Localized Probe and Detector Models

A localized detector is a finite quantum system with a specified coupling and readout, not a definition of particles. Its response becomes a controlled QFT observable only after the detector Hilbert space, coupling region, switching, smearing, field state, perturbative order, and measured probe effect have all been fixed. The analytic Gaussian calculation below separates vacuum switching noise from the response to a one-particle wavepacket. Because Gaussian profiles have tails, it is a finite-resolution spectroscopy benchmark rather than a certificate of exact compact localization.

Required background. Current correlators and response supplies the response-theory viewpoint. Relativistic causality supplies the support constraint. Use the protocol specification on operational locality throughout.

Chapter map. From a local coupling to a field instrument follows the complete measurement chain. The claim-validity table records which hypotheses license each conclusion, while three independent validity questions separates mathematical definition, causal implementation, and empirical interpretation.

The smeared two-level detector

Let the detector have ground state $|g\rangle$, excited state $|e\rangle$, and gap $\Omega>0$. In its rest frame,

$$\mu(t)=e^{i\Omega t}\sigma^+ +e^{-i\Omega t}\sigma^-.$$

For an inertial detector centered at the origin, take

$$H_I(t)=\lambda\chi_T(t)\mu(t)\Phi(F_L,t), \qquad \Phi(F_L,t)=\int d^3\mathbf x\,F_L(\mathbf x)\phi(t,\mathbf x).$$

The detector begins in $|g\rangle$ and is measured in the energy basis in an idealized outgoing limit. For the Gaussian switching used below, “outgoing” means after a declared tail tolerance because the interaction never vanishes exactly; a strictly local protocol replaces it by compact switching and smearing. At leading nontrivial order,

$$P_{g\to e}=\lambda^2\mathcal F_\rho(\Omega)+O(\lambda^4),$$ $$\mathcal F_\rho(\Omega)= \int dt\,dt'\, \chi_T(t)\chi_T(t')e^{-i\Omega(t-t')} W_{\rho,F_L}(t,t'),$$

where $W_{\rho,F_L}$ is the state-$\rho$ Wightman distribution smeared in both spatial arguments. This response-function construction, including why smooth switching makes the distributional pairing unambiguous, is set out in Louko and Satz 2006, § 2, Eqs. (2.1)–(2.5), PDF.

Use the four-dimensional massless field convention

$$\phi(t,\mathbf x)= \int\frac{d^3\mathbf k}{\sqrt{2(2\pi)^3k}} \left[ a(\mathbf k)e^{-ikt+i\mathbf k\cdot\mathbf x} +a^\dagger(\mathbf k)e^{ikt-i\mathbf k\cdot\mathbf x} \right],$$

with $[a(\mathbf k),a^\dagger(\mathbf q)]=\delta^3(\mathbf k-\mathbf q)$ and $k=|\mathbf k|$. Freeze the apparatus profiles to

$$F_L(\mathbf x)= \frac{e^{-|\mathbf x|^2/(2L^2)}}{(2\pi L^2)^{3/2}}, \qquad \chi_T(t)=e^{-t^2/(2T^2)}.$$

For the Fourier convention $\widehat f(\nu)=\int dt\,e^{i\nu t}f(t)$,

$$\widetilde F_L(\mathbf k)=e^{-L^2k^2/2}, \qquad \widehat\chi_T(\nu)=\sqrt{2\pi}\,T e^{-T^2\nu^2/2}.$$

Spatial smearing enters detector spectroscopy precisely through $\widetilde F_L$; see Martín-Martínez, Montero, and del Rey 2013, § III, Eqs. (13)–(16), PDF.

Exact benchmark: vacuum and one wavepacket

For the Minkowski vacuum, the angular integrals give the positive spectral expression

$$\mathcal F_0(\Omega) =\frac{T^2}{2\pi} \int_0^\infty dk\,k\, e^{-L^2k^2-T^2(k+\Omega)^2}.$$

Writing $A=L^2+T^2$ and $b=T^2\Omega$, this is also

$$\mathcal F_0(\Omega) =\frac{T^2e^{-T^2\Omega^2}}{4\pi A} \left[ 1-\frac{\sqrt\pi\,b}{\sqrt A} e^{b^2/A}\operatorname{erfc}\!\left(\frac b{\sqrt A}\right) \right].$$

This formula is an important sign check: a ground-state detector samples $\widehat\chi_T(\Omega+k)$, not $\widehat\chi_T(\Omega-k)$, in the vacuum term.

