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At the absolute wavenumber, the group velocity $d\omega /dk$ is zero. The corresponding accessible saddle therefore governs the disturbance seen at a fixed position. Its exponential growth rate isSaddle accessibility can be checked directly here. With $D=1+i{c}_d$, whose real part is positive, Fourier inversion gives the Green function of the linear complex Ginzburg-Landau equation,At fixed $x$ its exponential rate is $\mathrm{Im}{\omega }_0$, while along the packet centre $x=Ut$ it is $\mu$. For real wavenumbers of Fourier modes, the temporal growth rate is $\mathrm{Im}\omega (k)=\mu -k^2$, so the flow is temporally unstable exactly when $\mu >0$.

Consequently the classifications for a localized disturbance in this laboratory frame areFor convective hydrodynamic instability, a travelling wave packet amplifies but the response at each fixed position decays. For absolute hydrodynamic instability, that fixed-position response amplifies. The equality is the marginal absolute threshold: its exponential rate is zero and the impulse response has a $t^{-1/2}$ prefactor. For $\mu <0$ there is no temporal growth; $\mu =0$ is temporally marginal. When $U=0$ the convective window is empty. These conclusions use the infinite-line impulse problem; a general dispersion relation requires its own spatial-branch or saddle selection.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.

For electron charge $-e$ with $e>0$ in a constant electric field $\mathcal{E}$, the single-band semiclassical equations are $\hslash \dot{k}=-e\mathcal{E}$, $\dot{x}=v(k)$. Thus $k(t)=k_0-e\mathcal{E}t/\hslash$, and for $\mathcal{E}\ne 0$,These are Bloch oscillations of period $2\pi \hslash /(|e\mathcal{E}|a)$. The zero-field limit is $x-x_0=v(k_0)t$. This assumes a narrow packet and neglects scattering and interband transitions.

For many electrons obeying the Pauli exclusion principle, a partially filled band has available nearby states and can carry current under a displaced occupation distribution: it describes a metal within band theory. A completely filled band has cancelling velocities over the zone, so its current is zero; if separated from empty higher bands by a nonzero gap, it describes an insulator at sufficiently low temperature and weak electric field. With spin degeneracy this isolated band is filled at two electrons per site. The single-electron Hamiltonian alone does not specify the filling or a gap to a next band, so those extra physical assumptions are required for classifying a material.

Without self-gravity and for $k>0$,The phase velocity and group velocity areso

With self-gravity restored,The product changes sign atFor $k>k_{\mathrm{crit}}$, crests and a localized wave packet travel in the same radial direction. For $k<k_{\mathrm{crit}}$, the group velocity and phase velocity have opposite signs: the envelope and its wave energy propagate opposite to the individual crests. At $k_{\mathrm{crit}}$ the group velocity vanishes and neighboring wave components cause the packet to spread.

Consider first a scalar constant-coefficient Cauchy problem for a partial differential equation on the whole line. A translation-invariant spatial discretization is a convolution operator, so the discrete Fourier transform turns it into multiplication by its Fourier symbol. This reduction is exact because Fourier modes are the simultaneous generalized eigenfunctions of every translation-invariant stencil.

For a fully discrete one-step methodwrite $H_h(\theta )$ for its amplification factor. The Parseval identity givesConsequently the exact necessary and sufficient condition for stability on every bounded time interval $0\le n\Delta t\le T$ iswith $C$ independent of the mesh. Sufficiency follows directly from Parseval; necessity follows by choosing transformed initial data concentrated where the multiplier is largest. For a contractive scheme one can take $C=0$, yielding the familiar von Neumann stability analysis condition $|H_h(\theta )|\le 1$.

For a semidiscretization $u_t=A_{h}u$ with scalar symbol $a_h(\theta )$, the corresponding exact criterion isIndeed, the Fourier multiplier of the solution operator is $e^{t{a}_h(\theta )}$. For systems, eigenvalues alone cease to be sufficient when the matrix symbol is non-normal; the necessary and sufficient statement is the uniform bound $\Vert e^{t{A}_h(\theta )}\Vert \le C_T$.

