TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Norm-continuous finite channel semigroups are GKSL

Statement

Let $d \geq 1$ and let $T_{t} : M_{d} (C) \to M_{d} (C)$ , $t \geq 0$ measured in seconds, be a semigroup of finite channels, so $T_{0} = id$ and $T_{t + s} = T_{t} \circ T_{s}$ . Assume $t \mapsto T_{t}$ is continuous in the induced operator norm on the finite-dimensional space of linear maps $M_{d} (C) \to M_{d} (C)$ . 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 exponential series of The exponential series of a bounded operator. Then there is a unique linear generator $L$ , in $s^{- 1}$ , with $T_{t} = e^{t L}$ , and $L$ has the finite GKSL form in Finite GKSL generators and units with at most $d^{2} - 1$ jump matrices. Conversely, every such finite GKSL generator produces a norm-continuous channel semigroup, as in Finite GKSL generates CPTP semigroups.

Facts & Assumptions

Given: $d \geq 1$ , the finite channel definition, a norm-continuous semigroup $T_{t}$ with $T_{0} = id$ , and $ℏ > 0$ in the units of the GKSL definition.

[F1]

Finite completely positive trace-preserving maps: each $T_{t}$ is completely positive and trace-preserving.

[F2]

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

[F3]

Finite matrix spectral and state calculus: finite Hermitian matrices have real spectral decompositions and positive matrices have positive square roots.

[F4]

Complete finite Choi-Kraus equivalence: a finite map is completely positive exactly when its Choi matrix is positive, and every finite CP map has a Kraus form.

[F5]

Finite GKSL generates CPTP semigroups: every finite GKSL generator has a norm-continuous semigroup of channels for $t \geq 0$ , with the operator exponential defined by its norm-convergent series.

[F6]

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

[F7]

Riemann integration of continuous curves in a complete normed space, with the fundamental theorem of calculus and coefficient functionals: a continuous curve into a complete normed real vector space has a norm-Riemann integral, its indefinite integral differentiates to the curve, and bounded-interval integrals are additive.

[F8]

Finite matrix trace cyclicity and nondegeneracy: setting the rectangular dimensions equal gives square trace cyclicity.

[F9]

The exponential series of a bounded operator: the operator-norm exponential series satisfies $E (0) = I$ , $E^{'} (t) = A E (t) = E (t) A$ , and $E (t + s) = E (t) E (s)$ , with the stated quadratic remainder bound.

Proof

1.1givenF6F7algebra

Let $E = {End}_{C} (M_{d} (C))$ with its induced operator norm. It has finite complex dimension $d^{4}$ , so its underlying real normed space is complete by [F6]. For small $h > 0$ , set $A_{h} = h^{- 1} \int_{0}^{h} T_{s} d s$ , using [F7]. The integral norm bound in [F7] and norm continuity at zero give $∥ A_{h} - I ∥ \leq \sup_{0 \leq s \leq h} ∥ T_{s} - I ∥ \to 0$ ; hence $∥ I - A_{h} ∥ < 1$ for sufficiently small $h$ . The series $B_{N} : = \sum_{k = 0}^{N} (I - A_{h})^{k}$ is Cauchy in the finite-dimensional complete map space: the induced operator norm is submultiplicative, so $∥ (I - A_{h})^{k} ∥ \leq ∥ I - A_{h} ∥^{k}$ , and the geometric series converges to a limit $B$ . The telescoping identities $A_{h} B_{N} = B_{N} A_{h} = I - (I - A_{h})^{N + 1}$ pass to the limit, so $A_{h} B = B A_{h} = I$ and $A_{h}$ is invertible. The semigroup law gives $T_{t} A_{h} = h^{- 1} \int_{t}^{t + h} T_{u} d u$ . By additivity and the fundamental theorem [F7], this moving integral has derivative $(T_{t + h} - T_{t}) / h$ for $t > 0$ , and the same right derivative at $t = 0$ . Right multiplication by $A_{h}^{- 1}$ yields $T_{t}^{'} = T_{t} G_{h}$ , where $G_{h} = (T_{h} - I) A_{h}^{- 1} / h$ , with right derivative $G_{h}$ at zero. Since that right derivative is unique, $G_{h}$ is independent of all sufficiently small $h$ ; set $L = G_{h}$ .

