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Quantum statistics matter when thermal energy is comparable with a characteristic vibrational level spacing. The chain's natural angular frequency is ${\omega }_0=\sqrt{\kappa /m}=5.7471\times 10^{12}{\mathrm{s}}^{-1}$, givingFactors of order unity depend on which vibrational scale is used. The largest chain phonon angular frequency is $2{\omega }_0$, corresponding to $87.8\mathrm{K}$. An isolated two-atom bond has reduced mass $m/2$ and angular frequency $\sqrt{2}{\omega }_0$, corresponding to $62.1\mathrm{K}$. Thus quantum corrections become appreciable on the scale of several tens of kelvin, and all modes are classical only when $T$ is comfortably above the largest relevant vibrational scale. Below it, high-frequency phonons are progressively frozen out and equipartition fails.

For the isolated bond, the zero-temperature root-mean-square relative displacement isIt is small enough for a local bond expansion. The classical weakly anharmonic estimate in part d therefore has a useful parametric windowwhere the lower condition means comfortably above the vibrational scale, rather than a sharp threshold. The model's upper bound concerns local anharmonicity and binding, not a prediction of the melting temperature.

Use the long-wavelength linear dispersion relation $\omega \simeq v_L|\mathbf{k}|$ as an estimate. The phonon energy is thenThis extrapolates the measured low-wavevector speed of sound to the $X$ point, the first Brillouin zone boundary along $[100]$. It is not an exact phonon dispersion relation there: the discrete crystal structure generally bends the branch, and the initial slope alone cannot determine its boundary energy. In particular, $(2\pi /a,0,0)$ is not a reciprocal lattice vector of the face-centered cubic lattice; the smallest such axial vector is $(4\pi /a,0,0)$. There is therefore no reason for this acoustic phonon energy to vanish.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.

A neutron can exchange both energy and crystal momentum with a lattice vibration. Determine its incoming and outgoing wave vectors ${\mathbf{k}}_i,{\mathbf{k}}_f$ from neutron wavelengths and directions. Its nonrelativistic kinetic energy is $E_n={\hslash }^2{k}_n^2/(2m_n)$, so the measured transfers areFor phonon creation, momentum conservation reads $\mathbf{Q}=\mathbf{G}+\mathbf{q}$ and the neutron loses energy $\hslash {\omega }_{\nu }(\mathbf{q})$. For phonon annihilation the corresponding phonon energy is gained by the neutron. Reduce $\mathbf{Q}$ by a reciprocal lattice vector $\mathbf{G}$ to obtain the phonon wave vector, and trace a branch by repeating this measurement at several reduced wave vectors.

A triple-axis spectrometer provides a suitable arrangement: a monochromator selects the incident wavelength, the oriented sample defines the momentum transfer through the scattering angles, and an analyzer crystal selects the final wavelength before detection. A time-of-flight spectrometer instead uses pulse timing and flight paths to obtain the energy change. In either case instrumental energy and angular resolution must be separated from intrinsic line widths. Nuclear scattering from atomic displacements couples neutrons to phonons; the one-phonon intensity contains a polarization factor involving $\mathbf{Q}\cdot {\mathbf{e}}_{\nu }$, atomic scattering lengths, and thermal occupation factors. Intensities can therefore differ even at equivalent phonon frequencies.

The difference of the two wave-vector labels is $(1,0,-1)$, an allowed reciprocal lattice vector because $1+0-1$ is even. Bloch's theorem of the lattice dynamical matrix therefore givesfor corresponding phonon branches. Neither individual odd-sum point needs to be a reciprocal lattice point. It is their difference that establishes equivalence. Structure factors and neutron polarization factors can still change the observed intensities at those two scattering vectors, so equal frequencies do not require identical measured spectra peak by peak in intensity.

Subtract the reciprocal lattice vector $(0,0,8)$, leaving a reduced wave vector $(0.2,0,0)$ along $a$. A linear-acoustic estimate givesInterpret the quoted intrinsic width as the full width at half maximum in energy, $\Gamma =0.1$ meV. The corresponding energy-relaxation mean lifetime is $\tau =\hslash /\Gamma =6.58\times 10^{-12}$ s, giving mean free path $v_g\tau \simeq 19.7$ nm. If the given width instead denotes a half width, the inferred mean lifetime differs by a factor of two; the convention must be specified.

