System–Probe Scattering and Measurement Models

A local measurement should be a physical process, not an instantaneous projection imposed on an entire time slice. Prepare a probe, couple it to the field only inside a compact spacetime region, and read the outgoing probe: the resulting scattering map determines the outcome probabilities, the state disturbance, and the spacetime support of the intervention. This page derives that chain and then checks it in an exactly soluble scalar-field/qubit benchmark.

Required background. Operational locality supplies the support and record data that a relativistic protocol must declare.

Helpful background. Localized probe and detector models develops finite-dimensional probes and their field couplings.

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 place this construction within the full measurement protocol.

A compact coupling produces an instrument

Let $\mathcal A$ be the field algebra and $\mathcal B$ the probe algebra. The theories are uncoupled outside a compact interaction region $K$. Comparing the early-time and late-time identifications of the coupled theory with $\mathcal A\otimes\mathcal B$ gives a scattering automorphism

$$\Theta:\mathcal A\otimes\mathcal B\longrightarrow \mathcal A\otimes\mathcal B.$$

Prepare the probe in a state $\sigma$ and define the partial expectation

$$\eta_\sigma(A\otimes B)=\sigma(B)A.$$

If $E_r$ is a probe effect for outcome $r$, the corresponding field effect is

$$\varepsilon_\sigma(E_r) =\eta_\sigma\!\left(\Theta(\mathbf 1\otimes E_r)\right).$$

Indeed, for every field state $\omega$,

$$\omega\!\left(\varepsilon_\sigma(E_r)\right) =(\omega\otimes\sigma)\!\left(\Theta(\mathbf1\otimes E_r)\right),$$

so the induced effect reproduces the actual outgoing-probe probability. The same dynamics determines more than a POVM. Its unnormalized selective update is the functional

$$\mathcal J_r(\omega)(A) =(\omega\otimes\sigma)\!\left(\Theta(A\otimes E_r)\right),$$

with

$$\mathcal J_r(\omega)(\mathbf1) =\omega\!\left(\varepsilon_\sigma(E_r)\right).$$

When this number is nonzero, the posterior state is $\omega_r=\mathcal J_r(\omega)/\mathcal J_r(\omega)(\mathbf1)$. Summing over all outcomes gives the nonselective channel. These are the induced observable and pre-instrument of Fewster and Verch 2020, Eqs. (3.11)–(3.13) and (3.19)–(3.23), PDF.

Exact scalar-field/qubit benchmark

Consider a regulated real scalar field and a real test function $f\in C_0^\infty(M)$ with $\operatorname{supp}f\subset K$. Normalize the smeared observable

$$Q=\Phi(f)$$

to be dimensionless, and work in a representation in which $Q$ is self-adjoint. Prepare a qubit probe in $|0\rangle$, couple it through

$$U=\exp\!\left[-igQ\otimes\sigma_y\right],$$

and measure the probe in the $\{|0\rangle,|1\rangle\}$ basis. Functional calculus gives the two field Kraus operators exactly:

$$M_0=\langle0|U|0\rangle=\cos(gQ), \qquad M_1=\langle1|U|0\rangle=\sin(gQ).$$

The Schrödinger-picture branches and effects are therefore

$$\mathcal J_r(\rho)=M_r\rho M_r^\dagger, \qquad E_r=M_r^\dagger M_r.$$

Each branch is completely positive because it has a Kraus form. The exact normalization check is the operator identity

$$E_0+E_1=\cos^2(gQ)+\sin^2(gQ)=\mathbf1.$$

Now take a centered quasifree state whose normalized smearing has variance $v=\langle Q^2\rangle$. Its characteristic function is $\langle e^{itQ}\rangle=e^{-t^2v/2}$, hence

$$\begin{aligned} p_0&=\langle\cos^2(gQ)\rangle =\frac{1+e^{-2g^2v}}{2},\\ p_1&=\langle\sin^2(gQ)\rangle =\frac{1-e^{-2g^2v}}{2}. \end{aligned}$$

For the reproducible choice $v=1$ and $g=1/2$,

$$p_0=0.8032653299, \qquad p_1=0.1967346701, \qquad p_0+p_1=1.$$

The value $v=1$ is not an unstated field normalization: starting from any centered quasifree state and compact smearing with positive finite variance, rescale $f$ by $1/\sqrt{\langle\Phi(f)^2\rangle}$. This leaves its support unchanged and makes the benchmark dimensionless.

These probabilities are analytic; the displayed decimals are rounded to ten places and have no sampling or integration uncertainty. A numerical realization should separately report its floating-point error and its regulator or truncation error.

The nonselective Heisenberg channel also exposes locality. Write $W(h)=e^{i\Phi(h)}$ and use

$$[\Phi(f),\Phi(h)]=i\Delta(f,h)\mathbf1.$$

Since

$$\mathcal E^*(B) =\cos(gQ)B\cos(gQ)+\sin(gQ)B\sin(gQ) =\frac12\sum_{s=\pm1}e^{isgQ}Be^{-isgQ},$$

one obtains

$$\mathcal E^*(W(h)) =\cos\!\bigl(g\Delta(f,h)\bigr)W(h).$$

If $\operatorname{supp}h$ is causally disjoint from $K$, microcausality gives $\Delta(f,h)=0$, so $\mathcal E^*(W(h))=W(h)$. Complete positivity and causal support have now been checked separately rather than inferred from one another.

