research36 min studyLesson 446 in Physics

Normalize spin noise by the signal it preserves

Derive the Wineland squeezing parameter from collective-spin readout, including contrast loss, variance gain, and the distinction between noise reduction and useful sensitivity.

On this page

What you’ll understand

  • Normalize collective-spin fluctuations against a coherent-spin reference.
  • Calculate metrological squeezing using both transverse variance and mean spin.
  • Distinguish a variance improvement from standard-deviation and decibel gains.

A narrow distribution of population differences looks attractive, but a sensor also needs a signal that moves with the unknown parameter. Spin squeezing is useful when it reduces readout noise while preserving enough collective polarization to retain that response. The metrological squeezing parameter explicitly divides noise by signal slope. This makes it a more demanding diagnostic than comparing a variance against the projection noise of an uncorrelated ensemble.

Choose a collective-spin convention

Use dimensionless collective operators Jk=12∑jσk(j)J_k=\frac12\sum_j\sigma_k^{(j)} for N spin-one-half particles. A coherent-spin state polarized along x has mean ⟨Jx⟩=N/2\langle J_x\rangle=N/2, zero mean transverse components, and transverse variance N/4N/4. Physical angular momentum would multiply J by ℏ\hbar, but using dimensionless spins keeps the normalization explicit. The uncertainty relation between transverse components then constrains how much one noise direction can be compressed while maintaining a large mean spin.

A small phase rotation about y moves the mean population difference JzJ_z with slope whose magnitude is ∣⟨Jx⟩∣|\langle J_x\rangle| at the operating point. Linear error propagation for that readout divides the standard deviation of JzJ_z by this slope. For the coherent-spin reference the result is 1/N1/\sqrt N. This describes a small-angle calibrated estimator using collective population readout; it is not a claim about every possible measurement on the state or about an arbitrary large-angle likelihood.

Var⁡CSS(Jz)=N4,∣⟨Jx⟩CSS∣=N2,(Δϕ)2≈Var⁡(Jz)∣⟨Jx⟩∣2\operatorname{Var}_{\mathrm{CSS}}(J_z)=\frac N4,\qquad |\langle J_x\rangle_{\mathrm{CSS}}|=\frac N2,\qquad (\Delta\phi)^2\approx\frac{\operatorname{Var}(J_z)}{|\langle J_x\rangle|^2}
J is dimensionless and CSS denotes an x-polarized coherent-spin state. The phase relation is local error propagation for a y rotation read out through Jz; its sign depends on the rotation convention, while the squared sensitivity does not.↔ Scroll wide equations horizontally.

Include contrast in the squeezing parameter

The Wineland parameter ξR2\xi_R^2 is the phase variance inferred by this readout divided by the coherent-spin phase variance at the same N. Define contrast C=2∣⟨Jx⟩∣/NC=2|\langle J_x\rangle|/N and normalized transverse variance v=4Var⁡(Jz)/Nv=4\operatorname{Var}(J_z)/N. Then ξR2=v/C2\xi_R^2=v/C^2. A state with half the reference variance has v=1/2v=1/2, but it gains nothing if its remaining contrast is too small. Signal loss enters quadratically, so comparing only histograms of population noise can give the wrong experimental conclusion.

When the parameter is less than one, this readout offers sub-shot-noise small-angle sensitivity under the stated calibration. The reciprocal is a phase-variance gain; its square root is the standard-deviation gain. Squeezing is often reported in decibels, which use ten times the base-ten logarithm of the variance ratio. A fourfold variance gain is about six decibels and a twofold standard-deviation gain. State whether a plotted number is a ratio, its square root, or a logarithmic representation before comparing experiments.

ξR2=NVar⁡(Jz)∣⟨Jx⟩∣2=vC2,Gvar=1ξR2,GdB=−10log⁡10(ξR2)\xi_R^2=\frac{N\operatorname{Var}(J_z)}{|\langle J_x\rangle|^2}=\frac{v}{C^2},\qquad G_{\mathrm{var}}=\frac{1}{\xi_R^2},\qquad G_{\mathrm{dB}}=-10\log_{10}(\xi_R^2)
All ratios are dimensionless for fixed known N. The chosen z direction must be the squeezed readout quadrature transverse to the mean spin. Gvar compares local phase variances with the coherent-spin reference, assuming equivalent detection and calibration.↔ Scroll wide equations horizontally.

Match the squeezed quadrature to the experiment

Squeezing transfers uncertainty into another collective direction rather than removing the uncertainty relation. Imperfect pulse phases can rotate an enlarged noise direction into the measured component, especially when the anti-squeezing is large. Preparation may also reduce spin length, scatter particles, or take extra time. Detection noise adds to the measured population variance unless it is separately characterized. A gain quoted after subtracting detector noise describes a different operational quantity from the precision achieved by the complete readout chain.