On the whole lattice, a finite stencil defines a Laurent operator, whose operator norm is the essential supremum of its symbol. Restricting the same stencil to a half-line gives a Toeplitz operator. The Cauchy symbol condition remains necessary for a stable initial-boundary scheme because data localized far from the boundary behave like the whole-line problem for a finite time. It is not sufficient: the boundary closure can support growing modes or amplify incoming modes even when every interior Fourier mode is stable.

Boundary stability therefore requires a uniform estimate for the forced half-line recurrence. After a Laplace transform in time and a Fourier transform in tangential variables, one solves a normal-direction recurrence. The Uniform Kreiss--Lopatinskii condition requires the boundary equations to determine its decaying roots with a uniformly bounded inverse. In practical terms, one imposes one independent boundary condition for each incoming characteristic or numerical mode and none for outgoing modes. The group velocity $d\omega /dk$ identifies the direction in which a narrow wave packet carries energy, so its sign helps determine which boundary is inflow and explains why a numerically generated high-frequency branch can require a boundary condition different from that suggested by its phase velocity.

For a two-step method, the Fourier substitution produces an amplification polynomial of a multilevel finite difference scheme. Every root must lie in the closed unit disk, and unit-modulus roots must be simple, uniformly in the wavenumber. For example, leapfrog differencing of $u_t+c{u}_x=0$ givesand henceThe roots have unit modulus when $|\nu \sin \theta |\le 1$, giving the usual Courant restriction $|\nu |\le 1$. At the endpoint, a wavenumber with $|\nu \sin \theta |=1$ produces a repeated unit root and violates the uniform root condition; strict $|\nu |<1$ avoids this marginal linear growth. The second root is the familiar oscillatory computational mode, illustrating why a multilevel scheme requires its full amplification polynomial rather than a single multiplier.

A wave crest moves at the phase velocity, while the envelope of a narrow wave packet moves at the group velocity. Since $c_g>c_p$ for every nonzero wavenumber, the crests move more slowly than the packet.

For particle positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrödinger equation isWrite $|a{⟩}_i$ for the localized wave packet ${\psi }_{ai}$ and $d_{ab}=|x_{a1}-x_{b2}|$. Neglecting packet spreading and branch overlap, the initial product state evolves branchwise asup to phases generated independently on the two particles. These branch-dependent phases generally cannot be separated into one phase depending only on $a$ and one depending only on $b$, so the Newtonian gravitational potential energy creates gravitationally induced entanglement.

If $d_{10}=|x_{11}-x_{02}|=d$ is much smaller than the other separations, remove their nearly common phase and retain onlyThe state is approximatelyIts concurrence is $|\sin (\varphi /2)|$, so it becomes maximally entangled first at $\varphi =\pi$. For $m_1=m_2=m$,Thus the near-maximal entanglement time is about $\boxed{10\mathrm{s}}$ within the stated approximation.

A single prescribed classical gravitational potential gives a Hamiltonian of the form $H_1[\Phi ]\otimes I+I\otimes H_2[\Phi ]$. Its evolution factorizes as $U_1\otimes U_2$ and preserves every initial product state, so it cannot generate this entanglement. A semiclassical mean field sourced only by expectation values likewise gives each particle a local one-body potential and does not provide a quantum mediator carrying branch correlations.