Now prepare the normalized one-particle state

$$|1_h\rangle=a^\dagger(h)|0\rangle, \qquad h(\mathbf k)=\frac{e^{-k/\kappa}}{\sqrt{\pi\kappa^3}}, \qquad \int d^3\mathbf k\,|h(\mathbf k)|^2=1.$$

Its two-point function is

$$W_{1_h}(x,x')=W_0(x,x') +u_h(x)\overline{u_h(x')} +\overline{u_h(x)}u_h(x').$$

After applying the same detector filter to all three terms,

$$\mathcal F_{1_h}(\Omega) =\mathcal F_0(\Omega)+|\mathcal A_+|^2+|\mathcal A_-|^2,$$ $$\mathcal A_\pm= \sqrt{\frac2\pi}\,\frac{T}{\kappa^{3/2}} \int_0^\infty dk\,k^{3/2} \exp\!\left[ -\frac{k}{\kappa} -\frac{L^2k^2}{2} -\frac{T^2(\Omega\pm k)^2}{2} \right].$$

$\mathcal A_-$ is the resonant absorption contribution; $\mathcal A_+$ is the counter-rotating contribution. The sum of their modulus squares is nonnegative, but its magnitude measures mode matching to this detector, not a universal particle count.

Set

$$L=T=\Omega=1,\qquad \kappa=\frac23,\qquad \lambda=0.05.$$

$L$ and $T$ are measured in a common inverse-energy unit, while $\Omega$ and $\kappa$ are measured in the reciprocal energy unit. Direct quadrature gives

Quantity	Benchmark value
$\mathcal F_0$	$0.00503997619544$
$\mathcal A_+$	$0.0559440651971$
$\mathcal A_-$	$0.249217451417$
$\lvert\mathcal A_+\rvert^2+\lvert\mathcal A_-\rvert^2$	$0.0652390765214$
$\mathcal F_{1_h}$	$0.0702790527169$
$\lambda^2\mathcal F_0$	$1.25999404886\times10^{-5}$
$\lambda^2\mathcal F_{1_h}$	$1.75697631792\times10^{-4}$

An adaptive 50-digit integral on $0\le k<\infty$ and a uniform Simpson calculation with $10^6$ panels on $0\le k\le12$ agree in every quoted response coefficient to better than $3\times10^{-12}$. This is a quadrature check only. The physical probabilities still carry an $O(\lambda^4)$ perturbative remainder that has not been bounded here, so the table is a reproducible leading-order benchmark rather than an error-certified nonperturbative prediction.

Adversarial control: shrink time and space independently

The same vacuum formula exposes why “take the detector to a point and an instant” is not a single limit. Keep the peak of $\chi_T$ fixed. Along $L=rT$,

$$\lim_{T\to0}\mathcal F_0(\Omega;T,rT) =\frac{1}{4\pi(1+r^2)}.$$

Different fixed ratios $r$ give different answers. More sharply,

$$\lim_{T\to0}\lim_{L\to0}\mathcal F_0 =\frac1{4\pi}, \qquad \lim_{L\to0}\lim_{T\to0}\mathcal F_0 =0.$$

The first order removes spatial suppression before the temporal Fourier window broadens; the second turns off a peak-fixed interaction at finite spatial size before taking the point limit. Neither result is a universal point-detector probability. Changing what is held fixed—peak amplitude, integrated coupling, or $L^2$ norm—changes the limiting family again. Smooth-switching and sharp-switching limits are carefully distinguished in Satz 2007, § 3, especially Eq. (3.8), and § 4, PDF.

The strongest surviving claim is therefore finite-$L$, finite-$T$: the frozen detector distinguishes the chosen wavepacket from the vacuum at leading order. The adversarial test rules out a regulator-independent joint pointlike-and-sudden limit for this family.