2.1step 1.1F9algebra

For $t > 0$ , differentiating the product in step 1.1 gives $\frac{d}{d t} (T_{t} e^{- t L}) = T_{t} L e^{- t L} - T_{t} L e^{- t L} = 0$ ; [F9] supplies the derivative and initial value of the matrix exponential. Continuity at zero gives $T_{0} e^{0} = I$ , so $T_{t} e^{- t L} = I$ and $T_{t} = e^{t L}$ . Any generator of this semigroup must equal its right derivative at zero, so $L$ is unique.

2.2givenF1F3step 1.1algebra

Each $T_{t}$ preserves adjoints: it is positive by [F1], and every Hermitian matrix is a difference of two positive matrices by [F3]. Taking the right derivative at zero shows $L (X^{*}) = L (X)^{*}$ . Differentiating $Tr (T_{t} X) = Tr X$ also gives $Tr (L (X)) = 0$ for every $X$ . Consequently $J (L)$ is Hermitian: $J (L)^{*} = \sum_{i, j} E_{j i} \otimes L (E_{i j})^{*} = \sum_{i, j} E_{j i} \otimes L (E_{j i}) = J (L)$ after reindexing.

3.1givenF3F4F9step 2.1step 2.2algebra

Let $Ω = \sum_{i = 1}^{d} ∣ i ⟩ \otimes ∣ i ⟩$ . By [F4], $J (T_{t}) \geq 0$ , and step 2.1 together with the quadratic remainder in [F9] gives $J (T_{t}) = ∣ Ω ⟩ ⟨ Ω ∣ + t J (L) + o (t)$ . For every $v ⊥ Ω$ , dividing $⟨ v, J (T_{t}) v ⟩ \geq 0$ by $t > 0$ and taking $t ↓ 0$ gives $⟨ v, J (L) v ⟩ \geq 0$ . Put $e = Ω / \sqrt{d}$ , $Q = I - ∣ e ⟩ ⟨ e ∣$ , $M = Q J (L) Q$ , $z = Q J (L) e$ , and $a = ⟨ e, J (L) e ⟩ \in R$ . Then $M \geq 0$ , and the block decomposition relative to $C e \oplus e^{⊥}$ is $J (L) = M + ∣ e ⟩ ⟨ z ∣ + ∣ z ⟩ ⟨ e ∣ + a ∣ e ⟩ ⟨ e ∣$ .

4.1step 3.1F3F4algebra

By [F3], write $M = \sum_{b = 1}^{q} ∣ w_{b} ⟩ ⟨ w_{b} ∣$ , where $q = rank M \leq d^{2} - 1$ , absorbing each positive eigenvalue into its eigenvector. Define $L_{b}$ by $w_{b} = \sum_{i} ∣ i ⟩ \otimes L_{b} ∣ i ⟩$ , and define $K$ by $k : = \sum_{i} ∣ i ⟩ \otimes K ∣ i ⟩ = z / \sqrt{d} + (a / (2 d)) Ω$ . Direct expansion of $J (Θ) = \sum_{i j} E_{i j} \otimes Θ (E_{i j})$ gives $J (X \mapsto L_{b} X L_{b}^{*}) = ∣ w_{b} ⟩ ⟨ w_{b} ∣$ , $J (X \mapsto K X) = ∣ k ⟩ ⟨ Ω ∣$ , and $J (X \mapsto X K^{*}) = ∣ Ω ⟩ ⟨ k ∣$ . The latter two sum to $∣ e ⟩ ⟨ z ∣ + ∣ z ⟩ ⟨ e ∣ + a ∣ e ⟩ ⟨ e ∣$ . Equality of Choi matrices gives equality on every input matrix unit because its $(i, j)$ block is $Θ (E_{i j})$ ; linearity then gives equality on all matrices. Thus $L (X) = \sum_{b} L_{b} X L_{b}^{*} + K X + X K^{*}$ .

5.1givenF2F8step 4.1algebra

Taking traces in step 4.1 and using finite cyclicity [F8] gives $Tr (X (K + K^{*} + \sum_{b} L_{b}^{*} L_{b})) = 0$ for every $X$ . Testing on all matrix units forces $K + K^{*} + \sum_{b} L_{b}^{*} L_{b} = 0$ . Set $H = (i ℏ / 2) (K - K^{*})$ ; then $H = H^{*}$ and $K = - \frac{1}{2} \sum_{b} L_{b}^{*} L_{b} - i H / ℏ$ . Substitution into step 4.1 gives the finite GKSL form in [F2] with at most $d^{2} - 1$ jumps.

6.1F5∎

Conversely, [F5] gives a norm-continuous channel semigroup from every finite GKSL generator.

Norm-continuous finite channel semigroups are GKSL

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources