Thermal contractions and absorption contribution

PDF p. 12: Text before Eq. (S3.6), Eqs. (S3.6)–(S3.8)

Assume the Axiom of Choice, Countable Choice and Dependent Choice, and adopt the finite-mode thermal-bath, regularized photon-mode and real-polarization dipole-coupling postulates. For modes $j=1,\dots ,m$, $m\ge 1$, let ${\omega }_j>0$, $\beta >0$, and $\gamma >0$. Let $a_j$ and $a_j^{†}$ be the annihilation and creation operators, $\mathrm{\mid }\mathbf{n}⟩$ the occupation basis, and ${\rho }_B=Z^{-1}{e}^{-\beta H_B}$ the thermal state, where $H_B={\sum }_j{\omega }_j{a}_j^{†}{a}_j$ and $Z={\prod }_j(1-e^{-\beta {\omega }_j}{)}^{-1}$. Write $N_j=(e^{\beta {\omega }_j}-1{)}^{-1}$ and $⟨X{⟩}_{\beta }=\mathrm{Tr}(X{\rho }_B)$.

Use the maximal occupation-basis mode operators, identified with their finite-mode symmetric Fock realization as in the thermal-correlator expansion. For this correlator, the field operators are the extensions to $\mathcal{D}={\bigcap }_{j=1}^m(D(a_j)\cap D(a_j^{†}))$, with $B_{\alpha }=-{\gamma }^{-1\mathrm{/}2}{\sum }_j{P}_{\alpha j}[\kappa ({\omega }_j)a_j+\kappa ({\omega }_j{)}^{\ast }{a}_j^{†}]$ on $\mathcal{D}$. Its free conjugate has the same domain. The Gibbs-weighted two-field products are defined on the whole bath space and trace class by that expansion result.

For every occupation vector, the diagonal matrix elements of $a_j{a}_k$ and $a_j^{†}{a}_k^{†}$ vanish. The normal thermal contraction is $⟨a_j^{†}{a}_k{⟩}_{\beta }={\delta }_{jk}{N}_j$; its occupation sum factors into the selected mode's weighted number sum and the other modes' partition functions.

Let $P_{\alpha j}={\widehat{e}}_{\alpha }\cdot {ϵ}_j$ for real transverse polarization vectors ${ϵ}_j$, and let $\kappa ({\omega }_j)$ be the finite-mode coupling coefficient. For $B_{\alpha }=-d_0{E}_{\alpha }\mathrm{/}\sqrt{\gamma }$, the bath correlator is ${\mathcal{J}}_{\alpha \alpha }(t)=\mathrm{Tr}[{\tilde{B}}_{\alpha }(t)B_{\alpha }{\rho }_B]$. Its normal, or absorption, contribution for real $t$ is

${\mathcal{J}}_{\alpha \alpha }(t)\{a^{†}a\}=\frac{1}{\gamma }{\sum }_{j=1}^m{P}_{\alpha j}^2\mathrm{\mid }\kappa ({\omega }_j){\mathrm{\mid }}^2{N}_j{e}^{i{\omega }_{j}t}\mathrm{.}$

The article writes the corresponding continuum expression as $\int d^{3}k{\sum }_{\epsilon }[{\widehat{e}}_{\alpha }\cdot \epsilon (\mathbf{k}){]}^2\mathrm{\mid }\kappa (\omega ){\mathrm{\mid }}^2{e}^{i\omega t}\mathrm{/}(e^{\beta \omega }-1)$. This is recorded as its formal expression, not as a proved regulator-removal limit. With the bath-operator normalization above, the finite calculation retains $1\mathrm{/}\gamma$; absorbing it requires a consistent redefinition of the coupling or correlator.