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Selective Operations, Postselection, and State Update

Postselection keeps only one branch of a measurement record. It can expose strong conditional correlations, but the selected map is subnormalized and the outcome label is a physical classical resource. A remote observer cannot use that conditional ensemble before the label arrives, and a large value on a vanishingly rare branch is not a large deterministic yield. These distinctions are worked out below in an exact two-region Gaussian field-detector benchmark.

Required background. Local measurement instruments defines selective and nonselective operations.

Helpful background. Hypothesis testing and asymptotics supplies operational error criteria for comparing rare conditional ensembles.

Chapter map. Use the route from a local coupling to a field instrument, the chapter claim-validity table, and the three independent validity questions to track the branch map, its causal record, and its success rate separately.

For a discrete outcome xx, let Ix\mathcal I_x be CP and trace nonincreasing. Given an input state ρ\rho,

px=tr⁡Ix(ρ),ρx=Ix(ρ)pxp_x=\operatorname{tr}\mathcal I_x(\rho), \qquad \rho_x=\frac{\mathcal I_x(\rho)}{p_x}

when px>0p_x>0. The instrument normalization condition is

E=∑xIx\mathcal E=\sum_x\mathcal I_x

with E\mathcal E trace preserving. Thus Ix(ρ)\mathcal I_x(\rho), not ρx\rho_x alone, is the linear output of the physical branch; its trace remembers how often that branch occurs.

The record can be represented explicitly by a classical register XX:

ΩXQ=∑x∣x⟩ ⁣⟨x∣X⊗Ix(ρ).\Omega_{XQ} =\sum_x |x\rangle\!\langle x|_X\otimes\mathcal I_x(\rho).

An observer with access to X=xX=x uses ρx\rho_x. An observer without that register traces it out and obtains E(ρ)\mathcal E(\rho). For continuous or algebraic outcome spaces, posterior states require the corresponding measurable disintegration; Okamura and Ozawa 2016, Definition V.2 and Eq. (69), PDF states this normalization precisely.

First QFT benchmark: two spacelike detector records

Section titled “First QFT benchmark: two spacelike detector records”

Use a finite lattice regulator for a real scalar field and choose two cells whose compact detector-coupling regions are spacelike separated. Let fAf_A and fBf_B be the corresponding real smearings and set

QA=Φ(fA),QB=Φ(fB).Q_A=\Phi(f_A), \qquad Q_B=\Phi(f_B).

Microcausality gives [QA,QB]=0[Q_A,Q_B]=0. Prepare the two regulated field modes in a centered two-mode squeezed Gaussian state with squeezing

s=12log⁡3.s=\frac12\log3.

In the convention where a vacuum quadrature has variance 1/21/2, this state has

Var⁡(QA)=Var⁡(QB)=12cosh⁡(2s)=56,Cov⁡(QA,QB)=12sinh⁡(2s)=23.\operatorname{Var}(Q_A)=\operatorname{Var}(Q_B) =\frac12\cosh(2s)=\frac56, \qquad \operatorname{Cov}(Q_A,Q_B) =\frac12\sinh(2s)=\frac23.

Couple an independent Gaussian pointer of noise variance δ2=1/2\delta^2=1/2 to each smeared observable as on the local-instrument page. The raw pointer variances are then 5/6+1/2=4/35/6+1/2=4/3, while their covariance remains 2/32/3. After standardizing each readout by 4/3\sqrt{4/3}, call them XX and YY. Their exact covariance matrix is

Σ=(11/21/21).\Sigma= \begin{pmatrix} 1&1/2\\ 1/2&1 \end{pmatrix}.

Its eigenvalues 1/21/2 and 3/23/2 are positive, and its correlation coefficient is

r=2/34/3=12.r=\frac{2/3}{4/3}=\frac12.

The regulator, state parameter, smearing support, pointer noise, threshold, and standardization have all been declared, so the benchmark can be reproduced without fitting an unspecified covariance.

Threshold each pointer at zero:

x=+  ⟺  X>0,y=+  ⟺  Y>0.x=+\iff X>0, \qquad y=+\iff Y>0.

For a centered bivariate normal distribution with correlation rr,

Pr⁡(X>0,Y>0)=14+arcsin⁡r2π.\Pr(X>0,Y>0) =\frac14+\frac{\arcsin r}{2\pi}.

