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A wave packet in an energy band has group velocity $v={\hslash }^{-1}dE/dk$. The semiclassical force law $\hslash \dot{k}=F$ then gives $\dot{v}={\hslash }^{-2}(d^{2}E/d{k}^2)F$. The curvature therefore defines the one-dimensional effective mass by $1/m^{\ast }=E^{\prime \prime }/{\hslash }^2$. Here $E^{\prime \prime }(k)=A{a}^2\cos (ka)$, so at the band minimum $k=0$,It measures acceleration within the band, rather than a change in the electron's rest mass. At a band maximum the electron curvature mass can be negative; a missing electron is then described by a positively charged hole with a positive curvature mass.

Why degenerate perturbation theory is needed. If several unperturbed states have equal or nearly equal energy, a perturbation can mix them at leading order. In that subspace, diagonalize the perturbation matrix rather than using nondegenerate formulas with small energy denominators. The resulting eigenvectors identify the combinations selected by the coupling, and the eigenvalues give the leading splitting. Other remote states can then be included perturbatively.

For the linear combination of atomic orbitals description of molecular hydrogen, equivalent atomic $1s$ states on the two nuclei supply the nearly degenerate starting functions. In an orthonormal two-state model, equal diagonal energy $E_a$ and real hopping $t$ give $E_a\pm t$ and symmetric/antisymmetric combinations. For actual overlapping atomic functions with overlap $S$ and Hamiltonian coupling $h$, one instead solves a generalized eigenvalue problem:The bonding state has enhanced density between the nuclei and the antibonding state a separating node. Two opposite-spin electrons occupying the bonding molecular orbital describe the simplest closed-shell hydrogen-molecule bond. Electron-electron interaction and nuclear repulsion are needed to calculate a realistic total binding energy; the two-state splitting alone is not the full molecular energy.

In the nearly-free electron model, the analogous degenerate states are plane waves $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$ at a Bragg scattering. A Fourier coefficient $U_{\mathbf{G}}$ of the periodic potential couples them. At exact degeneracy $E_B$, diagonalizing their two-by-two matrix givesThe associated standing waves place electronic density differently relative to the ionic potential. These are the same gap-opening combinations illustrated in Q1(b). The labels of the two starting basis states survive as mixed orbital or plane-wave character, rather than independent crossing eigenstates.

Origin of broad and narrow bands. Spatially extended s/p-derived orbitals have substantial overlap and therefore large hopping amplitudes and broad bands. More localized d orbitals overlap less strongly and produce narrower bands. A multiorbital tight-binding calculation assigns the on-site orbital energies and hopping matrices, forms Bloch sums, and diagonalizes the resulting matrix at each wavevector. Different hopping ranges and orbital symmetry determine the bandwidths and allowed interband coupling. The two-band description retains one broad combination and one nearly flat d-derived combination; a real transition metal contains several d orbitals and more complicated dispersions.

Hybridized eigenvalues and eigenstates. Let $|{\psi }_{\mathbf{k}}⟩=a|{\varphi }_{\mathbf{k}}⟩+b|{\chi }_{\mathbf{k}}⟩$, in an orthonormal basis. Hermitian operator fixes the reverse coupling to be $V^{\ast }$. Projection onto the two components givesSolving this quadratic yields the band hybridization dispersionFor nonzero $V$, one eigenvector representation is $(a,b)\propto (V,E-E_1)$, followed by normalization. Thus the wavefunctions exchange broad-band and d-band character through the avoided crossing; the levels are equal mixtures at the uncoupled crossing.

At $V=0$, the parabola crosses the horizontal level at $k=\pm k_0$, with $k_0=\sqrt{2m{E}_d}/\hslash$. For small nonzero $|V|$, the curves avoid crossing with minimum direct separation $2|V|$. The lower branch follows the parabola below the crossing and flattens toward $E_d$ above it; the upper does the opposite. The lower branch satisfies $E_{-}<E_d$ everywhere and tends to $E_d$ from below as $k\to \infty$, while the upper branch always lies above $E_d$. This is the perfectly flat-band continuum approximation, with zone-boundary intersections excluded as requested.

