Thermal generator on the regularized beam-splitter output splits into the input generators

Let $F=\ell^2(\mathbb{N}^n,\mathbb{C})$ be the $n$-mode Fock space with number basis $(|k\rangle)_{k\in\mathbb{N}^n}$, total number $|k|=k_1+\dots+k_n$, annihilation operators $a_j$, and the diagonal weight $\hat N=1+\sum_ja_j^\dagger a_j$, i.e. $\hat N|k\rangle=(1+|k|)|k\rangle$. A state is a positive bounded operator on $F$ whose number-basis diagonal sums to $1$. For a state $\rho$ write $S(\rho)=-\operatorname{Tr}\rho\log\rho$, $E(\rho)=\sum_k|k|\,\rho_{kk}$ and $M_4(\rho)=\sum_k(1+|k|)^4\rho_{kk}$; $\rho$ has finite $m$-th moment if $\sum_k(1+|k|)^m\rho_{kk}<\infty$.

Fix $\eta,\delta,\kappa\in[0,1]$ and states $\tau_D,\tau_0$, and for states $\rho,\sigma$ put

where $\mathcal B_\theta$ is the beam-splitter output at transmissivity $\theta$. For a state $\mu$ with finite sixth moment, its canonical weighted columns are $x^\mu_m=\hat N^{3}\sqrt\mu\,|m\rangle$ $(m\in\mathbb{N}^n)$, so that $y_m=\hat N^{-3}x^\mu_m=\sqrt\mu|m\rangle$.

For $s\in\mathbb{R}$ and a bounded operator $Z$, the generator pairing is

with $\hat n_j=a_j^\dagger a_j$ (all vectors lie in $F$ because $\hat N^{2}a_j\hat N^{-3}$ etc. are bounded). Formally $\Gamma_s(x;Z)=\operatorname{Tr}\mu\,\mathcal L_s^*(\hat N^2Z\hat N^2)$, where $\mathcal L_s^*(K)=\sum_j(s+1)(a_j^\dagger Ka_j-\frac12\{\hat n_j,K\})+s(a_jKa_j^\dagger-\frac12\{a_ja_j^\dagger,K\})$ is the Heisenberg generator of the thermal-noise semigroup with mean photon number $s$.

For $\theta\in[0,1]$, a state $\nu$ with finite fourth moment and a bounded $T$, let $\mathcal C^{\theta}_{\nu}(T)$ and $\mathcal C'^{\theta}_{\nu}(T)$ be the weighted conditional operators: with $U_\theta$ the passive beam-splitter unitary of transmissivity $\theta$ on $\ell^2(\mathbb{N}^n\times\mathbb{N}^n)$, they are defined by

and likewise $\mathcal C'^\theta_\nu$ with $\sqrt\nu|m\rangle\otimes x$ (the known state in the first port). (In the file these are written with bounded weight ratios, as $\sum_mW_m^*(T\otimes\mathrm{Id})W_m$ for bounded $W_m$, so they are bounded operators; the displayed formula is their meaning.)

Theorem. Let $\rho,\sigma,\tau_D,\tau_0$ have finite sixth moments, let $a,b\in\mathbb{R}$, $d>0$, $r_0>0$ with

and assume $\tau_D$ and $\tau_0$ are the number-diagonal thermal states with mean photon numbers $d$ and $r_0$ per mode (matrix entries $\langle k|\tau_D|l\rangle=\delta_{kl}\prod_j\frac{1}{d+1}\bigl(\frac d{d+1}\bigr)^{k_j}$, and similarly for $\tau_0$ with $r_0$). Then $\omega=\omega(\rho,\sigma)$ has finite sixth moment, and for every bounded operator $B$

Formally: the thermal-noise generator with mean $r_0$, evaluated on the output, equals $(1-\kappa)$ times the sum of the generators with means $a$ and $b$ acting on the two inputs, transported through the Heisenberg duals of the two beam splitters. The compatibility condition on $r_0$ says that $r_0$ is the output mean photon number when the inputs are thermal with means $a,b$; the $\kappa\tau_0$ component, being thermal with mean $r_0$, does not contribute.

Formalization Note The parameters are fields of the structure Regularization n (R.output is $\omega$); the canonical columns are WeightedColumns.canonical (built from weightedRoot 3), $\Gamma_s$ is WeightedGenerator.pairing, and X.slice, X.sliceRight are $\mathcal C$, $\mathcal C'$ with the columns $\hat N^{2}\sqrt\nu|m\rangle$. The existence of the moment proof hm0 is part of the conclusion.