Switching, Smearing, and Detector Regularization

Switching and smearing are part of a detector observable, not disposable decorations. The switching function says when the coupling acts and with what amplitude; the spatial profile says which field modes reach the probe. Together they determine coupling support, frequency resolution, ultraviolet suppression, and even whether a pointlike or sudden limit exists. Smoothness and support answer different questions: smoothness controls high-frequency decay, while compact support licenses exact statements about spacetime separation. This page compares two normalized switching profiles against the exact free-field spectrum and then reverses the pointlike and sudden limits as an adversarial test.

Required background. Point splitting and Wick products explains local ultraviolet singularities. Test functions and distributional support supplies the correct pairing with fields. The response model is defined on localized detectors.

Helpful background. Zero modes, boundaries, and infrared effects helps separate long-distance sensitivity from switching ultraviolet noise.

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.

A response is a distributional pairing

For a stationary inertial detector coupled to the four-dimensional massless scalar vacuum, spatially smear with

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

The profile is normalized so that $\int d^3\mathbf x\,F_L(\mathbf x)=1$, and $L$ is its spatial width. It approaches a point profile as $L\to0$, but for every nonzero $L$ this Gaussian has tails at arbitrarily large distance. It therefore defines a convenient spectral benchmark, not an exactly compact interaction region. An exact locality claim would instead require compact spacetime support, while an approximate claim must state a tail or signaling tolerance.

If $\widehat\chi(\nu)=\int dt\,e^{i\nu t}\chi(t)$, the leading response coefficient is

$$\mathcal F_\chi(\Omega,L) =\int\frac{d^3\mathbf k}{2(2\pi)^3k} |\widetilde F_L(\mathbf k)|^2 |\widehat\chi(\Omega+k)|^2 =\frac1{4\pi^2}\int_0^\infty dk\,k\,e^{-L^2k^2} |\widehat\chi(\Omega+k)|^2.$$

This is the free Wightman distribution paired with the detector test function, written in a manifestly positive spectral form. Each factor has a distinct role. The measure $k\,dk$ is the massless four-dimensional density of modes after angular integration, $e^{-L^2k^2}$ is the squared spatial filter, and $|\widehat\chi(\Omega+k)|^2$ is the temporal filter. The plus sign is important: a ground-state detector exciting from the vacuum must supply both its positive gap $\Omega$ and a positive field-mode energy $k$. Finite switching broadens the frequency filter, but it does not turn this vacuum term into resonant absorption at $k=\Omega$. The general response and the role of smooth compact switching are developed in Louko and Satz 2006, § 2, Eqs. (2.1)–(2.5), PDF.

The ultraviolet behavior can be read before integration. A Gaussian spatial profile supplies exponential momentum suppression. A pointlike detector sets $\widetilde F_L\to1$ and leaves convergence entirely to the large-frequency decay of $\widehat\chi$.

Exact benchmark: Gaussian versus compact switching

Compare profiles with the same $L^2$ norm, so their overall coupling budget rather than their peak height is held fixed. This choice makes the two response coefficients comparable, but it is a convention rather than a universal definition of equal detector strength. Matching peak height or integrated switching area would produce different families and different numerical ratios.

$$\chi_G(t)= \frac{e^{-t^2/(2T^2)}}{\pi^{1/4}\sqrt T},$$ $$\chi_C(t)= \begin{cases} \dfrac{2}{\sqrt{3T}} \cos^2\!\left(\dfrac{\pi t}{2T}\right),&|t|\lt T,\\ 0,&|t|\ge T. \end{cases}$$

Both obey $\int dt\,|\chi(t)|^2=1$. The compact raised cosine is $C^1$ but not $C^\infty$ at its endpoints, an intentional intermediate case between a smooth bump and a step. Their Fourier transforms are

$$\widehat\chi_G(\nu) =\sqrt2\,\pi^{1/4}\sqrt T\, e^{-T^2\nu^2/2},$$ $$\widehat\chi_C(\nu) =\frac{2a^2}{\sqrt{3T}}\, \frac{\sin(\nu T)}{\nu(a^2-\nu^2)}, \qquad a=\frac{\pi}{T},$$

with the removable values $\widehat\chi_C(0)=2\sqrt{T/3}$ and $\widehat\chi_C(\pm a)=\sqrt{T/3}$. The apparent poles at $\nu=0$ and $\nu=\pm a$ are artifacts of the closed form: taking the corresponding limits gives the finite values just stated.

The Gaussian decays exponentially, whereas the compact $C^1$ profile decays as $|\nu|^{-3}$. This difference can be anticipated without doing the full transform. Integration by parts transfers powers of $\nu^{-1}$ to endpoint derivatives; every derivative that joins smoothly to zero removes another boundary contribution. The raised cosine and its first derivative vanish at $|t|=T$, but its second derivative jumps there, producing the cubic tail. A step already jumps in value and therefore has only a $|\nu|^{-1}$ tail. Compact support and spectral sharpness consequently involve a regularity tradeoff, not a simple choice between “good” and “bad” profiles.

Take $T$ as the time unit and set

$$\frac{L}{T}=1, \qquad \Omega T=1.$$

These are the two dimensionless parameters that control the benchmark. To see this explicitly, write $\chi(t)=T^{-1/2}f(t/T)$ and rescale momentum by $q=kT$. Then

$$\begin{aligned} z&=\Omega T,& r&=\frac LT,& q&=kT,\\ \mathcal F_\chi(\Omega,L) &=\frac1T\,\mathcal G_\chi(z,r),\\ \mathcal G_\chi(z,r) &=\frac1{4\pi^2}\int_0^\infty dq\,q\,e^{-r^2q^2} |\widehat f(z+q)|^2. \end{aligned}$$

Thus the reported numbers are values of the dimensionless combination $T\mathcal F_\chi$ at $r=z=1$, equivalently values of $\mathcal F_\chi$ in units of $T^{-1}$. Because the switchings have unit $L^2$ norm, $\mathcal F_\chi$ carries one power of energy and the coupling that converts it into a probability carries inverse-square-root energy dimension. Substitution into the exact spectral integral gives

Switching	Normalization	Dimensionless response $T\mathcal F_\chi$
Gaussian $\chi_G$	$\int dt\,\lvert\chi_G\rvert^2=1$	$0.00284350207080$
Compact raised cosine $\chi_C$	$\int dt\,\lvert\chi_C\rvert^2=1$	$0.0104731620986$

A 50-digit adaptive integral on $q\in[0,\infty)$ and a uniform Simpson calculation with $10^6$ panels on $0\le q\le12$ agree to better than $3\times10^{-12}$; the removable value at $q=\pi-1$ is inserted analytically for the compact profile. The compact profile gives about $3.68319$ times the response here because its Fourier tail places more weight in the positively weighted integral at $(\Omega+k)T=1+q\ge1$. That ratio is not universal: it changes with $L/T$, $\Omega T$, and the chosen normalization.

The larger compact-profile value does not mean that compact switching is intrinsically a more sensitive or more physical detector. It means only that, under this $L^2$ matching and at these two dimensionless ratios, more of its spectral weight lies in the frequency range sampled by the vacuum response. If the same two-level detector and coupling strength $\lambda$ are used, the leading probability is $P_{g\to e}=\lambda^2\mathcal F_\chi+O(\lambda^4)$, so the response ratio is also the ratio of the leading probabilities. The quoted quadrature agreement controls the numerical coefficient, not the omitted $O(\lambda^4)$ dynamics. Changing the gap or the strength convention can change the comparison without contradicting either calculation.

Gaussian finite-time detector responses and their limiting subtleties were worked out in Sriramkumar and Padmanabhan 1996, § 3(a), Eqs. (51)–(59), PDF. The benchmark above differs by keeping a finite spatial Gaussian and matching the $L^2$ norms of the two switchings, so every normalization factor is visible.

