Dressed Charges, Soft Sectors, and Infrared Information

A charged excitation in a massless gauge theory is inseparable from its long-range field. Bare Fock amplitudes are infrared divergent, inclusive calculations trace unresolved radiation, and dressed constructions attach coherent soft clouds to asymptotic charges. These are different descriptions of a measurement task. Information quantities become meaningful only after the measured algebra, detector response, dressing, and order of limits have been stated.

Required background. Superselection and accessible entanglement supplies sector-conditioned information; reference frames and asymmetry supplies relational charge resources.

Helpful background. Zero modes, boundaries, and infrared effects supplies the regulator boundary.

Chapter map. The overview gives the task-to-resource map, the comparison table, and the three gates for an operational claim.

Inclusive and dressed descriptions

For two hard scattering branches, write the outgoing state schematically as

$$|\Psi_{\rm out}\rangle =c_0|h_0\rangle|s_0\rangle +c_1|h_1\rangle|s_1\rangle.$$

Tracing the unobserved radiation gives

$$(\rho_{\rm hard})_{ij} =c_i c_j^*\langle s_j|s_i\rangle.$$

In the inclusive $S$-matrix treatment, the off-diagonal coefficient scales with the infrared regulator as $\lambda^{\Delta A+\Delta B}$. It survives $\lambda\to0$ precisely when the angle-dependent electromagnetic and gravitational currents of the two hard branches agree. Carney et al. 2017, Eqs. (9)–(12) and the current-matching statement following Eq. (12), pp. 3–4 establish this asymptotic result.

For one incoming and one outgoing massive charge in QED, two outgoing momenta $q$ and $q'$ give

$$\Delta A_{qq',p} =\frac{e^2}{8\pi^2} \left[ \frac{1}{\beta_{qq'}} \log\!\left(\frac{1+\beta_{qq'}}{1-\beta_{qq'}}\right)-2 \right],$$

where $\beta_{qq'}$ is their relative speed. The exponent is nonnegative and vanishes only for $q=q'$. This is Carney et al. 2017, Eq. (16), p. 4.

Faddeev–Kulish-type descriptions instead build coherent soft fields into the asymptotic charged states. Kulish and Faddeev 1970, §§ 2–3 construct the asymptotic dynamics, while Carney et al. 2018, § II, Eqs. (10)–(13), and § III derive the dressed hard density matrix and interpret the vanishing cloud overlap in the infinite-time limit.

These papers do not prove a universal finite-time detector formula. The inclusive calculation uses the infinite-time $S$-matrix, and Carney et al. 2017, final paragraph before the black-hole discussion explicitly leaves its dependence on that approximation open. A finite-time statement therefore needs a detector model rather than the replacement of a smooth response by a sharp lower-frequency cutoff.

A smooth finite-time response model

The following finite benchmark is a declared operational model, not a universal QED theorem. Let the detector distinguish radiation at angular frequency $\omega$ with weight

$$w_T(\omega):=1-e^{-(\omega T)^2}.$$

It is insensitive as $w_T(\omega)\sim(\omega T)^2$ for $\omega T\ll1$ and resolves modes smoothly for $\omega T\gg1$. Retaining the leading logarithmic soft spectrum, define the hard-coherence visibility

$$\eta(T,\lambda,\Delta E) :=\exp\!\left[ -\Delta A \int_{\lambda}^{\Delta E} \frac{d\omega}{\omega}\,w_T(\omega) \right].$$

The integral is elementary:

$$I(T,\lambda,\Delta E) =\log\frac{\Delta E}{\lambda} -\frac12 \left[ \operatorname{Ei}(-\Delta E^2T^2) -\operatorname{Ei}(-\lambda^2T^2) \right].$$

At fixed finite $T$, the regulator can be removed:

$$I_T(x) :=\lim_{\lambda\to0}I =\log x+\frac{\gamma_{\rm E}}{2} -\frac12\operatorname{Ei}(-x^2), \qquad x:=\Delta E T.$$

Thus the detector, not an artificial hard band $1/T<\omega<\Delta E$, supplies the operational resolution. For $x\ll1$, $I_T(x)=x^2/2+O(x^4)$; for $x\gg1$, $I_T(x)=\log x+\gamma_{\rm E}/2+O(e^{-x^2}/x^2)$.

Executed hard, soft, and inclusive benchmark

Take equal hard amplitudes and a checked QED exponent $\Delta A=0.10$. The hard reduced state in the branch basis is

$$\rho_{\rm hard} =\frac12 \begin{pmatrix} 1&\eta\\ \eta&1 \end{pmatrix}.$$

Three complementary quantities are then exact within the response model:

$$F_+^2 :=\langle+|\rho_{\rm hard}|+\rangle =\frac{1+\eta}{2},$$ $$P_{\rm guess}^{\rm soft} =\frac12\left(1+\sqrt{1-\eta^2}\right),$$

and $\operatorname{Tr}|\Psi_{\rm out}\rangle\langle\Psi_{\rm out}|^2=1$ for the inclusive hard-plus-soft state. The first is explicitly the squared fidelity with the coherent hard state; the second is the Helstrom success probability for equal-prior pure soft clouds.