Fixed particle number and the larger Fermi sphere. Denote the uncoupled broad-band Fermi radius by $k_0$ to distinguish it from the new one $K$. With two spin states, the broad-band number density at $V=0$ is $k_0^3/(3{\pi }^2)$. The stipulated equal population in the flat sector makes the total density twice that value. At zero temperature and any nonzero hybridization, all occupied states fit into the monotonic lower branch below $E_d$, while the upper branch is empty. ThereforeThis is the stated factor increase. It uses fixed total particle number, equal spin counting, an exactly flat unhybridized band and enough available states below any Brillouin-zone restriction. In generic pairs of dispersive bands both hybridized pockets can remain occupied; then Fermi-volume conservation in two hybridized electron bands fixes a sum of volumes rather than this one radius. The exactly flat band is degenerate at $V=0$, so its initial occupation must be specified, as it is here.

Fermi-velocity suppression. Set $\Delta (k)=E_1(k)-E_d$. The exact derivative of the lower branch isAt $k=K$, $E_1(K)=2^{2/3}{E}_d$ and $\Delta (K)=(2^{2/3}-1)E_d>0$. For $|V|\ll \Delta (K)$, the square bracket divided by two is $|V{|}^2/\Delta (K{)}^2+O(|V{|}^4/\Delta (K{)}^4)$. Since group velocity is ${\hslash }^{-1}dE/dk$ and the broad-band reference velocity is $v_F^{(1)}=\hslash k_0/m$,so the flat-band hybridization Fermi-velocity suppression isThe reference velocity is that of the original broad-band Fermi sphere, not the broad-band slope at the larger $K$; omitting the radius change loses the factor $2^{1/3}$.

The low-temperature qualification becomes more stringent as $V\to 0$: the new chemical potential approaches $E_d$ with $E_d-\mu \simeq |V{|}^2/\Delta (K)$, and the relevant thermal energy must be small compared with this hybridization scale. At fixed nonzero temperature the singular flat-band redistribution is not automatically resolved for arbitrarily small coupling. A small but finite narrow-band dispersion can also change the occupation topology.

Density of states and magnetism. For an isotropic three-dimensional branch with spin degeneracy two,A small Fermi velocity therefore produces a large density of states. The localized, partially filled d-derived bands in iron, cobalt and nickel place many states close to the Fermi level; broad s/p-derived conduction bands in copper, silver and gold coexist with filled d bands lying below it. This explains the substantially larger Fermi-level density of states of the former group within the band picture. An exact factor of ten requires the actual band parameters and is not fixed by the two-band toy model.

A high density of states lowers the kinetic-energy cost of spin polarization. A repulsive exchange interaction can then satisfy the Stoner criterion and stabilize ferromagnetism. With a single-spin density of states $g_s$ and an on-site interaction normalization, the criterion is $U{g}_s(E_F)>1$; with spin-summed $g=2g_s$, it is $Ug(E_F)>2$. The large d-state contribution favors this instability in Fe, Co and Ni. The noble metals do not satisfy the corresponding instability condition in the same ordinary state; this does not imply that they have no weak magnetic response whatsoever.

mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity isDifferentiating the velocity givesThis effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator $T_{\mathbf{a}}\psi (\mathbf{r})=\psi (\mathbf{r}-\mathbf{a})$. A lattice-periodic scalar potential satisfies $V(\mathbf{r}+\mathbf{a})=V(\mathbf{r})$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf{a}}]=0$. The lattice translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate energy subspace.

For a common eigenstate, write $T_{\mathbf{a}}\psi =t(\mathbf{a})\psi$. Unitarity gives $|t(\mathbf{a})|=1$, and the group law gives $t(\mathbf{a}+\mathbf{b})=t(\mathbf{a})t(\mathbf{b})$. On primitive Bravais vectors ${\mathbf{a}}_j$, choose $t({\mathbf{a}}_j)=e^{-i\mathbf{k}\cdot {\mathbf{a}}_j}$. Every lattice vector is an integer combination of these generators, henceThe vector $\mathbf{k}$ is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define $u_{n\mathbf{k}}(\mathbf{r})=e^{-i\mathbf{k}\cdot \mathbf{r}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in $\mathbf{k}$ and an identity term quadratic in $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Thenbecause opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore ${\hslash }^2{k}^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energyHermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at $\mathbf{k}=0$ enter the sum. The remainder is even because time-reversal symmetry gives ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

