Regulated soft coherent photon-count prediction

Finite-sector vectors lie in $D(N)$ and are dense by [F1]. For $\mathrm{\Psi },\mathrm{\Phi }\in D(N)$, the Cauchy–Schwarz inequality gives absolute convergence of ${\sum }_{n}n⟨{\mathrm{\Psi }}_n,{\mathrm{\Phi }}_n{⟩}_2$, and because each $n$ is real, $⟨N\mathrm{\Psi },\mathrm{\Phi }{⟩}_2={\sum }_{n}n⟨{\mathrm{\Psi }}_n,{\mathrm{\Phi }}_n{⟩}_2=⟨\mathrm{\Psi },N\mathrm{\Phi }{⟩}_2\mathrm{.}$ Thus $N$ is densely defined and symmetric.

For each Borel $B$, [F2] makes $E_N(B)$ the diagonal multiplier by the real indicator ${\mathbf{1}}_B(n)$. It is bounded with norm at most one, self-adjoint, and idempotent. The same coordinate formula gives $E_N(\mathrm{\varnothing })=0$, $E_N(\mathbb{R})=I$, and $E_N(B)E_N(C)=E_N(B\cap C)$.

Suppose $y\in D(N^{\ast })$ and write $z=N^{\ast }y$. Test [F4] on vectors supported in only sector $n$. Since $N$ acts there as $nI$, the identity for every $x_n\in H_s^{(n)}$ gives $z_n=n{y}_n$. Hence ${\sum }_n{n}^2\mathrm{\parallel }{y}_n{\mathrm{\parallel }}^2={\sum }_n\mathrm{\parallel }{z}_n{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $y\in D(N)$ and $N^{\ast }y=Ny$. Conversely, if $y\in D(N)$, the convergent sector sum in step 1.1 gives $⟨Nx,y{⟩}_2=⟨x,Ny{⟩}_2$ for every $x\in D(N)$; [F4] then gives $y\in D(N^{\ast })$ and $N^{\ast }y=Ny$. Thus $D(N^{\ast })=D(N)$ and $N=N^{\ast }$, which proves self-adjointness by [F5].

Let $(B_j{)}_{j\ge 0}$ be pairwise disjoint Borel sets with union $B$, and fix $\mathrm{\Psi }\in {\mathcal{F}}_s(H)$. Given $ϵ>0$, choose $M$ so that ${\sum }_{n>M}\mathrm{\parallel }{\mathrm{\Psi }}_n{\mathrm{\parallel }}^2<ϵ$. Among the finitely many labels $0,\dots ,M$, each label in $B$ belongs to one $B_j$; choose $J$ at least as large as all those finitely many indices. Then $E_N(B)\mathrm{\Psi }-{\sum }_{j=0}^J{E}_N(B_j)\mathrm{\Psi }$ has zero components through sector $M$, and its squared norm is at most ${\sum }_{n>M}\mathrm{\parallel }{\mathrm{\Psi }}_n{\mathrm{\parallel }}^2<ϵ$. This proves strong countable additivity. Together with step 1.2 it proves the PVM clauses in [F3]. The coordinate formula in [F2] shows this PVM retains exactly the sectors on which $N$ acts by their labels.

In a finite Borel partition $(B_a{)}_{a=1}^m$ of $\mathbb{R}$, distinct sets are disjoint, so step 1.2 gives $E_N(B_a)E_N(B_b)=0$ for $a\mathrm{\ne }b$. Append empty sets to this finite list to apply the countable-additivity result of step 2.2; since the union is $\mathbb{R}$ and $E_N(\mathbb{R})=I$, the finite projections sum to $I$.

For each $i$ and $\mathbf{n}\in S^2$, $p_i\cdot q=\omega E_i(1-{\mathbf{v}}_i\cdot \mathbf{n})\ge \omega E_i(1-\mathrm{\mid }{\mathbf{v}}_i\mathrm{\mid })>0$. Since there are finitely many massive terms, $K^{\mu }(\mathbf{n})={\sum }_i{s}_i{Q}_i(1,{\mathbf{v}}_i)\mathrm{/}(1-{\mathbf{v}}_i\cdot \mathbf{n})$; each denominator is bounded below by $1-\mathrm{\mid }{\mathbf{v}}_i\mathrm{\mid }>0$, so $K$ and its transverse components are bounded. For the measurable orthonormal frame $({ϵ}_1,{ϵ}_2,\mathbf{n})$, ${\sum }_r\mathrm{\mid }\mathbf{K}\cdot {ϵ}_r{\mathrm{\mid }}^2=\mathrm{\mid }\mathbf{K}-(\mathbf{K}\cdot \mathbf{n})\mathbf{n}{\mathrm{\mid }}^2=\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^2$. Thus the integrand is nonnegative and measurable, and compactness of $S^2$ gives $0\le C<\mathrm{\infty }$.

