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In the hydrogenic donor state approximation, the screened Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{ϵ}_{r}r)$ and the kinetic term uses $m^{\ast }=0.067m_e$. Rescaling the hydrogen atom gives the effective Bohr radius of a hydrogenic donor and the positive binding energyThereforeThe radius here is the exponential length of the ground-state wavefunction, not its root-mean-square radius. With $\psi (r)=(\pi a_b^3{)}^{-1/2}{e}^{-r/a_b}$, radial integration gives $⟨r^2⟩=3a_b^2$ and $⟨x^2+y^2⟩=2a_b^2$.

For dilute independent donors of number density $n$, a bound spin one-half electron has magnetic moment magnitude ${\mu }_s=|g|{\mu }_B/2$, with the Bohr magneton defined using the bare electron mass. Its weak-field Curie law response isFor the orbital response, expand the effective mass Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m^{\ast })$ in a uniform magnetic field, using $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. The spherical ground state has zero orbital angular momentum; the quadratic term raises its energy byDifferentiating this energy twice and using $B\simeq {\mu }_{0}H$ givesThe donor-only magnetic susceptibility cancels whenThe negative sign of $g=-0.45$ drops out of the linear magnetic susceptibility. This crossover is far below $E_b/k_B\simeq 63.6\mathrm{K}$, consistent with donor binding. It assumes sufficiently weak magnetic field, negligible exchange between donors and a thermal scale above any inter-donor splitting; sufficiently dilute donors can satisfy those assumptions. The background diamagnetism from part (b) is excluded from this cancellation.

A hydrogenic donor state has an exponentially decaying envelope $\psi (r)\propto e^{-r/a_b}$. Neighbouring donors at typical separation $r_d\sim n^{-1/3}$ therefore have exponentially small orbital overlap. The corresponding hopping integral is of order $E_b{e}^{-r_d/a_b}$, apart from algebraic factors and factors from the number of neighbours. Hybridizing many such localized states produces an impurity band with a bandwidth of the same scale:The order-one constant $\alpha$ absorbs details of the donor spacing and envelope. The exponential dependence expresses quantum tunnelling between weakly overlapping donor orbitals, rather than thermal excitation over the donor binding 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.

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.

Use the same normalization for the elastic energy and electronic energy density: if $\kappa {\delta }^2$ is an energy per unit cell, divide it by $a$ first; if it is already per length, use it directly. Write its coefficient per length as $K>0$ and put $A={\pi }^2{V}_0^2/(4a{E}_{\mathrm{ZB}})>0$. The small-distortion change in total energy density isNo minimization calculation is needed. For arbitrarily small nonzero $|\delta |$, the logarithm is negative with arbitrarily large magnitude. Hence the negative electronic Peierls energy gain, proportional to ${\delta }^2\ln (1/|\delta |)$, eventually exceeds any fixed positive quadratic elastic energy. There are distorted configurations with lower energy than the undistorted configuration for every finite $K$ and nonzero coupling. The ideal zero-temperature infinite-chain model therefore hasThe source's ${\delta }_{\mathrm{eq}}>0$ is correct when $\delta$ denotes a nonnegative distortion amplitude. If $\delta$ is signed, the energy density is even and the two signs describe equally stable patterns related by a lattice translation. For example, $V(x+a/2;\delta )=V(x;-\delta )$, so a negative equilibrium distortion has the same energy as the positive one. The zero-coupling case $V_0=0$ instead has its minimum at $\delta =0$. The argument assumes a coupled chain, complete lower-energy band filling and zero temperature; it does not by itself establish finite-temperature long-range order in an isolated one-dimensional system.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

The plateau in both source plots is approximately $10^{18.5}{\mathrm{m}}^{-3}$, soThese are graphical estimates, not high-precision fitted material parameters. To extract activation energies consistently with part (a), retain the temperature powers in the effective density of states factors.