Two atomic orbitals. In the linear combination of atomic orbitals approximation, use $|\psi ⟩=c_A|A⟩+c_B|B⟩$, where the two basis functions are localized atomic orbitals. Projecting $\hat{H}|\psi ⟩=E|\psi ⟩$ onto each orbital gives the generalized eigenvalue problemHere the scalar overlap in the matrix is $S=⟨A|B⟩$, while $H_{ab}=⟨a|\hat{H}|b⟩$. Normalization requires $c^{†}Sc=1$, with $S$ now denoting the full overlap matrix. For equivalent real orbitals with diagonal energy $E_a$, real hopping $t$, and real overlap $s$, the eigenstates and energies areFor the usual negative hopping convention, the symmetric combination is the lower bonding orbital and the antisymmetric combination is the higher antibonding orbital. In an orthonormal approximation, $s=0$ and the result reduces to a two-by-two Hermitian matrix eigenproblem.

Hybridized bands. At each $\mathbf{k}$, take the two Bloch basis states to be orthonormal, as implicit in the stated band model, and write ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The Hamiltonian isFor $|{\psi }_{\mathbf{k}}⟩=a_{\mathbf{k}}|{\varphi }_{\mathbf{k}}⟩+b_{\mathbf{k}}|{\chi }_{\mathbf{k}}⟩$, the eigenvalue condition isso $(E-{ϵ}_k)(E-E_d)-|V{|}^2=0$. Thus band hybridization givesFor nonzero $V$, a convenient explicit normalized pair of eigenstates isA phase convention can make $V$ real, but the eigenvalues depend only on its modulus. These are exact within the truncated two-band model, and approximate as a description of a material.

Dispersions and limits. For $V=0$, the parabola crosses the flat band atFor $V\ne 0$ the crossing is avoided, with a direct hybridization gap $2|V|$ at $k_0$. The lower and upper branches exchange orbital character through this avoided crossing. At the origin their energies areso for $|V|\ll E_d$ the small-$k$ expansions areAt large $k$, where ${ϵ}_k-E_d\gg |V|$,The lower branch approaches $E_d$ from below and becomes flat; the upper branch approaches the free-electron parabola from above. At zero coupling the unsorted orbital bands remain exactly flat and parabolic. Sorting them by energy instead gives $E_{-}=\min ({ϵ}_k,E_d)$ and $E_{+}=\max ({ϵ}_k,E_d)$, which cross at $k_0$ without mixing.

Fermi wavevector and velocity. A partially occupied perfectly flat band has chemical potential $E_d$. Thus before mixing, the occupied broad band has radius $k_0$ defined above. For spin degeneracy two, its particle density is $n_1=k_0^3/(3{\pi }^2)$; equal population of the flat sector gives total density $n=2n_1$.

For any nonzero mixing, $E_{-}(k)<E_d$ at finite $k$, while $E_{+}(k)>E_d$. Moreover $E_{-}(k)$ increases monotonically. At zero temperature and this fixed density, a single occupied sphere in the lower branch therefore accommodates all particles, and the upper branch is empty. Fermi-volume conservation in two hybridized electron bands givesThe new radius is about $26\%$ larger than the old broad-band radius. The strictly flat uncoupled band has no uniquely determined Fermi radius of its own; its prescribed population is what enters the count.

The radial group velocity is $v_{-}(k)={\hslash }^{-1}d{E}_{-}/dk$. Differentiating the exact dispersion givesAt $k_F$, put $\delta =2^{2/3}-1$, so ${ϵ}_{k_F}-E_d=\delta E_d$. Relative to the original broad-band Fermi velocity $v_F^{(1)}=\hslash k_0/m$,Expanding for $|V|\ll E_d$ gives the flat-band hybridization Fermi-velocity suppressionThe coefficient is about $3.65$. The enlarged occupied volume coexists with a small velocity because the new Fermi surface lies on the almost-flat part of the lower branch. This result assumes fixed particle density, sufficient available states before any Brillouin zone boundary, and temperature low enough to resolve energy differences of order $|V{|}^2/E_d$. The limit $V\to 0$ at a fixed nonzero temperature need not give this sharp redistribution. Giving the narrow band a finite dispersion can also leave both pockets occupied, so the result is specific to the ideal model.

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction electrons of charge $-e$, the isotropic Drude model giveswhere $\tau$ is the momentum relaxation time.

The positive electron mobility ${\mu }_e=|{\mathbf{v}}_d|/|\mathbf{E}|$ measures the ease of this drift in linear response. In the same approximation,Its units are ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$, while $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$. ThusFor an isotropic parabolic minimum this reduces to $E=E_c+{\hslash }^2{k}^2/(2m^{\ast })$ and $m^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}$. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.

Continue with $m=0$. Along a characteristic $x=x_0+t$, the chain rule givesThus the initial wave packet ${\psi }_R(0,x)=f(x)$ evolves as ${\psi }_R(t,x)=f(x-t)$. Similarly, along $x=x_0-t$, the left component is constant, givingA feature initially at $x_0$ therefore appears at $x_0+t$ for the right component and $x_0-t$ for the left component. The profiles translate without changing shape at the speed of light, which is 1 in these units. For positive-energy plane waves, the right component has $E=p>0$ and the left component has $E=-p>0$, another way to identify their propagation directions.

The conclusion concerns the massless Dirac equation. For $m\ne 0$, the components couple as in c, and a positive-energy packet has group velocity $p/\sqrt{p^2+m^2}$, rather than a universal speed $\pm 1$. For example, a massive positive-energy rest solution has $p=0$ and ${\psi }_R={\psi }_L$, so neither component alone describes an independently moving light-speed packet.

Assume a cold, unmagnetized, weakly collisional plasma, a small fractional bandwidth and $\omega \gg {\omega }_p$. The dispersion relation is ${\omega }^2=c^2{k}^2+{\omega }_p^2$, with ${\omega }_p^2=n_e{e}^2/({ϵ}_0{m}_e)$. The group velocity and propagation time areThe lower frequencies arrive later. Taking the magnitude of the frequency derivative over $\Delta \omega$ givesFor a nonuniform line of sight, replace $n_{e}L$ by the dispersion measure $\int n_{e}dl$. The light travels at the group velocity, not at the phase velocity.

Write $L_0\psi =-U\psi$ and use the outgoing inverse of $L_0$. Adding the homogeneous incoming wave yieldsFor the attractive delta potential, let $\eta =m\alpha /(k{\hslash }^2)$. The equation reduces to $\psi (x)=e^{ikx}+i\eta e^{ik|x|}\psi (0)$ and $\psi (0)=1/(1-i\eta )$. Separating the two half-lines givesBoth sides have the same group velocity, soThe same amplitudes satisfy continuity and the exact delta-potential derivative jump ${\psi }^{\prime }(0^{+})-{\psi }^{\prime }(0^{-})=-2m\alpha \psi (0)/{\hslash }^2$, providing an independent local check on the sign.

The phase velocity is the speed of a constant-phase surface, $v_{\mathrm{ph}}=\omega /k$. The group velocity is the envelope speed of a narrow packet, $v_{\mathrm{gr}}=d\omega /dk$, requiring a local dispersion relation rather than only a single monochromatic wave. In an isotropic transparent medium with $k=n(\omega )\omega /c$,For a vacuum wave, $n\equiv 1$ over the packet bandwidth, so both speeds are $c$ and $\boxed{v_{\mathrm{ph}}{v}_{\mathrm{gr}}=c^2}$. No superluminal signalling is implied by a phase speed in a dispersive material.

Strictly, the value $n({\omega }_0)=1$ at one frequency alone does not prove the product identity. For example $n(\omega )=1+q(\omega -{\omega }_0)/{\omega }_0$ gives $v_{\mathrm{ph}}({\omega }_0)=c$ but $v_{\mathrm{gr}}({\omega }_0)=c/(1+q)$. The vacuum interpretation, or an equivalent nondispersive assumption, is necessary.

