and the higher terms are negligible, for example $|V^{\prime \prime \prime }(a)\delta |\ll 3K$ and $|V^{\prime \prime \prime \prime }(a){\delta }^2|\ll 12K$. The relevant small quantity is the relative displacement of neighbouring atoms. Small bond excursions about a stable minimum give an amplitude-independent spring constant $K=V^{\prime \prime }(a)$. Large thermal excursions, bond breaking or appreciable anharmonicity invalidate the approximation.

With atomic displacement $u_j$, the harmonic potential energy is $\frac{1}{2}K{\sum }_j(u_{j+1}-u_j{)}^2$. Newton's second law givesFor a periodic chain of $N$ atoms, its normal modes are $u_j=U{e}^{i(kja-\omega t)}$, with $k=2\pi s/(Na)$ and one representative per Brillouin zone. Substitution yields the monatomic-chain dispersion relationA real general displacement is a superposition of these normal modes and their complex conjugates. The $k=0$ mode is a rigid translation. At small $|k|$, the acoustic phonon branch is linear, with speed of soundFor $0<k<\pi /a$, the group velocity is $a\sqrt{K/m}\cos (ka/2)$ and vanishes at the zone boundary. The dashed initial tangent in the sketch has slope $v$.

For the Lennard-Jones potential,Its minimum and harmonic spring constant areUsing the monatomic-chain dispersion relation,The maximum is at $|k|=\pi /a$. Taking the argon atomic mass as $m\simeq 39.95\mathrm{u}=6.63\times 10^{-26}\mathrm{kg}$ and $ϵ=1.602\times 10^{-21}\mathrm{J}$ givesThese are the stipulated one-dimensional nearest-neighbour model estimates; a three-dimensional argon crystal has additional neighbours and phonon polarizations.

Normalize the plane waves in a periodic crystal of area or volume $\mathcal{V}$. The Fourier coefficients of the periodic potential obeyTranslation of the integral by any lattice vector $\mathbf{R}$ leaves it invariant but multiplies it by $e^{-i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{R}}$. A nonzero result therefore requires $\mathbf{k}-{\mathbf{k}}^{\prime }$ to be a reciprocal lattice vector.

Near an isolated Bragg scattering degeneracy, ordinary nondegenerate perturbation theory fails because the free energies coincide. Retain the resonant plane waves $|\mathbf{k}⟩$, $|\mathbf{k}+\mathbf{G}⟩$ and apply degenerate perturbation theory. Absorb the constant $V_0$ into the energy origin. Since a real potential has $V_{-\mathbf{G}}=V_{\mathbf{G}}^{\ast }$, the projected Schrödinger equation has coefficient matrixA nonzero eigenstate requiresOff-resonant plane waves give higher-order energy shifts omitted in this leading nearly-free electron model.

Let $u=V_{10}<0$ andAt $X=(\pi /a,0)$ the two degenerate plane waves have momenta $(\pi /a,0)$ and $(-\pi /a,0)$ and coupling $u$. ThusTheir symmetric and antisymmetric combinations are the lower and upper states respectively. The local band gap is $2|u|$.

Put $w=V_{11}<0$. At $M=(\pi /a,\pi /a)$, the free energy is $2E_X$. The four cyclic combinations in the supplied secular equation give the four-plane-wave splitting at a square-zone cornerFor example, the normalized coefficient vectors in the source's cyclic basis are $(1,1,1,1)/2$, $(1,0,-1,0)/\sqrt{2}$, $(0,1,0,-1)/\sqrt{2}$, and $(1,-1,1,-1)/2$. Because $u<w<0$, these are ordered as written: a lower singlet, a central doublet and an upper singlet. The sketch puts the $X$ and $M$ energies on one common scale; its representative negative couplings are stated in the caption.

There are two electrons per primitive unit cell; a spin-degenerate energy band accommodates exactly that number. A semiconductor results when the first band is full and does not overlap the second. In the leading two-wave/four-wave approximation, compare the first-band maximum at $M$ with the second-band minimum at $X$:This is an indirect band gap; a nonzero local splitting at $X$ alone is insufficient. At the onset,With $a=0.3\mathrm{nm}$, ${\hslash }^2/(2m_e)=0.03810\mathrm{eV}\mathrm{n}{\mathrm{m}}^2$ gives $E_X\simeq 4.178\mathrm{eV}$, henceFor $V_{10}$ more negative than this value the estimated indirect band gap is positive; at equality the bands just touch. This is the onset estimate within the source's nearly-free electron model truncation, not an exact critical potential after coupling all plane waves.

The Brillouin function is odd, increases monotonically through zero, and approaches $\pm 1$ for $x\to \pm \infty$. Its small-$x$ slope is $(J+1)/(3J)$. For $x\gg 1$, the magnetic ions overwhelmingly occupy the largest field-aligned moment projection, so the magnetization reaches magnetic saturation:The sketch shows several $J$ values; all have the same limiting values, while their slopes differ.

The $1/x$ terms in the Brillouin function cancel. Keeping the linear terms givesConsequently,Use the dimensionless SI magnetic susceptibility $\chi =\partial M/\partial H_{\mathrm{applied}}$ and $B_{\mathrm{applied}}={\mu }_0{H}_{\mathrm{applied}}$, as in the source's molecular-field convention. The Curie law isIf susceptibility is instead defined as $\partial M/\partial B$, its coefficient is $C/{\mu }_0$ and is dimensionful. The SI convention above makes the subsequent Curie–Weiss law and $\lambda$ consistent.

For positive $\lambda$, a fluctuation of magnetization produces a Weiss molecular field in the same direction, reinforcing alignment. The mean-field self-consistency equation isLinearizing in the paramagnetic phase givesAt zero applied field, the right-hand side's slope at $M=0$ is $T_C/T$. Below $T_C$ this exceeds one, and the saturation of the Brillouin function allows stable nonzero solutions of opposite sign: spontaneous ferromagnetism. Above $T_C$, only the zero-field solution $M=0$ remains stable. The stated linear susceptibility is the paramagnetic branch for $T>T_C$; it must not be continued as a negative susceptibility below the transition.

Using the supplied iron parameters givesAn internal Weiss molecular field depends on temperature through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng{\mu }_{B}J\simeq 1.58\times 10^6\mathrm{A}{\mathrm{m}}^{-1}$:At and above $T_C$ the zero-applied-field magnetization vanishes in this model, so this is a saturation estimate, not the field at $T_C$.

Use the paper's stipulated face-centered cubic lattice, which has twelve nearest neighbours. The saturated one-ion alignment scale isDividing the effective field among twelve equivalent neighbours gives the per-bond exchange interaction scaleTo specify the convention, write the Heisenberg model as $\mathcal{H}=-\mathcal{J}{\sum }_{⟨ij⟩}{\mathbf{J}}_i\cdot {\mathbf{J}}_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g{\mu }_B{B}_e=12\mathcal{J}⟨J_z⟩$, so $\mathcal{J}=3k_B{T}_C/[12J(J+1)]\simeq 11.2\mathrm{meV}$ here. A convention with $-2{\mathcal{J}}^{\prime }$ instead has ${\mathcal{J}}^{\prime }=5.62\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal{J}{J}^2/2\simeq -67.4\mathrm{meV}$; the factor of one half avoids counting each bond twice.