On the semiconductor carrier freeze-out tail, put $x=1/T$. ThenFor example, the right-hand graph gives roughly ${\log }_{10}[n/({\mathrm{m}}^{-3})]\simeq 17.7$ at $x=0.10{\mathrm{K}}^{-1}$ and $13.0$ at $x=0.30{\mathrm{K}}^{-1}$. Subtract the known prefactor contribution before interpreting the slope:Thus $\boxed{E_i\simeq 9\mathrm{meV}}$, to the accuracy of the drawing. Ignoring the weak prefactor gives about $9.3\mathrm{meV}$, a similar rough estimate.

For the intrinsic branch, useThe left-hand graph gives $n(100\mathrm{K})\simeq 2.5\times 10^{21}{\mathrm{m}}^{-3}$ and $n(500\mathrm{K})\simeq 1.5\times 10^{24}{\mathrm{m}}^{-3}$, both far above $N_D$. ThereforeHere $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$. Omitting the $T^{3/2}$ prefactor over this wide temperature interval would instead give approximately $0.14\mathrm{eV}$, so it is not a negligible correction for the band-gap extraction from intrinsic carrier activation. The source graph warrants an approximate narrow band gap, rather than several significant figures.

There are three regimes. At the lowest temperatures, most donor impurities retain their electrons; thermal ionization gives semiconductor carrier freeze-out. The result from part (b) isThis produces the approximately straight descending low-temperature tail on the logarithmic scale plot against $1/T$. The activation energy in the exponential is $E_i/2$, because the chemical potential adjusts with donor ionization.

At intermediate temperatures, the donor impurities are essentially fully ionized while intrinsic excitation remains small. Charge neutrality gives $n\simeq N_D$, producing the extrinsic plateau, roughly from $20$ to $40\mathrm{K}$ in the figure.

At higher temperatures, excitation from the valence band to the conduction band creates electron-hole pairs. The intrinsic carrier concentration grows as $n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$ for constant parabolic band effective masses and constant band gap. With ionized donor impurities, charge neutrality and the semiconductor mass-action law giveThus $n\simeq n_i+N_D/2$ when $n_i\gg N_D$. This explains the renewed rapid rise and the eventual intrinsic branch; that branch is governed by the band gap, rather than the donor binding energy.

For a fixed open-shell electron configuration in the LS coupling regime, Hund's rules select the free-ion ground term in the following order:

The first two rules come from the Coulomb interaction between electrons within the partially filled subshell, including its exchange interaction. Parallel spins require an antisymmetric spatial wavefunction, reducing close approach and often lowering the repulsive Coulomb interaction energy. Among the allowed high-spin terms, the angular dependence of the Coulomb interaction generally favours the largest $L$; a simple magnetic alignment argument alone does not derive this second ordering rule.

The weaker spin-orbit coupling then splits a selected $L,S$ term. In LS coupling, its term-level energy is proportional toThe effective sign appropriate to less-than-half versus more-than-half filling gives the third rule. These rules assume the electrostatic term separation dominates spin-orbit coupling; strong crystal field splitting or other coupling regimes require further analysis.

For neutron energy loss, $E_f=30-20=10\mathrm{meV}$. Write $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$. The neutron dispersion $E={\hslash }^2{k}^2/(2m_n)$, or $E[\mathrm{meV}]=2.0721k[{\mathring{\mathrm{A}}}^{-1}{]}^2$, givesA triangle of these wavevectors exists ifWith $c=4.50\mathring{\mathrm{A}}$, the requested momentum transfers have $Q=(2\pi /c)|l|=1.3963|l|{\mathring{\mathrm{A}}}^{-1}$. ConsequentlyFor example $|l|=1$ is too small even for forward scattering at these unequal neutron energies, and $|l|=5$ exceeds the backscattering maximum. This identifies accessible reciprocal lattice points, not guaranteed nonzero intensity: a one-phonon structure factor contains the atomic scattering lengths, eigenvectors and factors $\mathbf{Q}\cdot {\mathbf{e}}_j$. A transverse eigenvector perpendicular to $\mathbf{Q}$ may be invisible, and basis interference can cause extinctions. The supplied geometry and cell constants do not specify that full intensity selection rule.

