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A lossless mode transformation maps a product of coherent states to another product of coherent states with transformed complex amplitudes. unitarity preserves the total mean particle number. It does not generally turn a fixed-number input into coherent states.

The coherent state satisfies $⟨a⟩=\alpha$, $⟨a^{†}⟩={\alpha }^{\ast }$ and $⟨a^{†}a⟩=|\alpha {|}^2$. The canonical commutation relation adds the vacuum term $⟨a{a}^{†}⟩=|\alpha {|}^2+1$. Thereforeand the coherent-state electric-field variance isHere $E_0$ is taken positive as the field-amplitude scale. The coherent state displaces the mean field without changing this vacuum optical quadrature noise.

A single-mode coherent state is an annihilation operator eigenstate:Its photon statistics are Poisson distribution: $⟨n⟩=\mathrm{Var}n=|\alpha {|}^2$. For optical quadratures $X=(a+a^{†})/2$, $P=(a-a^{†})/(2i)$, one has $[X,P]=i/2$ and $\mathrm{Var}X=\mathrm{Var}P=1/4$. Its field mean oscillates classically while retaining vacuum fluctuations.^[ref][ref]

A classical mixture of coherent states has a nonnegative Glauber-Sudarshan representation. A nonclassical state of light cannot be represented that way. Examples are a squeezed vacuum state, with quadrature variances $e^{-2r}/4$ and $e^{2r}/4$, and a nonvacuum number state, with zero photon number variance. Noise below the vacuum level in one quadrature, sub-Poisson number noise, or photon antibunching can witness nonclassicality. For $|n⟩$, $g^{(2)}(0)=1-1/n<1$; coherent light has $g^{(2)}(0)=1$. Thermal light has excess noise but is a classical mixture: noncoherent does not by itself mean nonclassical.

Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|{\varphi }_n⟩=|{\psi }_n⟩+\lambda |{\eta }_n⟩+O({\lambda }^2)$ and ${\mathcal{E}}_n=E_n+\lambda \Delta E_n+O({\lambda }^2)$. Fix phase and normalization at this order by $⟨{\psi }_n|{\eta }_n⟩=0$. The first-order Schrödinger equation isangular momentum projection onto $⟨{\psi }_n|$ gives $\Delta E_n=⟨{\psi }_n|H_1|{\psi }_n⟩$. angular momentum projection onto a different eigenstate givesCompleteness therefore yields the first-order nondegenerate perturbation theory resultNonzero energy denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding integral as well as any discrete sum.

For the oscillator in a static electric field, electrostatic potential energy is $H_1=-qϵx$ and $\omega =\sqrt{k/m}$. parity gives $⟨0|x|0⟩=0$, and the supplied Gaussian wavefunctions giveSince $E_0-E_1=-\hslash \omega$, the two-state first-order ground state isThus $⟨x⟩=2\eta x_{\mathrm{zp}}+O({ϵ}^2)$, and the induced electric dipole isIt is aligned with the field for either sign of $q$. The harmonic-oscillator electric polarizability is $q^2/k$. The position operator connects $|0⟩$ only to $|1⟩$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.

Complete the square to find the exact electric-field displacement of a quantum harmonic oscillator:The exact normalized ground wavefunction is ${\psi }_0(x-x_0)$, with energy $\hslash \omega /2-q^2{ϵ}^2/(2k)$. Its mean displacement is $x_0$, soThe first-order dipole equals the exact dipole, because the restoring force is exactly linear and translating a harmonic well does not change its width or frequency. This agreement of one expectation value does not make the truncated wavefunction exact. The Hellmann–Feynman theorem provides an independent check: $d=-\partial E_0/\partial ϵ$.

For perturbation theory to approximate the state, $|\eta |\ll 1$, or $|x_0|\ll 2x_{\mathrm{zp}}$. The exact translated ground state has the original-oscillator expansionIt is a coherent state with mean occupation ${\eta }^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The normalized two-state truncation of a displaced oscillator instead giveswhich incorrectly tends towards zero for large $|\eta |$ instead of following the displaced well. In position space the truncated polynomial times Gaussian also develops a node absent from the exact ground state. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.

A coherent state is a normalized eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding in the Fock state eigenbasis gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and wavefunction normalization fixesThe displacement operator therefore shifts the oscillator ground state in phase space. Distinct coherent states are not orthogonal:

The occupation probabilities form a Poisson distribution,For the quantum harmonic oscillator, the expectation values of position and momentum areThe fluctuations equal those of the ground state,Thus the position wavefunction is a displaced Gaussian wave packet that saturates the Heisenberg uncertainty principle.

