Single-pole thermal contour and exact detailed balance

Write $\sinh w=(e^w-e^{-w})\mathrm{/}2$ and $\mathrm{csch}w=1\mathrm{/}\sinh w$ where the denominator is nonzero. From [F5]–[F6], $\sinh (w+i\pi )=-\sinh w$, $\sinh (-w)=-\sinh (w)$, and $\sinh w=0$ exactly when $w\in i\pi \mathbb{Z}$. Thus $C(z)$ has poles only at $z=in\beta$ ($n\in \mathbb{Z}$). It is holomorphic on the stated strip: $\sinh$ is holomorphic by [F5] and the chain rule; where it is nonzero, its reciprocal is complex differentiable because $\frac{1\mathrm{/}\sinh (w+h)-1\mathrm{/}\sinh (w)}{h}=-\frac{\sinh (w+h)-\sinh (w)}{h\sinh (w+h)\sinh (w)}$ converges as $h\to 0$, using continuity from [F7]. The exponential addition law gives period $i\beta$, and oddness of $\sinh$ gives evenness of $C$. These are identities of the general analytic kernel; their distributional boundary interpretation follows once the boundary has been constructed.

Set $z_0=-i\beta$ and $\delta =z-z_0$. Since $b{z}_0=-i\pi$, $\sinh (bz)=-b\delta q(\delta )$, where $q(\delta )=\sinh (b\delta )\mathrm{/}(b\delta )$ for $\delta \mathrm{\ne }0$ and $q(0)=1$. Its power series $q(\delta )={\sum }_{m=0}^{\mathrm{\infty }}(b\delta {)}^{2m}\mathrm{/}(2m+1)!$ converges for all $\delta$ as a subseries of the complex exponential series. The termwise-differentiation theorem in [F7] makes $q$ holomorphic and its displayed series makes it even. Since $q(0)=1$, continuity from [F7] makes $q$ nonzero on a small disk; the reciprocal difference quotient using [F7] makes $q^{-2}$ holomorphic there, and its even series is $1+O({\delta }^2)$. Therefore $C(z)=-\frac{L}{(z-z_0{)}^2}+H(z),L=\frac{A}{b^2},$ where $H(z)$ is holomorphic at $z_0$, with no simple-pole term.

Fix $\omega \in \mathbb{R}\setminus \{0\}$ and $0<ϵ<\beta$. Integrate $e^{-i\omega z}C(z)$ around the clockwise rectangle with top from $-R-iϵ$ to $R-iϵ$ and bottom at imaginary part $-(\beta +ϵ)$. Its interior lies in $-2\beta <\mathrm{Im}z<0$ and contains only $z_0$. The holomorphic part $e^{-i\omega z}H(z)$ integrates to zero by [F3] on a slightly enlarged convex rectangle: its closure contains the contour, and convexity contracts the closed contour to a constant loop inside that open domain. For the double-pole term, the function $(e^{-i\omega z}-e^{-i\omega z_0})\mathrm{/}(z-z_0)$ extends holomorphically at $z_0$, so [F3] gives zero for its contour integral. To evaluate the remaining simple pole, write $a=\beta -ϵ>0$ and $c=ϵ>0$. Relative to $z_0$, the rectangle's vertices are $(-R,a),(R,a),(R,-c),(-R,-c)$. With ${\theta }_a=\arctan (a\mathrm{/}R)$ and ${\theta }_c=\arctan (c\mathrm{/}R)$, the continuous argument changes along its four clockwise sides by $-\pi +2{\theta }_a$, $-{\theta }_a-{\theta }_c$, $-\pi +2{\theta }_c$, and $-{\theta }_a-{\theta }_c$, respectively; their sum is $-2\pi$. The net change of $\log \mathrm{\mid }z-z_0\mathrm{\mid }$ is zero, so $\oint dz\mathrm{/}(z-z_0)=-2\pi i$ and $\oint e^{-i\omega z}(z-z_0{)}^{-1}dz=-2\pi i{e}^{-i\omega z_0}$. The function $e^{-i\omega z}\mathrm{/}(z-z_0)$ is single-valued and has continuous derivative on a neighborhood of the contour, which avoids $z_0$; by [F4] its derivative integrates to zero. Since $d\text{\hspace{0.17em}⁣}\left(\frac{e^{-i\omega z}}{z-z_0}\right)=(-i\omega \frac{e^{-i\omega z}}{z-z_0}-\frac{e^{-i\omega z}}{(z-z_0{)}^2})dz,$ we obtain $\oint e^{-i\omega z}(z-z_0{)}^{-2}dz=-2\pi \omega e^{-i\omega z_0}$.

The vertical-side integrals tend to zero as $R\to \mathrm{\infty }$: for $\mathrm{Re}z=R$, ${\mathrm{csch}}^2(bz)=4e^{-2bz}(1-e^{-2bz}{)}^{-2}=O(e^{-2bR})$ uniformly on the finite vertical interval, and the analogous formula with $e^{2bz}$ gives the same bound on $\mathrm{Re}z=-R$. The factor $e^{-i\omega z}$ is bounded there for fixed $\omega ,ϵ$. The bottom line is the top line translated by $-i\beta$; periodicity of $C$ and the Fourier phase multiply its oriented integral by $-e^{-\beta \omega }$. Combining this with step 2.1 gives $(1-e^{-\beta \omega })I_{ϵ}(\omega )=2\pi L\omega e^{-\beta \omega },$ where $I_{ϵ}(\omega )={\int }_{\mathbb{R}}{e}^{-i\omega (s-iϵ)}C(s-iϵ)ds$.