At a valley $K$, $\gamma (K)=1+2\cos (2\pi /3)=0$. For h-BN the inequivalent sublattice energies therefore leave the direct two-band gapGraphene's equivalent carbon sites instead have ${ϵ}_B={ϵ}_N$ in this model, so the two bands touch at the valleys. Near a valley, $|\gamma (K+\mathbf{q})|\simeq (\sqrt{3}a/2)|\mathbf{q}|$: graphene has approximately linear, gapless Dirac cones while h-BN has gapped band edges. The model panel in the original figure in solution displays this distinction.

First, undoped h-BN is an electrical insulator with thermally suppressed carriers, whereas neutral graphene is a zero-gap semimetal whose transport is readily changed by gating. Disorder, temperature and contacts still affect its measured conductivity. Second, h-BN suppresses interband absorption below its large gap and is transparent over much of the visible range, whereas graphene permits low-energy interband absorption unless occupations block it. A quantitative optical threshold in real h-BN also involves additional bands and strong excitonic effects; a two-orbital model should not be presented as a numerical prediction of its complete optical spectrum.

Angle-resolved photoemission spectroscopy illuminates a surface with photons of known energy and measures the kinetic energy and emission direction of the escaping electrons. Referencing the initial binding energy to the Fermi level givesHere $\Phi$ is the sample/analyser work-function reference appropriate to the calibrated energy scale, and $\theta$ is the emission angle from the surface normal. Parallel crystal momentum is conserved modulo a surface reciprocal lattice vector; measuring two angular coordinates maps the two in-plane momentum components.

Sweeping angle and energy traces the occupied spectral weight and its dispersion. In a quasi-two-dimensional sheet, no detailed free-electron model of the perpendicular final-state momentum is required to map the in-plane bands. Peaks reflect the occupied single-particle spectral function, multiplied by Fermi occupation and photoemission matrix elements and broadened by finite lifetimes and resolution. Ordinary photoemission does not directly measure empty conduction bands; a gap may need complementary information. A clean surface, vacuum and charge control are especially important for an insulating specimen.

Use the explicit convention $H=J_1{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j+J_2{\sum }_{⟨⟨ij⟩⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, counting every bond once and taking dimensionless spins of length $S$. On the simple cubic lattice there are six nearest and twelve face-diagonal next-nearest neighbours.

For type I all six nearest bonds are antiparallel and all twelve next-nearest bonds parallel, so $E_I/N=S^2(-3J_1+6J_2)$. Choose the parallel direction of type II as $z$. It has two parallel and four antiparallel nearest bonds, and four parallel $xy$ face diagonals but eight antiparallel $xz,yz$ diagonals. Thus $E_{II}/N=S^2(-J_1-2J_2)$ andConsequently, for the stated antiferromagnetic exchanges,This comparison is also consistent with the full classical spiral energy. With $u_i=\cos (q_{i}a)$,It is affine in each $u_i\in [-1,1]$, so some minimum occurs at a cube vertex, giving the competing collinear states above. At the boundary, the coefficient of $u_z$ vanishes for $u_x=u_y=-1$, producing the larger degeneracy $(\pi /a,\pi /a,q_z)$ and permutations. It is not an isolated two-state crossing. The first panel of the model figure below shows the competing energies.

