Manuscript Em03-H@8 · Em03-H-v8.tex · theorem · 91e9e3bd

Theorem: Signed branches and complete channels

Exact signed extraction and its reached space · lines 762–801 · proof included or linked

Em03-H@8#thm:branches

Writing $c = cos Θ (u)$ , $s = sin Θ (u)$ , boundary-first evolution is

U_{ext, m} (u) (∣ P ⟩_{B} ∣ \emptyset ⟩_{R}) = e^{- i Ω u ∣ P ∣} L \subseteq P \sum (- 1)^{I (L, P ∖ L)} c^{∣ L ∣} (- i s)^{∣ P ∣ - ∣ L ∣} \times ∣ L ⟩_{B} ∣ P ∖ L ⟩_{R} .

If $K_{L}$ is the bath coefficient, signed bath-first maps are

K_{L} ∣ P ⟩ = (- 1)^{∣ L ∣ (∣ P ∣ - ∣ L ∣)} K_{L} ∣ P ⟩ .

Consequently,

A_{u} (X) = L \sum K_{L} X K_{L}^{†},

A_{u}^{c} (X) = L, L^{'} \sum tr (K_{L} X K_{L^{'}}^{†}) ∣ L ⟩ ⟨ L^{'} ∣ .

Zero powers are interpreted as one at endpoints.

Proof. The active creator transformation is

U c_{j}^{†} U^{†} = e^{- i Ω u} (c c_{j}^{†} - i s b_{j}^{†}) .

Expand the increasing creator product. Transferred creators cross retained ones, giving the inversion sign. Equation (eq:regroup) supplies the additional access sign. Partial traces give the channels, and unitarity gives completeness.

Displayed formulas in this passage

unlabeled · line 766
$U_{ext, m} (u) (∣ P ⟩_{B} ∣ \emptyset ⟩_{R}) = e^{- i Ω u ∣ P ∣} L \subseteq P \sum (- 1)^{I (L, P ∖ L)} c^{∣ L ∣} (- i s)^{∣ P ∣ - ∣ L ∣} \times ∣ L ⟩_{B} ∣ P ∖ L ⟩_{R} .$
unlabeled · line 776
$K_{L} ∣ P ⟩ = (- 1)^{∣ L ∣ (∣ P ∣ - ∣ L ∣)} K_{L} ∣ P ⟩ .$
unlabeled · line 780
$A_{u} (X) = L \sum K_{L} X K_{L}^{†},$
unlabeled · line 783
$A_{u}^{c} (X) = L, L^{'} \sum tr (K_{L} X K_{L^{'}}^{†}) ∣ L ⟩ ⟨ L^{'} ∣ .$
unlabeled · line 793
$U c_{j}^{†} U^{†} = e^{- i Ω u} (c c_{j}^{†} - i s b_{j}^{†}) .$

Theorem: Signed branches and complete channels

Displayed formulas in this passage

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