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The commutator gives $a(a^{†}{)}^n=(a^{†}{)}^{n}a+n(a^{†}{)}^{n-1}$. Hence $a|n⟩=\sqrt{n}|n-1⟩$ and $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. Starting with the normalized vacuum, repeated application gives $⟨0|a^n(a^{†}{)}^n|0⟩=n!$, proving unit norm. The number operator then has eigenvalue $n$, soDifferent levels are orthogonal because the Hamiltonian is self-adjoint and their eigenvalues are distinct.

For the perturbation, the ground-state diagonal matrix element is zero, while $V|0⟩=\hslash \omega \sqrt{r!}|r⟩$. The nondegenerate second-order perturbation formula consequently givesFor $r=1$, $V|1⟩=\hslash \omega (|0⟩+\sqrt{2}|2⟩)$. The lower and upper intermediate levels contribute respectively $+{\lambda }^2\hslash \omega$ and $-2{\lambda }^2\hslash \omega$, giving $\Delta E_1=-{\lambda }^2\hslash \omega$ to second order.

For $r=1$ the exact completion of the square isThe shifted operators $b=a+\lambda$, $b^{†}=a^{†}+\lambda$ have the same commutator and are unitarily related to the original ones by a displacement operator. Their vacuum is the coherent state with amplitude $-\lambda$. Therefore every exact level shifts by $-{\lambda }^2\hslash \omega$.

The stability of monomial ladder-operator perturbations gives a necessary qualification to the request for a lowest level for every $r$. For $r>2$ and real nonzero $\lambda$, a coherent state of amplitude $R{e}^{i\theta }$ has energy expectation $\hslash \omega [R^2+1/2+2\lambda R^r\cos (r\theta )]$. Choose the phase so the last term is negative. As $R\to \infty$ the expectation tends to minus infinity, so the perturbed Hamiltonian has no true ground state. The displayed second-order answer is then the formal perturbative coefficient of the branch from the unperturbed vacuum, not an exact lowest eigenvalue. For $r=2$ stability requires $|\lambda |<1/2$; for $r=1$ the completed-square Hamiltonian is bounded below for every real $\lambda$.

Generalized position eigenstates satisfy $⟨x|x^{\prime }⟩=\delta (x-x^{\prime })$ and $\int |x⟩⟨x|dx=I$. Define the position-space wavefunction $\psi (x)=⟨x|\psi ⟩$. Completeness then gives $⟨\psi |\psi ⟩=\int |\psi (x){|}^{2}dx$. Position acts by multiplication, $⟨x|\hat{x}|\psi ⟩=x\psi (x)$. Using $⟨x|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipx/\hslash }$ and momentum completeness, differentiating the Fourier integral gives $⟨x|\hat{p}|\psi ⟩=-i\hslash {\psi }^{\prime }(x)$ on the momentum operator's domain.

For the unit-mass, unit-frequency oscillator defineSince $[H,a]=-\hslash a$, a simultaneous energy eigenstate and $a$-eigenstate would satisfy $[H,a]|{\psi }_{\alpha }⟩=0=-\hslash \alpha |{\psi }_{\alpha }⟩$, forcing $\alpha =0$. For $\alpha =0$, the energy is $\hslash /2$, the smallest possible because $⟨a^{†}a⟩=\Vert a\psi {\Vert }^2\ge 0$.

The position-space equation is $(x+\hslash {\partial }_x){\psi }_{\alpha }=\sqrt{2\hslash }\alpha {\psi }_{\alpha }$. Solving and normalizing, with $\alpha =u+iv$, gives up to a constant phaseIts squared modulus is a normalized Gaussian, so every complex $\alpha$ gives a normalizable coherent state.

For a creation eigenstate, $(x-\hslash {\partial }_x)\psi =\sqrt{2\hslash }\beta \psi$ instead gives $\psi =C\exp [x^2/(2\hslash )-\sqrt{2/\hslash }\beta x]$. Every nonzero such function grows quadratically exponentially at infinity and fails square integrability. Hence there are no nonzero normalizable eigenstates of $a^{†}$. This last symbol is $a^{†}$ in the original PDF, although the converted TeX drops its dagger.

