Condensed matter physics course of the University of Oxford (Physics Part B, B6, B3)

Paper code B3 in 2012–2013; B6 in 2014–2019 and 2021–2022.

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 76–80. The paper cover establishes 2022; the attachment filename uses 2021.

The ordered crystal structure consists of a primitive cubic lattice with the four-site atomic basisin fractional crystal coordinates. The drawing shows all boundary representatives of these sites. Eight corner gold chemical atoms, shared among eight unit cells each, contribute one gold chemical atom. Six face-centred copper chemical atoms, shared between two unit cells each, contribute three copper chemical atoms. Thuswhich gives the required stoichiometry. Although the unlabelled positions form a face-centered cubic lattice, a face-centering lattice translation exchanges gold and copper and therefore fails to preserve the ordered crystal structure. Its Bravais lattice is primitive cubic lattice.

For the reciprocal lattice vector ${\mathbf{G}}_{hkl}=(2\pi /a)(h,k,l)$, sum the atomic scattering factors with the atomic basis phase factors. Taking the positive phase convention gives the structure factorChanging the phase convention complex-conjugates the structure factor and leaves the diffraction wave intensity unchanged. Since the Miller indices are integers,If all three Miller indices have the same integer parity, all pairwise sums are even. Otherwise exactly one pairwise sum is even and two are odd. ConsequentlyThe corresponding structure factors are $f_{\mathrm{Au}}+3f_{\mathrm{Cu}}$ and $f_{\mathrm{Au}}-f_{\mathrm{Cu}}$. Reflections with mixed integer parity are superlattice reflections that reveal chemical ordering in alloys.

In elastic scattering, the incoming and outgoing wave vectors have equal magnitude $k_0=2\pi /\lambda$. Their mutual angle is $2{\theta }_{hkl}$. The scattering vector $\mathbf{q}={\mathbf{k}}_f-{\mathbf{k}}_i$ therefore has magnitudeThe elastic Bragg scattering condition is $\mathbf{q}={\mathbf{G}}_{hkl}$, givingEquivalently, $G_{hkl}=2\pi /d_{hkl}$ turns this relation into Bragg's law, $2d_{hkl}\sin {\theta }_{hkl}=\lambda$, with the reflection order incorporated into the Miller indices.

The primitive cubic lattice has $G_{hkl}=(2\pi /a)\sqrt{h^2+k^2+l^2}$. Combining this with Bragg's law gives the Bragg angleUsing $a=0.375\mathrm{nm}$ and $\lambda =0.154\mathrm{nm}$ yieldsThese are the full scattering angles; the Bragg angles themselves are half as large.

At each reflection, the argument of the atomic scattering factor isSubstitution into the given atomic scattering factor approximation givesThus the structure factors are $S_{110}=36.2620$ and $S_{111}=113.7340$. The powder diffraction multiplicity is $m_{110}=12$: choose which Miller index is zero in three ways and the signs of the other two in four ways. For $\{111\}$, the eight sign choices give $m_{111}=8$. With the specified instrumental and angular factors omitted, the integrated powder diffraction wave intensity is proportional to $m_{hkl}|S_{hkl}{|}^2$, soIt would be incorrect to use the same atomic scattering factors at the two different scattering vectors, or to omit the powder diffraction multiplicity.

The stoichiometry fixes the site probabilities to $1/4$ gold and $3/4$ copper. Every site in the average crystal structure therefore has the same atomic scattering factorThe average structure factor isAt $(110)$ the bracket vanishes, while at $(111)$ it is four. HenceThe $(111)$ powder diffraction multiplicity is unchanged, so its coherent Bragg diffraction wave intensity equals the ordered value under the stated equal-lattice constant approximation.

The zero refers to the coherent Bragg diffraction peak of the average crystal structure. A random alloy can still give occupational diffuse scattering at the same scattering vector: the average of the squared amplitude includes fluctuations as well as the square of the average amplitude. Such background is not a surviving $(110)$ Bragg diffraction peak.

Replace every coloured site in the drawing by an identical site carrying $\bar{f}$. The face-centering lattice translations now preserve the average crystal structure, whose Bravais lattice is a face-centered cubic lattice. One choice of primitive lattice translation vectors isA coherent Bragg diffraction peak requires $\mathbf{G}\cdot {\mathbf{a}}_j\in 2\pi \mathbb{Z}$ for all three primitive lattice translation vectors. Thus $k+l$, $h+l$, and $h+k$ must all be even, equivalent to all three Miller indices having the same integer parity. This is the face-centred cubic selection rule: $(110)$ is forbidden and $(111)$ is allowed. The restoration of these lattice translations explains the disappearance of the ordered superlattice reflections.

Specify the normalization of the alloy sublattice order parameter rather than assuming that its endpoint values uniquely define it. Let $p_c$ be the probability of gold occupation at a corner site and defineThe complementary copper probability is $3(1-\rho )/4$, so the corner-site atomic scattering factor isThe stoichiometry provides one gold chemical atom per four-site unit cell. If all three face sublattices are equivalent, their gold occupation probability $p_f$ obeys $p_c+3p_f=1$. ThereforeAt $\rho =1$ the corner and face sites carry pure gold and pure copper; at $\rho =0$ they have identical statistical compositions. These formulas describe the branch with $0\le \rho \le 1$ and equivalent face-site averages.

Use the same atomic basis phases as in the ordered structure factor, replacing each atomic scattering factor by its site average. For the reflection with equal integer parity,Conservation of the total composition removes the alloy sublattice order parameter from this amplitude. For the reflection with mixed integer parity,The coherent powder diffraction wave intensities consequently satisfyHere the lattice constant, atomic scattering factors, powder diffraction multiplicities, and omitted experimental factors are held fixed. The ratio of a superlattice reflection to a fundamental reflection can therefore measure the magnitude of the alloy sublattice order parameter.

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

Write $U(x)=-e\varphi (x)$ for the potential energy of an electron in an electrostatic potential $\varphi$, where $e>0$. With $G=2\pi /a$, the coupling in the nearly-free electron model is the Fourier coefficientIf a symbol $V(x)$ already denotes the electron potential energy, as in part (c), it replaces $U(x)$ without another factor $-e$.

At the Brillouin zone boundary the two resonant plane waves have unperturbed energy $E_{\mathrm{ZB}}$. Their Hamiltonian operator in the two-state basis has diagonal entries $E_{\mathrm{ZB}}$ and off-diagonal entries $V_G,V_G^{\ast }$; its eigenvalues are $E_{\mathrm{ZB}}\pm |V_G|$. ThusThis is the local band gap in the two-wave nearly-free electron model.

For zero coupling, $\sqrt{4y^2}=2|y|$, and the two folded energy bands areThey touch at $y=0$, where the underlying free plane wave branches cross. A nonzero Fourier coefficient replaces this crossing by an avoided crossing, with $E_{\pm }(0)=E_{\mathrm{ZB}}(1\pm ϵ)$. The lower energy band moves downward and the upper energy band upward. Both branches are smooth near $y=0$ for $ϵ>0$.

The figure uses the boundary-centred interval $-1\le y\le 1$ required in part (b). With the printed definition $y=1-ka/\pi$, this is the translated Brillouin zone $0\le k\le 2\pi /a$. The printed $0\le y\le 2$ instead corresponds to $-\pi /a\le k\le \pi /a$; the local two-wave formula must be folded appropriately before treating it as a dispersion over that entire zone. It must not be integrated over $0\le y\le 2$ as though it were the filled first energy band there. The subsequent calculation follows the explicit continuation prescribed in part (b).