Keep the once-per-bond Hamiltonian convention of part (a). Let each sublattice magnetization be its magnetic moment per total physical volume, so the density of B-sublattice spins is $1/(2a^3)$. Writing their moments as $\mathbf{\mu }=g{\mu }_B\mathbf{S}$ fixesThe exchange energy of one A spin is $6J_1{\mathbf{S}}_A\cdot ⟨{\mathbf{S}}_B⟩=-{\mathbf{\mu }}_A\cdot {\mathbf{B}}_{\mathrm{mf}}$, givingA convention using $2J_1$ per bond would double this coefficient; defining magnetization per sublattice volume instead changes its normalization. The definitions above make the total magnetization $\mathbf{M}={\mathbf{M}}_A+{\mathbf{M}}_B$.

For equal fixed sublattice magnitudes $M_0$, put ${\mathbf{M}}_A=M_0(\sin \theta ,0,\cos \theta )$ and ${\mathbf{M}}_B=M_0(\sin \theta ,0,-\cos \theta )$, with the external field along $+x$. The effective A-site field is $(B_{\perp }-\lambda M_0\sin \theta ,0,\lambda M_0\cos \theta )$. Its alignment with ${\mathbf{M}}_A$ requires $B_{\perp }=2\lambda M_0\sin \theta$, soThe last equality uses the externally imposed vacuum field $B_{\perp }={\mu }_0{H}_{\perp }$ and neglects demagnetizing corrections. Canting costs exchange energy but need not change the magnitude of either sublattice moment. The result holds before saturation and assumes the ordered axis is maintained when defining a transverse response.

Seek the steady forced displacement as $x=x_0{e}^{-i\omega t}$. The oscillator equation givesFor $n$ ion pairs per volume, the ionic polarization is $P_{\mathrm{ion}}=nqx$. Add the electronic polarization ${ϵ}_0({ϵ}_{\infty }-1)E$, assumed frequency-independent over this infrared range. The complex relative permittivity is thereforeFor a longitudinal free oscillation without external free charge, $\mathbf{k}\cdot \mathbf{D}=0$ and the longitudinal electric field requires ${ϵ}_r({\omega }_{\mathrm{LO}})=0$ in the lossless limit. HenceEliminating the microscopic oscillator strength givesHere ${\omega }_{\mathrm{LO}}$ denotes the undamped longitudinal frequency; with finite damping the zero is at a complex frequency. The sign of the imaginary term follows the stated $e^{-i\omega t}$ convention and gives positive absorptive $\mathrm{Im}{ϵ}_r$ for $\omega >0$, $\gamma >0$. At zero frequency the same model yields the Lyddane-Sachs-Teller relation ${ϵ}_r(0)/{ϵ}_{\infty }={\omega }_{\mathrm{LO}}^2/{\omega }_{\mathrm{TO}}^2$. The high-frequency limit here is above the phonon resonance but below electronic resonances, not arbitrarily high photon energy.

At normal incidence from vacuum the reflectivity iswith the passive branch of the complex refractive index. Below the optical resonance the real permittivity rises towards its pole and reflectance increases. Between the TO and LO frequencies, the lossless permittivity is negative; the wave is evanescent and the reflectivity is unity in that idealization. Finite damping rounds and reduces this reststrahlen band. Above the LO edge, positive permittivity returns and reflectivity can become very small when the refractive index is near one, before rising towards its electronic-background value.

Reading the surviving source graph to its available precision givesThese are band-edge estimates, not a high-precision damping fit. The minimum immediately above the band is not exactly the LO zero: for zero damping ${ϵ}_r=1$ occurs at ${\omega }^2=({ϵ}_{\infty }{\omega }_{\mathrm{LO}}^2-{\omega }_{\mathrm{TO}}^2)/({ϵ}_{\infty }-1)>{\omega }_{\mathrm{LO}}^2$.

The high-energy plotted tail near $90\mathrm{meV}$ has $R\simeq 0.22$. Ignoring residual phonon dispersion gives $n\simeq (1+\sqrt{R})/(1-\sqrt{R})=2.77$ and $ϵ\simeq 7.65$. Correcting with the finite-energy oscillator ratio givesThus a background of order $8$–$9$ is supported; equating the still-dispersive tail exactly to the asymptotic background would slightly underestimate it. The original model curves below use these approximate energies, without copying or fitting individual source data points.