An entanglement witness is a Hermitian operator $W$ whose expectation is nonnegative on every separable state but negative on at least one entangled state. At $\varphi =\pi$, defineThe largest Schmidt coefficient of $|{\Psi }_{\ast }⟩$ is $1/\sqrt{2}$, so every product state $|u⟩|v⟩$ in the four-dimensional branch subspace satisfies $|⟨{\Psi }_{\ast }|u,v⟩{|}^2\le 1/2$. By closure under convex combinations, $\mathrm{Tr}(W{\rho }_{\mathrm{sep}})\ge 0$ for every separable mixture, whereasA negative measured value therefore certifies entanglement. Under the assumptions that the masses began unentangled and interacted only through gravity, such certification would show that the mediator can transmit quantum coherence; it would be evidence against a purely classical gravitational channel and for the quantum nature of gravity.

Put $\tau =T/\rho >0$ and take the positive-frequency branch for $k>0$. The phase velocity and group velocity areSince the square root is increasing, minimizing $c_p$ is equivalent to minimizing $g/k+\tau k$. Its derivative is $-g/k^2+\tau$, which changes sign from negative to positive atThe phase speed diverges at both positive endpoints $k\downarrow 0$ and $k\to \infty$, so this stationary point is its global minimum. At $k_c$, $\tau k_c^2=g$, givingThe relative speed is especially transparent:A narrow wave packet moves with $c_g$, while its crests move with $c_p$. For $k<k_c$, $c_p>c_g$: crests move forward through the packet, from its rear towards its front. For $k>k_c$, $c_p<c_g$: crests drift backwards relative to the packet, although both absolute velocities are positive. At $k_c$ the leading-order relative drift vanishes. This is the minimum phase speed of a gravity-capillary wave.

For either normalized two-spinor ${\chi }^s$, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=p^{2}I$ givesThe Dirac equation alone does not fix $N$: a normalization convention is needed. In the usual covariant external-spinor convention ${\bar{u}}_r{u}_s=2m{\delta }_{rs}$, choose real positive $N=\sqrt{E+m}$. Then $u_r^{†}{u}_s=2E{\delta }_{rs}$ and ${\bar{u}}_s{\gamma }^{\mu }{u}_s=2p^{\mu }$. A second common convention is unit rest density, $\bar{u}u=1$, giving $N=\sqrt{(E+m)/(2m)}$ and density $E/m$. Both explain why the boosted density is not one particle per unit coordinate volume: the density is a four-current component and changes under a Lorentz transformation.

One may instead choose unit laboratory density, $N=\sqrt{(E+m)/(2E)}$, or box normalization $N=\sqrt{(E+m)/(2EV)}$ for one particle in volume $V$. An infinite plane wave is not normalizable to one particle over all space; a localized wave packet is required for that. Thus a unique $N$ cannot be inferred from the prompt without specifying which of these conventions is intended.

Let $\chi =(\sqrt{5}/3,2/3{)}^{\mathsf{T}}$, so ${\chi }^{†}\chi =1$. The Pauli identity $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^{2}I$ givesusing $E^2=|\mathbf{p}{|}^2+m^2$. Thus unit spinor probability density is obtained withAn arbitrary constant phase is immaterial. For the alternative Lorentz normalization $u^{†}u=2E$, choose $N=\sqrt{E+m}$. Neither convention makes an infinite plane wave square-integrable; a box-normalized state includes an additional $V^{-1/2}$, or one forms a normalizable wave packet.

Put $r=p/(E+m)$ and assume $p>0$. The common plane wave has unit modulus. The component norm isFor unit spinor probability density choosewhere $E^2=p^2+m^2$ was used. This is spinor normalization, not square normalization of a plane wave over infinite space; a box or wavepacket gives the corresponding spatial normalization. If the convention is instead ${\psi }^{†}\psi =2E$, use $N=\sqrt{(E+m)/10}$.

For momentum along $+z$, the positive-helicity components are the first and third, while the second and fourth have negative helicity. HenceBoth are independent of speed. In the source's chiral block order the exact left-chiral weight is $1/2+3p/(10E)$, tending to $4/5$ at high energy; helicity and chirality probabilities agree only in that limit.