For a long chain of length $L$ with periodic boundary conditions, each spin has $Ldk/(2\pi )$ quantum states in a wavenumber interval $dk$. Including two spin states, a full energy band contains $2L/a$ electrons. The state counting in a Bloch band therefore gives the energy densityThe integral of $1+y^2$ is $8/3$. For $ϵ>0$, the even square-root integral isFor example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end logarithm contributes $\ln ϵ$. HenceThe continuous zero-coupling limit is $\mathcal{U}(0)=2E_{\mathrm{ZB}}/(3a)$. This agrees with directly integrating the one-dimensional free electron model up to $k_F=\pi /a$ at two electrons per unit cell.

At every $y$ and every $ϵ>0$,so the lower energy band lies strictly below its zero-coupling counterpart. Its state counting in a Bloch band and complete occupation are unchanged. Subtracting the two energy densities directly givesThusEquality holds at $ϵ=0$. This is the downward displacement of the filled lower energy band in the sketch. The upward displacement of the empty upper energy band contributes nothing to the zero-temperature electronic energy density.

For $ϵ\to 0^{+}$, the Taylor expansions needed in the exact energy density areSubstitution yieldsThereforeHere Big O notation describes the neglected small-coupling terms, and the logarithm is natural. The nonanalytic ${ϵ}^2\ln ϵ$ term comes from the formerly degenerate quantum states near the Brillouin zone boundary; a uniform nondegenerate perturbation theory expansion of the integrand would fail there. This is the same mechanism as the Peierls energy gain.

Regard $V(x)$ as an electron potential energy, and take $G=2\pi /a$. Each cosine supplies half its coefficient to each of its two Fourier coefficients. The nonzero coefficients areThe Fourier coefficient connecting the two plane waves at $k=\pm \pi /a$ is $V_G$, so the two-wave band gap isAt $\delta =0$ the actual periodic potential has period $a/2$ and no Fourier harmonic at $G$. The crossing at the artificially halved Brillouin zone boundary then has zero band gap. Higher-order weak-potential corrections can change the proportionality coefficient at nonzero $\delta$; the displayed coefficient belongs to the given two-wave approximation.

There is also a small-displacement qualification in the source: the claimed positions of the potential energy minima are correct to first order in $\delta$, not exactly. Put $t=2\pi x/a$. ThenAt $\delta =0.01$ and $x/a=1/4-\delta =0.24$, this equals approximately $-2.802\times 10^{-3}$, rather than zero. For $V_0>0$, expanding about $x/a=1/4$ and $3/4$ gives minima at $1/4-\delta +O({\delta }^3)$ and $3/4+\delta +O({\delta }^3)$. This does not alter the leading Fourier coefficient or the linear opening of the band gap.

Without the Peierls distortion, adjacent chemical atoms are separated by $a/2$. The primitive unit cell contains one chemical atom and contributes one electron. A energy band with two spin states accommodates two electrons per primitive unit cell, so the first energy band is half filled. Its true first Brillouin zone extends to $\pm 2\pi /a$, while the Fermi points are at $\pm \pi /a$.

Describing the same chain with a doubled unit cell of length $a$ merely folds the half-filled energy band into two branches that touch at those Fermi points. The unbroken lattice translation by $a/2$ prevents an artificial band gap there. A Peierls distortion breaks that smaller lattice translation and supplies the missing Fourier coefficient at $2k_F=G$; the lower folded energy band can then be completely filled and separated from the upper energy band. Thus the zero-band gap result at $\delta =0$ agrees with the original one-electron band filling.

Define the dimensionless coupling $C=\pi |V_0|/E_{\mathrm{ZB}}$, so that $ϵ=C|\delta |$. Substitution into the energy density from part (b) gives, within the prescribed continuation of the two-wave nearly-free electron model,Its value at $\delta =0$ is understood by continuity. For $C|\delta |\ll 1$,Both electrons per doubled unit cell have been included. The fixed $2G$ harmonic affects the underlying energy bands beyond the two-wave approximation; the expression here is the energy density requested from part (b), rather than an exact solution for all harmonics of the full periodic potential.