Approximating the thermal average by this velocity and mean lifetime, and using the isotropic angular factor $⟨v_a^2⟩=v_g^2/3$, kinetic theory givesUsing a single purely $a$-directed effective channel without angular averaging gives $2.17$ W m${}^{-1}$ K${}^{-1}$, still far below the observed total $39$ W m${}^{-1}$ K${}^{-1}$. The isotropic estimate is roughly $54$ times smaller than the measured total.

The calculation is not a complete thermal-transport average. Electrons in this metallic compound can carry substantial heat, the supplied total heat capacity need not be entirely phononic, and one $7$ meV phonon need not represent all thermally active branches and mean lifetimes. At $20$ K, $k_{B}T\simeq 1.72$ meV, so that measured mode is relatively high in the thermal distribution; lower-energy modes can have much longer mean lifetimes. A measured phase velocity is also only an estimate of the group velocity if the branch is nonlinear.

electron-phonon interaction is a plausible damping mechanism in a phonon-mediated superconductor, but the measurement is above its $15$ K transition and probes the normal state. It cannot be attributed directly to decay into an already established superconducting condensate. Below the transition an energy gap can modify electron-hole damping. The weak-coupling scale $2\Delta \simeq 3.52k_B{T}_c\simeq 4.55$ meV is below the $7$ meV phonon energy, so even superconductivity need not prohibit all pair-breaking damping of this mode. Temperature-dependent spectra would distinguish that mechanism from defects or anharmonic phonon decay.

For a periodic single-electron Hamiltonian operator, Bloch's theorem givesTwo bands at the same $\mathbf{k}$ have the same translation eigenvalues but different cell-periodic eigenfunctions, generally different energies and different orbital/nodal character. They may be chosen orthogonal even at a band degeneracy. The basis choices are treated in the two scoped branches below.

For a slowly varying external potential $U(\mathbf{r})$, the semiclassical wavepacket Hamiltonian is ${ϵ}_b(\mathbf{k})+U(\mathbf{r})$. Hamilton's equation for the conjugate variables $\mathbf{r},\hslash \mathbf{k}$ gives $\hslash \dot{\mathbf{k}}=-\nabla U=\mathbf{F}$. Equivalently the electric perturbation changes the translation phase at the rate $q\mathbf{E}\cdot \mathbf{R}/\hslash$. This is the single-band acceleration theorem, requiring a slowly varying field and negligible interband tunnelling/scattering; it refers to crystal momentum, not free-space mechanical momentum.

Let $e>0$ and $K=e{E}_0/(\hslash \omega )$. Integrating from the specified initial condition $\mathbf{k}(0)=0$ gives, with the first field alone,With both components switched on,This is a circle centred at $(-K,0)$, not at the band minimum. Since $v_i=\hslash k_i/m_i^{\ast }$, integration givesThe switch-on drift in a circularly driven parabolic band accompanies unequal transverse scales: it is a stretched cycloid. In a frame moving with that drift, the oscillation is an ellipse with semiaxis ratio $m_y^{\ast }/m_x^{\ast }\gg 1$. A closed real-space ellipse would require different initial momentum. Indeed, after one period $x-x_0=-2\pi e{E}_0/(m_x^{\ast }{\omega }^2)\ne 0$, an explicit check against that interpretation.

Writing $\theta =\omega t$, the kinetic energy above the band minimum isFor $m_x^{\ast }<m_y^{\ast }$ it is largest at $\theta =\pi$, with value $2e^2{E}_0^2/(m_x^{\ast }{\omega }^2)$. Equating this to the optical-phonon energy gives the emission threshold estimateThis treats the optical branch as approximately dispersionless and assumes a momentum-allowed final state, a cold initial electron, and negligible phonon absorption. It is the amplitude of each Cartesian field component; the circular field has constant magnitude $E_0$. A steady periodic orbit centred at $\mathbf{k}=0$ would instead give twice this amplitude, but fails the stated switch-on condition.

