Quadrature Operators and the Uncertainty Relation

Quantizing a single mode of the electromagnetic field hands you the machinery of the harmonic oscillator: a lowering operator $\hat a$ and a raising operator $\hat a^\dagger$ with $[\hat a,\hat a^\dagger]=1$ , inherited here from aq-optics-field-quantization. Those ladder operators count photons perfectly, but they are not Hermitian, so no detector can read either one. The electric field is real and measurable, and this article asks the follow-up: which Hermitian combinations of $\hat a$ and $\hat a^\dagger$ are the field, and what does quantum mechanics forbid about them? The answer is the two quadrature operators, $\hat X_1=\tfrac12(\hat a+\hat a^\dagger)$ and $\hat X_2=\tfrac1{2i}(\hat a-\hat a^\dagger)$ — the real and imaginary parts of the amplitude, the components of the oscillation in phase and ninety degrees out of phase with a reference cosine. They are the optical twins of a mechanical oscillator's position and momentum, and inherit the same fatal feature: they do not commute, $[\hat X_1,\hat X_2]=i/2$ . That single limit is the seed crystal of quantum optics, forcing $\Delta X_1\,\Delta X_2\ge\tfrac14$ so that no state of light — not even the photon-free vacuum — can hold both field components perfectly still. The residual jitter $\Delta X_1=\Delta X_2=\tfrac12$ is the zero-point noise that beats an interferometer, drives spontaneous emission, and sets the standard quantum limit. The whole cast of quantum states is organized by how it distributes this unavoidable noise between the two quadratures, and because balanced homodyne detection reads a chosen quadrature directly, they are the working currency of the laboratory.

Key takeaways

The quadratures $\hat X_1,\hat X_2$ are the Hermitian real and imaginary halves of $\hat a$ — the dimensionless "position" and "momentum" of a field mode, and the in-phase and out-of-phase amplitudes of the measurable field.
They satisfy the canonical commutator $[\hat X_1,\hat X_2]=\tfrac{i}{2}$ , which forces the uncertainty relation $\Delta X_1\,\Delta X_2\ge\tfrac14$ — the floor on simultaneous knowledge of the two field components.
The vacuum and coherent states saturate it isotropically, $\Delta X_1=\Delta X_2=\tfrac12$ — the standard quantum limit.
Squeezed states saturate it anisotropically, $\Delta X_1=\tfrac12e^{-r}$ and $\Delta X_2=\tfrac12e^{+r}$ — one quadrature is pushed below the vacuum at the exact cost of the other, keeping the product at $\tfrac14$ .
Homodyne detection measures the rotated quadrature $\hat X_\theta$ set by the local-oscillator phase — what an experiment actually records.
Key idea: split the non-Hermitian annihilation operator into two Hermitian, measurable halves — the in-phase and out-of-phase field amplitudes. Their canonical commutator $[\hat X_1,\hat X_2]=i/2$ is the origin of all optical quantum noise: it makes the vacuum fluctuate and sets the floor $\Delta X_1\Delta X_2\ge1/4$ .

Where this lives today

Their most visible home is gravitational-wave astronomy. Advanced LIGO and Virgo read the phase quadrature of light returning from a four-kilometre arm, where the noise floor is the vacuum $\Delta X=\tfrac12$ Caves identified. Since 2019 both observatories inject squeezed vacuum into the dark port, cutting the noise by several decibels and, because detection rate scales as the cube of range, enlarging the volume of universe searched — extra black-hole mergers every run, bought with nonlinear crystals that reshape the vacuum noise ellipse. The same operators run quantum-state tomography: sweeping $\theta$ records $\hat X_\theta$ at every angle, and an inverse Radon transform reconstructs the Wigner function — how experimentalists photograph a squeezed or cat state. And in continuous-variable quantum information, encoding directly in $\hat X_1,\hat X_2$ enables deterministic teleportation and cluster-state computing, with two-mode squeezing supplying the EPR resource $\Delta(\hat X_1^A-\hat X_1^B)\to0$ .