$x=\Delta E T$	$I_T(x)$	Hard visibility $\eta$	$F_+^2$	$P_{\rm guess}^{\rm soft}$	Inclusive purity
$1$	$0.398300$	$0.960953$	$0.980476$	$0.638356$	$1$
$10$	$2.591193$	$0.771731$	$0.885865$	$0.817975$	$1$
$100$	$4.893778$	$0.613008$	$0.806504$	$0.895038$	$1$

Longer observation makes the two radiation records more distinguishable and reduces hard coherence, while the inclusive state remains pure. This table does not say that every infrared sector is a finitely readable classical label; it reports one specified response and one two-branch leading-log model.

As an independent finite mode-split check, take two coherent clouds with squared phase-space distance $\|\delta\alpha\|^2=4$ and assign one quarter of that distance to detector-resolved modes. The unresolved modes leave hard coherence

$$\eta_{\rm unres}=e^{-(3/4)4/2}=e^{-3/2}=0.223130,$$

while the detector records have overlap $\eta_{\rm det}=e^{-(1/4)4/2}=e^{-1/2}=0.606531$. Their product is the full-cloud overlap $e^{-2}=0.135335$, as required by factorization over independent coherent modes.

Executed resolution and dressing adversaries

Remove finite resolution. If the response is replaced by $w_T(\omega)=1$ all the way to the regulator, then

$$\eta_{\rm ideal}(\lambda) =\left(\frac{\lambda}{\Delta E}\right)^{\Delta A} \longrightarrow0.$$

The finite regulator limit of the operational model is lost. The strongest surviving statement is the asymptotic current-matching theorem: unequal currents have vanishing off-diagonal coefficients in that ideal limit. It does not determine a finite detector’s discrimination error.

Change the dressing-defined split. As a finite control, take two normalized cloud states with overlap $\eta=0.6$. The hard eigenvalues are $0.8$ and $0.2$, so

$$S_{\rm hard}=H_2(0.8)=0.721928\ \text{bits}.$$

A branch-dependent change of dressing that makes the recorded cloud states orthogonal gives $S_{\rm hard}=1$ bit under the newly declared trace. This does not change an inclusive prediction when state and observable algebra are transformed consistently; it changes the hard/soft subsystem prescription. Holding the old observable algebra fixed while changing only the state is a different physical preparation, not a harmless convention change.

The robust claim is therefore narrower than a “soft-photon entropy”: inclusive observables can be infrared safe, while hard-state coherence and entropy depend on the dressing, response, and traced algebra.

Common pitfalls

Using $1/T$ as an exact spectral wall. A finite switching function produces a Fourier response with tails. A sharp band is an additional model assumption.

Confusing orthogonality with accessible distinguishability. Orthogonal asymptotic representations and a finite detector’s error probability are statements about different algebras and limits.

Changing a dressing on only one side of a prediction. A representation change transforms states and observables together. A fixed partial trace can make a convention-dependent subsystem entropy appear physical.

Exercises

1. Positivity of the one-charge exponent

Show that $\Delta A_{qq',p}\ge0$ for $0\le\beta_{qq'}<1$, with equality only at $\beta_{qq'}=0$.

Solution

Use

$$\frac{1}{\beta}\log\frac{1+\beta}{1-\beta} =\frac{2\operatorname{artanh}\beta}{\beta} =2\sum_{k=0}^{\infty}\frac{\beta^{2k}}{2k+1}.$$

The first term is $2$ and every remaining term is nonnegative. The bracket defining $\Delta A$ is therefore nonnegative and is zero only when $\beta=0$, which for equal masses means equal momenta.

2. Remove the infrared regulator

Derive the finite expression for $I_T(x)$ from the regulated integral.

Solution

Since

$$\int\frac{d\omega}{\omega}e^{-\omega^2T^2} =\frac12\operatorname{Ei}(-\omega^2T^2),$$

the stated regulated result follows. For small positive $z$,

$$\operatorname{Ei}(-z)=\gamma_{\rm E}+\log z-z+O(z^2).$$

Substituting $z=\lambda^2T^2$ cancels the $\log\lambda$ term and gives

$$I_T(x)=\log x+\frac{\gamma_{\rm E}}2-\frac12\operatorname{Ei}(-x^2).$$

3. Information complementarity

For the two-branch pure state, show that the hard visibility and the optimal soft-cloud discrimination obey

$$\left(2P_{\rm guess}^{\rm soft}-1\right)^2+\eta^2=1.$$

Solution

The trace distance between two equal-prior pure states with overlap magnitude $\eta$ is $\sqrt{1-\eta^2}$. Helstrom’s formula therefore gives

$$2P_{\rm guess}^{\rm soft}-1=\sqrt{1-\eta^2}.$$

Squaring and adding $\eta^2$ proves the identity. It quantifies the same which-path/coherence tradeoff seen in the benchmark table.

References

• Carney, Daniel, Laurent Chaurette, Dominik Neuenfeld, and Gordon Walter Semenoff. “Dressed Infrared Quantum Information.” Physical Review D 97 (2018): 025007. DOI. Open PDF.
• Carney, Daniel, Laurent Chaurette, Dominik Neuenfeld, and Gordon Walter Semenoff. “Infrared Quantum Information.” Physical Review Letters 119 (2017): 180502. DOI. Open PDF.
• Kulish, P. P., and L. D. Faddeev. “Asymptotic Conditions and Infrared Divergences in Quantum Electrodynamics.” Theoretical and Mathematical Physics 4 (1970): 745–757. DOI.