For $N$ ion pairs per unit volume, a relative displacement $x$ produces ionic polarization $P=Nqx$. Solving the driven Lorentz oscillator model with the source's $e^{-i\omega t}$ convention givesThe constant ${ϵ}_{\infty }$ includes the fast electronic response: infinity here means well above the phonon band but below electronic resonances. Writing $\Delta ϵ={ϵ}_s-{ϵ}_{\infty }$ makes the numerator $\Delta ϵ{\omega }_{\mathrm{TO}}^2$. In an undamped longitudinal mode the bulk displacement field vanishes without an external free charge, so a nonzero electric field requires ${ϵ}_r({\omega }_{\mathrm{LO}})=0$. ThereforeThis is the Lyddane-Sachs-Teller relation for the single polar mode.

At normal incidence from vacuum the Fresnel equations giveBelow the transverse resonance the dielectric response increases, raising reflectivity. Between the transverse and longitudinal frequencies the undamped dielectric function is negative, giving an evanescent transmitted wave and nearly total reflection: the Reststrahlen band. Above the longitudinal edge it becomes small and positive, passing through ${ϵ}_r\simeq 1$ at the reflectivity minimum and eventually approaching its electronic background.

Reading the original graph to its limited precision gives $\hslash {\omega }_{\mathrm{TO}}\simeq 30\mathrm{meV}$ and $\hslash {\omega }_{\mathrm{LO}}\simeq 37\mathrm{meV}$, hence ${\omega }_{\mathrm{TO}}\simeq 4.6\times 10^{13}{\mathrm{s}}^{-1}$ and ${\omega }_{\mathrm{LO}}\simeq 5.6\times 10^{13}{\mathrm{s}}^{-1}$. Extrapolating the low-energy reflectivity to roughly $R(0)\simeq 0.25$ givesValues around $9$–$10$ and $5.5$–$6.5$ are appropriate to this graphical estimate. Neither the dip near $39\mathrm{meV}$ nor the finite-energy right endpoint represents the high-frequency asymptote; treating that endpoint as $R(\infty )$ would underestimate ${ϵ}_{\infty }$.

At $\omega ={\omega }_{\mathrm{TO}}$, ${ϵ}_r={ϵ}_{\infty }+iA$ with $A=\Delta ϵ{\omega }_{\mathrm{TO}}/\gamma$. Weak damping gives $A\gg {ϵ}_{\infty }$, so for the complex refractive index $n+i\kappa$,Squaring and substituting $A$ provesUse the reflectivity at the estimated transverse resonance, not automatically the plateau maximum. A representative reading $R_{\mathrm{TO}}\simeq 0.9$, with $\Delta ϵ\simeq 3.1$, gives $\gamma /{\omega }_{\mathrm{TO}}\simeq 0.0043$, $\gamma \simeq 2.0\times 10^{11}{\mathrm{s}}^{-1}$ andGraphical uncertainty in the transverse edge makes this only a few-picosecond estimate: readings $R_{\mathrm{TO}}\simeq 0.85$–$0.93$ give approximately $2$–$11\mathrm{ps}$. The displacement amplitude decays as $e^{-\gamma t/2}$ and has lifetime $2/\gamma$; phonon occupation/energy decays as $e^{-\gamma t}$ and has lifetime $1/\gamma$. Anharmonic forces allow an optical phonon to decay into two lower-energy phonons when energy and crystal momentum are conserved. Spontaneous decay can persist at low temperature; defects and electron–phonon scattering can add further linewidth.

In BCS theory, an electron distorts the positively charged lattice, and the retarded response can attract another electron. Within a phonon-energy shell around the Fermi surface, the effective attraction can overcome the screened Coulomb repulsion and destabilize the normal Fermi sea toward Cooper pairs. In conventional elemental superconductors the pairs predominantly join time-reversed states of opposite momentum and spin, forming a spin singlet.

In the weak-coupling approximation,Here $N(0)$ is the appropriate single-spin density of states and $V_{\mathrm{eff}}>0$ the effective attractive interaction in a consistent normalization. Both the phonon scale and the exponential coupling matter. A large Fermi-level density of states, often associated with partially filled narrow $d$ bands, can raise $T_c$; weak coupling or a small density of states can suppress it even in a good conductor. With force constants and coupling fixed, ${\omega }_D\propto M^{-1/2}$ gives the conventional isotope trend. This is not a universal monotonic ranking of elements by atomic mass: phonon stiffness, electron–phonon matrix elements, Coulomb effects and magnetism vary between elements. Magnetic pair breaking helps explain why some elements do not follow a naive density-of-states expectation.

A phonon is one quantum of a normal vibrational mode of a crystal. In the harmonic approximation, each mode with wavevector $\mathbf{q}$, branch index $s$ and angular frequency ${\omega }_s(\mathbf{q})$ is an oscillator with energies $\hslash {\omega }_s(n+1/2)$. A phonon changes the occupation $n$ by one; it is not a particular atom moving through the solid.

For a three-dimensional primitive cell containing $r$ atoms there are $3r$ branches. Three acoustic branches approach zero frequency as $\mathbf{q}\to 0$: at zero wavevector they represent rigid translations, with all basis atoms displaced together. Their small-wavevector frequencies usually grow linearly with wavevector. The remaining $3r-3$ optical branches have internal relative motion of basis atoms and generally nonzero zone-centre frequencies. The names reflect their coupling to radiation in suitable polar crystals, not a rule that every optical mode must be infrared active. Acoustic and optical branches can each have longitudinal or transverse polarizations.

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.

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.

In weak-coupling BCS theory, a useful scale isHere the phonon spectrum sets the attractive-interaction cutoff, $N(0)$ is the consistently defined Fermi-level density of states and $V_{\mathrm{eff}}$ the effective pairing attraction after repulsion is accounted for. Light hydrogen supports high vibrational frequencies; strong electron–phonon coupling and favourable metallic density of states can raise $T_c$. Pressure changes bonding, phase stability, electronic structure and phonon frequencies. It is not simply an independent multiplier in this expression. In a real strongly coupled hydride, Eliashberg theory uses the full spectral function and a screened Coulomb pseudopotential; the elementary exponential is a mechanism guide rather than a controlled quantitative fit to $203\mathrm{K}$.

Measure a four-terminal resistance that falls below the instrumental bound, and test that a magnetic field suppresses the transition. Independently detect diamagnetic screening and, where feasible, field-cooled flux expulsion or an AC susceptibility signal with controlled cell background. A resistance drop alone can reflect contact changes or a filamentary path; a consistent magnetic response is important evidence for superconductivity. A reproducible isotope shift upon replacing hydrogen by deuterium tests the involvement of lattice motion, although it is not by itself proof of superconductivity.

Extreme-pressure samples are microscopic and enclosed in a diamond-anvil cell. Reliable separate electrical contacts, homogeneous pressure, pressure calibration during cooling, chemical stability and the tiny magnetic signal relative to the anvils and gasket are practical difficulties. Structural identification matters because a sulfur-rich hydrogen compound formed under pressure need not retain the starting stoichiometry.

Historical correction: the original primary report associates the $203\mathrm{K}$ transition with pressures near $155\mathrm{GPa}$, while roughly $90\mathrm{GPa}$ is the metallization scale, not that same transition point. The printed pairing of $203\mathrm{K}$ with $90\mathrm{GPa}$ should not be treated as an established measured pair. The statement about the highest conventional transition temperature is also a historical statement in this 2018 examination.^[ref]

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.

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.

In inelastic scattering, the probe transfers energy as well as momentum to the specimen. A neutron, X-ray or electron can create or annihilate a phonon, while electronic or magnetic excitations provide other channels. Measuring the incident and outgoing energies resolves the transfer $\hslash \omega$; their wavevectors determine momentum transfer $\hslash \mathbf{Q}$. In a crystal, phonon momentum is conserved modulo a reciprocal lattice vector. Resolved spectra therefore probe excitation dispersions and lifetimes, rather than only average atomic positions.

Thermal diffuse scattering arises from thermally populated lattice displacements and their correlations. The equilibrium lattice produces sharp Bragg peaks, but displacements redistribute part of the scattering into diffuse intensity. To one-phonon order a mode contribution contains a displacement projection and thermal occupation factors of the formwith atomic scattering weights, phases and geometrical factors understood. Stokes and anti-Stokes energy-resolved components carry $n_B+1$ and $n_B$ respectively; the displayed combination describes their summed scale. At high temperature, $n_B\simeq k_{B}T/(\hslash \omega )$, so soft modes and their polarization strongly influence the pattern. Zero-point motion also contributes at low temperature. An energy-integrating diffraction image can record this inelastic intensity as a diffuse background without resolving its small energy transfers. Static disorder gives additional diffuse scattering, so diffuse intensity alone does not establish a thermal mechanism.

