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For physical spin ladder operators, $S_j^{\pm }=S_j^x\pm i{S}_j^y$. Operators at distinct sites commute. Expanding the two products cancels their cross terms and givesAdding the remaining component provesFor spin one-half, $S^{+}|\downarrow ⟩=\hslash |\uparrow ⟩$, $S^{-}|\uparrow ⟩=\hslash |\downarrow ⟩$, and raising an up spin or lowering a down spin gives zero. These formulas restore the physical $\hslash$ factors implicit in the dimensionless ladder hint.

Let $|n⟩$ abbreviate the state with the only down spin at site $n$, keeping $|{\psi }_0⟩$ for the fully up state. If $n$ lies on neither end of bond $(j,j+1)$, that bond contains two up spins. The ladder terms vanish, while $S_j^z{S}_{j+1}^z$ gives ${\hslash }^2/4$. ThusFor a down spin at $j$, the $z$ product is instead $-{\hslash }^2/4$, and $S_j^{+}{S}_{j+1}^{-}$ moves it to $j+1$ with amplitude ${\hslash }^2$. For a down spin at $j+1$, the other ladder product moves it back. ThereforeEvery bond in $|{\psi }_0⟩$ is aligned, so summing its $N$ bond energies givesThis is indeed a ground-state energy: a pair of spin-one-half particles has ${\mathbf{S}}_j\cdot {\mathbf{S}}_{j+1}$ eigenvalues ${\hslash }^2/4$ and $-3{\hslash }^2/4$, so each ferromagnetic bond is bounded below by $-J/4$, simultaneously attained by the fully aligned state. The isotropic Heisenberg model has a whole maximum-total-spin ground multiplet, not a unique all-up ground state.

In the single-excitation subspace, only the two bonds adjacent to the down spin differ from the aligned background. Their diagonal energy increase is $J$, and each neighboring hop has amplitude $-J/2$. Hence, with periodic site indices,This is an exact finite-dimensional hopping problem. Choose the active one-site translation convention $T|n⟩=|n+1⟩$. A periodic plane-wave coefficient pattern requires $e^{iq(n+N)a}=e^{iqna}$, soThere are $N$ independent such wavevectors, for example one set of $N$ consecutive integers $\ell$ in a Brillouin zone. The normalized one-magnon Bloch states on a ring are $|q⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{iqna}|n⟩$. Reindexing with the periodic coefficient condition gives $T|q⟩=e^{-iqa}|q⟩$ and hence $T^N|q⟩=|q⟩$.

There is a logical qualification to the condition stated in the paper. On an $N$-site ring $T^N=I$ already holds for every state. In particular, the finite sum written with an arbitrary real $q$ also returns to itself after $N$ translations; that condition alone does not quantize $q$. The displayed allowed values are the ones for which the ansatz is a single one-site translation eigenstate, equivalently a consistently periodic plane wave. This is the additional condition needed for the subsequent energy-eigenstate argument. Other $q$ values generally describe superpositions of the allowed momentum modes.

Apply the hopping Hamiltonian operator and shift the summation index using periodic boundary condition. The coefficients from $|n-1⟩$ and $|n+1⟩$ are respectively $e^{iqa}$ and $e^{-iqa}$ times the original coefficient. Thusand the nearest-neighbour ferromagnetic magnon dispersion isThe unnormalized state in the question has the same energy. Distinct allowed momenta are orthogonal, since $N^{-1}{\sum }_n{e}^{i(q-q^{\prime })na}={\delta }_{q{q}^{\prime }}$ over one set of representatives.

For large $N$ the sampled wavevectors become dense, and the requested sketch is the periodic curve $[E(q)-E_0]/J=1-\cos (qa)$ over $-2\pi <qa<2\pi$. It has a zero at $q=0$ and tends to the equivalent zero at either open endpoint; maxima $2J$ occur at $q=\pm \pi /a$. Near any equivalent zero,using $q$ measured from that zero. The low-momentum ferromagnetic magnon is quadratic, with group velocity ${\hslash }^{-1}dE/dq=(Ja/\hslash )\sin (qa)$. The $q=0$ state is proportional to the total spin lowering operator acting on $|{\psi }_0⟩$, so it belongs to the same ground multiplet and has exactly zero excitation energy. Extending the plot beyond one reciprocal period repeats states rather than adding new independent modes. The lattice spacing $a$ is the nearest-neighbor chord length specified in the paper; $Na$ is the translation length in this lattice description, not necessarily the continuous circumference of a circular arrangement.

Taking the Hermitian adjoint gives ${\phi }_{\mathbf{k}}=c_k{\psi }_{\mathbf{k}}+s_k{\psi }_{-\mathbf{k}}^{†}$. For nonzero wavevectors, the canonical commutation relations implyWhen the Kronecker delta is nonzero, $k=q$, so the hyperbolic Pythagorean identity $c_k^2-s_k^2=1$ givesFor the two annihilation operators, the only potentially nonzero terms areHere $\mathbf{q}=-\mathbf{k}$ implies equal magnitudes and hence equal $c,s$, because ${\theta }_k$ depends only on $|\mathbf{k}|$. ThusThe corresponding commutators of two creation operators vanish by the Hermitian adjoint. This verifies that the Bogoliubov transformation preserves the full bosonic canonical commutation relations.

Put ${ϵ}_k={\hslash }^2{k}^2/(2m)>0$ for $\mathbf{k}\ne 0$ and $g=u\rho$, using $\rho =N/L^3$. Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state isTogether with part (d), the fixed-mean-number Bogoliubov variational energy isFor repulsive contact interaction, $u>0$, each ${ϵ}_k+g>g>0$. Using the double-angle hyperbolic-function identities,soThe stationary point therefore satisfiesThis is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersionAt the displayed solution ${\theta }_k^{\ast }$, $\cosh 2{\theta }_k^{\ast }=({ϵ}_k+g)/E_k$ and $\sinh 2{\theta }_k^{\ast }=g/E_k$. HenceThe hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from $\theta =0$ isIndeed $E_k<{ϵ}_k+g$. More directly, for small positive $\theta$,The pair vacuum expectation value lowers the interaction energy at first order in $\theta$, whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.

Two qualifications delimit this result. First, the paper does not state the sign of $u$. More generally a finite stable minimum exists only when ${ϵ}_k+g>|g|$ for every retained nonzero mode; repulsion ensures this. For example, with $g=-{ϵ}_k$ in some mode, $f_k(\theta )={ϵ}_k\sinh (2\theta )/2$ tends to $-\infty$ as $\theta \to -\infty$, so an unconditional minimum claim for arbitrary $u$ is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.

Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large $k$, $E_k-{ϵ}_k-g=-g^2/(2{ϵ}_k)+O({ϵ}_k^{-2})$, whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.

The first-order excitation amplitude is the Fourier transform of the temporal pulse envelope evaluated at the Rabi detuning. Two rectangular time windows therefore give exactly the mathematical structure of double-slit interference: each window gives a sinc function amplitude, and their separation gives a relative quantum phase $\delta T$. The transition probability is a sinc function squared envelope multiplied by ${\cos }^2(\delta T/2)$.

Pulse duration corresponds to slit width, pulse-start separation to slit-center separation, and Rabi detuning to the transverse wavevector sampled on the screen. The narrow fringes measure the long separation, while the broad envelope measures the short interaction duration. As in spatial interference, adding probabilities instead of coherent probability amplitudes would lose the fringes. Randomizing the relative pulse phase averages the cosine squared to $1/2$.

Differentiating the tight binding dispersion produced by the Hückel method givesHenceUnlike a free particle, the band effective mass depends on wavevector and can change sign. At an inflection point $\cos (ka)=0$, the inverse mass vanishes and the curvature-based mass diverges; a quadratic approximation then ceases to describe the dispersion locally. For the usual $\beta <0$ convention, the bottom at $k=0$ has positive mass and the top at $k=\pi /a$ has negative electron mass. The constant $\alpha$ has no effect on curvature.

Let $\mathcal{A}$ be the total two-dimensional area. There are $\mathcal{A}/(2\pi {)}^2$ wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk givesFor areal number density $n_e=N_e/\mathcal{A}$ this yieldsThe disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

For the free electron model, the Fermi energy is the highest occupied one-electron energy at zero temperature, measured relative to the bottom of the free-electron band. The occupied wavevectors fill a disk in two dimensions or a sphere in three dimensions. The radius is the Fermi wavevector $k_F$, withEach allowed wavevector has two spin states. At nonzero temperature, the Fermi-Dirac distribution rounds the occupation boundary; the zero-temperature Fermi energy remains a useful density-defined scale and should not be defined as the thermal energy $k_{\mathrm{B}}T$.

The energy-band widths are the difference between their extrema:For the separated bands, the top of the $s$ band and bottom of the $p$ band are both at $k=\pi /a$. HenceIt is a direct band gap at the zone boundary, because both limiting states have the same wavevector. This result uses the opposite signs of the $s$ and axial-$p$ hoppings.

With mixing and $\gamma =|{\beta }_{sp}|$, the equal-on-site-energy result isFor the regime $0<\gamma <t$ considered in part (iii), the crossing becomes an avoided crossing with gap $4\gamma$ at $k=\pi /(2a)$. The lower band has its maximum there and the upper band its minimum. Their orbital characters exchange smoothly across this region, with equal $s/p$ weights at its centre. The lower-right panel gives the sketch.

The zero-mixing limit recovers the crossing of case (1). A useful special limiting case is $\gamma =t$, where the square root is independent of wavevector and both bands are flat at $E=\pm 2t$. This flat-band limit is outside the strict $\gamma <t$ assumption made in part (iii).

For $0<\gamma <t$, the positive branch ranges from $2\gamma$ to $2t$, and the negative branch from $-2t$ to $-2\gamma$. ThereforeThe smallest separation is between the upper-band minimum and lower-band maximum at the same wavevector $k=\pi /(2a)$:It is a direct band gap at the interior avoided crossing. The zero-mixing limit has zero gap; the width is not $4|\beta |$ because these are the mixed energy-ordered branches, each turning around at the middle of the plotted interval.

The Debye model treats a three-dimensional solid as an isotropic elastic continuum with three independent acoustic phonon polarizations. It assumes harmonic normal modes, a linear dispersion relation $\omega =vq$, and equilibrium Bose-Einstein statistics with zero phonon chemical potential. Real anisotropy and nonlinear dispersion are replaced by one effective Debye sound speed. A cutoff restores the finite number of atomic vibrational degrees of freedom; the continuum itself would otherwise have infinitely many modes.

For sample volume $V$, the wavevector state density is $V/(2\pi {)}^3$ per polarization. Counting a sphere givesThere are $3N$ modes for $N$ atoms. HenceThe associated phonon density of states isIf longitudinal and transverse sound speeds differ, use $3/v^3=1/v_L^3+2/v_T^3$ for the low-frequency count. The Debye frequency is an effective mode-counting cutoff, not a claim that every actual branch terminates at exactly that frequency.

For area $A$, allowed wavevectors have density $A/(2\pi {)}^2$. Include two electron spin states and fill a disk of radius $k_F$ at zero temperature. ThenThe free electron model has parabolic dispersion $E={\hslash }^2{k}^2/(2m_e)$, with its minimum chosen as zero, soThese are the Fermi wavevector and Fermi energy for a spin-degenerate two-dimensional free electron gas. An imposed spin polarization or additional valley degeneracy would change the state count.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

A phonon is one quantum of a collective harmonic normal mode of a crystal. A mode labelled by wavevector and branch has energies $E_m=\hslash \omega (m+1/2)$, with bosonic occupation statistics. The crystal momentum $\hslash \mathbf{k}$ is defined modulo a reciprocal lattice vector; it is not generally the mechanical momentum of one atom. The phonon dispersion relation assigns its frequency and polarization to the wavevector and branch.

Inelastic neutron scattering can measure this relation in a single crystal. Determine the incident neutron energy and wavevector, scatter at a chosen angle, and analyze the outgoing energy by time of flight or a spectrometer analyzer. Energy and crystal-momentum conservation giveThe signs distinguish phonon creation and annihilation with consistent momentum convention. Repeating measurements at different momentum transfers maps peaks in the dynamic scattering response into phonon branches. The measured intensity depends on occupation, scattering factors and the projection of the mode polarization along the momentum transfer; a weak or missing peak need not mean that the corresponding mode does not exist.

Write $q=ka$ and use $2\sqrt{C_1/M}$ as the frequency unit. The nearest-neighbour chain and its next-nearest-neighbour extension haveThese are each a single acoustic branch, repeated over reciprocal-lattice periods. There is no added optical branch merely because a longer-range spring is introduced. The cutoff $C_n=0$ for $|n|>1$ specifies the first chain as nearest-neighbour despite the source's repeated word in its description.

For small $|q|$, ${\omega }_1\simeq a\sqrt{C_1/M}|k|$ and ${\omega }_2\simeq a\sqrt{(C_1+4C_2)/M}|k|$. ThusThe maximum is not necessarily at the zone edge. Put $x={\sin }^2(q/2)$, so $0\le x\le 1$. The second squared dimensionless frequency isIt reaches its maximum at $x=3/4$, or $q=\pm 2\pi /3$ in the first zone. The first chain peaks at $q=\pm \pi$. ThereforeAt $q=\pm \pi$ both chains have the same frequency, but the second chain has already passed its interior maximum. The next-nearest-neighbour monatomic-chain frequency maximum is therefore crucial to a correct sketch. Periodic repetition gives the requested curves for all wavevectors.

Use periodic boundaries in a box of side $L$ and include two spin states for each wavevector. One orbital state occupies $(2\pi /L{)}^d$ in wavevector space. Let $v_d$ be the volume of the unit ball in dimension $d$, namely $2,\pi ,4\pi /3$. Filling the ball of radius $k_F$ givesSince $E_F={\hslash }^2{k}_F^2/(2m)$,The density $n$ means number per length, area or volume as appropriate; it is not assigned three-dimensional units in every case.

For the same spin-summed states, the cumulative number per $d$-dimensional volume below energy $E$ is $n(E)=2v_d(2mE/{\hslash }^2{)}^{d/2}/(2\pi {)}^d$. Differentiating yields $g(E)=(d/2)n(E)/E$. Evaluating at the filling energy,In one, two and three dimensions the density of states varies respectively as $E^{-1/2}$, a constant and $E^{1/2}$. The normalization counts both signs of momentum and both spin states exactly once. A single-spin density of states or extra valley degeneracy would require a different counting convention, not an extra unexplained factor in these constants.

Inelastic neutron scattering measures both momentum and energy exchanged with the crystal. Comparing incident and outgoing neutron wavevectors gives $\mathbf{Q}={\mathbf{k}}_i-{\mathbf{k}}_f$, while their energy difference gives a created or annihilated phonon energy $\hslash \omega$. Crystal momentum conservation identifies the reduced phonon wavevector modulo a reciprocal-lattice vector. Varying scattering geometry maps dispersion curves and polarization-dependent intensities across a substantial part of the Brillouin zone.

Inelastic X-ray scattering provides a complementary energy- and momentum-resolved probe, particularly when small samples or high-pressure environments favor photon beams. Raman and infrared spectroscopy measure optical-frequency resonances subject to different symmetry selection rules. Photon momentum at these frequencies is usually small on the Brillouin-zone scale, so these methods mostly sample near the zone center rather than an arbitrary phonon wavevector. Brillouin scattering and ultrasonic measurements probe low-frequency acoustic motion and infer sound velocities or elastic constants.

Heat-capacity measurements constrain the thermally weighted vibrational density of states and its low-frequency behavior, but do not by themselves reconstruct a unique wavevector-resolved dispersion. Elastic diffraction and temperature-dependent displacement factors also constrain mean-square motion, while energy-resolved inelastic measurements distinguish individual excitations directly. A monatomic scalar chain has only the acoustic branch derived below; it should not be assigned an optical branch merely because Raman spectroscopy exists for more complicated crystals.

The right spring changes length by $u_{n+1}-u_n$ and pulls atom $n$ to the right with force $C(u_{n+1}-u_n)$. The left spring contributes $-C(u_n-u_{n-1})$. Adding these restoring forces and applying Newton's second law givesUniform displacement makes both spring extensions vanish, as translational invariance requires. Search for a normal mode $u_n=u_0{e}^{i(kna-\omega t)}$. Substitution yieldsOne representative first Brillouin zone is $-\pi /a\le k\le \pi /a$. This is one positive-frequency acoustic branch: near zero wavevector, $\omega \simeq a\sqrt{C/m}|k|$, so the sound speed is $v_s=a\sqrt{C/m}$. Opposite signs of $k$ label opposite propagation directions, not a second polarization branch.

The model assumes small displacements, harmonic nearest-neighbor forces, one scalar displacement per atom and uniform equilibrium spacing. It excludes damping, anharmonic expansion and transverse polarizations. In a three-dimensional monatomic crystal there would normally be three acoustic polarizations; that counting cannot be imported into this explicitly one-dimensional scalar model.

For a long periodic chain of length $\mathcal{L}=Na$, neighboring allowed signed wavevectors differ by $2\pi /\mathcal{L}$. Counting both signs once gives a state density per length $dk/\pi$ on the positive half-zone. Therefore the density per angular-frequency interval and per length isIt is zero outside this band, starts at $g(0)=1/(\pi v_s)=7.958\times 10^{-5}$ s/m and has an integrable inverse-square-root Van Hove singularity as $\omega \to {\omega }_m^{-}$. The state-count normalization isThus the scalar chain has one vibrational mode per atom, with no extra factor of two for the already counted propagation directions. If the horizontal variable were ordinary frequency $\nu$, the density would be $g_{\nu }(\nu )=2\pi g_{\omega }(2\pi \nu )$. A finite chain has discrete lines instead of an actual infinite peak; the smooth divergent curve is the large-chain continuum description.