A wave packet is a localized superposition of nearby wavenumbers. Its rapid carrier crests have phase velocity $v_p=\omega /k$; the slowly varying envelope translates at the group velocity. Dispersion means that $\omega (k)$ is not simply a constant times $k$, so different carrier components have different phase velocities. Expand around $k_0$:The integral is an envelope depending on $x-{\omega }^{\prime }(k_0)t$, so $v_g=d\omega /dk$ at the central frequency. Neglected curvature causes packet spreading. For a deep-water gravity wave, $\omega =\sqrt{gk}$, hence $v_p=\sqrt{g/k}$ andFor crest crossings through a dispersive wave packet, an instantaneous group containing six crests has length approximately $L=6\lambda$. A stationary ship takes $L/v_g$ to encounter the group; its crest-arrival frequency is $v_p/\lambda$. ConsequentlyIn the envelope frame the ship moves backwards at $v_g$ while crests move forwards at $v_p-v_g=v_g$. Their relative speed is $2v_g$, explaining why counting a snapshot and counting crossings differ.

To estimate deep-water swell distance from period-dependent arrival times, assume a localized storm emits both periods at the same time and place, propagation is in deep water, and currents and refraction can be neglected. If its distance is $D$, the arrival lag obeysWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $D=15g(86400)/\pi \simeq 4.05\times 10^6\mathrm{m}$, about 4050 km. The corresponding travel times are three and four days. A moving or long-lived storm with frequency-dependent emission times would invalidate the unique distance inference.

identical particles have the same intrinsic properties. Exchanging their bookkeeping labels cannot distinguish a physical preparation or change a measurable prediction. This is stronger than merely failing to track classical trajectories. For ordinary bosons and fermions, the many-particle wavefunction has respectively bosonic exchange symmetry or fermionic exchange symmetry under every particle exchange operator, acting on spatial and spin coordinates together. The Spin-statistics theorem connects integer spin to bosons and half-integer spin to fermions; this connection uses relativistic quantum-field assumptions, rather than following from the nonrelativistic Schrödinger equation alone.

For one-particle states $u,v$ satisfying orthonormality, the two alternatives form a quantum superposition:Its probability density contains an exchange interference term, absent from an incoherent mixture of labelled alternatives. The antisymmetric expression vanishes when both one-particle states are identical: one complete one-particle state can contain at most one fermion. This Pauli exclusion principle explains electron shell filling, the electronic Fermi sea and degeneracy pressure. Opposite spins are different one-particle states, so two electrons may share an atomic orbital. Spatially separated wave packets with negligible overlap can often be treated as distinguishable for local measurements; the underlying exchange symmetry remains.

In thermal equilibrium, occupation of a level of energy $ϵ$ obeysbosons can accumulate macroscopically in one state, giving Bose-Einstein condensation; fermions instead fill successive states up to the Fermi energy at zero temperature. When occupation is dilute, both expressions reduce to the Maxwell-Boltzmann distribution. The identical-particle classical partition function contains $1/N!$, avoiding spurious entropy from permutations and the Gibbs paradox. Exchange effects become conspicuous when the thermal de Broglie wavelength is comparable to the interparticle spacing.

An operational test is the Hong-Ou-Mandel effect. Down-converted photon pairs enter separate ports of a balanced beam splitter; indistinguishable alternatives for one photon per output cancel. Scanning the relative arrival delay restores coincidences away from overlap. Matching polarization, spectral filtering and spatial alignment controls distinguishability; measuring delayed accidental coincidences controls background. The observed dip tests two-photon interference, and its finite depth must not automatically be interpreted as a failure of bosonic symmetry. The original apparatus varied beam-splitter position and subtracted a separately measured accidental background.^[ref]

The free particle dispersion relation is $\hslash \omega ={\hslash }^2{k}^2/(2m)$. HenceThe phase velocity follows constant-phase surfaces of a plane wave. The group velocity follows the envelope of a narrow-band wave packet and equals its classical velocity $p/m$. Dispersion can spread the packet; the phase speed is not the particle's transport velocity. In three dimensions ${\mathbf{v}}_g=\hslash \mathbf{k}/m$ and the phase speed quoted is along $\mathbf{k}$.