The spatial translation operator is $U(\alpha )=e^{-i\alpha \hat{p}/\hslash }$. From $[\hat{x},\hat{p}]=i\hslash$ and the preceding central-commutator result,Thus $\hat{x}U|x⟩=U\hat{x}|x⟩+\alpha U|x⟩=(x+\alpha )U|x⟩$, consistent with $U|x⟩=|x+\alpha ⟩$. Since $⟨x|U=⟨x-\alpha |$, the translated wavefunction is $\boxed{\psi (x-\alpha )}$.

For the quantum harmonic oscillator, the creation and annihilation operators give $\hat{p}=i\sqrt{m\hslash \omega /2}(a^{†}-a)$. Put $s=\alpha \sqrt{m\omega /(2\hslash )}$. The central commutator of $s{a}^{†}$ and $-sa$ is $s^2$, soAs $a|0⟩=0$ and $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$,Taking position wavefunctions provesThe coefficient squares sum to $e^{-s^2}{\sum }_n{s}^{2n}/n!=1$, checking normalization of this coherent state.

Write $\Delta =\Omega -\omega$. Usingthe interaction-picture perturbation is

The initial field is the coherent state with amplitude $-1$:Hence the initial amplitudes of both $|e,n⟩$ and $|g,n⟩$ areOnly $|g,n+1⟩$ contributes at first order to the amplitude of $|e,n⟩$. Since $A|n+1⟩=\sqrt{n+1}|n⟩$,But $\sqrt{n+1}{c}_{g,n+1}(0)=-c_{e,n}(0)$, soThe free Schrödinger-picture phase does not change the probability. Within this first-order perturbative approximation,At $t=\pi /(\Omega -\omega )$ the cosine is zero, so this probability vanishes to the stated perturbative order.

Orthogonality of the energy eigenstates givesUsing the lowering-operator action from part (a) and shifting the summation index,Thus this coherent state is a normalized eigenstate of $A$ with eigenvalue $z$.

The displacement operator is unitary, so $\Vert |\alpha ⟩\Vert =1$. By coherent state, $A|\alpha ⟩=\alpha |\alpha ⟩$, hence $\Vert A|\alpha ⟩\Vert =|\alpha |$.

Coherent state evolution givesThe condition from part (d) becomes

A coherent pump has a Poisson distribution of photon numbers. Second-harmonic conversion couples a pump number state $|n⟩$ to a state with two fewer pump photons with a matrix element proportional to $\sqrt{n(n-1)}$. Different number components therefore evolve at different rates, and the pump becomes correlated with the harmonic mode. Its reduced state is generally neither pure nor exactly Poisson coherent. An undepleted classical-pump approximation suppresses precisely this backaction; it cannot prove that the exact finite-depletion quantum state stays coherent. Vacuum input or zero interaction are trivial exceptions.

Let $A=a^{†2}$, $B=a^2$ and $N=a^{†}a$. The oscillator algebra gives $[A,B]=-4N-2$, $[N,A]=2A$ and $[N,B]=-2B$. Substituting into the definitions proves closure:The factors depend on the stated normalization: $K_3$ has an extra factor one-half relative to $N+1/2$. The Robertson uncertainty relation consequently givesFor a coherent state with $n=|\alpha {|}^2$, normal ordering givesThus $⟨K_1⟩=\mathrm{Re}{\alpha }^2$ and $⟨K_2⟩=-\mathrm{Im}{\alpha }^2$. In either variance, the fourth-power and $n^2$ terms cancel against the square of the mean, leavingThe variance product is exactly $4⟨K_3{⟩}^2$, so the coherent state saturates the stated bound.

Amplitude-squared squeezing in either of these axes meansThe other variance must then rise sufficiently to respect the product inequality. A rotated field-squared quadrature can reveal squeezing in another direction; the criterion concerns the declared quadrature and reference, not merely a reduction in mean optical intensity.

For stationary photodetection, the normalized correlation in second-order optical coherence is $g^{(2)}(\tau )=⟨E^{-}(t)E^{-}(t+\tau )E^{+}(t+\tau )E^{+}(t)⟩/⟨E^{-}{E}^{+}{⟩}^2$. Normal ordering expresses the absorption of two photons. In a single mode,A coherent state has $g^{(2)}(0)=1$. A single-mode thermal field has $g^{(2)}(0)=2$ and bunches, while a number state with $n>0$ gives $1-1/n$; a single-photon state has zero ideal coincidence probability at zero delay. Antibunching, a zero-delay suppression relative to nearby delays, is distinct from simply reporting an intensity variance. Multimode averaging and finite detector response reduce observed contrast.

