Soft coherent radiation, resolution and inclusive cancellation

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$: every sequence $(X_n{)}_{n\in \mathbb{N}}$ of nonempty sets admits a function $s$ with $s(n)\in X_n$ for every $n$. Use the following SI constants and meanings: $c>0$ is the speed of light in vacuum, in m s${}^{-1}$; $\mathrm{\hslash }>0$ is the reduced Planck constant, in J s; $e>0$ is the magnitude of the elementary charge, in C; and ${\mu }_0>0$ is the vacuum permeability, in H m${}^{-1}$, used here as a positive parameter without assigning it a numerical value. Set $g=e\sqrt{\frac{{\mu }_{0}c}{\mathrm{\hslash }}}>0,$ which is dimensionless. These are the adopted QFT SI conventions.

Let $I$ be a finite, possibly empty, set of leg labels. For each $i\in I$, choose an energy $E_i>0$ in J and a physical velocity ${\mathbf{v}}_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\in {\mathbb{R}}^3$ in m s${}^{-1}$ with $\mathrm{\mid }{\mathbf{v}}_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{\mid }<c$. Write $v_i={\mathbf{v}}_{i,\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}c\in {\mathbb{R}}^3$, so $v_i$ is dimensionless and $\mathrm{\mid }{v}_i\mathrm{\mid }<1$. Also choose a dimensionless charge number $Q_i\in \mathbb{R}$ (the assigned physical charge is $e{Q}_i$ in C) and an orientation ${\eta }_i\in \{-1,1\}$, where $-1$ denotes an incoming leg and $+1$ an outgoing leg. Assume charge balance ${\sum }_{i\in I}{\eta }_i{Q}_i=0$. For four-vectors $a=(a^0,\mathbf{a}),b=(b^0,\mathbf{b})\in {\mathbb{R}}^4$, write $h=\mathrm{diag}(1,-1,-1,-1)$ and $h(a,b)=a^0{b}^0-\mathbf{a}\cdot \mathbf{b}$ for the algebraic contraction with signature $(+,-,-,-)$ used in this model.

For $n\in S^2=\{x\in {\mathbb{R}}^3:\mathrm{\mid }x\mathrm{\mid }=1\}$ and angular frequency $\nu >0$ in s${}^{-1}$, with associated model photon energy $\mathrm{\hslash }\nu$ in J, put $p_i=E_i(1,v_i)$ and $q=\mathrm{\hslash }\nu (1,n)$ in ${\mathbb{R}}^4$. On $\mathcal{Q}=\{\mathrm{\hslash }\nu (1,n):\nu >0,n\in S^2\}\subset {\mathbb{R}}^4,$ define the prescribed current $J:\mathcal{Q}\to {\mathbb{R}}^4$ and angular coefficient $K:S^2\to {\mathbb{R}}^4$ by $J(q)={\sum }_{i\in I}{\eta }_i{Q}_i\frac{p_i}{h(p_i,q)},K(n)={\sum }_{i\in I}{\eta }_i{Q}_i\frac{(1,v_i)}{1-v_i\cdot n}\mathrm{.}$ Because $\mathrm{\mid }{v}_i\cdot n\mathrm{\mid }\le \mathrm{\mid }{v}_i\mathrm{\mid }<1$, every denominator $1-v_i\cdot n$ is strictly positive. The data make $h(p_i,q)=E_i\mathrm{\hslash }\nu (1-v_i\cdot n)>0$ and $J(q)=K(n)\mathrm{/}(\mathrm{\hslash }\nu )$. The balance condition gives $h(J(q),q)=0$; this is the conservation convention for this prescribed eikonal current in the model. Here “prescribed eikonal current” means exactly the displayed rational map $J$ on these model photon energy four-vectors; no scattering-amplitude factorization or approximation is asserted.

Let $A\subseteq S^2$ be a Borel angular acceptance region. The sphere $S^2\subset {\mathbb{R}}^3$ is a compact embedded smooth hypersurface, and $d\mathrm{\Omega }$ denotes its Euclidean chart surface measure: for a finite regular injective $C^1$ chart cover $X_j:V_j\to S^2$, where each $V_j$ is open in ${\mathbb{R}}^2$, with $C^1$ coordinate transitions and a subordinate ambient partition of unity ${\chi }_j$ whose supports are compactly contained in the chart neighborhoods, its integral on a nonnegative Borel function $F$ is ${\int }_{S^2}Fd\mathrm{\Omega }={\sum }_j{\int }_{V_j}({\chi }_{j}F)(X_j(y))\sqrt{\det (D{X}_j(y{)}^{T}D{X}_j(y))}dy\mathrm{.}$ This is the chart-independent induced surface integral. Write $K(n)=(K^0(n),\mathbf{K}(n))$ and define $K_{\perp }(n)=(I_3-n{n}^T)\mathbf{K}(n)$, where $I_3$ is the $3\times 3$ identity matrix. Put $b_0(n)=\frac{g{K}_{\perp }(n)}{\sqrt{2(2\pi {)}^3}},b(n)={\mathbf{1}}_A(n)b_0(n),C={\int }_{S^2}\mathrm{\mid }b(n){\mathrm{\mid }}^{2}d\mathrm{\Omega }(n)\mathrm{.}$ The earlier current theorem proves $b_0\in L^2(S^2,d\mathrm{\Omega };{\mathbb{R}}^3)$ for these finite balanced subluminal legs; since $A$ is Borel, $b\in L^2$ and $0\le C\le {\int }_{S^2}\mathrm{\mid }{b}_0{\mathrm{\mid }}^{2}d\mathrm{\Omega }<\mathrm{\infty }$. Thus $b$ and $C$ are dimensionless.

