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Lifted null-branch matrix elements and normalization

PDF p. 16: Eqs. (S4.10)–(S4.12)

Statement

Adopt the finite lifted model, the hypotheses of its null-projection and uniform-decay results, and the Axiom of Choice, Countable Choice and Dependent Choice inherited there. In particular, the finite physical space has orthonormal basis ∣J,m⟩ with J=0,…,N and m=−J,…,J; the initial density operator ρ0 is supported on the block J0∈{1,…,N}. A density operator is positive semidefinite with trace one. Let H be the model Hamiltonian with real lift Δ(J,m)=Δ0Jm and rate γ>0. Put V=−12∑δLδ†Lδ, Knull=I+τ(−iH+V), B=Knullρ0Knull†, and rmm′=⟨J0,m∣ρ0∣J0,m′⟩. Write amm′=Δ0J0(m−m′) and [ρ0,H]=ρ0H−Hρ0.

As τ↓0 at fixed model parameters, i[ρ0,H]mm′=−iamm′rmm′, Bmm′=[1+(−iamm′−γ)τ]rmm′+O(τ2), and Tr⁡B=1−γτ+O(τ2). For sufficiently small τ≥0, Tr⁡B>0 and the normalized null output ρnull=B/Tr⁡B satisfies [ρnull]mm′=rmm′[1−iamm′τ/(1−γτ)]+O(τ2)=rmm′(1−iamm′τ)+O(τ2). The displayed quotient expression is a first-order formula with a remainder, not an asserted exact finite-time identity.

The exact channel followed by the same null projection has the same displayed first-order matrix-element, trace and normalized-state expansions for sufficiently small τ. Here each O(τ2) is bounded in absolute value by a constant times τ2 near zero, with the model and initial state fixed; no uniform estimate over varying model parameters is claimed.

Facts & Assumptions

[F1]

In the lifted model, the Hamiltonian eigenvalue on ∣J0,m⟩ is Em=ω0J0(J0+1)+Δ0J0m, so Em−Em′=amm′. The diagonal phase-data convention uses differences of diagonal entries. General bounded diagonal lift for fast-checking analysis Calibrated diagonal lifts and transition phase differences

[F2]

Under the benchmark and its inherited system, bath, spectral and choice hypotheses, V=−γI/2 on every excited block. Normalized jumps give a scalar total decay operator

[F3]

The null-projection result gives B=ρ0+τ(i[ρ0,H]+Vρ0+ρ0V)+O(τ2) in spectral norm, and the exact projected channel output equals B+O(τ2) in that norm. The conditional output is obtained by dividing by its positive trace. Null-syndrome state and first-order expansion

[F4]

Scalar expansions with nonzero constant denominator can be divided through the desired order, retaining an O(τ2) remainder at fixed parameters. Finite exponential and smooth quotient remainder control

Proof

Given: The lifted model, supported density operator and checking interval in the statement, with the assumptions in [A1].

Goal: Derive the first-order unnormalized and normalized null-branch formulas for the approximate and exact channels.

1.1F1F2F3A1algebra

Since H is diagonal, [ρ0,H]mm′=(Em′−Em)rmm′=−amm′rmm′. On the support of ρ0, [F2] gives Vρ0+ρ0V=−γρ0. Substitution into [F3] yields the stated matrix-element expansion. A spectral-norm O(τ2) bound also bounds each matrix element between unit basis vectors.

2.1givenstep 1.1algebra

Summing the finitely many diagonal entries of step 1.1 gives Tr⁡B=(1−γτ)Tr⁡ρ0+O(τ2)=1−γτ+O(τ2): the commutator contribution vanishes on the diagonal. Thus the trace tends to one and is positive for sufficiently small nonnegative τ. The denominator 1−γτ is also nonzero there.

3.1F4step 1.1step 2.1algebra

By [F4], division by the trace gives [ρnull]mm′=rmm′(1−iamm′τ)+O(τ2). Moreover τ/(1−γτ)=τ+O(τ2), proving the equivalent displayed quotient form with its remainder. This argument does not require rmm′ to be nonzero.

4.1F3F4step 2.1step 3.1algebra∎

By [F3], the exact projected output differs from B by O(τ2) in spectral norm. In finite dimension this also changes each entry and the trace by O(τ2). Its trace consequently tends to one, so it is positive sufficiently near zero; applying [F4] once more gives the same normalized expansion. All constants here may depend on the fixed finite model and state.

Depends on

Used by · 1

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