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Start from electrons moving in the periodic equilibrium ionic potential. Displace ion $l$ by ${\mathbf{u}}_l$ and expand its contribution as $V(\mathbf{r}-{\mathbf{R}}_l-{\mathbf{u}}_l)\simeq V(\mathbf{r}-{\mathbf{R}}_l)-{\mathbf{u}}_l\cdot \nabla V(\mathbf{r}-{\mathbf{R}}_l)$. Quantizing the harmonic normal modes expresses ${\mathbf{u}}_l$ as a sum of creation and annihilation operators with displacement scale $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ are boson operators for lattice modes. The matrix element $M_{\mathbf{q}}$ contains the derivative of the ionic potential, phonon polarization, displacement normalization and electronic wavefunction overlaps. With phases chosen appropriately, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ ensures a Hermitian Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “Fröhlich Hamiltonian” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar semiconductor. A harmonic lattice, small displacements and a weak enough coupling to organize perturbation theory are the underlying approximations.

The Fröhlich canonical transformation removes real linear phonon operators from the electron-only effective problem to first order. Choose an anti-Hermitian $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe commutator describes virtual emission and absorption of a phonon. normal ordering yields an electron-electron term as well as one-electron energy renormalizations and phonon terms. Retain the former, absorb single-particle shifts into ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, so total momentum and the spins are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real phonons requires more care than this second-order construction.

For $|\Delta |<{\Omega }_{\mathbf{q}}$ the denominator is negative: virtual phonon exchange is attractive in this low-energy channel. Physically, an electron changes the ionic density and a second electron can gain energy by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every energy transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and superconductivity is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition temperature.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le {\xi }_{\mathbf{k}}\le D$, where ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. Measure the added-pair energy relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding eigenvalue equation, its eigenvalue must instead be $E_{\mathrm{abs}}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.

Write $g={\gamma }^2>0$ and $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis is the stated coefficient formula in the consistent Fermi-level energy convention. Its even dependence on $\mathbf{k}$ produces an even orbital state with a spin singlet. The constant $C$ is fixed up to an overall phase by normalization; for a bound state, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$:Here $\rho (E_F)$ counts orbital states for one spin species, as appropriate to the unrestricted $\mathbf{k}$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total density of states is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi =g\rho (E_F)$ as the paper's dimensionless coupling, inversion yieldsThe first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding energy is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.

Cooper's problem asks whether two electrons added above an inert Fermi sea bind despite Pauli exclusion of lower states. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. Its wavefunction draws on many time-reversed momentum states and becomes dominated by a narrow region near the Fermi surface at weak coupling; the corresponding pair is spatially extended. This establishes a Cooper instability of the unpaired reference state.^[ref] The full superconducting ground state needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. BCS theory supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.^[ref]

For the Bogoliubov spectrum, define ${ϵ}_k={\hslash }^2{k}^2/(2m)$ and $E_k=\sqrt{{ϵ}_k({ϵ}_k+2{\mu }_0)}$. For repulsive interactions and ${\mu }_0>0$,Thus the low-momentum excitations are gapless phonons and the high-momentum branch approaches free-particle curvature. The crossover occurs when kinetic and interaction energy are comparable. The Landau criterion follows because an obstacle moving at velocity $v$ can supply energy $\mathbf{v}\cdot \mathbf{p}$ to an excitation. HereThis is the homogeneous phonon kinematic threshold, not a guarantee that every perturbation dissipates immediately above it.

In two dimensions, $n=|\psi {|}^2$ has units of inverse area and ${\hslash }^2/m$ has units of energy times area. Therefore $\tilde{g}$ is dimensionless. Neglecting the kinetic term in the Gross–Pitaevskii equation, with positive coupling, giveswith $n=0$ outside the radius in the Thomas–Fermi approximation for a condensate. Multiplication givesThe sharp edge is an approximation; the neglected gradient term rounds it over a healing length scale.

Using the local sound-speed criterion for circular condensate stirring, the moving beam has speed $v=\Omega D$. The local interaction chemical potential is ${\mu }_0-V(D)=m{\omega }_r^2(R^2-D^2)/2$, soThe onset equality is $\Omega D=c(D)$ and the permitted excitation condition isThis uses the beam speed as the local relative-flow scale. A finite hole can distort the surrounding flow or nucleate vortices, so the result is the requested local phonon estimate, not a universal necessary condition for all experimental stirring.