Time evolution under $H=\hslash \omega (a^{†}a+1/2)$ preserves this form:The center follows the classical oscillator trajectory while the packet width stays constant. Its mean energy is $\hslash \omega (N+1/2)$; it is generally a superposition of energy eigenstates, rather than a state of definite energy. For $N\gg 1$, the relative occupation fluctuation is $1/\sqrt{N}$, explaining the classical appearance of a highly excited coherent state.

For the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

The initial quantum state is a coherent state. Since $a|\alpha ⟩=\alpha |\alpha ⟩$ and the state is normalized, its mean occupation number is $N=⟨a^{†}a⟩=|\alpha {|}^2$. Each quantum harmonic oscillator energy eigenstate acquires the phase $e^{-i\omega (n+1/2)t}$, soThe prefactor is a global phase and does not change an expectation value. Inverting the definitions of the annihilation operator and creation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. If $\alpha =\sqrt{N}{e}^{i\phi }$, thenFor a classical harmonic oscillator with amplitude $A$, the conserved energy is $E=m{\omega }^2{A}^2/2$. Setting $E=\hslash \omega N$ gives exactly the same amplitude, $A=\sqrt{2\hslash N/(m\omega )}$. Thus these coherent-state oscillator expectation values follow the classical orbit.

The full quantum mean energy is nevertheless $⟨H⟩=\hslash \omega (N+1/2)$: the extra zero-point energy belongs to the nonzero position and momentum uncertainties required by the Heisenberg uncertainty relation, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A coherent state follows a classical mean trajectory while retaining finite position and momentum uncertainties.

Normalize the specified state by $⟨0|0⟩=1$ and write $⟨A⟩=⟨0|A|0⟩$. Its zero mode is a coherent state, so any normally ordered product of zero-mode creation operators and annihilation operators can be evaluated using ${\psi }_0\mapsto \sqrt{N_0}$ and ${\psi }_0^{†}\mapsto \sqrt{N_0}$. In particular $⟨({\psi }_0^{†}{)}^2{\psi }_0^2⟩=N_0^2$, rather than $N_0(N_0-1)$, which would apply to a zero-mode number state.

Sort the momentum-space contact-interaction Hamiltonian by its number of zero-mode factors. Four such factors give $u{N}_0^2/(2V)$. Three zero-mode factors and one nonzero momentum violate momentum conservation, so no such term survives. With two zero-mode factors there are three possibilities:

The sums over $\mathbf{k}\ne 0$ include both members of each opposite-momentum pair, consistently with the overall factor $1/2$. Terms with one zero-mode factor contain three nonzero-mode factors; their expected values vanish in the centered multimode squeezed vacuum. Indeed each $\psi$ is linear in $\phi ,{\phi }^{†}$, and an odd product has zero vacuum expectation value. Terms with no zero-mode factors are independent of $N_0$ at fixed ${\theta }_k$. ThereforeTo evaluate these vacuum expectation values, use the inverse Bogoliubov transformation and its Hermitian adjoint:Since ${\phi }_{\mathbf{k}}|0⟩=0$, the only nonzero two-factor vacuum expectation value is $⟨{\phi }_{\mathbf{k}}{\phi }_{\mathbf{q}}^{†}⟩={\delta }_{\mathbf{k},\mathbf{q}}$. The bosonic squeezed-vacuum contractions are consequentlyIn particular the three requested quantities are $-\sinh {\theta }_k\cosh {\theta }_k$, $-\sinh {\theta }_k\cosh {\theta }_k$ and ${\sinh }^2{\theta }_k$, respectively. The minus sign in the pair vacuum expectation values comes from the inverse Bogoliubov transformation and is essential to lowering the interaction energy. Substitution yields

The coherent state in the zero mode has mean occupation number $N_0$. The other occupation numbers follow from part (c), giving exactlyImpose the fixed mean particle number $⟨\hat{N}⟩=N$ by choosingHere $D$ is the quantum depletion of a Bose condensate. The approximation requires $0\le D\ll N$, so $N_0$ remains positive and macroscopically large. The specified coherent state and multimode squeezed vacuum are not exact eigenstates of the total number operator; $N$ in this variational construction is their prescribed mean particle number.

