Coherent state

$|\alpha ⟩=D(\alpha )|0⟩$ is normalized, satisfies $A|\alpha ⟩=\alpha |\alpha ⟩$, and evolves to a phase times $|\alpha e^{-i\omega t}⟩$.

A normalized nonnegative probability measure over coherent states produces a classical mixture. The measure may contain Dirac delta distributions, so a pure coherent state is included. Conditioning on $\alpha$ gives Poisson distribution photon statistics; hence $\mathrm{Var}n=⟨n⟩+\mathrm{Var}(|\alpha {|}^2)\ge ⟨n⟩$. Each optical quadrature has variance at least its vacuum value because mixing adds nonnegative variance of conditional means. These inequalities provide simple nonclassicality witnesses.

The diagonal coherent state representation writes a single-mode density matrix in the form shown. Its weight may be a generalized distribution rather than an ordinary nonnegative probability density. A nonclassical state of light has no nonnegative measure of this form. This use of $P$ differs from the Wigner distribution; a nonnegative Wigner function does not by itself certify classicality.

A normalized bosonic field coherent state with $\bar{N}=\int |\psi {|}^2$ has $⟨\hat{\Psi }(\mathbf{r})⟩=\psi (\mathbf{r})$. For a normal-ordered contact-interacting Hamiltonian, its energy expectation is $\int [{\hslash }^2|\nabla \psi {|}^2/(2m)+V|\psi {|}^2+g|\psi {|}^4/2]d^{3}r$. Minimizing at fixed mean particle number gives the stationary Gross–Pitaevskii equation. This mean-field state has number fluctuations and does not include the pair correlations responsible for quantum depletion.

The normalized coherent state has number amplitude $e^{-|\alpha {|}^2/2}{\alpha }^n/\sqrt{n!}$, hence the displayed Poisson law with mean and variance $|\alpha {|}^2$. A product of coherent modes has factorized number probabilities.

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.

Expand both states in the Fock basis and sum the exponential series to obtain the displayed overlap. In particular $⟨\alpha |-\alpha ⟩=e^{-2|\alpha {|}^2}$. Distinct coherent states are nonorthogonal at finite separation; dropping this overlap can give incorrect normalization and small-amplitude limits.

The displacement operator translates oscillator phase space. It is unitary and obeys $D(\alpha {)}^{†}aD(\alpha )=a+\alpha$ and $D(\alpha )|0⟩=|\alpha ⟩$, a coherent state. The Baker--Campbell--Hausdorff formula with a scalar commutator givesThe conjugation on ${\alpha }^{\ast }$ ensures that the exponent is anti-Hermitian for arbitrary complex displacement. For real $\alpha =\sqrt{m\omega /(2\hslash )}{x}_0$, this becomes the spatial translation $e^{-i{x}_{0}p/\hslash }$.

Suddenly shifting a quantum harmonic oscillator equilibrium from zero to a signed position $x_0$ leaves the old ground state unchanged at the instant of the quantum quench. The new annihilation operator is $a^{\prime }=a-\lambda$, with $\lambda =\sqrt{m\omega /(2\hslash )}{x}_0$, so $a^{\prime }|0⟩=-\lambda |0⟩$. In the translated energy eigenstates $|n^{\prime }⟩=e^{-ip{x}_0/\hslash }|n⟩$, the expansion isThe sign follows by translating the old vacuum by $-x_0$ before expanding it in the old Fock states. A Baker--Campbell--Hausdorff formula with central commutator separates the resulting exponential into creation and annihilation parts. Evolution with the shifted Hamiltonian preserves this coherent state, with $\alpha (t)={\alpha }_0{e}^{-i\omega t}$. Its mean occupation is $|{\alpha }_0{|}^2$ and its mean energy above the new potential minimum is $\hslash \omega (|{\alpha }_0{|}^2+1/2)$. Omitting $(-\lambda {)}^n$ in favour of $|\lambda {|}^n$ changes the state when $x_0>0$.

A normalized oscillator coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$, so its mean occupation number is $N=|\alpha {|}^2$. Under the quantum harmonic oscillator Hamiltonian operator $\hslash \omega (a^{†}a+1/2)$, its evolution is $|\alpha (t)⟩=e^{-i\omega t/2}|\alpha e^{-i\omega t}⟩$. Inserting $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ and $p=-i\sqrt{m\hslash \omega /2}(a-a^{†})$ givesThese mean values obey the classical harmonic oscillator equations. The displacement amplitude is $\sqrt{2\hslash N/(m\omega )}$, whose classical orbit has energy $\hslash \omega N$. The full quantum mean energy has the additional zero-point energy $\hslash \omega /2$. Indeed the variances are $\hslash /(2m\omega )$ and $m\hslash \omega /2$, and their kinetic and potential contributions sum to $\hslash \omega /2$.

For an unnormalized bosonic coherent state, the measure is $d^2\psi /\pi$ over the complex plane. Gaussian moments $\int d^2\psi e^{-|\psi {|}^2}{\psi }^n{\bar{\psi }}^m/\pi =n!{\delta }_{nm}$ recover the identity in the Fock state basis. For a fermionic coherent state, use independent Grassmann variables and an ordered Berezin integral with $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$. The coefficient of $-\bar{\psi }\psi$ in the weighted projector is the identity. Products of these one-mode measures give the finite-mode result.

An unnormalized bosonic coherent state is $|\psi ⟩=e^{\psi a^{†}}|0⟩$ for a complex label $\psi$. The canonical commutation relation gives $a|\psi ⟩=\psi |\psi ⟩$ and overlap $⟨\psi |{\psi }^{\prime }⟩=e^{\bar{\psi }{\psi }^{\prime }}$. Multiplication by $e^{-|\psi {|}^2/2}$ produces a normalized coherent state. The unnormalized form makes the coherent-state resolution of identity and thermal time slicing particularly simple.

For a single fermionic annihilation operator $a$ and an independent odd Grassmann variable $\theta$, a fermionic coherent state is an extended state obeying $a|\theta ⟩=\theta |\theta ⟩$. With $\{a,\theta \}=0$ one may take $|\theta ⟩=\exp (-\theta a^{†})|0⟩$. Its overlap is exponential in the Grassmann labels. A resolution of identity with a Berezin integral gives the time-slice construction of a fermionic path integral. These formal eigenstates use Grassmann-valued coefficients.

The ordinary trace of an even fermionic operator requires $-\psi$ in the bra of its fermionic coherent state kernel. In one mode the diagonal matrix elements give $⟨-\psi |A|\psi ⟩=A_{00}-\bar{\psi }\psi A_{11}$, whose Gaussian Berezin integral is $A_{00}+A_{11}$. An untwisted bra gives $A_{00}-A_{11}$, the supertrace. This distinction produces the antiperiodic coherent-state thermal boundary conditions.