Photon number divergence can coexist with finite soft energy

Put $\lambda =C\log (\mathrm{\Delta }\mathrm{/}\epsilon )$. By [F2], the count distribution is $p_m=e^{-\lambda }{\lambda }^m\mathrm{/}m!$. Reindexing the nonnegative series gives $\mathbb{E}{N}_{(\epsilon ,\mathrm{\Delta })}=e^{-\lambda }{\sum }_{m\ge 1}\frac{m{\lambda }^m}{m!}=\lambda e^{-\lambda }{\sum }_{j\ge 0}\frac{{\lambda }^j}{j!}=\lambda \mathrm{.}$ This calculation also gives zero mean when $C=0$ by [F9].

If $C>0$, fix $\mathrm{\Delta }$ and set $\kappa =\mathrm{\Delta }\mathrm{/}c$ and $\delta =\epsilon \mathrm{/}c$. The Definition's profile $f_{\epsilon ,\mathrm{\Delta }}$ is exactly the supplier profile with these cutoffs and the same transverse angular function $b$; its hypotheses hold because $\kappa >0$ and $C_b=C>0$. Applying [F10] shows that ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}$ has no strong Fock-space limit as $\epsilon \downarrow 0$.

If $C>0$, then for every integer $q\ge 1$, whenever $\epsilon <\mathrm{\Delta }{2}^{-q}$, $\log (\mathrm{\Delta }\mathrm{/}\epsilon )>q\log 2\ge q\mathrm{/}2$, since $\log 2={\int }_1^{2}dt\mathrm{/}t\ge 1\mathrm{/}2$ by [F7]. Hence $C\log (\mathrm{\Delta }\mathrm{/}\epsilon )\to \mathrm{\infty }$ as $\epsilon \downarrow 0$. For $C=0$ the count and its mean remain zero.

Under the isometry in [F5], multiplication by $\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ and by the frequency-band indicator commutes pointwise with $P_T(\mathbf{k})$, so the scalar sector multiplier of [F5] restricts to the transverse Fock space. On the support of $f_{\epsilon ,\mathrm{\Delta }}$, $0<\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }<\mathrm{\hslash }\mathrm{\Delta }$. By [F5], the squared norm of the energy-operator image is bounded by $\mathrm{\parallel }{H}_{\gamma }{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}{\mathrm{\parallel }}^2\le (\mathrm{\hslash }\mathrm{\Delta }{)}^2{e}^{-\lambda }{\sum }_{m\ge 0}\frac{m^2{\lambda }^m}{m!}=(\mathrm{\hslash }\mathrm{\Delta }{)}^2({\lambda }^2+\lambda )<\mathrm{\infty }\mathrm{.}$ The identity follows by writing $m^2=m(m-1)+m$ and shifting indices in the two exponential series [F8], with [F9] covering $\lambda =0$. Thus ${\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}\in D(H_{\gamma })$.

On sector $m$, the energy multiplier is the sum of the $m$ one-particle energies. Product tensor pairings make each summand contribute $⟨f_{\epsilon ,\mathrm{\Delta }},\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }{f}_{\epsilon ,\mathrm{\Delta }}⟩\mathrm{\parallel }{f}_{\epsilon ,\mathrm{\Delta }}{\mathrm{\parallel }}^{2(m-1)}$. Including the coherent-state factor and the sector normalization gives $⟨{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }},H_{\gamma }{\mathrm{\Psi }}_{\epsilon ,\mathrm{\Delta }}⟩=e^{-\lambda }{\sum }_{m\ge 1}\frac{m⟨f_{\epsilon ,\mathrm{\Delta }},\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }{f}_{\epsilon ,\mathrm{\Delta }}⟩{\lambda }^{m-1}}{m!}=⟨f_{\epsilon ,\mathrm{\Delta }},\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }{f}_{\epsilon ,\mathrm{\Delta }}⟩,$ where $e^{-\lambda }{\sum }_{m\ge 1}m{\lambda }^{m-1}\mathrm{/}m!=1$ by [F8]. Polar coordinates [F6], $\nu =cr$, and the definition of $C$ give $⟨f_{\epsilon ,\mathrm{\Delta }},\mathrm{\hslash }c\mathrm{\mid }\mathbf{k}\mathrm{\mid }{f}_{\epsilon ,\mathrm{\Delta }}⟩={\int }_A\mathrm{\mid }b(n){\mathrm{\mid }}^{2}d\mathrm{\Omega }(n){\int }_{\epsilon \mathrm{/}c}^{\mathrm{\Delta }\mathrm{/}c}\mathrm{\hslash }cdr=\mathrm{\hslash }C(\mathrm{\Delta }-\epsilon )\mathrm{.}$ It is bounded by $\mathrm{\hslash }C\mathrm{\Delta }$ and tends to that finite value as $\epsilon \downarrow 0$.

If $C=0$, [F1] gives $\mathrm{\parallel }b{\mathrm{\parallel }}_{L^2}^2=0$, so $b$ is the zero $L^2$ equivalence class and hence $b=0$ almost everywhere. The same radial norm calculation in [F2] gives $\mathrm{\parallel }{f}_{\epsilon ,\mathrm{\Delta }}{\mathrm{\parallel }}^2=C\log (\mathrm{\Delta }\mathrm{/}\epsilon )$. For $C>0$ this is unbounded as $\epsilon \downarrow 0$ by step 2.1. If $f_{\epsilon ,\mathrm{\Delta }}$ had a norm limit $f$, choose a sequence ${\epsilon }_j\downarrow 0$ with $\mathrm{\parallel }{f}_{{\epsilon }_j,\mathrm{\Delta }}-f\mathrm{\parallel }<1$ for all sufficiently large $j$. The Hilbert-space triangle inequality would then give $\mathrm{\parallel }{f}_{{\epsilon }_j,\mathrm{\Delta }}\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }+1$, contradicting the divergence in step 2.1. For $C=0$ the displacement is the zero $L^2$ vector at every cutoff by [F2].