The physical picture

Freeze one mode and picture its complex amplitude as a phasor turning at frequency $\omega$ . The horizontal axis is $\hat X_1$ , the cosine amplitude; the vertical axis is $\hat X_2$ , the sine amplitude. A classical field is a sharp arrow of definite length and phase; a quantum field cannot be, because its tip may not sit at a precise point in both directions at once. Every state is instead a fuzzy blob, and the whole subject is its shape, size, and position. The area of the blob has a floor. Squash it thin in one direction and it must bulge in the perpendicular one to keep the area above the commutator's limit — Heisenberg's principle drawn in phase space. Push the noise below the vacuum in the quadrature you care about (squeezing) and you pay in the conjugate one you ignore. The first figure maps this: the vacuum disk, the same disk shoved out to a bright amplitude (a coherent state), and the thinned ellipse of a squeezed state.

Figure

Quadrature phase space X₁ versus X₂ every state of light is a noise blob of area at least π/4 the vacuum is a disk of radius 1/2 a coherent state is that disk pushed out to amplitude α and a squeezed state is an ellipse thinned to ΔX₁ = 1/4 at the cost of ΔX₂ = 1

Now let the phasor turn. A detector sees the blob projected onto the direction the field currently points, traced out in time. For a coherent state the blob is a symmetric disk, so this projected noise has the constant width $\Delta X=\tfrac12$ at every phase — a uniform band hugging a sine wave. For a squeezed state the blob is an ellipse, so as the reference angle sweeps the noise breathes: it thins to $\Delta X=\tfrac14$ on the squeezed axis and swells to $\Delta X=1$ a quarter cycle later. The second figure shows this directly, and choosing when in the cycle to read the field is what homodyne detection does with the local-oscillator phase.

Figure

The measured field E(t) ∝ X₁ cos ωt + X₂ sin ωt carries a noise band for a coherent state its width is the constant vacuum value ΔX = 1/2 at every phase for a phase-squeezed state it breathes thinning to ΔX = 1/4 where the field is read and swelling to ΔX = 1 a quarter cycle away

The core equations

The definition is the whole move. Split the non-Hermitian $\hat a$ into its Hermitian real and imaginary parts, each separately observable and together carrying exactly the information $\hat a$ did.

\hat X_1=\tfrac12\left(\hat a+\hat a^\dagger\right),\qquad \hat X_2=\tfrac{1}{2i}\left(\hat a-\hat a^\dagger\right) \qquad (1)

where: $\hat a,\hat a^\dagger$ — annihilation and creation operators, $[\hat a,\hat a^\dagger]=1$ ; $\hat X_1$ — in-phase (amplitude) quadrature, Hermitian; $\hat X_2$ — out-of-phase (phase) quadrature, Hermitian. Physically: $\hat X_1$ is how much cosine and $\hat X_2$ how much sine the field carries, and inverting gives $\hat a=\hat X_1+i\hat X_2$ . Because $\hat a$ and $\hat a^\dagger$ do not commute, neither do the halves. The commutator turns out to be a pure number — no operators survive — which makes the resulting uncertainty state-independent.

[\hat X_1,\hat X_2]=\tfrac{i}{2} \qquad (2)

where: the right side is a $c$ -number, so $\langle[\hat X_1,\hat X_2]\rangle=\tfrac{i}{2}$ in every state. Physically: this is the optical $[\hat x,\hat p]=i\hbar$ , the quadratures playing position and momentum for the mode. Feed that commutator into the general Robertson inequality and the floor on the noise product drops out. This is the central inequality of the whole subject, quoted here and derived below.

\Delta X_1\,\Delta X_2\ge\tfrac14 \qquad (3)

where: $\Delta X_i=\sqrt{\langle\hat X_i^2\rangle-\langle\hat X_i\rangle^2}$ — standard deviation of quadrature $i$ . Physically: sharpen one field component and the conjugate one blurs; the product never dips below one-quarter, and states hitting equality are minimum-uncertainty. The names "in-phase" and "out-of-phase" are literal. Rebuilding the field from the mode operators shows $\hat X_1$ and $\hat X_2$ are exactly its cosine and sine amplitudes.

\hat E(t)=E_0\left(\hat X_1\cos\omega t+\hat X_2\sin\omega t\right) \qquad (4)

where: $E_0=\sqrt{\hbar\omega/\varepsilon_0 V}$ — single-photon field amplitude; $\omega$ — mode frequency; $V$ — quantization volume. Physically: the detector reads $\hat X_1$ at the cosine and $\hat X_2$ a quarter-period later — the field sampled 90° apart. A homodyne experiment need not sample at the cosine or the sine. The local oscillator can pick any reference angle $\theta$ , and the measured observable is the rotated quadrature interpolating between $\hat X_1$ and $\hat X_2$ .

