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advanced32 min studyLesson 472 in Physics

Separate coherent amplitude from quadrature squeezing

Compare Poissonian coherent-state counts with anisotropic quantum noise and track vacuum admixture under loss.

Before you begin
On this page

The question for today

What you’ll understand

  • Derive coherent-state number probabilities from its ladder-operator eigenvalue.
  • Compute squeezing variances under the declared half-unit vacuum convention.
  • Explain why loss degrades squeezing and why quadrature squeezing does not always imply sub-Poisson counts.

A stable mean optical amplitude does not remove photon-count fluctuations. A coherent state combines a classical-looking mean field with irreducible vacuum quadrature noise and Poisson number statistics. A squeezed state changes that noise geometry by reducing one quadrature variance and increasing its conjugate. These are statements about different measurements. We will derive their normalizations and examine how ordinary optical loss mixes in vacuum, so a claim of reduced noise always names its quadrature, phase reference, and detection conditions.

A ladder eigenstate has a Poisson distribution

Define a coherent state as an eigenstate of the annihilation operator with complex eigenvalue alpha. Expand it in number states and apply the lowering relation. The resulting coefficient recurrence gives powers of alpha divided by square root of factorial; normalization supplies an exponential prefactor. Squaring the coefficients yields a Poisson photon-number distribution whose mean and variance are both the squared magnitude of alpha. A large amplitude makes relative number fluctuations small, not absolute fluctuations zero.

The real and imaginary parts of alpha determine the two quadrature means, while their variances remain one half in our convention. Different coherent amplitudes are not mutually orthogonal, although their overlap becomes small when separated widely in phase space. A number state and a coherent state with the same mean energy can therefore have very different count distributions and phase-sensitive measurement statistics. Describing both only by their mean photon number discards the feature that distinguishes them.

a^∣α⟩=α∣α⟩,∣α⟩=e−∣α∣2/2∑n=0∞αnn!∣n⟩,P(n)=e−∣α∣2∣α∣2nn!\hat a|\alpha\rangle=\alpha|\alpha\rangle,\qquad |\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle,\qquad P(n)=e^{-|\alpha|^2}\frac{|\alpha|^{2n}}{n!}
α and n are dimensionless. A coherent state has ⟨n⟩=Var(n)=|α|² and quadrature means √2 Reα and √2 Imα, with variance 1/2 in each.↔ Scroll wide equations horizontally.

Squeezing preserves a commutator, not both noise levels

Choose a real squeezing parameter and an orientation so the X quadrature is compressed. The canonical transformation scales X by an exponential and P by its reciprocal, preserving their commutator. Applied to vacuum, it produces one variance below one half and the other above it. Their product remains one quarter for a pure minimum-uncertainty squeezed vacuum. Reducing both below the same vacuum reference would contradict the declared canonical commutation relation.

Squeezed vacuum has zero coherent amplitude but a nonzero mean photon population. Its photon-number fluctuations are not generically sub-Poissonian; quadrature squeezing and reduced number variance are distinct properties. A displaced squeezed state can exhibit amplitude squeezing and reduced count noise for appropriate displacement and squeezing phase, but the phase relationship matters. The reference phase in balanced homodyne detection selects which quadrature is observed, so an unqualified statement that the beam is less noisy is incomplete.

S^(r)=er2(a^2−a^†2),VX=12e−2r,VP=12e2r,VXVP=14,nˉsq=sinh⁡2r\hat S(r)=e^{\frac r2(\hat a^2-\hat a^{\dagger2})},\qquad V_X=\frac12e^{-2r},\qquad V_P=\frac12e^{2r},\qquad V_XV_P=\frac14,\qquad \bar n_{\mathrm{sq}}=\sinh^2r
r is real and dimensionless, S acts on vacuum, and squeezing axes are chosen as X and P. The variances use [X,P]=i and vacuum variance 1/2; displaced or rotated states need corresponding mean and axis transformations.↔ Scroll wide equations horizontally.

Loss adds a vacuum contribution

Model a passive attenuator as a beam splitter coupling the selected mode to an independent vacuum mode. The output quadrature is the square-root transmission times the input quadrature plus the complementary vacuum contribution. Variances add because the two inputs are uncorrelated. A vacuum input remains vacuum; a squeezed variance moves toward one half under loss. This statement includes the quantum noise needed to preserve the output commutator, which a deterministic rescaling alone would lose.

Rotating a homodyne phase selects a different linear combination of X and P, so a squeezed state can appear quiet or noisy depending on that phase. The vector lab can isolate how projection onto a chosen phase axis changes a coherent mean while its magnitude stays fixed. Its arrow has no quantum noise ellipse. Compute variances and loss separately from the displayed geometry, and do not interpret a rotating mean projection as generation or measurement of squeezing.

a^out=η a^+1−η v^,Vout=ηVin+(1−η)12,0≤η≤1\hat a_{\mathrm{out}}=\sqrt\eta\,\hat a+\sqrt{1-\eta}\,\hat v,\qquad V_{\mathrm{out}}=\eta V_{\mathrm{in}}+(1-\eta)\frac12,\qquad 0\le\eta\le1
η is power transmission, v an independent vacuum mode, and V a quadrature variance in dimensionless units. This is a passive linear-loss channel with no thermal background, technical noise, or detector inefficiency beyond the modeled transmission.↔ Scroll wide equations horizontally.

Work it through

Compute a squeezed-vacuum noise budget

A pure squeezed vacuum has r=ln2 with its squeezed axis along X. Find both variances, their product, and mean photon number. Then pass it through a loss channel with power transmission η=1/2.

  1. Exponentiating gives e^(2r)=4 and e^(−2r)=1/4. Thus VX=1/8 and VP=2; their product is 1/4, the minimum in the stated convention.

  2. The hyperbolic sine is (2−1/2)/2=3/4, so mean photon number is 9/16. A zero mean field has not made this the zero-photon vacuum.

  3. At half transmission, the X variance is (1/2)(1/8)+(1/2)(1/2)=5/16. The P variance is (1/2)(2)+1/4=5/4. Their product is 25/64, larger than the pure-state minimum because the attenuated state is mixed.

The result

Initially VX=1/8, VP=2 and mean occupation=9/16. Half transmission gives variances 5/16 and 5/4; squeezing survives but its purity and noise advantage decrease.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Project a coherent mean onto a phase axis

For fresh comparisons Reset and set BOTH angle 0°, magnitude 2; then Reset and set angle 90°, magnitude 2. Predict horizontal mean projections 2 and 0. The vector model shows only a coherent-mean analogy; it has no uncertainty ellipse, bosonic state, or squeezing readout.

Project a coherent mean onto a phase axisPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSelecting a phase-sensitive projection is different from changing quantum variance.

Pause & reason

Check your understanding

Which statement is true for a pure squeezed vacuum with nonzero r?

Your turn

Try it yourself

A coherent state has α=1+i. Find its mean number, number variance, quadrature means, and probability of zero photons. Would a binomial lab with p=1/2 and n=4 reproduce its full count distribution?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use |α|² and the Poisson zero-count probability; compare the supports of the two distributions.

Compare with the solution

Mean number and variance are both 2; both quadrature means are √2. The zero-count probability is e⁻²≈0.1353. A four-trial binomial has the same mean but variance 1 and cannot produce more than four counts. It therefore does not reproduce the coherent state’s Poisson distribution, whose support extends to every nonnegative integer.

Keep the idea

Take it with you

  • A coherent state has Poisson photon counts and vacuum-level quadrature variance.
  • Squeezing redistributes conjugate noise while preserving the commutator.
  • Passive loss adds vacuum noise and reduces observable squeezing.

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