In Raman scattering, the photon transfers $\hslash \omega =\hslash ({\omega }_i-{\omega }_s)$ and $\mathbf{q}={\mathbf{k}}_i-{\mathbf{k}}_s$ to a collective excitation. Changing scattering geometry selects different $q$ and probes its dispersion.

The open circles rise and then flatten near $45\mathrm{meV}$: they trace the lower transverse phonon polariton branch. It is strongly dispersive on its photon-like side, then becomes predominantly TO-phonon-like and approaches the TO energy at larger wavevector. The open triangles stay near $50\mathrm{meV}$ and are the nearly nondispersive longitudinal optical phonon. Longitudinal motion creates its own polarization field and is shifted above the transverse resonance. It does not hybridize with a propagating transverse photon in the same way, and it should not be misidentified as the upper transverse polariton branch, which rises with wavevector. Raman tensor and polarization selection rules determine which excitation is visible.

Scale qualification: the source horizontal label appears to print $10^{-4}\mathrm{c}{\mathrm{m}}^{-1}$. Taken literally, one plotted unit would have vacuum photon energy only $\hslash c\times 0.01{\mathrm{m}}^{-1}\simeq 2\times 10^{-6}\mathrm{meV}$, inconsistent with the observed tens of meV and the dispersive polariton. The physically compatible scale is of order $10^4\mathrm{c}{\mathrm{m}}^{-1}$; the qualitative branch identification does not require silently treating the printed negative exponent as correct. The original model figure in solution explicitly labels its calculated wavevector scale.

The displayed horizontal variable is ${\mu }_{0}H$ in tesla. The source formulas written with a flux quantum actually have the dimensions of magnetic flux density, so define $B_{c1}={\mu }_0{H}_{c1}$ and $B_{c2}={\mu }_0{H}_{c2}$ when applying them:From the onset of curvature away from the low-field line in panel A, $B_{c1}$ is roughly $0.06\mathrm{T}$ with uncertainty of order $0.01\mathrm{T}$. Panel B's branches approach the normal background over approximately $2$–$3\mathrm{T}$; use $B_{c2}\simeq 2.5\mathrm{T}$ as a coarse central estimate. Resolving irreversible loop closure from the true thermodynamic upper field, and correcting demagnetization, would require better data and geometry. These readings support scales, not precise material constants.

The upper field givesThe leading-log critical-field inversion starts by dividing the two field formulas to eliminate $\xi$:The large-$\kappa$ solution is $\kappa \simeq 6.15$, henceThe penetration-depth range is an indicative graph/model scale, not a propagated confidence interval. There is a mathematical ambiguity worth making explicit: $f(\kappa )=\ln \kappa /(2{\kappa }^2)$ rises from zero at one, peaks at $\sqrt{e}$, then falls. The same central ratio also gives $\kappa \simeq 1.055$ and $\lambda \simeq 12.1\mathrm{nm}$, so the printed restriction $\kappa >1$ alone does not select a unique solution. The leading-log lower-field formula is intended for appreciably large $\kappa$, which motivates the $\kappa \simeq 6$ branch; near one it needs the omitted core-energy correction. Even the larger-branch estimate has systematic uncertainty from that approximation.

$\lambda$ is the London penetration depth, the magnetic screening scale, while $\xi$ is the superconducting coherence length, the recovery scale of the superconducting order parameter after spatial suppression. In the mixed state the complex order parameter vanishes in vortex cores of radius of order $\xi$, recovers between separated cores and winds in phase by $2\pi$ around each singly quantized vortex. Currents and magnetic field spread over scale $\lambda$; vortex density grows with the mean internal field according to $n_v{\Phi }_0=\overline{B}$. Near $H_{c2}$ cores overlap and the order-parameter amplitude becomes small throughout the sample. A schematic core profile appears in solution.