\hat X_\theta=\tfrac12\left(\hat a\,e^{-i\theta}+\hat a^\dagger e^{+i\theta}\right)=\hat X_1\cos\theta+\hat X_2\sin\theta \qquad (5)

where: $\theta$ — local-oscillator phase selecting the measured component. Physically: $\theta=0$ reads $\hat X_1$ , $\theta=\pi/2$ reads $\hat X_2$ ; sweeping $\theta$ projects the blob onto every direction — the raw data of tomography. Finally, the quadratures rebuild the photon-number operator. Doing so exposes the zero-point offset that never goes away — the bridge back to energy and the vacuum.

\hat n=\hat a^\dagger\hat a=\hat X_1^2+\hat X_2^2-\tfrac12 \qquad (6)

where: $\hat n$ — photon-number operator; the $-\tfrac12$ is the zero-point term that puts $\tfrac12\hbar\omega$ into $\hat H=\hbar\omega(\hat n+\tfrac12)$ . Physically: the squared radius of the blob is photons plus one half, so even $n=0$ has $\langle\hat X_1^2+\hat X_2^2\rangle=\tfrac12$ — a nonzero blob at the origin.

Derivation AThe uncertainty relation from the canonical commutator

The goal is to earn both equation (2) and equation (3) from one fact, $[\hat a,\hat a^\dagger]=1$ , and see why the bound holds for every state of light rather than any special one. We compute the quadrature commutator from the definitions, feed it into the Robertson inequality — the general variance bound for any two Hermitian observables — and read off $\tfrac14$ . The payoff is state-independence: the commutator collapses to a pure number, so no preparation can dodge the floor.

Step 1. Substitute the definitions into the commutator: $[\hat X_1,\hat X_2]=\big[\tfrac12(\hat a+\hat a^\dagger),\,\tfrac1{2i}(\hat a-\hat a^\dagger)\big]=\tfrac1{4i}\big([\hat a,-\hat a^\dagger]+[\hat a^\dagger,\hat a]\big)$ . Why: commutators are bilinear; the like-operator brackets $[\hat a,\hat a]$ and $[\hat a^\dagger,\hat a^\dagger]$ vanish, leaving only the two cross terms.

Step 2. Evaluate the cross terms with $[\hat a,\hat a^\dagger]=1$ : since $[\hat a,-\hat a^\dagger]=-1$ and $[\hat a^\dagger,\hat a]=-1$ , we get $[\hat X_1,\hat X_2]=\tfrac1{4i}(-1-1)=\tfrac{-2}{4i}=\tfrac{i}{2}$ . Why: this is equation (2), inheriting its value from the single bosonic relation — rescale the quadratures and this number changes, the root of the convention traps below.

Step 3. Invoke the Robertson inequality for Hermitian $\hat A,\hat B$ : $\Delta A\,\Delta B\ge\tfrac12|\langle[\hat A,\hat B]\rangle|$ , proven by applying the Cauchy–Schwarz inequality to the vectors $(\hat A-\langle\hat A\rangle)|\psi\rangle$ and $(\hat B-\langle\hat B\rangle)|\psi\rangle$ . Why: this is the rigorous general form of Heisenberg's principle, converting a commutator into a bound on measurable spreads.

Step 4. Insert the quadrature commutator: because $\langle[\hat X_1,\hat X_2]\rangle=\tfrac{i}{2}$ regardless of the state, $\Delta X_1\,\Delta X_2\ge\tfrac12\cdot\tfrac12=\tfrac14$ . Why: the expectation of a $c$ -number is that number, so no state can lower the bound — equality requires a Gaussian with no $\hat X_1$ – $\hat X_2$ correlation, met by the vacuum, coherent, and axis-aligned squeezed states.

Go deeper · Derivation B · 1 minVacuum fluctuations and the number-operator identity

Derivation BVacuum fluctuations and the number-operator identity

The aim is to compute the zero-point noise, $\Delta X_1=\Delta X_2=\tfrac12$ , and prove the identity $\hat X_1^2+\hat X_2^2=\hat n+\tfrac12$ tying the quadratures back to photon number and energy. These are the quantitative heart of "the vacuum is not empty": with no photons the field still carries a definite, calculable jitter — the standard quantum limit every squeezing experiment is scored against. We use only $\hat a|0\rangle=0$ and $[\hat a,\hat a^\dagger]=1$ .