Perturbative error is an operator check

It is tempting to replace the exact Kraus operators by

$$M_0^{(2)}=\mathbf1-\frac{g^2Q^2}{2}, \qquad M_1^{(1)}=gQ.$$

They reproduce the unitary expansion through the advertised low order, but their completeness relation is

$$\bigl(M_0^{(2)}\bigr)^\dagger M_0^{(2)} +\bigl(M_1^{(1)}\bigr)^\dagger M_1^{(1)} =\mathbf1+\frac{g^4Q^4}{4}.$$

Thus the truncated branches are not exactly trace preserving. In the Gaussian benchmark with $v=1$, $\langle Q^4\rangle=3$; at $g=1/2$ the mean normalization excess is

$$\frac{g^4\langle Q^4\rangle}{4}=\frac{3}{64}=0.046875.$$

Quoting only the formal order $O(g^4)$ would hide a five-percent error in this example. A perturbative instrument must report a state or energy domain on which the residual is controlled, or be completed to an exact CP, normalized map.

Causal factorization and its failure control

For two compactly supported couplings, let

$$G_j=g_j\Phi(f_j)\otimes\sigma_y^{(j)}, \qquad U_j=e^{-iG_j}.$$

If the supports are causally disjoint, $\Delta(f_1,f_2)=0$ and $[G_1,G_2]=0$. The two dilations commute, so the resulting instruments can be composed in either order. This is the finite benchmark counterpart of the causal-factorization theorem Fewster and Verch 2020, PDF, Eq. (3.28) and Theorem 3.5.

Move the second coupling into causal contact with the first. If $\Delta(f_1,f_2)=\kappa\ne0$, then

$$[G_1,G_2] =ig_1g_2\kappa\,\sigma_y^{(1)}\sigma_y^{(2)},$$

and ordered dependence is generically present. A special probe state or readout may accidentally hide it, so the adversarial check should compare the full induced maps or a separating set of field observables, not one outcome probability.

Moving only the readout after the coupling has ended does not move the field intervention: it changes when the classical outcome becomes available. Attempting to interpret a readout before the probe has reached an outgoing region instead breaks the scattering protocol itself. The distinction between an observer’s local marginal and joint postselection after records are brought together is made explicit in Bostelmann, Fewster, and Ruep 2021, § VI.B and § VI.E, especially Eqs. (21)–(26) and (36)–(37), PDF.

Scope and limitations

The benchmark is exact within a regulated representation and for the declared compact smearing. It does not show that every abstract POVM or CP map has a compactly supported QFT realization. On a general von Neumann algebra, a CP instrument needs the normal extension property to arise from a measuring process; local extension also uses hypotheses such as the split property Okamura and Ozawa 2016, Definition III.3 and Theorems III.4 and VI.2, PDF. Nor does localization mean absence of backreaction: observables in the causal future of $K$ can be disturbed.

Exercises

1. Reproduce the probe probabilities

Starting from $M_0=\cos(gQ)$ and $M_1=\sin(gQ)$, derive $p_0$ and $p_1$ for a centered Gaussian $Q$ of variance $v$.

Solution

Use $\cos^2z=(1+\cos2z)/2$ and $\sin^2z=(1-\cos2z)/2$. The Gaussian characteristic function gives

$$\langle\cos(2gQ)\rangle =\operatorname{Re}\langle e^{2igQ}\rangle =e^{-2g^2v}.$$

Substitution yields $p_0=(1+e^{-2g^2v})/2$ and $p_1=(1-e^{-2g^2v})/2$. Their sum is one independently of the state, as required by the operator completeness identity.

2. Diagnose the truncated instrument

Verify the completeness residual of $M_0^{(2)}$ and $M_1^{(1)}$. Why can it not be repaired by renormalizing every output state after the calculation?

Solution

Expanding the two positive terms gives

$$\left(\mathbf1-\frac{g^2Q^2}{2}\right)^2+g^2Q^2 =\mathbf1+\frac{g^4Q^4}{4}.$$

Dividing each final state by its trace would make the evolution nonlinear in the input state, so it would no longer be a quantum channel. Here the residual is positive, so adding another Kraus branch would only increase the excess. A valid repair must modify the existing operators—by keeping a consistent higher-order expansion or returning to the exact sine and cosine—so that the operator sum, not merely one state-dependent expectation, equals $\mathbf1$.

3. Separate coupling support from record time

Let $h$ be spacelike separated from $\operatorname{supp}f$. Show that the nonselective channel fixes $W(h)$. Would delaying the probe readout change this result?

Solution

Microcausality gives $\Delta(f,h)=0$. The exact channel formula then gives

$$\mathcal E^*(W(h))=\cos(0)W(h)=W(h).$$

Delaying a readout performed after the coupling changes only the spacetime point at which the classical record is available. The already completed system–probe coupling, and hence the nonselective field channel and its support, are unchanged.

References

• 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.