Common pitfalls

Treating the apparatus profile as a harmless regulator. $F_L$ and $\chi_T$ determine which field modes reach the probe. Changing either one changes the measurement.

Calling every excitation “absorption.” Vacuum excitation with finite switching arises from the counter-rotating spectral window. The resonant $\Omega-k$ term appears only when the state supplies a one-particle contribution.

Reporting a perturbative number as an exact probability. Numerical quadrature can be extremely accurate while the omitted $O(\lambda^4)$ dynamics remains uncontrolled. These are different uncertainties.

Exercises

1. Derive the vacuum spectral integral

Starting from the mode expansion, show that the frozen Gaussian detector has the stated $\mathcal F_0(\Omega)$.

Solution

The vacuum Wightman function contributes one annihilation and one creation mode. Smearing and switching give

$$\mathcal F_0 =\int\frac{d^3\mathbf k}{2(2\pi)^3k} |\widetilde F_L(\mathbf k)|^2 |\widehat\chi_T(\Omega+k)|^2.$$

Using $d^3\mathbf k=4\pi k^2dk$, $|\widetilde F_L|^2=e^{-L^2k^2}$, and $|\widehat\chi_T|^2=2\pi T^2e^{-T^2(\Omega+k)^2}$ gives

$$\mathcal F_0 =\frac{T^2}{2\pi}\int_0^\infty dk\,k e^{-L^2k^2-T^2(\Omega+k)^2}.$$

2. Normalize the wavepacket

Verify that $h(\mathbf k)=e^{-k/\kappa}/\sqrt{\pi\kappa^3}$ has unit norm and compute its mean momentum.

Solution

Radial integration gives

$$\int d^3\mathbf k\,|h|^2 =\frac4{\kappa^3}\int_0^\infty dk\,k^2e^{-2k/\kappa} =1.$$

Similarly,

$$\langle k\rangle =\frac4{\kappa^3}\int_0^\infty dk\,k^3e^{-2k/\kappa} =\frac{3\kappa}{2}.$$

For $\kappa=2/3$, the mean momentum is $1$, matched to the gap $\Omega=1$.

3. Explain positivity of the state correction

Why is $\mathcal F_{1_h}-\mathcal F_0$ nonnegative in this benchmark?

Solution

The one-particle correction to the Wightman function is the sum of two rank-one kernels, $u_h\overline u_h$ and $\overline u_h u_h$. Pairing each with the detector test function produces a modulus square. Hence

$$\mathcal F_{1_h}-\mathcal F_0 =|\mathcal A_+|^2+|\mathcal A_-|^2\ge0.$$

This argument is specific to the one-particle-versus-vacuum comparison and the linear detector coupling; it is not a general ordering theorem for arbitrary field states.

4. Reverse the shrinking limits

Derive the two iterated limits of $\mathcal F_0$.

Solution

At fixed $T$, first set $L=0$ and then rescale $q=Tk$. As $T\to0$,

$$\mathcal F_0\longrightarrow \frac1{2\pi}\int_0^\infty dq\,q e^{-q^2} =\frac1{4\pi}.$$

At fixed $L>0$, the integral tends to $\int_0^\infty dk\,k e^{-L^2k^2}=1/(2L^2)$ while its prefactor is $T^2/(2\pi)$, so the result tends to zero. Taking $L\to0$ afterwards leaves zero.

References

• Louko, J., and Satz, A. (2006). “How Often Does the Unruh–DeWitt Detector Click? Regularisation by a Spatial Profile.” Classical and Quantum Gravity 23, 6321–6344. DOI. Open PDF.
• Martín-Martínez, E., Montero, M., and del Rey, M. (2013). “Wavepacket Detection with the Unruh–DeWitt Model.” Physical Review D 87, 064038. DOI. Open PDF.
• Satz, A. (2007). “Then Again, How Often Does the Unruh–DeWitt Detector Click If We Switch It Carefully?” Classical and Quantum Gravity 24, 1719–1732. DOI. Open PDF.