Step 2. Expand the square: $\langle0|\hat X_1^2|0\rangle=\tfrac14\langle0|(\hat a+\hat a^\dagger)^2|0\rangle=\tfrac14\langle0|(\hat a^2+\hat a\hat a^\dagger+\hat a^\dagger\hat a+\hat a^{\dagger2})|0\rangle$ . Why: of the four terms only $\hat a\hat a^\dagger$ returns the state to $|0\rangle$ ; the rest annihilate or leave it orthogonal.

Step 3. Evaluate the surviving term: $\langle0|\hat a\hat a^\dagger|0\rangle=\langle0|(\hat a^\dagger\hat a+1)|0\rangle=1$ , so $\Delta X_1^2=\tfrac14\cdot1=\tfrac14$ and $\Delta X_1=\tfrac12$ . Why: symmetry gives the same for $\hat X_2$ , so the vacuum is an isotropic disk of radius $\tfrac12$ with saturating product $\tfrac14$ .

Step 4. Assemble the number identity: $\hat X_1^2+\hat X_2^2=\tfrac14(\hat a+\hat a^\dagger)^2-\tfrac14(\hat a-\hat a^\dagger)^2=\tfrac12(\hat a\hat a^\dagger+\hat a^\dagger\hat a)=\hat a^\dagger\hat a+\tfrac12$ , hence $\hat n=\hat X_1^2+\hat X_2^2-\tfrac12$ . Why: the cross terms cancel and the leftover $\tfrac12$ is the zero-point energy in disguise.

Go deeper · Derivation C · 1 minWhy a Fock state has isotropic but super-minimal noise

Derivation CWhy a Fock state has isotropic but super-minimal noise

This explains a result that surprises most students: the number state $|n\rangle$ has perfectly sharp photon number yet is nowhere near minimum-uncertainty, with $\Delta X_1\Delta X_2=\tfrac14(2n+1)$ growing without bound. Sharp photon number means completely random phase, so $|n\rangle$ is not a blob but a ring, sharing its noise equally between the quadratures. We use the number identity and the rotational symmetry of $|n\rangle$ .

Step 3. Split it by symmetry: $|n\rangle$ is invariant under phase-space rotations (its Wigner function is a ring), so the total shares equally, $\langle\hat X_1^2\rangle=\langle\hat X_2^2\rangle=\tfrac12(n+\tfrac12)=\tfrac14(2n+1)$ . Why: definite photon number has no preferred quadrature, forcing isotropy $\Delta X_1=\Delta X_2=\tfrac12\sqrt{2n+1}$ .

Step 4. Form the product: $\Delta X_1\Delta X_2=\tfrac14(2n+1)\ge\tfrac14$ , with equality only at $n=0$ . Why: for $n\ge1$ it sits far above the floor despite sharp photon number — exact energy costs all knowledge of phase, spreading both components.

Limiting cases

Case	Result	Meaning
Vacuum / coherent state	$\Delta X_1=\Delta X_2=\tfrac12,\ \ \Delta X_1\Delta X_2=\tfrac14$	Minimum-uncertainty, isotropic disk. The standard quantum limit — the yardstick for all optical noise.
Squeezed vacuum	$\Delta X_1=\tfrac12 e^{-r},\ \ \Delta X_2=\tfrac12 e^{+r}$	Product still $\tfrac14$ ; one quadrature beats the vacuum by $e^{-r}$ at the cost of the other. $r$ is the squeeze parameter.
Number (Fock) state $\vert n\rangle$	$\Delta X_1=\Delta X_2=\tfrac12\sqrt{2n+1}$	Isotropic but not minimum-uncertainty for $n\ge1$ ; product $\tfrac14(2n+1)$ . Random phase spreads the blob into a ring.
Thermal state, mean $\bar n$	$\Delta X_1=\Delta X_2=\tfrac12\sqrt{2\bar n+1}$	Same isotropic form as Fock but a Gaussian blob; classical thermal noise piles on top of the zero-point floor.