Use $\hslash =c=1$ and the mostly-plus Minkowski metric. Then the source's Klein-Gordon equation is $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. For a plane wave $\varphi =C{e}^{-i\omega t+i\mathbf{k}\cdot \mathbf{x}}$, substitution givesThus a real propagating wave vector requires $|\omega |\ge m$, or $|E|\ge m{c}^2$ on restoring units. The positive-energy branch has a rest-energy threshold $m{c}^2$. At $|\omega |=m$, $\mathbf{k}=0$; frequencies below the mass threshold give evanescent spatial dependence rather than real-wave-vector propagation. Both signs of $\omega$ occur in the classical equation; a real scalar field combines complex-conjugate modes. Its group velocity is $\mathbf{k}/\omega$ and is subluminal, though the positive-branch phase speed exceeds one when $m>0$.

For a static spherically symmetric scalar field, the equation outside the source isSetting $\psi =r\varphi$ gives ${\psi }^{\prime \prime }-m^2\psi =0$, soDecay at infinity selects $B=0$, the Yukawa potential. It is singular at the origin and describes a point source distributionally; a source-free solution regular at the origin and decaying at infinity must be zero. For $m=0$, $\varphi =A/r+B$, and decay sets the constant $B$ to zero.

Rigid walls require zero normal velocity, so ${\varphi }_x=0$ at $x=0,L$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables gives spatial modes $\cos (m\pi x/L)\cos (n\pi y/b)$, for nonnegative integers $m,n$. The wave equation therefore givesThe $(0,0)$ spatially constant mode has zero frequency and is not an ordinary positive-frequency acoustic oscillation.

A spatially uniform boundary forcing excites the $n=0$ transverse mode. Put $\varphi =X(x)\cos \omega t$. Then $X^{\prime \prime }+(\omega /c_0{)}^{2}X=0$ and $X^{\prime }(L)=0$, hence $X=C\cos [\omega (L-x)/c_0]$. The condition $X^{\prime }(0)=a$ fixes $C$ and givesArbitrary free oscillations can be superposed. The denominator vanishes at longitudinal resonances. Near them the lossless steady harmonic response ceases to be a uniformly small-amplitude approximation; damping from viscosity, heat transfer or wall losses regularizes the response, and large amplitudes can require nonlinear equations of motion or a more accurate moving-boundary model. At exact resonance a switched-on ideal forcing produces secular growth rather than a bounded harmonic particular solution.

For the semi-infinite duct expand $F(y)={\sum }_{n\ge 0}{F}_n\cos (n\pi y/b)$. Each axial mode hasFor a propagating mode a particular solution is ${\varphi }_n=(F_n/k_n)\cos (n\pi y/b)\cos (k_{n}x-\omega t)$; differentiating at zero gives the prescribed $F_n\cos (n\pi y/b)\sin \omega t$. A nonzero higher transverse coefficient propagates when $|\omega |>n\pi c_0/b$.

The outgoing radiation condition selects $c_0^2{k}_n/\omega >0$, its positive axial group velocity. Thus $\boxed{\omega k_n>0}$. This sign is selected by the absence of waves incident from infinity, not by the differential equation alone. If $F$ has a nonzero mean, its $n=0$ component can propagate even below every higher-mode cutoff.

For $F=\cos (\pi y/b)$ and $0<|\omega |<\pi c_0/b$, put $\kappa =\sqrt{(\pi /b{)}^2-{\omega }^2/c_0^2}$. The decaying solution isIt is evanescent and carries no mean energy to infinity. These mode and sign conditions are summarized by the radiation condition for a waveguide mode.

Along $x=Vt+x_0$, the phase is $t\Phi (k)+k{x}_0$, with $\Phi (k)=kV-\omega (k)$. Graphically, a line of slope $V$ intersects the dispersion curve transversely except where it is tangent. On a transverse region, neighboring wavenumbers acquire order-one phase differences over a width $\Delta k\sim t^{-1}$, so cancellation leaves an $O(t^{-1})$ contribution. More formally, integration by parts usesand gives that order when the amplitude and its derivative are integrable and ${\Phi }^{\prime }$ is bounded away from zero on the region.