For the loss-driven onset of condensate stirring, first integrate the density:At fixed coupling and trap frequency, ${\mu }_0\propto N^{1/2}$ and $R\propto N^{1/4}$. Hence $R(t)=R(0)e^{-t/(4\tau )}$ while $D$ stays fixed. For $\Omega ={\omega }_r/\sqrt{2}$, threshold requires $D/R=1/\sqrt{2}$. Starting at $D/R(0)=1/2$,At onset, $N=N_0/4$, $R=R(0)/\sqrt{2}$ and the beam is still inside the cloud. Excitations become kinematically allowed just beyond this equality, provided the stipulated adiabatic regime of the Thomas–Fermi approximation for a condensate remains valid.

Interpret the box as a hard-wall cube. Its one-particle levels are ${ϵ}_{n_x{n}_y{n}_z}={\pi }^2{\hslash }^2(n_x^2+n_y^2+n_z^2)/(2m{L}^2)$ for positive integers, giving $E_0=3{\pi }^2{\hslash }^2/(2m{L}^2)$. Counting lattice points in a positive octant and differentiating the large-radius count gives the supplied continuum density of states. Expressing its prefactor through $E_0$ yieldsHere $\mathcal{A}$ denotes the dimensionless constant requested in this question, to distinguish it from the unrelated hyperfine constant of question 1. Periodic boundary conditions would have zero ground energy and would not permit this parametrization by a nonzero $E_0$.

Separate the ground-state occupation from excited particles. The finite-box ground-energy convention in Bose condensation makes the chemical potential approach the ground energy from below as that occupation becomes macroscopic. In the thermodynamic continuum approximation one shifts the energy zero to the ground state and sets the limiting chemical potential to zero in the excited-state integral. The difference between the exact shifted density and its $\sqrt{ϵ}$ bulk form gives finite-size corrections; at the transition $k_B{T}_c/E_0\sim N^{2/3}\gg 1$. ThusThe identity follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating each term. The excited states have a finite saturation capacity. Setting that capacity equal to $N$ determines condensation:Equivalently $k_B{T}_c=(2\pi {\hslash }^2/m)[n/\zeta (3/2){]}^{2/3}$, with $n=N/L^3$.

At fixed $N$, below $T_c$ the chemical potential is pinned to its limiting value andAbove $T_c$, $N^{\prime }\simeq N$ and there is no macroscopic $N_0$; the chemical potential adjusts rather than allowing the saturation expression to exceed the available particles. At fixed $T$, put $N_c=N_{\max }^{\prime }(T)$. For $N<N_c$, $N^{\prime }\simeq N$ and $N_0$ is not macroscopic; for $N>N_c$, $N^{\prime }=N_c$ saturates while $N_0=N-N_c$ grows linearly. This is particle saturation in a homogeneous Bose gas; the sharp threshold is a thermodynamic-limit approximation to finite-box occupations.

The interaction scale compares differently with confinement and transition energies:Thus the stated range makes interactions dominate the ground-state confinement energy while remaining small on the thermal scale controlling $T_c$. This is an interaction-dominated ground state with nearly ideal condensation temperature. It also gives the dilute-gas parameter $n{a}^3=N(a/L{)}^3\ll 1$. A contact model has $g=4\pi {\hslash }^{2}a/m$ and $\mu \simeq gn$; the question's $E_I$ suppresses order-one factors, and the uniform ground energy per particle is $gn/2$, rather than exactly $gn$. Weak interactions can shift $T_c$ slightly, but their dimensionless strength $a{n}^{1/3}\ll 1$ makes that a small relative effect. In a hard-wall box, boundary healing layers do not overturn bulk interaction dominance when their thickness is small compared with $L$.