The Hanbury Brown and Twiss effect is measured by splitting the light onto two detectors and histogramming coincidences versus delay. Background counts, detector timing, dead time and afterpulsing need calibration. Two detectors avoid confusing the single detector's inability to register arbitrarily close counts with real antibunching. First-order interference and second-order correlations contain different information: weak first-order coherence does not imply an absence of intensity bunching.^[ref]

For $X=(a+a^{†})/2$ and $Y=(a-a^{†})/(2i)$, $[X,Y]=i/2$, so $\Delta X\Delta Y\ge 1/4$. Vacuum and coherent states have variance $1/4$ in either optical quadrature. quadrature squeezing means that some rotated optical quadrature has variance below this reference, with compensating noise elsewhere. A reduced second moment about zero is not the correct test when the field's mean is displaced.

For $S(\zeta )=\exp [({\zeta }^{\ast }{a}^2-\zeta a^{†2})/2]$ and $\zeta =r{e}^{i\theta }$, the quiet principal axis of the optical squeezed vacuum is at angle $\theta /2$ under the squeeze-operator convention used above, andThe pure Gaussian quantum state saturates the uncertainty product. Displacement moves its centre without changing its covariance ellipse; squeezed thermal states instead have a larger uncertainty product. The optical squeezed vacuum has zero mean field, even photon-number parity and mean occupation ${\sinh }^{2}r$; it need not have sub-Poissonian photon statistics. number squeezing and quadrature squeezing are different conditions.

Degenerate parametric amplification generates single-mode squeezing; a two-mode squeezed vacuum instead has correlated occupations and joint quadrature squeezing. Either individual mode of an ideal two-mode state is thermal after the other is ignored, not individually squeezed. balanced homodyne detection measures the noise versus local oscillator phase. Reference phase drift admixes the anti-squeezed optical quadrature, while electronic noise and detector imbalance need separate checks.

For an optical loss channel with efficiency $\eta$ and vacuum entering the unused port, optical squeezing degradation by loss givesThe linear-loss limit follows directly by writing the output field as $\sqrt{\eta }a+\sqrt{1-\eta }v$ with independent vacuum mode $v$. Loss therefore places a floor under measured squeezing. Squeezed input can improve phase-sensitive measurements when aligned with the measured optical quadrature; it does not reduce every noise source or both conjugate uncertainties at once.

An ideal cross-Kerr interaction between two single modes is $H_{\mathrm{int}}=\hslash \chi n_1{n}_2$, with $n_j=a_j^{†}{a}_j$. The number operators commute with the interaction, so this coupling changes phases without transferring photons between the modes. To implement the specified positive conditional phase, usewhere $n_{2V}=|V⟩⟨V|$ within the one-photon polarization subspace. The horizontal branch has zero interacting occupation. Since $e^{i\Theta n}|\alpha ⟩=|e^{i\Theta }\alpha ⟩$, the specified conditional rotation follows. This is an ideal lossless single-mode model; an actual travelling-wave medium's bandwidth, loss and noise are not established by writing this Hamiltonian.

Use $X_1=(a+a^{†})/2$ and $X_2=(a-a^{†})/(2i)$. A coherent state has centre $(\mathrm{Re}\alpha ,\mathrm{Im}\alpha )$ and variance $1/4$ in either optical quadrature. Multiplication of $\alpha$ by $e^{i\Theta }$ rotates that centre anticlockwise through $\Theta$, keeping the coherent state noise disk unchanged. For real positive $\alpha$ the two centres are $(\alpha ,0)$ and $(\alpha \cos \Theta ,\alpha \sin \Theta )$.

Normalize the initial polarization of an electromagnetic wave qubit. The output of the medium isThe joint state is generally an entangled state. The overlap magnitude is $|⟨\alpha |\beta ⟩|=e^{-2|\alpha {|}^2{\sin }^2(\Theta /2)}$. The state factorizes when the amplitudes coincide, while nearly distinguishable amplitudes make the polarization nearly maximally entangled with the optical mode. Before any polarization conditioning, tracing out mode 2 gives the mixture $(|\alpha ⟩⟨\alpha |+|\beta ⟩⟨\beta |)/2$, with no interference between branches.