Common trapThe number 1/2 for vacuum noise is a law of nature you can quote across textbooks

It is not — it is a convention, and mixing conventions is the commonest way to get squeezing numbers wrong. Here $\hat X_1=\tfrac12(\hat a+\hat a^\dagger)$ gives $[\hat X_1,\hat X_2]=i/2$ , vacuum noise $\tfrac12$ , and bound $\tfrac14$ . Other books set $\hat X_1=\tfrac1{\sqrt2}(\hat a+\hat a^\dagger)$ , giving $[\hat X_1,\hat X_2]=i$ , vacuum noise $\tfrac1{\sqrt2}$ , and bound $\tfrac12$ ; still others normalize the vacuum variance to $1$ . The physics is identical but the bare numbers are not, so before comparing an author's "shot-noise limit" check whether their vacuum sits at $\tfrac12$ , $\tfrac1{\sqrt2}$ , or $1$ . Decibels are convention-proof; absolute variances are not.

Worked examples

1 · How many decibels of squeezing, and what does it cost? (the real number)

A gravitational-wave detector injects squeezed vacuum and observes 6 dB of noise reduction in the measured quadrature relative to the vacuum. Find the squeeze parameter $r$ , the two quadrature standard deviations, verify the uncertainty product, and state the penalty paid in the conjugate quadrature.

Step 1. Relate decibels to $r$ . Squeezing is quoted as a noise-power ratio to the vacuum, and the squeezed variance $\Delta X_1^2=(\tfrac12 e^{-r})^2=\tfrac14 e^{-2r}$ lies a factor $e^{-2r}$ below the vacuum $\tfrac14$ . In decibels, $\mathrm{dB}=-10\log_{10}(e^{-2r})=8.686\,r$ , so $r=6/8.686=\mathbf{0.691}$ .

Step 2. The squeezed (measured) quadrature: $\Delta X_1=\tfrac12 e^{-0.691}=\mathbf{0.251}$ , versus the vacuum $\tfrac12$ . Its variance is a factor $0.251$ of the vacuum's, i.e. $-10\log_{10}(0.251)=6.0$ dB — the check closes.

Step 3. The conjugate quadrature and product: $\Delta X_2=\tfrac12 e^{+r}=\tfrac12(1.996)=\mathbf{0.998}$ , essentially doubled — $+6$ dB of excess noise — while $\Delta X_1\Delta X_2=0.251\times0.998=\mathbf{0.250}=\tfrac14$ , still minimum-uncertainty. The lesson: 6 dB gained where you look is paid with 6 dB lost where you do not, so squeezing only helps an experiment that reads one well-chosen quadrature. r = 0.69, ΔX₁ = 0.25, ΔX₂ = 1.00, product = 1/4.

2 · The shot-noise floor of a bright coherent beam

A laser is prepared in a coherent state $|\alpha\rangle$ with mean photon number $\bar n=|\alpha|^2=100$ and real amplitude, so $\alpha=10$ . Find the mean and noise of the amplitude quadrature, the signal-to-noise ratio of a quadrature measurement, and the relative amplitude noise.

Step 1. Mean amplitude quadrature: for $|\alpha\rangle$ , $\langle\hat X_1\rangle=\mathrm{Re}\,\alpha=\mathbf{10}$ , $\langle\hat X_2\rangle=0$ . A coherent state is the vacuum blob displaced to $\alpha$ , so its noise is unchanged: $\Delta X_1=\Delta X_2=\tfrac12$ .

Step 2. Signal-to-noise ratio of reading $\hat X_1$ : $\mathrm{SNR}=\langle\hat X_1\rangle/\Delta X_1=10/0.5=\mathbf{20}$ . The measured field amplitude stands twenty vacuum-noise widths above the fuzz.

Step 3. Relative amplitude noise. The mean scales as $|\alpha|=\sqrt{\bar n}$ while the noise stays fixed at $\tfrac12$ , so the fractional noise is $\Delta X_1/\langle\hat X_1\rangle=0.5/10=\mathbf{0.05}=1/(2\sqrt{\bar n})$ and the intensity shot noise scales as $1/\sqrt{\bar n}$ . Brighter beams are quieter relatively — the shot-noise limit, which squeezing sidesteps by shrinking the $\tfrac12$ rather than growing $\bar n$ . ⟨X₁⟩ = 10, ΔX₁ = 1/2, SNR = 20, relative noise = 0.05.