Write $P={\sum }_{\mathbf{k}\ne 0}{s}_k{c}_k$. Substituting into the retained part of the interaction energy before truncating givesThe last two terms quadratic in $D,P$ have no explicit large-$N$ factor when the Bogoliubov transformation parameters are held fixed, and are discarded along with the nonzero-mode quartic contribution already omitted in part (c). This is the same truncation that underlies the number-conserving Bogoliubov quadratic Hamiltonian. Thus, defining the physical number density correctly as $\rho =N/V=N/L^3$,Replacing $N_0$ by $N$ at the outset would give the wrong coefficient $2{\sinh }^2{\theta }_k$: expanding the leading condensate interaction energy supplies the necessary subtraction.

The printed definition $\rho =N/L$ is an error. In three dimensions the contact interaction coupling $u$ has units of energy times volume, so $uN/L$ does not even have units of energy. There is a direct counterexample with no depletion: when ${\theta }_k=0$, the specified coherent state has exactly $⟨H_{\mathrm{int}}⟩=u{N}^2/(2L^3)$. In fixed length and energy units, $L=2$, $u=1$ and $N=100$ give $625$, whereas the printed $u\rho N/2$ with $\rho =N/L$ gives $2500$. The requested formula and the minimization in part (e) hold with $\rho =N/L^3$.

Orthogonality of the number states makes the coherent state norm an exponential series:Using $\hat{a}|n⟩=\sqrt{n}|n-1⟩$ and reindexing $m=n-1$,It is not an eigenstate of the creation operator. Indeed its supposed eigenvalue would have to be ${\alpha }^{\ast }$, from its expectation, whereasThis nonzero residual holds even at $\alpha =0$.

The photon number probabilities of a coherent state form a Poisson distribution with mean $\mu =|\alpha {|}^2$. Alternatively normal ordering gives $⟨\hat{n}(\hat{n}-1)⟩=⟨{\hat{a}}^{†2}{\hat{a}}^2⟩=|\alpha {|}^4$. ThusThe last ratio is defined only for nonzero $\alpha$. At large $|\alpha |$ the absolute fluctuation grows but its size relative to the mean becomes negligible. This is the deterministic intensity limit within quantum optics; it does not assert the disappearance of vacuum-scale quadrature noise.

The annihilation operator eigenvalue property givesThis is a classical monochromatic electric field: the modulus of $\alpha$ fixes the wave amplitude, while its argument fixes the wave phase relative to the chosen mode convention. Direct normal ordering also gives $(\Delta E{)}_{\alpha }^2=|\mathcal{E}{|}^2$, independent of $\alpha$. Hence the fluctuation divided by the peak mean-field amplitude is $1/(2|\alpha |)$, small in the bright coherent state limit. A relative fluctuation evaluated exactly at a field zero is not a useful classicality test.

Choose the normalization constant real and positive; its otherwise arbitrary phase is physically immaterial. Orthogonality of the Fock states givesActing with the annihilation operator and shifting the summation index givesThus a coherent state is an annihilation-operator eigenket. It is not also an eigenket of the creation operator: a common eigenket would make $[\hat{a},{\hat{a}}^{†}]$ annihilate it, contradicting $[\hat{a},{\hat{a}}^{†}]=I$. Taking the adjoint of the annihilation-operator equation instead gives the distinct, valid bra relationWith $\hat{n}={\hat{a}}^{†}\hat{a}$, the photon number expected value is $\mu =|\alpha {|}^2$. Normal ordering ${\hat{n}}^2=({\hat{a}}^{†}{)}^2{\hat{a}}^2+{\hat{a}}^{†}\hat{a}$ gives $⟨n^2⟩={\mu }^2+\mu$. ThereforeAt large mean photon number the relative fluctuations vanish, giving a classical-field limit even though quantum fluctuations remain. For the vacuum $\mu =0$, the relative ratio itself is undefined.

The Born rule applied to the number-state coefficient gives the Poisson distributionIt depends on the single parameter $\mu =⟨n⟩$, and has equal expected value and variance, consistent with the preceding moments.

For a Fock state $|n⟩$, raising or lowering the number makes an orthogonal state, so the optical quadrature has $⟨\hat{X}⟩=0$. Expanding its square givesThe first two terms have zero diagonal matrix elements, while the last two contribute $n+1$ and $n$. In particular, the vacuum retains quadrature variance $1/2$.

Let the initial field be the normalized coherent state $|z⟩={\sum }_{n=0}^{\infty }{c}_n|n⟩$, where $c_n=e^{-|z{|}^2/2}{z}^n/\sqrt{n!}$. The atom is again initially excited. Linearity of the Schrödinger equation givesThe different photon numbers have different Rabi frequencies, so the state generally becomes entangled between atom and field. It cannot in general be replaced by one Rabi-oscillating atomic state multiplied by an unchanged coherent state.