At r=1/2r=1/2, arcsin⁡r=π/6\arcsin r=\pi/6, so the complete joint distribution is

Pr⁡(+,+)=Pr⁡(−,−)=13,Pr⁡(+,−)=Pr⁡(−,+)=16.\begin{aligned} \Pr(+,+)&=\Pr(-,-)=\frac13,\\ \Pr(+,-)&=\Pr(-,+)=\frac16. \end{aligned}

Postselecting the AA outcome x=+x=+ succeeds with

p+=12,p_+=\frac12,

and gives the record-conditioned probability

Pr⁡(y=+∣x=+)=Pr⁡(+,+)Pr⁡(x=+)=1/31/2=23.\Pr(y=+\mid x=+) =\frac{\Pr(+,+)}{\Pr(x=+)} =\frac{1/3}{1/2}=\frac23.

By contrast, the statistic available at BB without the AA record is

Pr⁡B(y=+)=Pr⁡(+,+)+Pr⁡(−,+)=13+16=12.\Pr_B(y=+) =\Pr(+,+)+\Pr(-,+) =\frac13+\frac16=\frac12.

This is the required three-way comparison: the unconditional probability is 1/21/2; the subensemble selected locally at AA has probability 2/32/3; and the remote observer still sees 1/21/2 until the outcome record arrives. The conditional change comes from the initial field correlation, not from a signal.

The fractions are exact and therefore have no model-integration uncertainty. Empirical repetitions would add multinomial sampling uncertainty, which must be reported rather than folded into the postselection effect.

Why the remote marginal cannot use an unavailable label

Section titled “Why the remote marginal cannot use an unavailable label”

Let EyBE_y^B be a detector effect in region BB. The joint probability is

p(x,y)=tr⁡ ⁣[EyBIxA(ρ)].p(x,y)=\operatorname{tr}\!\left[E_y^B\mathcal I_x^A(\rho)\right].

Before receiving xx, observer BB must sum over it:

pB(y)=∑xp(x,y)=tr⁡ ⁣[EyBEA(ρ)].\begin{aligned} p_B(y) &=\sum_xp(x,y)\\ &=\operatorname{tr}\!\left[E_y^B\mathcal E_A(\rho)\right]. \end{aligned}

For a properly localized, nonselective intervention in spacelike region AA,

EA∗(EyB)=EyB,\mathcal E_A^*(E_y^B)=E_y^B,

so pB(y)p_B(y) equals its value when AA does nothing. Selective state updates can nevertheless change conditional predictions because EyBE_y^B may be correlated with the induced AA effect. This exact distinction follows from the probe pre-instrument formulas in Fewster and Verch 2020, Eqs. (3.19), (3.23)–(3.26), and Theorem 3.4, PDF.

Operational postselection requires access to both records and therefore occurs only in their joint causal future. Bostelmann, Fewster, and Ruep 2021, PDF, Eq. (36) and the discussion immediately following it makes this record requirement explicit. Communicating the label does not retroactively alter the earlier BB data; it enables the parties to sort those data into conditional subensembles later.

Rare-event amplification and its failure control

Section titled “Rare-event amplification and its failure control”

Use the same standardized Gaussian readouts and postselect the rarer event X>tX>t. Write

ϕ(t)=e−t2/22π,Φ‾(t)=∫t∞ϕ(x) dx.\phi(t)=\frac{e^{-t^2/2}}{\sqrt{2\pi}}, \qquad \overline\Phi(t)=\int_t^\infty\phi(x)\,dx.

The success probability is pt=Φ‾(t)p_t=\overline\Phi(t). For a bivariate normal pair with correlation rr, E(Y∣X=x)=rx\mathbb E(Y\mid X=x)=rx, so

E(Y∣X>t)=rϕ(t)Φ‾(t).\mathbb E(Y\mid X>t) =r\frac{\phi(t)}{\overline\Phi(t)}.

The conditional mean grows as rtrt for large tt, but its signed contribution per original trial is

pt E(Y∣X>t)=rϕ(t)⟶0.p_t\,\mathbb E(Y\mid X>t)=r\phi(t)\longrightarrow0.

For the declared r=1/2r=1/2 and t=3t=3,

p3=0.0013498980,E(Y∣X>3)=1.6415493,p3E(Y∣X>3)=0.0022159242.\begin{aligned} p_3&=0.0013498980,\\ \mathbb E(Y\mid X>3)&=1.6415493,\\ p_3\mathbb E(Y\mid X>3)&=0.0022159242. \end{aligned}

One selected record therefore requires about 1/p3≃7411/p_3\simeq741 original trials on average. The conditional mean is larger than the unconditional mean zero, yet the per-trial contribution tends to zero as the threshold is raised. This is the adversarial limit required for any amplification or resource claim. For a nonlinear resource monotone, replace the signed mean by that task’s success-weighted rate and include repetitions, energy, duration, record storage, classical communication, and estimation uncertainty.

The tail probability and the two following decimals are analytic Gaussian values rounded to the shown digits. A Monte Carlo reproduction should attach confidence intervals to both the success rate and the conditional mean; the rare branch makes the latter uncertainty large at fixed total trial count.

Compare a heralded protocol with a deterministic one at equal total preparations, not equal retained samples. Report the success rule before examining the data, all failed trials, and the confidence interval after selection. If the choice of threshold or branch is made after looking at outcomes, selection bias becomes an additional inference problem.

The exact benchmark assumes a quasifree state, Gaussian pointer noise, compact smearings, and a regulator under which the pointer instruments are defined. It does not claim that pointlike projectors or instantaneous global collapses are local operations. Interacting fields, non-Gaussian detectors, and continuum limits require their own supported instrument, energy accounting, and error estimates. The causal statement concerns the nonselective remote marginal; it does not say that spacelike field correlations vanish.

For r=1/2r=1/2, use the Gaussian quadrant formula to find all four sign probabilities and verify both marginals.

Solution

Because arcsin⁡(1/2)=π/6\arcsin(1/2)=\pi/6,

Pr⁡(+,+)=14+112=13.\Pr(+,+)=\frac14+\frac1{12}=\frac13.

Sign symmetry gives Pr⁡(−,−)=1/3\Pr(-,-)=1/3. Each marginal is symmetric and therefore equals 1/21/2, so

Pr⁡(+,−)=Pr⁡(−,+)=12−13=16.\Pr(+,-)=\Pr(-,+)=\frac12-\frac13=\frac16.

The four probabilities sum to one, and summing either row or column gives 1/21/2.

Show that a trace-preserving instrument at AA cannot change the probability of a spacelike BB effect when its nonselective dual fixes that effect.

Solution

Sum the joint probabilities over the unavailable outcome:

∑xp(x,y)=tr⁡[EyBEA(ρ)]=tr⁡[ρEA∗(EyB)].\sum_xp(x,y) =\operatorname{tr}[E_y^B\mathcal E_A(\rho)] =\operatorname{tr}[\rho\mathcal E_A^*(E_y^B)].

Locality gives EA∗(EyB)=EyB\mathcal E_A^*(E_y^B)=E_y^B, hence the result is tr⁡(ρEyB)\operatorname{tr}(\rho E_y^B), exactly the probability without the intervention. No corresponding equality is required branch by branch.

At r=1/2r=1/2 and t=3t=3, calculate the conditional mean, the per-trial contribution, and the expected trials per success from ϕ(3)=0.0044318484\phi(3)=0.0044318484 and Φ‾(3)=0.0013498980\overline\Phi(3)=0.0013498980.

Solution

The conditional mean is

120.00443184840.0013498980=1.6415493.\frac12\frac{0.0044318484}{0.0013498980} =1.6415493.

Multiplication by the success probability cancels the tail denominator:

p3E(Y∣X>3)=12ϕ(3)=0.0022159242.p_3\mathbb E(Y\mid X>3) =\frac12\phi(3)=0.0022159242.

Finally, 1/p3=740.8…1/p_3=740.8\ldots, so approximately 741741 preparations are needed per selected record. Reporting only 1.64154931.6415493 would omit the dominant operational cost.

  • Bostelmann, H., Fewster, C. J., and Ruep, M. H. (2021). “Impossible Measurements Require Impossible Apparatus.” Physical Review D 103, 025017. DOI. Open PDF.
  • Fewster, C. J., and Verch, R. (2020). “Quantum Fields and Local Measurements.” Communications in Mathematical Physics 378, 851–889. DOI. Open PDF.
  • Okamura, K., and Ozawa, M. (2016). “Measurement Theory in Local Quantum Physics.” Journal of Mathematical Physics 57, 015209. DOI. Open PDF.

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