Packet and group index. A wave packet superposes nearby wavevectors to form a localized envelope. The phase velocity is $v_p=\omega /k$; for a narrow packet with weak distortion the envelope velocity is $v_g=d\omega /dk$. Since $k=2\pi fn(f)/c$,For the stipulated linear index, $n_g=1.5+\tau (2f-f_0)$, not simply $n(f)$.

bandwidth and speed spread. A pulse lasting $t_p$ has Fourier bandwidth of order $1/t_p$, so $\Delta f\sim 10^{10}\mathrm{Hz}$; an exact width coefficient depends on pulse shape and width convention. At the carrier,Across a full bandwidth $\Delta f$,Higher-frequency components have the lower group velocity. For a band symmetric about $f_0$, the endpoint velocities are approximately $v_g(f_0)\pm 370\mathrm{m}{\mathrm{s}}^{-1}$.

dispersion-limited digital-fibre link length. Differential group delay accumulates asThe pulse spacing is $t_r=10^{-9}\mathrm{s}$. Pulse overlap and intersymbol interference become serious when the dispersive spread approaches that spacing, givingA conservative full-width estimate $t_p+\Delta t_{\mathrm{disp}}\lesssim t_r$ gives about 45 km; this refinement is not a universal sharp boundary because pulse shape, chirp and detection tolerance are unspecified. Attenuation and modal or waveguide dispersion could impose further limits, but are not quantified by the supplied index law.

digital optical pulse regeneration. Each stage must recover and regenerate the data with fresh, short pulses and correct timing, as well as restore signal level. Amplifying the already broadened waveform alone does not remove accumulated dispersion; explicit dispersion compensation is an alternative to full regeneration.

For a monochromatic wave, a constant phase $kx-\omega t$ moves at phase velocity $v_p=\omega /k$. For a narrow wave packet around $k$, expansion $\omega (k+\Delta k)\simeq \omega (k)+(d\omega /dk)\Delta k$ shows that its envelope moves at group velocity $v_g=d\omega /dk$. The envelope interpretation requires a sufficiently narrow spectrum and weak distortion; it is not a general identification with a causal signal front. Since $\omega =k{v}_p$ and $\lambda =2\pi /k$,For the gravity-capillary wave, put $s=\sigma /\rho$ and take $k>0$. Direct differentiation givesThus $v_g=v_p$ when $s{k}_{\ast }^2=g$, orAt this wavelength the gravity and surface-tension restoring terms contribute equally. Also $d(v_p^2)/dk=-g/k^2+s=0$ with positive second derivative: this is the minimum phase speed of a gravity-capillary wave, $v_{p,\min }=(4g\sigma /\rho {)}^{1/4}=0.214\mathrm{m}{\mathrm{s}}^{-1}$. Locally the carrier crests and narrow envelope travel together. The medium has not become nondispersive: $d^2\omega /d{k}^2=k_{\ast }{d}^2{v}_p/d{k}^2\ne 0$ there, so a finite wave packet can still spread. In the gravity-dominated limit $v_g/v_p\to 1/2$; in the capillary-dominated limit it tends to $3/2$.

Differentiating the expectation value and using ${\partial }_t|\psi ⟩=-i\hat{H}|\psi ⟩/\hslash$ and ${\partial }_t⟨\psi |=i⟨\psi |\hat{H}/\hslash$ givesThe second term means explicit operator time dependence, distinct from the state's evolution.