For each fixed $0<ϵ<\beta$, $s\mapsto C(s-iϵ)$ has no real pole, is bounded on compact intervals, and decays exponentially at both ends, so it is in $L^1(\mathbb{R})$. If ${\omega }_n\to \omega$, the integrands converge pointwise and are bounded by $\mathrm{\mid }C(s-iϵ)\mathrm{\mid }\in L^1$; [F8] therefore makes its Fourier transform $J_{ϵ}(\omega )=\int e^{-i\omega s}C(s-iϵ)ds$ continuous. Since $I_{ϵ}=e^{-\omega ϵ}{J}_{ϵ}$, step 3.1 yields, for $\omega \mathrm{\ne }0$, $J_{ϵ}(\omega )=2\pi L{e}^{\omega ϵ}\frac{\omega }{e^{\beta \omega }-1}\mathrm{.}$ Continuity forces the value at $\omega =0$ to be $2\pi L\mathrm{/}\beta$; no point value at zero is left arbitrary.

Define $S(\omega )=2\pi L\omega \mathrm{/}(e^{\beta \omega }-1)$ with its continuous value $2\pi L\mathrm{/}\beta$ at zero. Continuity bounds $S$ on $[-1,1]$. For $\omega \ge 1$, $\mathrm{\mid }S(\omega )\mathrm{\mid }\le C\omega e^{-\beta \omega }$, while for $\omega \le -1$, $\mathrm{\mid }S(\omega )\mathrm{\mid }\le C\mathrm{\mid }\omega \mathrm{\mid }$ because $1-e^{\beta \omega }\ge 1-e^{-\beta }$. Thus $\mathrm{\mid }S(\omega )\mathrm{\mid }\le C(1+\mathrm{\mid }\omega \mathrm{\mid })$ globally. For any Schwartz test $\phi$, with seminorms $p_{m,n}(\phi )={\sup }_{\omega }\mathrm{\mid }{\omega }^m{\phi }^{(n)}(\omega )\mathrm{\mid }$, split at $\mathrm{\mid }\omega \mathrm{\mid }=1$: the integral of $(1+\mathrm{\mid }\omega \mathrm{\mid })\mathrm{\mid }\phi (\omega )\mathrm{\mid }$ is bounded by a constant times $p_{0,0}(\phi )+p_{3,0}(\phi )$, since the tail is bounded by $2p_{3,0}\mathrm{\mid }\omega {\mathrm{\mid }}^{-2}$. Hence $S$ defines a tempered distribution. For $0<ϵ\le \beta \mathrm{/}2$, $e^{\omega ϵ}S(\omega )$ is bounded by $C(1+\mathrm{\mid }\omega \mathrm{\mid })$ for $\omega \le 0$ and by $C(1+\omega )e^{-\beta \omega \mathrm{/}2}$ for $\omega \ge 0$. These bounds make the Schwartz-test product integrable and give an integrable majorant independent of $ϵ$; [F8] yields $J_{ϵ}\to S$ in the tempered-distribution sense. To construct the boundary without assuming a chiral specialization, apply Fourier inversion [F2] to a Schwartz test $\phi$ and use $C_{ϵ}\in L^1$ and $\widehat{\phi }\in L^1$ to justify Fubini [F9]: $⟨C_{ϵ},\phi ⟩=\frac{1}{2\pi }{\int }_{\mathbb{R}}{J}_{ϵ}(\omega )\widehat{\phi }(-\omega )d\omega \mathrm{.}$ The same majorant and [F8] give the limit $⟨C_{A,b},\phi ⟩:=\frac{1}{2\pi }{\int }_{\mathbb{R}}S(\omega )\widehat{\phi }(-\omega )d\omega \mathrm{.}$ This is continuous in the Schwartz topology by the seminorm bound above and continuity of the Schwartz Fourier map in [F2]; hence it defines a tempered distribution and $C_{ϵ}\to C_{A,b}$ on every Schwartz test. Angular Fourier inversion gives $\widehat{\widehat{\phi }}(-\omega )=2\pi \phi (\omega )$, so the displayed formula applied to $\widehat{\phi }$ yields $⟨{\widehat{C}}_{A,b},\phi ⟩=\int S(\omega )\phi (\omega )d\omega$. Thus the general boundary has exactly the stated spectrum.

For $\omega >0$, the numerator and denominator of $S_{A,b}(\omega )$ are positive; for $\omega <0$ both are negative, and its zero limit $2\pi L\mathrm{/}\beta =2A\mathrm{/}b$ is positive. Direct algebra gives $S_{A,b}(-\omega )=e^{\beta \omega }{S}_{A,b}(\omega )$. The analytic period and reflection from step 1.1 give $C(s-i\beta +iϵ)=C(s+iϵ)=C(-s-iϵ)$; reflection preserves Schwartz tests, so step 5.1 gives the stated tested boundary relation. For the chiral specialization $A={\kappa }^2\mathrm{/}(16\pi )$, $b=\kappa \mathrm{/}2$, [F1] identifies the same analytic regularizations with the compact-test boundary $C_{\kappa }$. Uniqueness of their distributional limit makes it the restriction of $C_{A,b}$, giving its tempered extension. Substituting $L=1\mathrm{/}(4\pi )$ and $\beta =2\pi \mathrm{/}\kappa$ yields the original chiral spectrum, zero value and ratio. The units follow because the phase $\omega s$ is dimensionless and Fourier transformation integrates once in the coordinate. These are two-point mathematical deductions only; neither the contour proof nor the strip identity establishes a full $C^{\ast }$-algebra KMS state or an exact detector transition probability.