Smooth, compact, and sharp are different claims

The compact raised cosine is sufficient for this finite benchmark, but $C_0^\infty$ switchings are the clean default for distribution theory. A standard smooth bump,

$$b_T(t)= \begin{cases} C_T\exp\!\left[-\dfrac1{1-(t/T)^2}\right],&|t|\lt T,\\ 0,&|t|\ge T, \end{cases}$$

has exact compact support and a Fourier transform that falls faster than every inverse power. Its transform is normally evaluated numerically. Notice that “smooth” alone is not enough: the constant function is smooth but does not switch the detector off. The useful hypothesis here is smooth compact support, or in this Minkowski spectral calculation a sufficiently rapidly decreasing Schwartz profile. The Gaussian is Schwartz and spectrally convenient, but its nonzero tails still prevent an exact finite-support locality certificate.

By contrast, the rectangular window $\chi_{\mathrm{step}}(t)=\mathbf1_{[-T,T]}(t)$ has

$$\widehat\chi_{\mathrm{step}}(\nu) =\frac{2\sin(\nu T)}{\nu}.$$

In the pointlike limit $L=0$, the response integrand behaves as $\sin^2[(k+\Omega)T]/k$ at large $k$ and is logarithmically divergent. A finite $L$ hides this divergence behind $e^{-L^2k^2}$; removing $L$ exposes it again. This is why a finite numerical answer at one nonzero $L$ is not evidence that the pointlike limit exists. Smooth compact switching produces a finite, covariantly defined pointlike response, while a controlled sharp limit requires separate treatment; see Satz 2007, § 3, especially Eq. (3.8), and § 4, PDF. Spatial-profile regularization and its physical detector-frame interpretation are analyzed in Schlicht 2004, § 4, Eqs. (17)–(20), PDF.

Adversarial control: reverse the idealizations

Use the peak-fixed Gaussian family

$$\chi_T^{\mathrm{peak}}(t)=e^{-t^2/(2T^2)}.$$

Its vacuum response is

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

The $L^2$-normalized Gaussian used in the benchmark is $\chi_T^{(2)}=\pi^{-1/4}T^{-1/2}\chi_T^{\mathrm{peak}}$, hence

$$\mathcal F^{(2)}_{T,L} =\frac{1}{\sqrt\pi\,T} \mathcal F^{\mathrm{peak}}_{T,L}.$$

The normalization factor grows as the duration shrinks. Peak fixing makes the integrated switching area proportional to $T$, whereas $L^2$ fixing raises the peak as $T^{-1/2}$ to keep the squared norm constant. The two families therefore represent different experimental controls even though both are described informally as “short pulses.”

Now reverse the limits:

$$\lim_{T\to0}\lim_{L\to0}\mathcal F^{(2)}_{T,L} =+\infty, \qquad \lim_{L\to0}\lim_{T\to0}\mathcal F^{(2)}_{T,L} =0.$$

For the first order, the pointlike limit leaves $\mathcal F^{\mathrm{peak}}\to1/(4\pi)$ as $T\to0$, and the $L^2$ normalization multiplies it by $1/(\sqrt\pi T)$. For the reverse order, at fixed $L$ one has

$$\mathcal F^{(2)}_{T,L} \sim\frac{T}{4\pi^{3/2}L^2}\longrightarrow0.$$

The two idealizations do not commute. The inner limit is the operation performed first: the left expression removes spatial smearing before shortening the pulse, while the right expression shortens the pulse at fixed nonzero spatial size. Once the first operation has produced a divergence or zero, the outer limit cannot undo it. More importantly, the verdict depends on what the apparatus holds fixed. Peak-fixed switching gives finite but path-dependent limits; $L^2$-fixed switching gives zero in one order and a divergence in the other. Neither family licenses a profile-independent “instantaneous point detector.”

A reproducible error record

For every reported response, record:

1. the formulas and normalization of $\chi$ and $F$;
2. exact supports or a declared tail tolerance;
3. the field state, dimension, mass, and $i0$ prescription;
4. the detector gap, trajectory, coupling, and readout;
5. integration variables, domains, cutoffs, and quadrature error;
6. perturbative truncation separately from numerical error;
7. the order and path of every $T\to0$, $L\to0$, long-time, or continuum limit.