3 · Is a Fock state a minimum-uncertainty state?

Take the number state $|3\rangle$ , which has perfectly known photon number. Compute its quadrature standard deviations and uncertainty product, and compare with the vacuum.

Step 1. Isotropy gives $\Delta X_1=\Delta X_2=\tfrac12\sqrt{2n+1}$ ; with $n=3$ , $\Delta X_1=\Delta X_2=\tfrac{\sqrt7}{2}=\mathbf{1.323}$ — over two and a half times the vacuum's $\tfrac12$ .

Step 2. Product: $\Delta X_1\Delta X_2=\tfrac14(2\cdot3+1)=\tfrac74=\mathbf{1.75}$ , seven times the floor $\tfrac14$ . So despite zero uncertainty in photon number, $|3\rangle$ sits far above the bound — emphatically not minimum-uncertainty. ΔX₁ = ΔX₂ = √7/2 ≈ 1.32, product = 7/4 = 1.75, seven times the vacuum floor.

Physics of the portrait

Origins · 1 minCarlton Caves and how this was discovered

Carlton Caves

b. 1950 · quantum limits of optical measurement and the birth of squeezed-light interferometry

Carlton Morley Caves is the theorist who turned the quadrature-uncertainty relation into a design principle for the world's most sensitive instruments. Trained under Kip Thorne at Caltech in the 1970s, he worked on the quantum limits of gravitational-wave antennas when it was unclear whether those limits were fundamental or merely technical. His 1980 and 1981 papers answered it unexpectedly. Analyzing a Michelson interferometer, he showed the limiting noise enters not through the bright laser port but through the empty port — the vacuum fluctuations of the quadratures, leaking in and beating the signal, so the standard quantum limit is set by the vacuum's $\Delta X_1\Delta X_2=\tfrac14$ . The consequence followed: replace that dark-port vacuum with squeezed vacuum, its phase-quadrature noise below $\tfrac12$ , and the interferometer beats the naive limit without breaking the uncertainty relation. Published years before squeezed light existed, this is the charter of every squeezing-enhanced detector running today.

History

1927

— Heisenberg formulates the uncertainty principle for position and momentum

1929

— Robertson casts the principle as a bound set by the commutator of any two observables

1976

— Yuen introduces the two-photon coherent (squeezed) states that beat symmetric noise

1981

— Caves shows squeezed vacuum in the dark port beats the interferometer standard quantum limit

1985

— Slusher and co-workers make the first observation of squeezed light by four-wave mixing

2019

— Advanced LIGO and Virgo run with routine squeezed-vacuum injection

Going further

Texts. Loudon, The Quantum Theory of Light, 3rd ed., Ch. 4–5 (quadratures, fluctuations, squeezing) · Gerry and Knight, Introductory Quantum Optics, Ch. 2–3 (ladder operators, coherent and squeezed states) · Walls and Milburn, Quantum Optics, Ch. 2 and Ch. 7 (squeezing, homodyne detection) · Saleh and Teich, Fundamentals of Photonics, 2nd ed., Ch. 12 · Scully and Zubairy, Quantum Optics, Ch. 2 · Caves, Phys. Rev. D 23, 1693 (1981), the founding squeezed-interferometer paper.

Provenance. The convention here is $\hat X_1=\tfrac12(\hat a+\hat a^\dagger)$ , giving $[\hat X_1,\hat X_2]=i/2$ , vacuum noise $\tfrac12$ , and bound $\tfrac14$ , as in Loudon (3rd ed.) and Gerry and Knight. A verifier should recompute: (1) the commutator $i/2$ from $[\hat a,\hat a^\dagger]=1$ ; (2) the vacuum variance $\langle0|\hat X_1^2|0\rangle=\tfrac14$ via $\langle0|\hat a\hat a^\dagger|0\rangle=1$ ; (3) the identity $\hat X_1^2+\hat X_2^2=\hat n+\tfrac12$ ; (4) the Fock variance $\tfrac14(2n+1)$ , so $|3\rangle$ gives product $\tfrac74$ ; (5) the conversion $\mathrm{dB}=8.686\,r$ , so 6 dB gives $r=0.69$ , $\Delta X_1=0.25$ , $\Delta X_2=1.00$ . Squeezed values $\tfrac12 e^{\mp r}$ follow Walls and Milburn Ch. 7; check the normalization against Saleh and Teich Ch. 12 before comparing absolute variances.

Prerequisites. Quantization of the Electromagnetic Field · Fock States and the Photon Number Basis · Vacuum State and Zero-Point Energy. Leads to. Coherent States · Squeezed States of Light · Phase-Space Quasiprobability Distributions · Photon Statistics and the Mandel Q Parameter.

Ask the tutor: Ask me to squeeze a coherent state step by step and watch its noise ellipse deform while the product $\Delta X_1\Delta X_2$ stays pinned at exactly one-quarter.

Are the quadratures really just position and momentum for light?

Mathematically, yes — that is the whole point. A field mode is a harmonic oscillator, and $\hat X_1,\hat X_2$ play $\hat x$ and $\hat p$ , with $[\hat X_1,\hat X_2]=i/2$ for $[\hat x,\hat p]=i\hbar$ . But they are not the position or momentum of anything material — there is no little mass sloshing back and forth. They are the two amplitudes of the electromagnetic oscillation itself: how much cosine and how much sine the field carries. The analogy is exact enough that every oscillator result carries straight over, which is why quantum optics inherited a finished formalism the day it was born.

Why is the vacuum noise a problem if there are no photons?

Because "no photons" does not mean "no field". The identity $\hat n=\hat X_1^2+\hat X_2^2-\tfrac12$ says the mean-square field in the vacuum is $\tfrac12$ , not zero — the blob sits at the origin but has finite size. Any measurement reading a quadrature sees a jitter $\Delta X=\tfrac12$ even with the laser off, and in an interferometer this leaks in through the dark port and beats the signal. That is the standard quantum limit — not a flaw a better detector removes but a consequence of the commutator, escaped only by reshaping the vacuum into a squeezed state.

What is the difference between the vacuum, a coherent state, and a squeezed state?

All three are Gaussian blobs; the vacuum and coherent states are both minimum-uncertainty. The vacuum is a symmetric disk of radius $\tfrac12$ at the origin. A coherent state is that same disk displaced to amplitude $\alpha$ — identical noise, nonzero mean field, the "most classical" light a laser makes. A squeezed state deforms the disk into an ellipse, pushing the noise below $\tfrac12$ in one quadrature and above it in the other while keeping the area minimal. Squeezed light is genuinely non-classical and the only way to read one field component more quietly than the vacuum permits.

How does homodyne detection actually measure a quadrature?

You overlap the signal with a strong local oscillator of phase $\theta$ on a 50:50 beam splitter and detect both output ports. Subtracting the photocurrents cancels the oscillator's own intensity and leaves a difference current proportional to $\langle\hat X_\theta\rangle$ , with fluctuations $\Delta X_\theta$ . The oscillator acts as phase reference and amplifier, lifting the tiny quadrature signal into a macroscopic current. Turning $\theta$ from $0$ to $\pi/2$ swings the observable from $\hat X_1$ to $\hat X_2$ , and a full sweep traces the noise ellipse — the workhorse of quantum-state tomography.

Does squeezing let you violate the Heisenberg uncertainty principle?

No — it saturates the relation, it never breaks it. A squeezed state keeps the product $\Delta X_1\Delta X_2$ pinned at exactly $\tfrac14$ , the minimum allowed. Squeezing merely redistributes the unavoidable noise: it moves fluctuations out of the quadrature you measure and into the one you ignore, so you gain sensitivity in $\hat X_1$ only by accepting more in $\hat X_2$ . If the experiment reads a single quadrature — an interferometer, a phase measurement — this is a pure win; if you needed both equally, squeezing buys nothing net.

Why does a Fock state have such large quadrature noise if its photon number is exact?

Because photon number and phase are conjugate, just as the two quadratures are. A state of definite photon number has completely undefined phase, so it is not a localized blob but a ring of radius $\sqrt{n+\tfrac12}$ smeared over all angles. Projecting that ring onto either axis gives a broad spread, so $\Delta X_1=\Delta X_2=\tfrac12\sqrt{2n+1}$ grows with $n$ and the product $\tfrac14(2n+1)$ climbs far above the floor. Knowing the energy perfectly costs all knowledge of phase, which reappears as quadrature noise; only the phaseless vacuum $n=0$ sits at the minimum.