A forward-biased p-n junction injects electrons from its n-type semiconductor and electron holes from its p-type semiconductor into an active region. A direct band gap allows radiative transitions without a compensating phonon, so stimulated emission can compete effectively with nonradiative processes. At thermal equilibrium there is one Fermi level; forward injection instead establishes separate electron and valence-electron quasi-Fermi levels $F_c$ and $F_v$. Sufficient injection makes conduction states more occupied than the corresponding valence states, producing optical gain.

Two parallel cleaved facets supply an optical resonator. At normal incidence their approximate intensity reflectance follows the Fresnel equations:A resonant normal mode grows when its round-trip amplification exceeds its losses. For length $L$, material optical gain $g$, overlap factor $\Gamma$, internal intensity loss ${\alpha }_i$ and facet reflectances $R_1,R_2$,The pump supplies the emitted energy; the facets provide feedback and let some output escape. Positive optical gain alone does not establish laser threshold.

There is a genuine periodicity mismatch in the source. Write $x_s=sb+u_s$. The intended periodic displacements in a harmonic chain satisfy $u_N=u_0$, equivalently $x_N=x_0+Nb$ for unwrapped positions. The printed $x_N=x_0$ instead makes the potential $\frac{1}{2}m{\lambda }^2{\sum }_s(x_{s+1}-x_s{)}^2+\frac{1}{2}Nm{\lambda }^2{b}^2$, with equilibrium at equal positions, not $x_s=sb$. The following lattice derivation uses periodic displacements, as required by the requested position operator.

Put $k_r=2\pi r/(Nb)$, $r=0,\dots ,N-1$, and Fourier transform the displacements and momenta with normalization $N^{-1/2}$. The spring quadratic form is diagonal, with factor $|e^{i{k}_{r}b}-1{|}^2=4{\sin }^2(k_{r}b/2)$. Thus each nonzero mode is a harmonic oscillator withCreation and annihilation operators obey $[a_r,a_{r^{\prime }}^{†}]={\delta }_{r{r}^{\prime }}$. Inverting the oscillator-coordinate expansion givesThe conjugate momentum is $p_s=-i\sqrt{m\hslash /(2N)}{\sum }_{r\ne 0}\sqrt{{\omega }_r}(a_r{e}^{i{k}_{r}sb}-a_r^{†}{e}^{-i{k}_{r}sb})$. Fourier orthogonality diagonalizes the Hamiltonian as ${\sum }_{r\ne 0}\hslash {\omega }_r(a_r^{†}{a}_r+1/2)$. The normalized oscillator vacuum is annihilated by every $a_r$; applying creation operators gives number states, each added phonon contributing $\hslash {\omega }_r$. The omitted zero mode is a free centre-of-mass coordinate, which must be fixed or treated separately rather than assigned an oscillator vacuum. Accordingly the reduced displacement commutator is $[u_s,p_t]=i\hslash ({\delta }_{st}-1/N)$.

For $iq{u}_s$, separate creation and annihilation parts. Their commutator is a scalar, so the supplied Baker--Campbell--Hausdorff formula and vacuum expectations giveSumming the geometric lattice phases yieldsThe ratio is interpreted by continuity at its removable singularities. The interference factor has its maximum $N^2$ at $q=2\pi n/b$, where all lattice phases coincide. The source's Fourier-sum hint should read $r\equiv 0(\mathrm{mod}N)$, rather than $r=Nb$.

The claim about the full matrix element also needs qualification: its Debye–Waller factor envelope varies with $q$. The absolute global maximum is $|M(0){|}^2=N^2$; at nonzero reciprocal-lattice points the value is $N^2{e}^{-2W(2\pi n/b)}<N^2$. Moreover the derivative of the envelope there is nonzero, whereas that of the interference factor is zero, so those points are not exact local maxima of the full finite-chain intensity. The reciprocal-lattice condition maximizes the interference factor, which is the usual constructive-scattering interpretation of the requested result.