Agreement between two algorithms at the same unresolved cutoff is not an independent continuum check. Numerical error, perturbative error, and model error answer different questions: refining a quadrature can reduce the first, but cannot bound omitted powers of the coupling or justify replacing a compact apparatus by a Gaussian-tailed one. A publishable result belongs either to a finite instrument or to a limit whose normalization and convergence have been demonstrated.

Common pitfalls

Calling compact support smooth. The raised cosine used above is compact and $C^1$, but it is not a test function in $C_0^\infty$. Regularity class and support are separate properties.

Comparing profiles with different strength conventions. Matching peaks, integrated coupling, or $L^2$ norm produces different families. State the invariant quantity before quoting a profile comparison.

Taking a numerical plateau for a continuum limit. A fixed grid can stabilize while the unresolved Fourier tail still dominates. Vary the momentum cutoff and resolution independently.

Exercises

1. Normalize both switching profiles

Verify $\int dt\,|\chi_G|^2=\int dt\,|\chi_C|^2=1$.

Solution

For the Gaussian,

$$\int dt\,|\chi_G|^2 =\frac1{\sqrt\pi T}\int_{-\infty}^{\infty}dt\,e^{-t^2/T^2} =1.$$

For the compact profile, use $\int_{-T}^{T}dt\,\cos^4(\pi t/2T)=3T/4$:

$$\int dt\,|\chi_C|^2 =\frac4{3T}\frac{3T}{4}=1.$$

2. Derive the compact Fourier transform

Show that the raised cosine has the stated transform and determine its large-$|\nu|$ decay.

Solution

Inside the support, $\cos^2(\pi t/2T)=[1+\cos(\pi t/T)]/2$. Integrating the two exponentials gives

$$\int_{-T}^{T}dt\, \cos^2\!\left(\frac{\pi t}{2T}\right)e^{i\nu t} =\frac{a^2\sin(\nu T)}{\nu(a^2-\nu^2)}.$$

Multiplying by $2/\sqrt{3T}$ gives $\widehat\chi_C$. Away from the removable points, the numerator is bounded and the denominator grows as $\nu^3$, so $\widehat\chi_C=O(|\nu|^{-3})$.

3. Diagnose the step-window divergence

Set $L=0$ and show why the rectangular switching has a logarithmically divergent response in four dimensions.

Solution

At large $k$, $\Omega+k\sim k$ and

$$k|\widehat\chi_{\mathrm{step}}(\Omega+k)|^2 \sim\frac{4\sin^2[(k+\Omega)T]}{k}.$$

The fixed phase $\Omega T$ does not change the average of $\sin^2[(k+\Omega)T]$, which is $1/2$ over many oscillations, so the displayed spectral integrand contributes $2\log\Lambda+O(1)$. Restoring the prefactor in $\mathcal F_\chi$ gives $\log\Lambda/(2\pi^2)+O(1)$. Oscillation does not make this nonnegative integral conditionally convergent.

4. Reverse the normalized limits

Starting from $\mathcal F^{(2)}_{T,L}=\mathcal F^{\mathrm{peak}}_{T,L}/(\sqrt\pi T)$, derive the two iterated limits.

Solution

If $L\to0$ first, rescaling $q=Tk$ gives

$$\mathcal F^{\mathrm{peak}}_{T,0}\longrightarrow\frac1{4\pi},$$

so division by $\sqrt\pi T$ diverges. If $T\to0$ first at fixed $L$, then

$$\mathcal F^{\mathrm{peak}}_{T,L} \sim\frac{T^2}{4\pi L^2},$$

and therefore $\mathcal F^{(2)}_{T,L}\sim T/(4\pi^{3/2}L^2)\to0$. Taking $L\to0$ afterwards leaves the already obtained 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.
• 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.
• Schlicht, S. (2004). “Considerations on the Unruh Effect: Causality and Regularization.” Classical and Quantum Gravity 21, 4647–4660. DOI. Open PDF.
• Sriramkumar, L., and Padmanabhan, T. (1996). “Response of Finite-Time Particle Detectors in Non-Inertial Frames and Curved Spacetime.” Classical and Quantum Gravity 13, 2061–2079. DOI. Open PDF.