Orthogonal frequency bands in the coherent soft model have independent Poisson counts

For every leg and every $n\in S^2$, $1-v_i\cdot n\ge 1-\mathrm{\mid }{v}_i\mathrm{\mid }>0$. The finite sum defining $K(n)$ is therefore continuous and bounded on $S^2$. Charge balance gives $h(K(n),(1,n))={\sum }_i{\eta }_i{Q}_i=0$, so $K^0(n)=n\cdot \mathbf{K}(n)$ and $K_{\perp }(n)$ is the transverse component. The current theorem [F10] uses the same $K_{\perp }$, $g$, and $d\mathrm{\Omega }$ with energy variable $\omega =\mathrm{\hslash }\nu$ and proves that the full-sphere coefficient $C_0$ is finite. Here $b={\mathbf{1}}_A{b}_0$, so $C={\int }_A\mathrm{\mid }{b}_0{\mathrm{\mid }}^{2}d\mathrm{\Omega }\le C_0$; the Borel restriction remains transverse. Thus the angular coefficient is square-integrable and $C$ is finite and nonnegative.

For $\mathbf{k}=rn\mathrm{\ne }0$, the coefficient $b(n)$ is perpendicular to $n$, so $\mathbf{k}\cdot f_{\epsilon ,\mathrm{\Lambda }}(\mathbf{k})=0$; hence $f_{\epsilon ,\mathrm{\Lambda }}\in {\mathcal{H}}_T$ once its $L^2$ norm is finite. The Borel cutoff and Borel angular coefficient make the squared norm integrand nonnegative and Borel. Applying [F6] and then $\nu =cr$ gives $\mathrm{\parallel }{f}_{\epsilon ,\mathrm{\Lambda }}{\mathrm{\parallel }}^2={\int }_A\mathrm{\mid }b(n){\mathrm{\mid }}^{2}d\mathrm{\Omega }(n){\int }_{\epsilon \mathrm{/}c}^{\mathrm{\Lambda }\mathrm{/}c}\frac{dr}{r}=C{\int }_{\epsilon }^{\mathrm{\Lambda }}\frac{d\nu }{\nu }=C\log \frac{\mathrm{\Lambda }}{\epsilon }<\mathrm{\infty }\mathrm{.}$ For each band define $f_j(\mathbf{k})={\mathbf{1}}_{\{\nu (\mathbf{k})\in B_j\}}{f}_{\epsilon ,\mathrm{\Lambda }}(\mathbf{k})$. The same change of variables gives $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}^2=C{\int }_{B_j}d\nu \mathrm{/}\nu ={\lambda }_j$. A singleton endpoint has zero one-dimensional Lebesgue measure, since it is covered by intervals of arbitrarily small length; therefore any endpoint convention gives the same integral. Thus the band components belong to pairwise orthogonal closed subspaces of ${\mathcal{H}}_T$, and their squared norms sum to $\lambda =C\log (\mathrm{\Lambda }\mathrm{/}\epsilon )$.

In the $m$-particle sector, $\mathrm{\parallel }{f}^{{\otimes }_{s}m}\mathrm{/}\sqrt{m!}{\mathrm{\parallel }}^2={\lambda }^m\mathrm{/}m!$. By [F8] and [F9], their sum over $m\ge 0$ is $\exp (\lambda )<\mathrm{\infty }$. Therefore $ϵ(f_{\epsilon ,\mathrm{\Lambda }})$ is a Fock vector of squared norm $\exp (\lambda )$, and the prescribed scalar $\exp (-\lambda \mathrm{/}2)$ normalizes ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Lambda }}$ to norm one.

Put $M={\sum }_j{m}_j$. The symmetrized tensors with different band occupancy tuples are orthogonal because some tensor factor then lies in orthogonal frequency subspaces. For $x=f_1^{\otimes m_1}\otimes \cdots \otimes f_r^{\otimes m_r}$, the projection formula [F2] gives $\mathrm{\parallel }{P}_{s,M}x{\mathrm{\parallel }}^2=⟨x,P_{s,M}x⟩=\frac{{\prod }_j{m}_j!}{M!}{\prod }_j\mathrm{\parallel }{f}_j{\mathrm{\parallel }}^{2m_j}\mathrm{.}$ Indeed, in the permutation sum only permutations within each band block have nonzero inner product with $x$; there are ${\prod }_j{m}_j!$ such permutations, each contributing ${\prod }_j\mathrm{\parallel }{f}_j{\mathrm{\parallel }}^{2m_j}$. Expanding $f^{\otimes M}=({\sum }_j{f}_j{)}^{\otimes M}$, the component with occupancy $(m_1,\dots ,m_r)$ is $\frac{M!}{{\prod }_j{m}_j!}{P}_{s,M}x$.

The squared norm of the normalized coherent-state projection onto occupancy $(m_1,\dots ,m_r)$ is consequently $\exp (-\lambda )\frac{M!}{{\prod }_j{m}_j{!}^2}\mathrm{\parallel }{P}_{s,M}x{\mathrm{\parallel }}^2=\exp \text{\hspace{0.17em}⁣}(-{\sum }_j{\lambda }_j){\prod }_j\frac{{\lambda }_j^{m_j}}{m_j!}={\prod }_j(\exp (-{\lambda }_j)\frac{{\lambda }_j^{m_j}}{m_j!})\mathrm{.}$ Each factor sums to one by [F8], so this is a joint probability law that factors into Poisson distributions of means ${\lambda }_j$. If $C=0$, all ${\lambda }_j=0$; [F9] makes the only nonzero outcome the all-zero tuple. The marked frequency intensity is $\mathrm{\mid }b(n){\mathrm{\mid }}^{2}d\mathrm{\Omega }(n)d\nu \mathrm{/}\nu$ because $\mathrm{\mid }f(rn){\mathrm{\mid }}^2{d}^3\mathbf{k}=\mathrm{\mid }b(n){\mathrm{\mid }}^{2}d\mathrm{\Omega }(n)d\nu \mathrm{/}\nu$ under $\nu =cr$.