At $k_{B}T\ll \mu$, thermally occupied Bogoliubov modes have $E_k\simeq \hslash c_{s}k$, with $c_s=\sqrt{\mu /m}$. There is one acoustic branch for this spinless gas. Its thermal energy determines the phonon heat capacity of a dilute Bose gas:The zero-temperature interaction energy does not itself contribute to this leading temperature derivative at fixed $N,V$. This bulk power law also requires $k_{B}T\gg \hslash c_s/L$ so that the phonon spectrum may be treated as continuous; a strictly finite box eventually has discrete-level activated behaviour.

For the noninteracting condensed Bose gas, the ideal Bose-gas heat capacity below condensation follows by integrating $ϵg(ϵ)f(ϵ)$:Its limiting value at $T_c^{-}$ is about $1.926N{k}_B$. At very low temperature the interacting phonon curve is suppressed relative to the ideal curve, has slope three on logarithmic axes, and crosses over towards particle-like $T^{3/2}$ behaviour around $k_{B}T\sim \mu$. Since $\mu \ll k_B{T}_c$, most of the interval below $T_c$ is approximately ideal at leading weak-coupling order. The figure marks a qualitative extrapolation near the transition: a zero-temperature Bogoliubov spectrum does not determine precise interacting critical heat capacity or its small transition shift. The illustrative fixed-$\mu$ interpolation is not claimed as a controlled critical-region calculation.

Write ${ϵ}_k={\hslash }^2{k}^2/(2m)$, $E_k=\hslash \omega$, and take $\mu >0$. The Bogoliubov spectrum is gapless and positive for $k\ne 0$. Its small- and large-momentum limits areThe low-energy excitations are collective sound modes; high-energy excitations approach particles with a mean-field energy offset. With healing length $\xi =\hslash /\sqrt{2m\mu }$ and $q=|k|\xi$,The crossover occurs at $q$ of order one, not at an additional energy gap.

The critical velocity of a homogeneous condensate follows because the Landau criterion compares an excitation energy with the energy that flow can release by transferring its momentum. Creating a mode is energetically possible when $E_k-\hslash \mathbf{k}\cdot \mathbf{v}\le 0$. For this isotropic spectrum,The infimum is the long-wavelength speed of sound. This tests energetic stability against the stated excitations; it is not a calculation of a dissipation rate or a universal vortex-nucleation threshold.

For an external potential varying slowly on the healing scale, the Thomas–Fermi approximation for a condensate neglects density-gradient kinetic energy in the stationary Gross–Pitaevskii equation. In occupied regions it givesHere $\rho$ is the appropriate local number density; in the ring it is a linear density and $g$ is the one-dimensional coupling. This is the local-density sound speed of a condensate. The formula applies where $\mu >V$, not in an empty region where its square root would be imaginary.

Normalize the condensate order parameter to the atom number:up to an arbitrary constant phase. A wavefunction normalized to one instead has amplitude $1/\sqrt{2\pi R}$, with $N|\psi {|}^2$ then providing density. Acting with $L_z=-i\hslash {\partial }_{\theta }$ gives angular momentum $n\hslash$ per atom and $Nn\hslash$ in total for the uniform ring. The superfluid velocity isThe interaction chemical potential is ${\mu }_0=gN/(2\pi R)$; a flowing uniform state also has the constant kinetic contribution ${\hslash }^2{n}^2/(2m{R}^2)$ to its total chemical potential. Its speed of sound is set by the interaction term. Applying $|v|<\sqrt{{\mu }_0/m}$ givesFor a uniform ring condensate, this is the continuum phonon criterion used here. Quantization of the available excitation wavelengths in a very small ring can modify a finite-size threshold.

For the nonrotating equilibrium with a raised left half, let ${\rho }_L$ and ${\rho }_R$ be the bulk linear densities. Within the same Thomas–Fermi approximation for a condensate,When both halves are occupied, normalization fixesand thereforeThese expressions require $V_0<2{\mu }_0$. Beyond that threshold, extending them would give a negative left density and is unphysical. Instead the left bulk empties, all $N$ atoms occupy a segment of length $\pi R$, andAt equality the left bulk just reaches zero density. kinetic energy rounds the interface and produces tails beyond the Thomas–Fermi approximation for a condensate, so an exactly empty half is the idealized bulk result.

A nonconstant $V(\theta )$ breaks rotational symmetry:Consequently angular momentum is no longer a good quantum number, and an eigenfunction need not be $e^{in\theta }$ times a constant amplitude. Nevertheless, while the order parameter is nonzero around the whole ring, single-valuedness requiresThe integer is a phase winding of a condensate, not an assertion that the inhomogeneous state remains an angular-momentum eigenstate. A superfluid phase slip can change it if the amplitude vanishes locally.

Conservation of particles gives ${\partial }_t\rho +{\partial }_s(\rho v)=0$. For steady density the one-dimensional particle current $j=\rho v$ is constant. Since the two halves have equal lengths,The current bottleneck in an inhomogeneous condensate occurs because a lower-density region has faster flow at the same current; it also has a smaller speed of sound. This makes the raised half the limiting region.

For $V_0=2{\mu }_0/3$ the equilibrium approximation givesThe winding condition and equal lengths requireThus the left-hand sound-speed test isThe right-hand condition is only $n^2<3m{R}^2{\mu }_0/{\hslash }^2$, so it is less restrictive.

This local sound-speed estimate for ring-flow stability uses the step-potential equilibrium in a ring condensate densities together with the local sound-speed test, as suggested by the preceding approximations. A quantitatively self-consistent finite-current solution also includes $m{v}^2/2$ and the quantum pressure of a condensate in the stationary Gross–Pitaevskii equation, allowing flow to change the density. Smooth steps suppress reflection but do not alone remove those terms. The $3/8$ result is the stated local-density estimate; it should not be promoted to an exact stability theorem for every finite interacting ring.

There are three useful temperature regimes.

At low temperature, dopant freeze-out keeps most donor electrons bound. Warming ionizes donors and increases conductivity through the rising mobile-electron concentration. For an uncompensated dilute donor system, negligible holes and spin degeneracy two, charge neutrality with partial donor ionization givesDeep in freeze-out, $n\ll N_D$, soThe activation scale is set by the shallow donor binding, rather than the whole band gap.

In the extrinsic regime, donors are largely ionized and $n\simeq N_D$ is nearly constant. The temperature dependence of conductivity then mainly follows the electron mobility. Increasing phonon scattering can make conductivity fall as temperature rises, even though donors remain ionized; impurity scattering can give a different mobility trend at lower temperature.

At sufficiently high temperature, thermally generated electron–hole pairs overwhelm the dopant population. The intrinsic, gap-activated increase in conductivity returns. The boundaries depend on donor concentration, compensation and scattering, so no one exponential describes the entire doped-sample curve.

A direct band gap has its conduction band minimum and valence band maximum at the same crystal momentum. Near-edge absorption or emission can therefore conserve momentum with the electron and photon alone, since optical photon momentum is small on the Brillouin-zone scale. An indirect band gap has its extrema at different wavevectors and normally needs a phonon or another momentum-transfer process in addition to the photon.

Direct-gap materials are usually preferred for efficient light emission, because the radiative transition avoids the extra phonon-assisted process. Their stronger near-edge absorption also permits thinner photovoltaic absorbers. Indirect-gap materials can nevertheless make excellent solar cells: sufficiently thick optical paths, light trapping, long carrier lifetimes and suitable material quality can compensate for weaker absorption. The direct/indirect distinction alone does not rank every device.

This momentum distinction and the difference between bound exciton lines and the free-pair continuum are treated together in the primary effective-mass absorption theory.^[ref]

At cryogenic temperatures, the intrinsic carrier concentration is exponentially small, while a low phonon population permits relatively sharp band-edge and exciton spectral features. Near room temperature, thermal electron–hole generation increases strongly:This generally raises intrinsic conductivity, although increased phonon scattering tends to lower the individual carrier mobilities. With a gap of $1.24$ eV, the material remains far from a degenerate metal at room temperature.

For optical structure, compare $k_{\mathrm{B}}T\simeq 25.9$ meV at 300 K with the exciton binding energies. The $n=2$ binding of $9.56$ meV is substantially smaller, so excited bound states are readily thermally disrupted and broadened. The ground binding of $38.24$ meV is larger but only about $1.48k_{\mathrm{B}}T$: the ground exciton need not disappear completely, yet strong thermal ionization and broadening can weaken its sharp line. A comparison with $k_{\mathrm{B}}T$ is a useful scale argument, not an exact bound-pair fraction independent of density or lifetime.