The commutator product rule follows by adding and subtracting $\hat{B}\hat{A}\hat{C}$:For the mathematical induction step, let $\hat{D}=[\hat{A},\hat{B}]$ and use $[\hat{A},\hat{D}]=0$. The case $n=1$ is immediate. If $[{\hat{A}}^m,\hat{B}]=m\hat{D}{\hat{A}}^{m-1}$, thenThus the claimed identity holds for every positive integer. Commutation of $\hat{B}$ with $\hat{D}$ is not needed for this particular step. This is the central-commutator differentiation rule. Termwise application to a convergent power series, on a common domain where the operations are justified, gives

For position and momentum, $[\hat{x},\hat{p}]=i\hslash \hat{I}$. Hence $[{\hat{p}}^2,\hat{x}]=-2i\hslash \hat{p}$, $[V(\hat{x}),\hat{x}]=0$ and $[V(\hat{x}),\hat{p}]=i\hslash V^{\prime }(\hat{x})$. Substitution gives Ehrenfest theorem:The mean momentum changes by the mean force. It does not follow for a general nonlinear Potential energy that $⟨V^{\prime }(\hat{x})⟩=V^{\prime }(⟨\hat{x}⟩)$: a closed classical trajectory for the mean position requires an additional condition, such as a quadratic Potential energy or a sufficiently localized wave packet.

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

Ehrenfest theorem relates the dynamics of quantum expectation values to operator commutators. For a normalized state obeying $i\hslash |\dot{\psi }⟩=\hat{H}|\psi ⟩$ and a possibly time-dependent observable $\hat{A}$, differentiate $⟨A⟩=⟨\psi |\hat{A}|\psi ⟩$. Using the adjoint Schrödinger equation for the bra givesThusThe identity assumes a common domain on which the required products and derivatives are defined.

For $H=p^2/(2m)+V(x)$, the canonical commutation relation gives $[x,H]=i\hslash p/m$ and $[p,H]=-i\hslash V^{\prime }(x)$. ThereforeThese resemble Hamilton's equations and Newton's second law, but generally $⟨V^{\prime }(x)⟩\ne V^{\prime }(⟨x⟩)$. If the state is localized about $\overline{x}=⟨x⟩$, a local expansion givesThe classical trajectory equation for the mean is accurate when the force changes little across the wave packet. For a quadratic harmonic potential, $V^{\prime }$ is linear and this closure is exact for every state with the relevant finite moments: $m{d}^2⟨x⟩/d{t}^2=-m{\omega }^2⟨x⟩$. This exact classical evolution of the centroid does not eliminate its quantum position uncertainty or make every state a point particle.

The correspondence principle is broader: quantum predictions recover classical ones in an appropriate limit, such as large action compared with $\hslash$ and observations that cannot resolve individual quantum levels. It is not the assertion that every large quantum number automatically makes every possible state classical. Delocalized superpositions and interference can remain significant unless the physical preparation and measurement justify a classical description.

A spectral example is useful. For a quantum harmonic oscillator, neighboring levels differ by $\hslash \omega$, giving transition frequency $\omega /(2\pi )$, equal to its classical oscillation frequency. A coherent state makes the connection more concrete: its mean position and momentum oscillate classically while the finite-width Gaussian profile retains its quantum uncertainty. For hydrogenic Bohr levels $E_n=-A/n^2$, transitions between nearby large $n$ have $\Delta E/\hslash \simeq 2A/(\hslash n^3)$, which equals the classical circular orbital angular frequency in that model. Ehrenfest's theorem supplies an exact expectation-value bridge; a classical trajectory follows only when the force can also be evaluated at the mean position.

Superpose two nearby harmonic waves. The identityseparates a rapid carrier from a slow beat envelope. A constant point of the envelope moves at $\Delta \omega /\Delta k$, which tends toas the components approach one another. Equivalently, expand a narrow wave packet spectrum around $k_0$: $\omega (k)={\omega }_0+v_g(k-k_0)+\cdots$. Factoring the carrier out of its Fourier transform representation leaves an envelope depending on $x-v_{g}t$. This interpretation requires a narrow spectrum and little attenuation or distortion; higher derivatives of the dispersion relation can change the envelope shape. The carrier phase velocity is $v_p=\omega /k$ and need not equal $v_g$.