For the one-charge rest-to-velocity current, put $\mu =\mathbf{n}\cdot \widehat{\mathbf{z}}$. Its spatial current is $\mathbf{K}=v\widehat{\mathbf{z}}\mathrm{/}(1-v\mu )$, hence $\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^2=v^2(1-{\mu }^2)\mathrm{/}(1-v\mu {)}^2$. Integrating over azimuth and then setting $x=1-v\mu$ gives $I(v):={\int }_{-1}^1\frac{v^2(1-{\mu }^2)}{(1-v\mu {)}^2}d\mu =\frac{1}{v}{\int }_{1-v}^{1+v}(\frac{v^2-1}{x^2}+\frac{2}{x}-1)dx=\frac{1}{v}{[\frac{1-v^2}{x}+2\log x-x]}_{1-v}^{1+v}=\frac{2}{v}\log \frac{1+v}{1-v}-4.$ Therefore $C(v)=\frac{g_{\mathrm{e}\mathrm{m}}^2}{2(2\pi {)}^3}(2\pi I(v))=\frac{\alpha }{\pi }(\frac{1}{v}\log \frac{1+v}{1-v}-2)\mathrm{.}$ For $f(v)=\log ((1+v)\mathrm{/}(1-v))-2v$, $f(0)=0$ and $f^{\mathrm{\prime }}(v)=2v^2\mathrm{/}(1-v^2)>0$ on $(0,1)$, so $C(v)>0$.

The fixed-current, charge-balance, transverse-polarization, and cutoff hypotheses match [L1]. The current formula simplifies to $J(q)=K(\mathbf{n})\mathrm{/}\omega$, and step 1.1 identifies the polarization sum with $\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^2$, exactly the angular convention of [L1]. This is a conditional use of the inspected draft carrier at its actual interface.

For each spatial direction $n$ where the prescribed real polarization frame is orthonormal, define $R_n:{\mathbb{C}}^2\to n^{\perp }\subset {\mathbb{C}}^3$ by $R_{n}a={\sum }_r{a}_r{ϵ}_r(n)$. Direct expansion gives $\overline{R_{n}a}\cdot R_{n}b={\sum }_r{\bar{a}}_r{b}_r$. For every complex transverse vector $v$, its real and imaginary parts are in the real plane spanned by the two frame vectors, so $v={\sum }_r({ϵ}_r\cdot v){ϵ}_r$; these coefficients give the inverse. The frame is measurable, hence $Rh(\mathbf{q})=R_{\mathbf{q}\mathrm{/}\mathrm{\mid }\mathbf{q}\mathrm{\mid }}h(\mathbf{q})$ is a measurable unitary from ${\mathfrak{h}}_E$ onto [L1]'s ${\mathbb{C}}^3$ transverse energy space, preserving the weighted integral pairing; its inverse has components ${ϵ}_r\cdot v$ and is also measurable. Composing $R$ with the unitary $U$ of [L1] gives precisely the displayed $U_E$ with its unchanged factor $\mathrm{\hslash }c\mathrm{/}\sqrt{(2\pi {)}^{3}2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}$. Because $J\cdot {ϵ}_r=-\mathbf{K}\cdot {ϵ}_r\mathrm{/}\omega$, the prescribed amplitude satisfies $R{h}_{\lambda ,\mathrm{\Lambda }}=-g_{\mathrm{e}\mathrm{m}}{K}_{\perp }\mathrm{/}\omega$ on the cutoff shell. With $g>0$ from [F2], put $\sigma =-g_{\mathrm{e}\mathrm{m}}\mathrm{/}g\in \{-1,+1\}$. Thus $U_E{h}_{\lambda ,\mathrm{\Lambda }}=\sigma f_{\lambda }$ in [L1]'s notation, and its exact squared norm is $N_{\gamma }$. This sign does not change any sector norm or energy: on sector $n$ it multiplies the coherent tensor by ${\sigma }^n$, and this direct-sum unitary preserves the energy domain and commutes with the sector energy multipliers. Consequently [L1]'s coherent-sector and energy formulas apply to the normalized vector in this Statement. These uses require only its displayed mathematical interfaces, not its separate infrared clause.

By [F5]–[F6], $P_n$ is an orthogonal projection and $I-P_n$ is its orthogonal complement. The two Kraus operators $P_n$ and $I-P_n$ are bounded and satisfy $P_n^{\ast }{P}_n+(I-P_n{)}^{\ast }(I-P_n)=P_n+(I-P_n)=I$, so they form a normalized finite projective instrument. The state from step 3.1 is a unit vector. Put ${\rho }_{\lambda ,\mathrm{\Lambda }}:=T_{{\mathrm{\Psi }}_{\lambda ,\mathrm{\Lambda }},{\mathrm{\Psi }}_{\lambda ,\mathrm{\Lambda }}}$, with $T_{\mathrm{\Psi },\mathrm{\Psi }}x=\mathrm{\Psi }⟨\mathrm{\Psi },x{⟩}_2$; by the unit-vector case in [A3], this is the density-operator input to the Born postulate. Applying its outcome formula to the first Kraus operator $P_n$ gives $p_n=\mathrm{tr}({\rho }_{\lambda ,\mathrm{\Lambda }}{P}_n)=\mathrm{\parallel }{P}_n{\mathrm{\Psi }}_{\lambda ,\mathrm{\Lambda }}{\mathrm{\parallel }}^2$. This uses the finite two-outcome partition for each singleton event and assumes no countably infinite Kraus instrument.