Thermal lattice expansion and electron–phonon interactions can also shift a real material's gap and optical edge. The given fixed tight-binding parameters contain no temperature dependence, so they do not determine the magnitude or even a universal sign of that extra gap shift. Within the stated model, the dependable trends are more thermally generated carriers and less distinct excitonic structure on warming.

At equilibrium, electrons initially diffuse from the n side and holes from the p side. Recombination leaves ionized donors and acceptors in a depletion region, producing an internal electric field and band bending. The Fermi level is constant across the equilibrium p-n junction. With p material on the left, both band edges lie higher on the p side, as in the left panel.

Forward bias, with the p side positive relative to n, reduces the injection barrier. Electrons and holes enter the junction region and can recombine radiatively, emitting photons with energy close to the band gap. Under bias, separate electron and hole quasi-Fermi levels replace a single equilibrium Fermi level.

Both GaN polymorphs have a direct band gap, so a conduction-band electron and valence-band hole can recombine without a large crystal-momentum change. Silicon and germanium have indirect band gaps, requiring phonon assistance for near-edge radiative recombination, which makes efficient band-edge light emission much harder. A direct gap enables the process; defects, carrier confinement and competing nonradiative channels still determine device efficiency.

Lattice vibrations contribute a phonon heat capacity. At temperatures well below the Debye temperature, its leading contribution is proportional to $T^3$, whereas the electronic part is proportional to $T$:The lattice term can therefore become important rapidly as temperature rises, even if the electronic term dominates sufficiently close to zero. At higher temperature the lattice contribution approaches the Dulong-Petit law value $3R$ per mole of atoms within the harmonic approximation. This is not a claim that the low-temperature $T^3$ law holds at all temperatures.

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.

The FCC conventional unit cell has four atoms, so the cube containsatoms, neglecting surface corrections. A phonon is one excitation quantum of a vibrational normal mode; there can be many quanta per mode, and zero-point motion does not add half a phonon to the thermal population.

For the low-temperature thermal phonon number, integrate the Bose-Einstein distribution against the Debye model phonon density of states:At $10\mathrm{K}$, ${\Theta }_D/T=21.5$, so extending the upper limit to infinity has negligible effect at the precision required. The supplied integral yieldsThis is the thermally excited phonon number, not the number $3N$ of available modes.

The continuum approximation fails when the thermally occupied wavelengths become comparable with the cube size $L$. From the Debye temperature,A lowest nontrivial acoustic frequency is of order $v/L$; a standing-wave estimate ${\omega }_{\min }\sim \pi v/L$ corresponds to $\hslash {\omega }_{\min }/k_B\sim 0.25\mathrm{K}$. Periodic boundary conditions instead give $2\pi v/L$ and $0.50\mathrm{K}$. The precise spectrum depends on mechanical boundary conditions and branch speeds, neither of which is specified.

This finite-crystal phonon freeze-out requires summing discrete normal modes. For a finite positive lowest frequency, $N_{\mathrm{ph}}\sim g_{\min }{e}^{-\hslash {\omega }_{\min }/(k_{B}T)}$, rather than extrapolating the bulk $T^3$ law indefinitely. Rigid translations and rotations of an isolated crystallite are not internal thermal phonons. Surface acoustic modes can modify the intermediate crossover, but do not justify inventing a unique boundary spectrum.

The harmonic phonon contribution to the internal energy isFor $T\gg {\Theta }_D$, every retained mode has $x=\hslash \omega /(k_{B}T)\ll 1$. Expanding $1/(e^x-1)=1/x-1/2+O(x)$ cancels the explicit half quantum and gives $k_{B}T$ per mode to leading order. ThereforeEquivalently, differentiating before integrating givesThe integrand tends to $x^2$, and its integral is $({\Theta }_D/T{)}^3/3$, producing the same Dulong-Petit law. In classical language, the equipartition theorem assigns $k_{B}T/2$ each to the kinetic and elastic parts of every harmonic normal mode. Fixed zero-point energy does not contribute to the heat capacity.

Substitute the traveling normal mode $u_j=u\exp [i(kja-\omega t)]$. Combining the opposite-neighbour phases givesso the positive-frequency branch isFor positive spring constants this is a stable acoustic branch. At long wavelengths,The next-nearest-neighbour bond spans twice the distance, hence contributes the factor four. Relative to $G=0$, positive $G$ increases the sound speed by $\sqrt{1+4G/K}$. The chain still has one acoustic branch and no optical branch.

The group speed near $k=0$ and the maximum frequency need not change in the same way. The zone-edge value is always ${\omega }_0(\pi /a)=2\sqrt{K/m}$, because the next-nearest-neighbour sites then move together. For $G>K/4$ the frequency maximum moves inside the zone, with $\cos (ka)=-K/(4G)$; the small-$k$ sound speed nevertheless keeps increasing.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

An electron losing energy to an acoustic phonon has, in a normal process, ${\mathbf{k}}^{\prime }={\mathbf{k}}_F-{\mathbf{k}}_{\mathrm{ph}}$, with possible addition of a reciprocal-lattice vector for an Umklapp process. The energy transfer is at most of order $\hslash {\omega }_D=k_B{\theta }_D$, much smaller than $E_F=k_B{T}_F$. Hence the magnitudes change very little even though the momentum direction can change substantially. At $T\sim {\theta }_D\ll T_F$, thermal vacancies near the Fermi surface allow emission into some lower-energy states; exactly at zero temperature emission into an occupied Fermi sea would be blocked by the Pauli exclusion principle.

Using $k^{\prime }=(1-\delta )k_F$ and the free-electron dispersion, energy conservation givesSince $\delta \ll 1$, the bracket is $2\delta +O({\delta }^2)$. Rearranging yieldsEquivalently $\delta \simeq \hslash c{k}_{\mathrm{ph}}/(2E_F)$. For a Debye-cutoff phonon in the preceding model, $\delta \simeq {\theta }_D/(2T_F)=1/480=0.00208$, confirming the approximation. Near-elastic scattering constrains the radial distance in wave-vector space, not the scattering angle alone.

The electron-acoustic recoil length compares electronic recoil with the acoustic quantum energy associated with the same momentum. For transferred wave number $q$,The dispersions cross at $q=1/\lambda$; there the reduced electron de Broglie wavelength is $\lambda$ and its free-particle group velocity is $2c$. Thus $\lambda$ is the length scale at which an electron's localization/recoil energy and a corresponding acoustic excitation energy become comparable. At $q\lambda \gg 1$ the electron recoil energy is large compared with the phonon energy; this explains the separation of electronic and acoustic energy scales when lattice wave numbers exceed $1/\lambda$.

Formally it is half the Reduced Compton wavelength with the speed of light replaced by the speed of sound. That is an analogy of scales, not a relativistic electron mass or a pair-production threshold: sound does not replace the limiting speed in electron dynamics. Also it is a reduced-wavelength convention; an ordinary de Broglie wavelength $2\pi /q$ at the crossing is $2\pi \lambda$. No scattering lifetime appears, so $\lambda$ is not an electron mean free path.

The Debye model treats the crystal as an isotropic elastic continuum with independent harmonic vibrational modes. There are three acoustic polarizations, approximated by linear dispersions and a common effective speed. A cutoff is chosen to retain $3N$ modes for $N$ atoms. The model neglects anharmonic interactions and the detailed high-wavevector dispersion. In a multi-atom primitive cell, optical branches require additional treatment; copper has one atom per primitive cell even though its conventional cubic cell contains four atoms.

A frequency interval contains, per volume,Each phonon mode has Bose-Einstein distribution occupation $(e^{\hslash \omega /(k_{B}T)}-1{)}^{-1}$. The fixed zero-point energy contributes no heat capacity. Its thermal energy density isFor $T\ll {\theta }_D$, extending the upper limit to infinity gives $u_{\mathrm{ph}}=(3{\pi }^4/5)n{k}_B{T}^4/{\theta }_D^3$. ThereforeThe cubic law reflects the three-dimensional acoustic mode density proportional to ${\omega }^2$. It is a low-temperature limit, not a valid extrapolation through the classical high-temperature regime.

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.