At a nondegenerate tangent point $k_{\ast }$, ${\Phi }^{\prime }(k_{\ast })=0$, equivalently $\boxed{V={\omega }^{\prime }(k_{\ast })}$, and $\Phi (k)=\Phi (k_{\ast })-{\omega }^{\prime \prime }(k_{\ast })(k-k_{\ast }{)}^2/2+\cdots$. Coherent wavenumbers now occupy width $O(t^{-1/2})$, yielding the stationary-phase contributionThese are the packets whose group velocity follows the observer. The displayed decay rates require the usual regularity and nondegeneracy hypotheses; $A\to 0$ alone is not a complete analytic hypothesis, and a vanishing amplitude or degenerate stationary point changes the leading order.

For deep water take a velocity potential satisfying Laplace's equation below the undisturbed surface and decaying with depth. A Fourier mode has vertical factor $e^{|k|z}$. The linearized kinematic and dynamic surface conditions are ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ at $z=0$. ThusRelease from rest gives ${\overline{\eta }}_t(k,0)=0$, so inversion of the Fourier transform givesEvenness of the initial profile makes its transform even and the reconstructed surface even; the same formula also works without that symmetry when the usual reality condition holds.

Split the cosine into its two time phases. Along the moving observer, their stationary conditions are $\pm {\omega }^{\prime }(k)=V$, where $\omega (k)=\sqrt{g|k|}$. Since $|{\omega }^{\prime }(k)|=\sqrt{g}/(2\sqrt{|k|})$, for $V\ne 0$,These are the possible dominant stationary-phase packets, provided the initial Fourier amplitude at the selected wavenumber is nonzero. For $V=0$ there is no finite nonzero stationary wavenumber; the formula must not be applied by division by zero.

Write $\omega$ for the nonzero forcing frequency, to distinguish it from the vertical velocity $w$. Put $k_z=\pi /h$ and $Z(z)=\cos (\pi (2z+h)/(2h))=-\sin (k_{z}z)$. Then $Z(0)=Z(h)=0$ and $Z^{\prime \prime }=-k_z^{2}Z$, so it meets the rigid-wall conditions. Substituting $w=\mathrm{Re}(f(x)e^{-i\omega t}Z(z))$ into the scalar wave equation givesorThe radiation sign for an inertial wave must exclude any disturbance arriving from $x=+\infty$. This means an radiation condition for propagating waves, or bounded decay for an evanescent wave.

For $0<|\omega |<2|\Omega |$, put $q=k_z\sqrt{4{\Omega }^2/{\omega }^2-1}$. The solutions are oscillatory. The inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$; its horizontal group velocity is $c_{g,x}=-\omega k_x/(k_x^2+k_z^2)$. Positive energy transport into the channel therefore requires $k_x=-\mathrm{sgn}(\omega )q$. The solution isFor a positive frequency, the phase travels towards negative $x$ while the group velocity travels towards positive $x$. Choosing the outgoing sign from the phase velocity alone would give the wrong result.

For $|\omega |>2|\Omega |$, set $\kappa =k_z\sqrt{1-4{\Omega }^2/{\omega }^2}$. Excluding the exponentially growing solution givesThis disturbance decays away from the wavemaker and does not propagate horizontally. At the cutoff the ODE degenerates to $f^{\prime \prime }=0$; zero frequency is not a nonzero forced harmonic mode with this vertical shape.

There is a second printed inconsistency: the stated velocity $u\mathbf{i}+w\mathbf{k}$, if taken literally with no $y$ dependence and no transverse velocity, cannot support a nontrivial rotating inertial wave. Its transverse momentum equation would give $2\Omega u=0$, then continuity and the rigid walls would force $w=0$. The scalar solutions above are the intended inertial-wave solutions when two-dimensional means dependence on $x,z$ while retaining the transverse component $v$. For a mode, $-i\omega v+2\Omega u=0$ explicitly supplies the generally nonzero transverse velocity. Thus the scalar equation can be solved as requested, but the full vector field needs this correction to the printed two-component description.