A localized quantum particle is described by a wave packet: a coherent quantum superposition of momentum components, rather than a single delocalized plane wave. The Born rule assigns the position probability density $|\psi (x,t){|}^2$. A useful normalized initial Gaussian wave packet isIts mean position and momentum are $x_0$ and $p_0$, and its standard deviations satisfy ${\sigma }_x(0)={\sigma }_0$, ${\sigma }_p=\hslash /(2{\sigma }_0)$. This unchirped initial state saturates the Heisenberg uncertainty relation.

For a free particle, the momentum-space Schrödinger equation multiplies each amplitude by $e^{-i{p}^{2}t/(2m\hslash )}$. The inverse Fourier transform is a Gaussian integral, giving the free Gaussian wave packet evolutionOne way to obtain this expression explicitly is to write the initial momentum amplitude as proportional to $\exp [-{\sigma }_0^2(p-p_0{)}^2/{\hslash }^2-ip{x}_0/\hslash ]$, multiply by the free evolution phase and complete the square in $p$. Taking the modulus squared removes the position-dependent phase and leavesThe expectation value follows uniform classical motion while the density spreads. A narrower initial packet needs a wider momentum distribution and therefore spreads more rapidly. At long times its width grows linearly with $|t|$; its variance grows quadratically, unlike the linear variance growth of ordinary diffusion. The probability continuity equation preserves total probability despite the decreasing peak height. The mean momentum and the mean energy $(p_0^2+{\sigma }_p^2)/(2m)$ remain constant, since the momentum populations do not change.

The same result has a wave interpretation. For a free nonrelativistic particle $\omega (k)=\hslash k^2/(2m)$, so the group velocity is $d\omega /dk=\hslash k/m$, equal to $p_0/m$ at the packet center. The curvature $d^2\omega /d{k}^2=\hslash /m$ makes different momentum components travel with different group velocities, producing dispersion. The phase velocity is $\omega /k=\hslash k/(2m)$ for this energy convention and does not describe the motion of the envelope. The evolving quadratic spatial phase is essential to the spreading; a density snapshot alone does not specify the full quantum state.

A freely spreading Gaussian wave packet remains a pure Gaussian state but develops position-momentum correlations. Its symmetrized covariance is $C_{xp}=⟨\{x-⟨x⟩,p-⟨p⟩\}⟩/2=t{\sigma }_p^2/m$. ConsequentlyEquality in the Robertson-Schrödinger uncertainty relation survives at every time, whereas the simpler product ${\sigma }_x{\sigma }_p$ equals $\hslash /2$ only initially. A chirped Gaussian wave packet can already have $C_{xp}(0)\ne 0$. More generally, $x(t)=x(0)+p(0)t/m$ gives ${\sigma }_x(t{)}^2={\sigma }_x(0{)}^2+2t{C}_{xp}(0)/m+t^2{\sigma }_p(0{)}^2/m^2$. A sufficiently negative initial covariance permits initial contraction before subsequent spreading.

A potential energy changes this behavior. A quadratic potential preserves the Gaussian form because the evolution of position and momentum is linear. For a quantum harmonic oscillator, the Gaussian covariance in a harmonic potential isA coherent state has ${\sigma }_x^2=\hslash /(2m\omega )$, ${\sigma }_p^2=m\hslash \omega /2$ and zero centered covariance. Its width is constant, and its center follows $x_c(t)=x_0\cos \omega t+p_0\sin \omega t/(m\omega )$. In oscillator notation its coherent state label rotates as $\alpha (t)=\alpha (0)e^{-i\omega t}$, apart from the common ground-energy phase of the state vector. A squeezed coherent state generally has a width that breathes at $2\omega$. Thus being Gaussian does not imply either free spreading or constant width; the Hamiltonian operator and initial covariance determine the result.