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Finite GKSL generates CPTP semigroups

Statement

Let $L$ be a finite GKSL generator on $M_{d} (C)$ , $d \geq 1$ , with the units and conventions of Finite GKSL generators and units. For a bounded operator $A$ on a complex Banach space, write $e^{t A} : = \sum_{n = 0}^{\infty} t^{n} A^{n} / n!$ for the operator-norm convergent series of The exponential series of a bounded operator. Then $T_{t} = e^{t L}$ is a finite channel for every $t \geq 0$ . The family satisfies $T_{0} = id$ , $T_{t + s} = T_{t} \circ T_{s}$ , and is norm-continuous. Time is measured in seconds; no negative-time channel or physical bath limit is asserted.

Facts & Assumptions

Given: $d \geq 1$ , $ℏ > 0$ , $H = H^{*}$ , a finite family $L_{a}$ , and the resulting $L$ of Finite GKSL generators and units.

[F1]

Finite GKSL generators and units: $L (X) = - \frac{i}{ℏ} [H, X] + \sum_{a} (L_{a} X L_{a}^{*} - \frac{1}{2} {L_{a}^{*} L_{a}, X})$ .

[F2]

Finite completely positive trace-preserving maps: for $d, e \geq 1$ , positivity sends every matrix $X \geq 0$ to a positive matrix. Complete positivity requires this after every finite ancilla extension; trace preservation separately requires equality of traces for every matrix input, without a positivity restriction on that input.

[F3]

Complete finite Choi-Kraus equivalence: finite Kraus maps are completely positive, and their Kraus adjoint-square sum characterizes trace preservation.

[F4]

Channel duals and norm closure: finite-dimensional completely positive maps are closed under induced-norm limits; finite channels are closed under such limits, composition, and finite convex combination.

[F5]

Finite matrix trace cyclicity and nondegeneracy: the finite matrix trace is cyclic.

[F6]

Every finite-dimensional normed space is Banach: every finite-dimensional real or complex normed space is complete.

[F7]

The exponential series of a bounded operator: on a complex Banach space the operator exponential series converges absolutely in operator norm, differentiates to $A e^{t A} = e^{t A} A$ , and satisfies $e^{(t + s) A} = e^{t A} e^{s A}$ ; it also has the stated quadratic remainder bound.

Proof

1.1givenF1F6F7algebra

Put $D = \sum_{a} L_{a}^{*} L_{a}$ , $K = - i H / ℏ - D / 2$ , $N (X) = K X + X K^{*}$ , and $J (X) = \sum_{a} L_{a} X L_{a}^{*}$ . Then $L = N + J$ . Left multiplication by $K$ and right multiplication by $K^{*}$ commute; expanding their norm-convergent exponential series and using the binomial formula gives $e^{t N} (X) = e^{t K} X e^{t K^{*}}$ . Termwise adjunction gives $e^{t K^{*}} = (e^{t K})^{*}$ for real $t$ . For every ancilla dimension $r$ , the amplification is conjugation by $I_{r} \otimes e^{t K}$ , hence preserves positivity. Thus $e^{t N}$ is completely positive.

2.1step 1.1F3F4F6F7algebra

The map $J$ is completely positive by [F3], and every iterate $J^{n}$ is completely positive by composition. For $t \geq 0$ , the series $e^{t J} = \sum_{n = 0}^{\infty} t^{n} J^{n} / n!$ converges in finite-dimensional map norm; its partial sums are completely positive, so [F4] implies that $e^{t J}$ is completely positive.

3.1step 1.1step 2.1F4F7algebra

By [F7], for $h \geq 0$ and each bounded linear map $A$ , $∥ e^{h A} - I - h A ∥ \leq \frac{1}{2} h^{2} ∥ A ∥^{2} e^{h ∥ A ∥}$ . Applying this to $N, J, N + J$ shows that for fixed $t$ , $∥ e^{h N} e^{h J} - e^{h (N + J)} ∥ \leq C_{t} h^{2}$ for $0 \leq h \leq t$ , with finite $C_{t}$ . Set $h = t / n$ . The telescoping identity for the difference of the $n$ th powers and the bound $∥ e^{s A} ∥ \leq e^{s ∥ A ∥}$ from [F7] show $∥ {(e^{t N / n} e^{t J / n})}^{n} - e^{t (N + J)} ∥ \leq n C_{t} (t / n)^{2} e^{t (∥ N ∥ + ∥ J ∥)} ⟶ 0.$ Each product on the left before taking the limit is completely positive by step 1.1 and step 2.1; [F4] therefore proves that $e^{t L}$ is completely positive.

4.1givenF2F5F7step 3.1algebra∎

Cyclicity of finite trace [F5] gives $Tr (L (X)) = 0$ for every $X$ : the commutator trace vanishes, and each $Tr (L_{a} X L_{a}^{*})$ cancels $\frac{1}{2} Tr (L_{a}^{*} L_{a} X + X L_{a}^{*} L_{a})$ . By [F7], the norm-convergent exponential series differentiates term-by-term on bounded time intervals, so $\frac{d}{d t} Tr (e^{t L} X) = Tr (L (e^{t L} X)) = 0$ . Therefore $Tr (e^{t L} X) = Tr X$ . The exponential law [F7] gives $T_{t + s} = T_{t} \circ T_{s}$ , and the series gives norm continuity. Together with complete positivity from step 3.1, [F2] makes every $T_{t}$ a channel.

Finite GKSL generates CPTP semigroups

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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