For the one-particle space, write $x\cdot y={\sum }_{j=1}^3{x}_j{y}_j(x,y\in {\mathbb{R}}^3),a\cdot b={\sum }_{j=1}^3{a}_j{b}_j(a,b\in {\mathbb{C}}^3),$ where the second dot product is the complex-bilinear extension. The fiber Hermitian pairing is $\bar{a}\cdot b={\sum }_j\overline{a_j}{b}_j$, with norm $\mathrm{\mid }a{\mathrm{\mid }}^2={\sum }_j\mathrm{\mid }{a}_j{\mathrm{\mid }}^2$. Set $\mathcal{H}=L^2({\mathbb{R}}^3,d^3\mathbf{k};{\mathbb{C}}^3),⟨u,v{⟩}_2={\int }_{{\mathbb{R}}^3}\overline{u(\mathbf{k})}\cdot v(\mathbf{k})d^3\mathbf{k},$ Here $d^3\mathbf{k}$ is Lebesgue measure, and this pairing is second-variable-linear. For $\mathbf{k}\mathrm{\ne }0$, put $\widehat{\mathbf{k}}=\mathbf{k}\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ and $P_T(\mathbf{k})=I_3-\widehat{\mathbf{k}}{\widehat{\mathbf{k}}}^T$, the Euclidean orthogonal projection onto the fiber plane perpendicular to $\mathbf{k}$; assign any value at $\mathbf{k}=0$. Its multiplier is an orthogonal projection on $\mathcal{H}$, so its range is closed. The transverse one-particle Hilbert space is ${\mathcal{H}}_T=\mathrm{Ran}{M}_{P_T}=\{u\in \mathcal{H}:\mathbf{k}\cdot u(\mathbf{k})=0\text{ for a.e. }\mathbf{k}\mathrm{\ne }0\},$ where $M_{P_T}$ is multiplication by $P_T$, and the real Euclidean dot product is extended complex-bilinearly in the transverse condition. This is the transverse one-particle space used in this model.

For $m\ge 1$, use the iterated Hilbert tensor completion ${\mathcal{H}}_T^{\widehat{\otimes }m}$, with product pairing on elementary tensors. Here $S_m$ is the permutation group and $U_{\pi }(u_1\otimes \cdots \otimes u_m)=u_{{\pi }^{-1}(1)}\otimes \cdots \otimes u_{{\pi }^{-1}(m)}$. Set $P_{s,m}=\frac{1}{m!}{\sum }_{\pi \in S_m}{U}_{\pi },{\mathcal{H}}_s^{(m)}=\mathrm{Ran}{P}_{s,m};$ put $P_{s,0}=I_{\mathbb{C}}$ and ${\mathcal{H}}_s^{(0)}=\mathbb{C}$. The symmetric bosonic Fock space is ${\mathcal{F}}_s({\mathcal{H}}_T)=\{({\mathrm{\Psi }}_m{)}_{m\ge 0}:{\mathrm{\Psi }}_m\in {\mathcal{H}}_s^{(m)},{\sum }_{m\ge 0}\mathrm{\parallel }{\mathrm{\Psi }}_m{\mathrm{\parallel }}^2<\mathrm{\infty }\},$ with componentwise pairing and vacuum $1\in \mathbb{C}$ in sector zero. For $f\in {\mathcal{H}}_T$, define the exponential-vector map $ϵ:{\mathcal{H}}_T\to {\mathcal{F}}_s({\mathcal{H}}_T)$ by $ϵ(f)={⨁}_{m=0}^{\mathrm{\infty }}\frac{P_{s,m}(f^{\otimes m})}{\sqrt{m!}}\mathrm{.}$ Here $f^{\otimes 0}=1$ in the vacuum sector. Since $f^{\otimes m}$ is already symmetric, $P_{s,m}(f^{\otimes m})=f^{\otimes m}$ for $m\ge 1$; write this tensor as $f^{{\otimes }_{s}m}$.

