Coherent states and squeezing

Nothing requires a non-Hermitian operator to have eigenvectors, but nothing forbids it either. Put |α⟩ = Σ cₙ|n⟩ into a|α⟩ = α|α⟩ and use a|n⟩ = √n |n−1⟩: matching the coefficient of |n⟩ gives √(n+1) c_(n+1) = α cₙ, so cₙ = αⁿ c₀/√(n!). Normalisation Σ|cₙ|² = |c₀|² e^(|α|²) = 1 then fixes c₀ = e^{(−|α|²/2)}. Every complex α gives exactly one normalised solution, so the spectrum of a is the whole complex plane and each eigenvalue is non-degenerate. Now run the same argument for a†. Since a†|ψ⟩ = Σₙ cₙ √(n+1)|n+1⟩ has no |0⟩ component at all, the |0⟩ line of a†|β⟩ = β|β⟩ reads 0 = β c₀, the |1⟩ line then gives c₀ = β c₁, and the whole vector collapses to zero. a† has no normalisable eigenvector, and the asymmetry is the ladder's own: lowering terminates at |0⟩, raising never terminates.

The same state has a constructive definition. D(α) = exp(α a† − α* a) has an anti-Hermitian exponent, so it is unitary. Because [α a†, −α* a] = |α|² is a c-number, the Baker–Campbell–Hausdorff identity e^(A+B) = e^A e^B e^{(−[A, B]/2)} splits it into D(α) = e^{(−|α|²/2)} e^{(α a†)} e^{(−α* a)}. Act on the vacuum: the rightmost factor is the identity there because a|0⟩ = 0, and e^{(α a†)}|0⟩ = Σₙ (αⁿ/n!)(a†)ⁿ|0⟩ = Σₙ (αⁿ/√(n!))|n⟩, reproducing the Fock sum exactly. The operator statement is the useful one: [a, α a† − α* a] = α is a number, so the commutator series for D†(α) a D(α) terminates after one term at a + α. A displacement adds a constant to a. The means move to ⟨X⟩ = √2 Re α and ⟨P⟩ = √2 Im α; every variance, and every commutator, is untouched. Draw it and the vacuum disc has slid to a new centre without changing shape.

Project onto the Fock basis: P(n) = |⟨n|α⟩|² = e^(−|α|²)|α|⁽²ⁿ⁾/n!, a Poisson distribution of parameter |α|². The moments come from the eigenvalue property with no sum at all. ⟨n⟩ = ⟨α|a†a|α⟩ = |α|², and normal-ordering gives ⟨n²⟩ = ⟨α|a†(a†a + 1)a|α⟩ = |α|⁴ + |α|², so (Δn)² = |α|² and Δn = |α|. The Fano factor (Δn)²/⟨n⟩ is exactly 1. That is the definition of shot noise, and it is why a perfectly stabilised laser still has a noise floor. A 1 ms sample of a 1 mW beam at 633 nm carries n̄ = 3.2 × 10¹² photons, so Δn = 1.8 × 10⁶ and the fractional noise is 5.6 × 10⁻⁷ — small, but not removable by improving the source. A Fano factor below 1, sub-Poissonian light, is impossible for any classical field and demands a state outside this family.

Evolve with U(t) = e^(−iHt/ħ) = e^(−iωt/2) e^(−iωtN). On the Fock sum it multiplies the |n⟩ amplitude by e^(−inωt), and since that amplitude is αⁿ/√(n!), the entire effect is α → α e^(−iωt): U(t)|α⟩ = e^(−iωt/2)|α e^(−iωt)⟩. A coherent state stays coherent, and its label runs clockwise on a circle of radius |α|. Writing x = √(ħ/2mω)(a + a†) then gives ⟨x⟩(t) = 2√(ħ/2mω)|α| cos(ωt − φ) and ⟨p⟩(t) = −√(2mħω)|α| sin(ωt − φ): the classical orbit, exactly, with no spreading, because Δx = √(ħ/2mω) at every instant. Do not generalise this. Ehrenfest's equations close on the means only when the force is linear in x; add a quartic term and ⟨V′(x)⟩ ≠ V′(⟨x⟩), the phase-space disc shears into a crescent, and the state stops being coherent, on a time set by the size of the quartic term.

To get below the vacuum floor you need an operator quadratic in a, not linear. S(ξ) = exp[½(ξ* a² − ξ a†²)] with ξ = r e^(iθ) is again unitary, and its action is a Bogoliubov transformation, S†(ξ) a S(ξ) = a cosh r − a† e^(iθ) sinh r, mixing a with a† instead of adding a number to it. Take θ = 0 and the dimensionless quadratures rescale cleanly: X → X e^(−r), P → P e^(+r). The variances become e^(−2r)/2 and e^(+2r)/2, so their product stays at ¼ and the ellipse has exactly the vacuum disc's area. In decibels the squeezed quadrature sits 8.686 r below vacuum. The cost is real: S(ξ)|0⟩ carries ⟨n⟩ = sinh²r photons and only even Fock components, because a†² creates them in pairs. Ten decibels means r = 1.151 and 2.03 photons, and it takes a two-photon process — parametric down-conversion, or a spring constant modulated at 2ω — since no linear drive can produce it.

Two coherent states are never orthogonal. ⟨β|α⟩ = exp(−½|α|² − ½|β|² + β*α), so |⟨β|α⟩|² = e^{(−|α − β|²)}: at |α − β| = 2 that is e⁻⁴ = 1.8%, at 3 it is 1.2 × 10⁻⁴, and it never reaches zero. Yet the set still resolves the identity, (1/π)∫d²α |α⟩⟨α| = 1 with d²α = d(Re α)d(Im α). A linearly dependent family that resolves unity is overcomplete: any one |α⟩ can be expanded in the others, so expansion coefficients are not unique and phase space is described by quasi-distributions (Q, Wigner, P) rather than by amplitudes. What an experiment reads also depends on the phase it measures at. Homodyne detection measures the rotated quadrature X_φ = (a e^(−iφ) + a† e^(iφ))/√2, whose variance in a θ = 0 squeezed state is ½(e^(−2r)cos²φ + e^(2r)sin²φ): get the local-oscillator phase 5° wrong and 10 dB of squeezing reads as 7.6 dB.