The hysteresis is not the reversible equilibrium magnetization: defects pin vortices, so entry and exit leave different spatial flux distributions. Macroscopic currents oppose changes in flux, trapped vortices give remanence, and depinning and creep depend on history and temperature. Surface barriers can also contribute. The loop width probes an irreversible current scale, but its conversion to current density additionally needs sample geometry and an appropriate critical-state model.

The cosmological principle permits a Friedmann-Lemaître-Robertson-Walker metric. The Friedmann equation, with vanishing cosmological constant, isFor separately conserved pressureless matter, the cosmological continuity equation gives ${\bar{\rho }}_m={\bar{\rho }}_{m0}{a}^{-3}$: its rest energy is conserved while its physical volume grows as $a^3$. Define the present critical density and matter density parameter byEvaluating the Friedmann equation today gives $-k=H_0^2(1-\Omega )>0$. Thus the model has negative spatial curvature of an FLRW universe; the second contribution to $H^2$ represents that spatial curvature of an FLRW universe, rather than another material fluid. Taking the expanding branch givesHere $0<\Omega <1$. At early times, $a\ll \Omega /(1-\Omega )$, matter domination gives $\dot{a}\simeq H_0\sqrt{\Omega }{a}^{-1/2}$, hence, measuring cosmic time from the Big Bang,At late times the $a^{-2}$ term dominates the Friedmann equation. This curvature-dominated universe has $\dot{a}\to H_0\sqrt{1-\Omega }$ and thereforeThe late expansion approaches the coasting Milne model. The early limit belongs to this idealized matter-only model; including radiation in cosmology would alter the sufficiently early evolution.

The four contributions have distinct physical origins. The intrinsic photon temperature perturbation is ${\delta }_{\gamma }/4$: a local blackbody radiation energy density satisfies ${\rho }_{\gamma }\propto T^4$, so its first-order fractional perturbation is four times the fractional temperature perturbation. The endpoint cosmological Newtonian potential term ${\psi }_{\ast }$ is the gravitational redshift between the emitting plasma and the observer; escaping a negative potential well lowers the measured photon energy. Together with the intrinsic contribution it produces the ordinary Sachs-Wolfe effect. The term $-\hat{\mathbf{n}}\cdot {\mathbf{v}}_{\ast }$ is the Doppler CMB anisotropy, caused by the plasma's line-of-sight peculiar velocity at last scattering, with the source's direction convention.

Finally, $2\int d\chi {\psi }^{\prime }$ is the Integrated Sachs-Wolfe effect. A photon passing through a static potential well gains energy on entry and loses it on exit, with no net shift. When the potential evolves while the photon crosses, the two shifts fail to cancel. The factor two is the sum of the temporal and spatial scalar-potential contributions when anisotropic stress is negligible, so the two potentials are equal. The prime is a partial derivative at fixed comoving coordinate; it is not the total derivative along the light ray.

For a low-cosmological redshift galaxy sample, the intrinsic, endpoint-potential and last-scattering Doppler CMB anisotropy terms are localized near $z\simeq 1100$, whereas the galaxy density contrast is measured much nearer us. In the Limber approximation their radial kernels have no overlap, so their leading cross correlations vanish. The late Integrated Sachs-Wolfe effect samples the same nearby evolving potentials as the galaxies, giving the desired late-time CMB-matter cross correlation. The early Integrated Sachs-Wolfe effect near matter-radiation equality is likewise radially separated.

This is the leading linear, separated-kernel description intended here. The primary Cosmic microwave background anisotropy and late density contrast share primordial initial conditions, so their exact large-angle correlation need not be identically zero outside the Limber approximation. Additional secondary effects, such as the Sunyaev-Zeldovich effect, can also correlate with galaxies in real maps; they are absent from the stipulated linear expression.