For fixed angular-frequency cutoffs $0<\epsilon <\mathrm{\Delta }<\mathrm{\Lambda }$, let $\nu (\mathbf{k})=c\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ in s${}^{-1}$, with $\mathbf{k}$ in m${}^{-1}$. The one-particle coherent displacement is the transverse mode function $f_{\epsilon ,\mathrm{\Lambda }}(\mathbf{k})={\mathbf{1}}_{\{\epsilon <\nu (\mathbf{k})<\mathrm{\Lambda }\}}\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^{-3\mathrm{/}2}\times b\text{\hspace{0.17em}⁣}\left(\frac{\mathbf{k}}{\mathrm{\mid }\mathbf{k}\mathrm{\mid }}\right)(\mathbf{k}\mathrm{\ne }0),f_{\epsilon ,\mathrm{\Lambda }}(0)=0.$ The function $f_{\epsilon ,\mathrm{\Lambda }}$ has units m${}^{3\mathrm{/}2}$ in ${\mathcal{H}}_T$; the associated free one-photon energy is $\mathrm{\hslash }\nu (\mathbf{k})=\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ in J. Define its normalized coherent vector by ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Lambda }}=\exp \text{\hspace{0.17em}⁣}(-\frac{\mathrm{\parallel }{f}_{\epsilon ,\mathrm{\Lambda }}{\mathrm{\parallel }}^2}{2})ϵ(f_{\epsilon ,\mathrm{\Lambda }})\mathrm{.}$ For a finite partition $B_1,\dots ,B_r$ of $(\epsilon ,\mathrm{\Lambda })$ into disjoint Borel intervals, let $H_j=\{u\in {\mathcal{H}}_T:u(\mathbf{k})=0\text{ a.e. unless }\nu (\mathbf{k})\in B_j\}$. Let $P_j$ be multiplication on ${\mathcal{H}}_T$ by ${\mathbf{1}}_{\{\nu (\mathbf{k})\in B_j\}}$. This bounded self-adjoint idempotent preserves transversality because it is scalar-valued, so it is an orthogonal projection. Its range is exactly $H_j$, making $H_j$ closed. For $j\mathrm{\ne }k$, disjointness gives $P_j{P}_k=0$, hence the ranges are orthogonal. For a tuple $(m_1,\dots ,m_r)\in {\mathbb{N}}_0^r$, put $M={\sum }_j{m}_j$. Its occupation subspace is the closed span in sector ${\mathcal{H}}_s^{(M)}$ of the vectors $P_{s,M}\left({⨂}_{j=1}^r{⨂}_{\mathrm{\ell }=1}^{m_j}{u}_{j,\mathrm{\ell }}\right),u_{j,\mathrm{\ell }}\in H_j,$ where the empty tensor for $M=0$ is the vacuum $1\in \mathbb{C}$. It has exactly $m_j$ factors in $H_j$ and no factor outside ${⨁}_j{H}_j$. Define $N(B_j)$ to have value $m_j$ on this subspace. The model assigns the tuple probability equal to the squared norm of the projection of ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Lambda }}$ onto it. Its resolution parameter $\mathrm{\Delta }$ divides $(\epsilon ,\mathrm{\Lambda })$ into the unresolved band $(\epsilon ,\mathrm{\Delta })$ and the resolved band $[\mathrm{\Delta },\mathrm{\Lambda })$. The threshold frequency is assigned to the resolved band; its exact assignment does not change any $L^2$ vector or probability because a single frequency has zero one-particle measure. This is a finite-cutoff free coherent-state model; it does not posit a detector response or an interacting QED scattering law.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

QFT example units, metric, gauge, and pairing conventions

The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$

Schwartz/distribution spaces and Hilbert Fock conventions

Positive transverse photon one-particle space

Free transverse-photon energy multiplier

Surface integration on compact C1 hypersurfaces

Agreement with the existing polar sphere measure

Soft-current angular norm and exact cutoff coherent model

Dereziński, §§5.3–5.6

PTB Quantum Optics, Lecture 5

The well-definedness of $C$, the transverse $L^2$ vector $f_{\epsilon ,\mathrm{\Lambda }}$, the normalized exponential vector, and the band counting probabilities is established in Orthogonal frequency bands in the coherent soft model have independent Poisson counts ↗. The exact formulas are conditional on the stated finite leg data and cutoffs. They introduce no empirical photon counts, no virtual corrections, and no claim of full Bloch–Nordsieck or KLN cancellation.

