Manuscript Em04-A@2 · Em04-A-v2.tex · theorem · f5b28abc

Theorem: Exact vacuum-plus-single-excitation reduction

Physicality and exact reduction · lines 345–381 · proof included or linked

Em04-A@2#thm:reduction, Em04-A@2#eq:S, Em04-A@2#eq:reduction

For the specified model, let $S_{0}$ be any density matrix on $C^{N}$ , and take $ρ (0) = V S_{0} V^{†}$ . There is a unique density matrix $S (t)$ satisfying

S ˙ = - i [h, S] - 2 γ_{d} (S - Δ S), S (0) = S_{0} .

Writing $q (t) = e^{- γ_{r} t}$ , the full state is exactly

ρ (t) = (1 - q (t)) ∣Ω ⟩ ⟨ Ω∣ + q (t) V S (t) V^{†} .

In particular,

R = q S, Tr R = q, Tr (ρ N_{exc}) = q .

The default preparation corresponds to $S_{0} = ∣0 ⟩ ⟨ 0∣$ .

Proof. Tracing (eq:R) gives $(Tr R)^{'} = - γ_{r} Tr R$ , with initial trace one, hence $Tr R = q$ . Equation (eq:p) gives $p = 1 - q$ . For finite $t$ , $q > 0$ , and substituting $R = q S$ gives (eq:S). This conditional equation is itself GKSL:

2 γ_{d} (Δ S - S) = k \sum D [2 γ_{d} ∣ k ⟩ ⟨ k ∣] (S) .

Thus it has a unique physical solution. The reconstructed state satisfies the full equation and initial condition, so uniqueness proves the identity.

Displayed formulas in this passage

Em04-A@2#eq:S · line 350
$S ˙ = - i [h, S] - 2 γ_{d} (S - Δ S), S (0) = S_{0} .$
Em04-A@2#eq:reduction · line 355
$ρ (t) = (1 - q (t)) ∣Ω ⟩ ⟨ Ω∣ + q (t) V S (t) V^{†} .$
unlabeled · line 361
$R = q S, Tr R = q, Tr (ρ N_{exc}) = q .$
unlabeled · line 374
$2 γ_{d} (Δ S - S) = k \sum D [2 γ_{d} ∣ k ⟩ ⟨ k ∣] (S) .$

Theorem: Exact vacuum-plus-single-excitation reduction

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