The expected photon energy is the energy-coordinate expectation from [L1], transferred to the adopted free Hamiltonian $H_{\gamma }$ by its unitary map and [A2]. The cutoffs are energies in joules, so $C(\mathrm{\Lambda }-\lambda )$ has unit joule.

Assume $C>0$ for the nonconvergence claim. The Segal–Weyl supplier [F7] requires a separable complex Hilbert space; [F8] verifies this for ${\mathfrak{h}}_T$, and $u_j\in {\mathfrak{h}}_T$ by [L1]. Let ${\lambda }_j=\mathrm{\Lambda }{2}^{-j}$ for $j\ge 1$, and set $u_j=U_E{h}_{{\lambda }_j,\mathrm{\Lambda }}$ and $V_j=W(-i\sqrt{2}{u}_j)$. The difference $u_{j+1}-u_j$ is the same current amplitude restricted to the shell ${\lambda }_{j+1}<\omega <{\lambda }_j$. Applying [L1]'s cutoff norm formula to that shell gives $\mathrm{\parallel }{u}_{j+1}-u_j{\mathrm{\parallel }}^2=C\log ({\lambda }_j\mathrm{/}{\lambda }_{j+1})=C\log 2$. The explicit exponential-vector formula in [F7] gives ${\mathrm{\Psi }}_{{\lambda }_j,\mathrm{\Lambda }}=V_j\mathrm{\Omega }$ for either sign of the amplitude. Unitarity and the Weyl product law in [F7] give $V_j^{\ast }{V}_{j+1}=e^{i{\theta }_j}W(-i\sqrt{2}(u_{j+1}-u_j))$ for some real phase ${\theta }_j$. Applying the vacuum expectation formula in [F7] yields $\mathrm{\mid }⟨{\mathrm{\Psi }}_{{\lambda }_j,\mathrm{\Lambda }},{\mathrm{\Psi }}_{{\lambda }_{j+1},\mathrm{\Lambda }}{⟩}_2\mathrm{\mid }=e^{-\mathrm{\parallel }{u}_{j+1}-u_j{\mathrm{\parallel }}^2\mathrm{/}2}=e^{-C\log 2\mathrm{/}2}<1.$

The projection $P_n$ retains exactly sector $n$ by [F5]. The sector formula from [L1] gives $P_n{\mathrm{\Psi }}_{\lambda ,\mathrm{\Lambda }}=e^{-N_{\gamma }\mathrm{/}2}(U_E{h}_{\lambda ,\mathrm{\Lambda }}{)}^{{\otimes }_{s}n}\mathrm{/}\sqrt{n!}$. The symmetric tensor norm is multiplicative and $U_E$ preserves the one-particle norm, so $p_n=e^{-N_{\gamma }}{N}_{\gamma }^n\mathrm{/}n!$.

The exponential series gives ${\sum }_{n\ge 0}{p}_n=e^{-N_{\gamma }}{e}^{N_{\gamma }}=1$ and, by reindexing its nonnegative terms, ${\sum }_{n\ge 0}n{p}_n=e^{-N_{\gamma }}{N}_{\gamma }{e}^{N_{\gamma }}=N_{\gamma }$. The zero-sector term is $e^{-N_{\gamma }}$, including $N_{\gamma }=0$ under $0^0=1$.

For unit vectors $x,y$, expansion of $\mathrm{\parallel }x-\zeta y{\mathrm{\parallel }}^2$ and choice of $\zeta$ to align their inner product give ${\inf }_{\mathrm{\mid }\zeta \mathrm{\mid }=1}\mathrm{\parallel }x-\zeta y{\mathrm{\parallel }}^2=2-2\mathrm{\mid }⟨x,y{⟩}_2\mathrm{\mid }$. Step 4.3 therefore gives the fixed positive squared adjacent-ray distance $2(1-e^{-C\log 2\mathrm{/}2})$ along ${\lambda }_j$. A projective-ray limit would make these adjacent distances tend to zero, a contradiction. Ordinary Fock-norm convergence is the special case without phase adjustment and is also impossible. The energy from step 4.2 tends to the finite value $C\mathrm{\Lambda }$; nonconvergence is not caused by divergent mean energy.