For $\bar{n}=|z{|}^2\gg 1$, the Poisson distribution of photon number has width $\sqrt{\bar{n}}\ll \bar{n}$. Initially the atom therefore undergoes nearly classical Rabi oscillations at angular frequency approximately $2g\sqrt{\bar{n}}$. However, the atomic inversion is a sum over slightly different frequencies:The components dephase, causing collapse of the visible Rabi oscillations on a time scale of order $1/g$; a Gaussian photon-number approximation gives an initial envelope approximately $e^{-g^2{t}^2/2}$. Their discrete phases later rephase, giving revivals nearThe revival estimate follows by requiring the phase difference of neighbouring photon-number components to advance by $2\pi$. These Jaynes-Cummings collapse and revival effects arise from atom-field entanglement and photon-number discreteness. Large amplitude makes the initial classical description accurate over many Rabi periods, but does not eliminate quantum collapse and revival at all later times.

Four complementary classical features of a large-amplitude coherent state are:

These properties identify a particular classical oscillation, rather than a state of definite photon number. Absolute quantum fluctuations remain even when their relative size is small.

For one electromagnetic normal mode, $a$ and $a^{†}$ are the annihilation operator and creation operator, with commutator $[a,a^{†}]=I$. Their action on normalized number states isThey remove or add one photon, whose energy is $\hslash \omega$. The vacuum state $|0⟩$ is normalized and obeys $a|0⟩=0$. It has no photons, but retains zero-point energy $\hslash \omega /2$ for the quantum harmonic oscillator Hamiltonian $H=\hslash \omega (a^{†}a+1/2)$. The coherent state $|\alpha ⟩=D(\alpha )|0⟩$ is a displaced vacuum state.

Set $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$. Their commutator isIt commutes with both operators, so the stated Baker--Campbell--Hausdorff formula terminates:In particular $e^{-{\alpha }^{\ast }a}|0⟩=|0⟩$. Expanding the remaining exponential and using $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$ gives the number state expansionThe orthonormality of the number states givesThis is a Poisson distribution with parameter $\mu =|\alpha {|}^2$. Its probabilities sum to $e^{-\mu }{e}^{\mu }=1$. The phase of $\alpha$ affects the coherent state's relative probability amplitudes but not its photon number probabilities. At $\alpha =0$ the distribution is concentrated at $n=0$.

The number operator is $\hat{N}=a^{†}a$. The number state expansion gives $a|\alpha ⟩=\alpha |\alpha ⟩$ by shifting the summation index; taking the adjoint gives $⟨\alpha |a^{†}={\alpha }^{\ast }⟨\alpha |$. Thus the expectation value isFor the second moment, use $a{a}^{†}=a^{†}a+I$:Its expectation value in the coherent state is $⟨{\hat{N}}^2⟩=|\alpha {|}^4+|\alpha {|}^2$. ThereforeThe variance equals the mean, exactly as for the Poisson distribution found in b. For $\mu >0$ the fractional standard deviation is $1/\sqrt{\mu }$, the usual photon shot noise scaling. For the vacuum state, both the mean photon number and its variance vanish; this does not imply zero optical quadrature fluctuations. At $\mu =0$ the fractional fluctuation is undefined.

For the normalized coherent state, $⟨a⟩=\alpha$ and $⟨a^{†}⟩={\alpha }^{\ast }$. The dimensionless electric field and magnetic field optical quadratures consequently have expectation valuesThe physical electric field and magnetic field strengths have the corresponding proportionality factors and mode functions. In free evolution replace $\alpha$ by $\alpha e^{-i\omega t}$ to recover the classical oscillation.

Expanding the squared operators cancels the $a^2$ and $(a^{†}{)}^2$ terms:On the other hand, $(\alpha +{\alpha }^{\ast }{)}^2+[-i(\alpha -{\alpha }^{\ast }){]}^2=4|\alpha {|}^2$. HenceThe difference is the sum of the variances, not an extra mean field. Indeed $\mathrm{Var}(\mathcal{E})=\mathrm{Var}(\mathcal{B})=1$, independent of $\alpha$. The commutator $[\mathcal{E},\mathcal{B}]=2iI$ requires ${\sigma }_{\mathcal{E}}{\sigma }_{\mathcal{B}}\ge 1$, and the coherent state attains equality. The residual two units are the vacuum fluctuations of the two optical quadratures.