Include the normalization factors in the half-wave plate map: $|H⟩\mapsto (|H⟩+|V⟩)/\sqrt{2}$ and $|V⟩\mapsto (-|H⟩+|V⟩)/\sqrt{2}$. The transformed joint state isConditioning on a vertical result and setting $\Theta =\pi$ therefore preparesThis pure optical cat state is the conditional output; it is not the unconditioned reduced state immediately after the Kerr medium. The measurement erases the polarization label, allowing the two amplitudes to interfere. The calculated optical Wigner functions below display the resulting fringes; their negative regions are quasiprobability, not negative detection probabilities. These illustrative plots choose $\alpha =1.15$: the rotation panel uses $\Theta =2\pi /3$, while the cat panels use the conditional Kerr rotation $\Theta =\pi$ and three different relative phases $\phi$.

The coherent-state overlap follows by expanding in the Fock basis:Thus $s=⟨\alpha |-\alpha ⟩=e^{-2|\alpha {|}^2}$ andThe postselection probability for Kerr-cat preparation includes both the normalized input qubit and the wave plate. At $\alpha =0,\varphi =\pi$ the vertical branch has zero probability and no normalized conditional state. For the remaining calculation take $\alpha$ real, as required.

Using $a|\pm \alpha ⟩=\pm \alpha |\pm \alpha ⟩$, put $D=1+s\cos \varphi$. Directly evaluating diagonal and cross terms givesThe last quantity is the mean photon number in the conditional cat, which differs from the input coherent state mean ${\alpha }^2$. Since $a{a}^{†}=n+1$,However, its first moment iswhich is generally nonzero. The original PDF's assertion of a zero mean for arbitrary $\varphi$ is inconsistent with its stated pure state. Consequently its displayed expression is the second moment, not the general variance. The phase-dependent cat-state quadrature variance isThe printed result does become the variance for $\sin \varphi =0$, notably the even and odd cat states. It could also describe an ensemble mixing equal weights of $+\varphi$ and $-\varphi$: that different state's mean vanishes while its second moment is unchanged. Such phase averaging is not part of the specified pure-state preparation.

For nonzero finite $\alpha$, $0<s<1$ and $D>0$. The correct quadrature squeezing condition in this quadrature is thereforeFor a representative phase in $-\pi <\varphi \le \pi$, this is $|\varphi |<\arccos (-e^{-2{\alpha }^2})$. If instead one applies the printed second moment as a variance, the resulting condition is $\cos \varphi >0$, or $|\varphi |<\pi /2$ modulo $2\pi$. That is sufficient, but not necessary, for the actual pure state's squeezing. For example, at $\varphi =\pi /2$, $⟨X_2^2⟩=1/4$ while $\mathrm{Var}{X}_2=1/4-{\alpha }^2{e}^{-4{\alpha }^2}<1/4$. This explicitly tests the difference.

For a fixed phase other than $\pi$ modulo $2\pi$, expandAs the input mean number ${\alpha }^2\to 0$, the normalized state tends to vacuum, $⟨X_2⟩\sim -\alpha \tan (\varphi /2)$ andThis expansion is nonuniform near the odd-cat phase. At exactly $\varphi =\pi$ the vacuum terms cancel and the normalized odd coherent-state superposition tends to $|1⟩$, giving $\mathrm{Var}{X}_2\to 3/4$, while its postselection probability tends to zero. The even coherent-state superposition at $\varphi =0$ instead contains only even photon numbers and tends smoothly to vacuum.

At large input mean number, $s\to 0$, the two coherent state branches become nearly orthogonal, and $\mathrm{Var}{X}_2\to 1/4$ for every fixed phase. The mean tends to zero as well. This does not make the state an incoherent mixture: phase-space interference still distinguishes the ideal quantum superposition, although its contribution to this low-order moment is exponentially small. The loss of coherence in an optical cat state arises when ordinary loss reveals the branch to the environment and suppresses that interference, with factor $e^{-2(1-\eta ){\alpha }^2}$ for transmissivity $\eta$.

For the especially simple even cat,It starts at $1/4$, falls to about $0.1108$ at $|\alpha |\simeq 0.7995$, and returns towards $1/4$ from below. The odd-cat curve decreases from $3/4$ towards $1/4$ from above. Other fixed-phase curves have the small- and large-amplitude limits derived above; for negative $\cos \varphi$, squeezing ends when $e^{-2{\alpha }^2}=-\cos \varphi$. In balanced homodyne detection, centre each quadrature distribution on its measured mean before inferring its variance.

Writing $a=cx+idp$, where $c=\sqrt{m\omega /(2\hslash )}$ and $d=1/\sqrt{2m\omega \hslash }$, the canonical commutation relation givesConsequently $[a^{†}a,a]=-a$ and $\boxed{[H,a]=-\hslash \omega a}$. If $a|{\varphi }_0⟩$ were nonzero, this commutator would make it an energy eigenstate with energy $E_0-\hslash \omega$, contradicting the lowest-energy property. Therefore $a|{\varphi }_0⟩=0$.

The commutator product rule and induction giveActing on the vacuum and shifting the sum then shows that the unnormalized bosonic coherent state $|\tilde{\psi }⟩=e^{\alpha a^{†}}|0⟩$ satisfies $a|\tilde{\psi }⟩=\alpha |\tilde{\psi }⟩$.

The displayed series lacks its wavefunction normalization factor. Since $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$, its squared norm is ${\sum }_n|\alpha {|}^{2n}/n!=e^{|\alpha {|}^2}$. The normalized coherent state needed for the stated expectation values isUse $⟨a⟩=\alpha$, $⟨a^{†}⟩={\alpha }^{\ast }$, $⟨a^{†}a⟩=|\alpha {|}^2$ and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$. For $x=x_0(a+a^{†})$, $x_0=\sqrt{\hslash /(2m\omega )}$,The displacement changes the mean position but not its ground-state uncertainty. Unnormalized bras and kets give the same physical expectation values only after division by their squared norm.

In the small-angle approximation the physical pendulum is a harmonic oscillator with $\omega =2\pi /T$. Its classical energy is $E=m{\omega }^2{A}^2/2=3.55\times 10^{-3}\mathrm{J}$. Equating this to $(n+1/2)\hslash \omega$ givesAn oscillating bob is not a single stationary energy eigenstate. An ideal coherent state has a Poisson distribution of oscillator occupations centred on this mean, with relative spread $1/\sqrt{\bar{n}}$, exceptionally small here. This quantum occupation scale of a classical oscillator fixes the excitation scale without requiring the real, damped clock to be a pure coherent state.

Hamiltonian and commutators. Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ and define $c_0=\sqrt{m\omega /(2\hslash )}$. Expanding the product of the given creation and annihilation operators,Therefore the quantum harmonic oscillator Hamiltonian operator isSimilarly,Using $[AB,C]=A[B,C]+[A,C]B$ gives

Lowering a normalized energy eigenstate. The commutator impliesFor $n>0$, nondegeneracy of this one-dimensional energy level means $\hat{a}|{\varphi }_n⟩=c_n|{\varphi }_{n-1}⟩$. Its norm is fixed by the number operator:Choosing successive eigenstate phases in the conventional way makes $c_n=\sqrt{n}$, soFor $n=0$, the same norm is zero, hence $\hat{a}|{\varphi }_0⟩=0$. A nonzero lowered ground state would have energy below the lowest allowed level, and positivity of ${\hat{a}}^{†}\hat{a}$ forbids that possibility.

Coherent-state eigenvalue. Applying the annihilation operator term by term, with the $n=0$ contribution zero, givesThusThe coherent state is normalized because $e^{-|\alpha {|}^2}{\sum }_n|\alpha {|}^{2n}/n!=1$. Its annihilation operator is not Hermitian, so the eigenvalue $\alpha$ may be complex.

Time evolution. Under the time-independent Hamiltonian operator, each energy component acquires its own phase:Since $|\alpha e^{-i\omega t}|=|\alpha |$, the normalization prefactor is unchanged. HenceThe state remains a coherent state, with its complex parameter rotating at the oscillator's angular frequency; the displayed extra factor is a global phase from the zero-point energy.

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.

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.