Fix the finite-leg current data and cutoffs of the resolved-soft model. For every finite partition $B_1,\dots ,B_r$ of $(\epsilon ,\mathrm{\Lambda })$ into disjoint Borel frequency intervals, the coherent state ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Lambda }}$ has independent band counts. For each $j$, $N(B_j)$ has Poisson distribution with mean ${\lambda }_j=C{\int }_{B_j}\frac{d\nu }{\nu },$ where $\nu$ is angular frequency in s${}^{-1}$. Thus ${\lambda }_j=C\log (b_j\mathrm{/}{a}_j)$ for an interval with endpoints $0<a_j<b_j$; endpoint conventions do not matter because $d\nu \mathrm{/}\nu$ assigns a singleton zero measure. For $C=0$ every band count is zero. The model's marked frequency density is the one-particle coherent intensity $\mathrm{\mid }b(n){\mathrm{\mid }}^{2}d\mathrm{\Omega }(n)d\nu \mathrm{/}\nu$.

Dereziński, §§5.3–5.6

PTB Quantum Optics, Lecture 5

In the defined finite-cutoff coherent model, let $0<\epsilon <\mathrm{\Delta }<\mathrm{\Lambda }$ and $C\ge 0$. Let $N_u$ and $N_r$ be the counts in the unresolved band $(\epsilon ,\mathrm{\Delta })$ and resolved band $[\mathrm{\Delta },\mathrm{\Lambda })$, respectively. The exclusive no-emission probability is $\mathbb{P}(N_u=0,N_r=0)=\exp \text{\hspace{0.17em}⁣}(-C\log \frac{\mathrm{\Lambda }}{\epsilon })\mathrm{.}$ For every $n\in {\mathbb{N}}_0$, $\mathbb{P}(N_u=n,N_r=0)=\exp \text{\hspace{0.17em}⁣}(-C\log \frac{\mathrm{\Lambda }}{\epsilon })\frac{{(C\log \frac{\mathrm{\Delta }}{\epsilon })}^n}{n!}\mathrm{.}$ Summing over all unresolved counts gives the finite-resolution inclusive probability $\mathbb{P}(N_r=0)=\exp \text{\hspace{0.17em}⁣}(-C\log \frac{\mathrm{\Lambda }}{\mathrm{\Delta }}),$ which is independent of the lower cutoff $\epsilon$. This is an exact identity in the defined free coherent model; it does not assert real–virtual cancellation in interacting QED or a detector's empirical response.

Dereziński, §§5.3–5.6

Srednicki, Chapter 26

The sketch records only the interval partition used above; the proof does not rely on the diagram. The boundary frequency $\mathrm{\Delta }$ is assigned to the resolved interval, and its zero measure is why this convention does not change the calculation.

In the defined coherent model, put $f_{\epsilon ,\mathrm{\Delta }}={\mathbf{1}}_{\{\epsilon <\nu (\mathbf{k})<\mathrm{\Delta }\}}{f}_{\epsilon ,\mathrm{\Lambda }}$ and ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}=\exp (-\mathrm{\parallel }{f}_{\epsilon ,\mathrm{\Delta }}{\mathrm{\parallel }}^2\mathrm{/}2)ϵ(f_{\epsilon ,\mathrm{\Delta }})$, the normalized coherent Fock vector on the unresolved band $(\epsilon ,\mathrm{\Delta })$. For $C\ge 0$ its mean band count and free-photon energy expectation are ${\mathbb{E}}_{{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}}{N}_{(\epsilon ,\mathrm{\Delta })}=C\log \frac{\mathrm{\Delta }}{\epsilon },{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}\in D(H_{\gamma }),⟨{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }},H_{\gamma }{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}⟩=\mathrm{\hslash }C(\mathrm{\Delta }-\epsilon ),$ where $H_{\gamma }=\mathrm{\hslash }cd\mathrm{\Gamma }(\mathrm{\mid }\mathbf{k}\mathrm{\mid })$ and $\nu =c\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ is angular frequency. If $C>0$, the mean number diverges as $\epsilon \downarrow 0$, while the energy expectation remains bounded by $\mathrm{\hslash }C\mathrm{\Delta }$ and tends to $\mathrm{\hslash }C\mathrm{\Delta }$. The one-particle displacement has $\mathrm{\parallel }{f}_{\epsilon ,\mathrm{\Delta }}{\mathrm{\parallel }}^2=C\log (\mathrm{\Delta }\mathrm{/}\epsilon )$; when $C>0$, it has no limit in one-particle Hilbert norm as $\epsilon \downarrow 0$. If $C=0$, then $b=0$ almost everywhere and $f_{\epsilon ,\mathrm{\Delta }}$ is the zero one-particle vector at every cutoff, so it converges in that norm to zero. This is an exact free coherent-state model statement; it asserts no empirical photon statistics, no interacting QED limit, and no implication about convergence of a different physical state.

Dereziński, §§5.3–5.6

PTB Quantum Optics, Lecture 5

Srednicki, Chapter 26