For spin-one-half ensembles, metrological squeezing below one witnesses useful particle correlations under the stated particle-count and collective-spin assumptions. The converse is false: some useful non-Gaussian entangled states have little mean spin and are poorly described by this particular ratio. Quantum Fisher information can recognize advantages available through other measurements. Treat the squeezing parameter as a direct diagnostic for a specified collective readout, and retain the full likelihood or a broader information analysis when the mean-spin linear calibration becomes inadequate.

A fourfold variance gain

An ensemble has N=100N=100, dimensionless mean spin ∣⟨Jx⟩∣=40|\langle J_x\rangle|=40, and readout variance Var⁡(Jz)=4\operatorname{Var}(J_z)=4. Assume an ideal small-angle readout without added detector noise. Find contrast, normalized variance, squeezing, and phase precision relative to the coherent-spin reference.

  1. The coherent-spin reference has mean spin 5050 and variance 2525. The actual contrast is C=2(40)/100=0.8C=2(40)/100=0.8, and normalized noise is v=4(4)/100=0.16v=4(4)/100=0.16.

  2. Include the reduced response: ξR2=0.16/(0.8)2=0.25\xi_R^2=0.16/(0.8)^2=0.25. Equivalently, 100(4)/402=1/4100(4)/40^2=1/4. The variance gain is therefore four.

  3. The phase standard deviation is 4/40=0.050 rad\sqrt4/40=0.050\,\mathrm{rad}, compared with 1/100=0.10 rad1/\sqrt{100}=0.10\,\mathrm{rad} for the reference. The gain is two in standard deviation, or 10log⁡104≈6.0210\log_{10}4\approx6.02 decibels in variance.

The result

This state gives useful metrological squeezing under the stated readout: phase standard deviation 0.050 rad and a fourfold variance improvement. Reporting only the noise suppression would overstate the gain because contrast is 0.8.

Explore the model

Interactive lab

Interactive lab · sampling

Establish the unsqueezed population baseline

Reset and set p = 0.5 and n = 100. The exact sample-mean variance is 0.0025; multiplying it by n squared gives count variance 25, equal to coherent-spin Jz variance because Jz = K − n/2. Reset and set p = 0.5 and n = 25 to predict count variance 6.25. The lab contains independent Bernoulli outcomes, with no squeezed correlations or adjustable contrast; playback only highlights exact outcomes.

Establish the unsqueezed population baselineExact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Current readings: Mean sample proportion: 0.5; Sample-mean variance: 0.0125; Uniform CDF error bound: ≤ 0.1062. All parameters are adjustable with the labeled controls below.Exact distribution of the mean of 20 independent Bernoulli variables with success probability 0.5; playback highlights outcome 0.00.250.50.75100.0580.1150.1730.23Sample mean · k / nProbability massNormal bin masses are an approximationk = 0 · exact probability = 9.54e-7
Exact binomial massNormal interval massHighlighted outcome / mean
Mean sample proportion
0.5
Sample-mean variance
0.0125
Uniform CDF error bound
≤ 0.1062
t = 0 s

Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). 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 forProjection noise supplies a reference variance, while useful squeezing also requires the signal slope.

Check your understanding

An ensemble has normalized transverse variance v = 0.25 and contrast C = 0.25. What does the Wineland ratio predict for this readout?

Try it yourself

Keep N=100N=100 and Var⁡(Jz)=4\operatorname{Var}(J_z)=4, but reduce ∣⟨Jx⟩∣|\langle J_x\rangle| to 1010. Compute the metrological squeezing ratio and local phase standard deviation. Explain why the noise is still below the reference while the sensitivity is worse.

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

Give me a hint

The population variance is unchanged, but the denominator in phase error propagation has changed.

Compare with the solution

The ratio becomes ξR2=100(4)/102=4\xi_R^2=100(4)/10^2=4. The phase standard deviation is 4/10=0.20 rad\sqrt4/10=0.20\,\mathrm{rad}, twice the coherent-spin reference 0.10 rad0.10\,\mathrm{rad}. Its variance of four is still smaller than the reference variance of twenty-five, but its phase-response slope is only ten instead of fifty. The loss of response outweighs the noise suppression. Another measurement might access information unavailable to this mean-spin readout; the squeezing ratio alone does not optimize over all measurements.

Take it with you

  • Normalize spin operators and the coherent-spin reference explicitly.
  • Metrological squeezing divides normalized noise by contrast squared.
  • Reduced transverse variance is useful only when adequate phase response and readout survive.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.