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Finite completely positive trace-preserving maps

1 · Prerequisites

2 · Summary

This page develops finite-dimensional quantum channels as complex-linear, completely positive and trace-preserving maps on matrix algebras. A local matrix-unit lemma proves rectangular trace cyclicity and nondegeneracy wherever those identities are needed. The Choi matrix uses the unnormalized input-reference $\otimes$ output convention, and trace preservation is distinct from unitality. It proves the finite Choi–Kraus equivalence, trace duality, norm closure and trace-norm contraction for arbitrary matrix inputs.

The generator results use finite matrices, explicit SI units, and continuous coefficients on compact time intervals. They give the exact finite GKSL channel-semigroup class and time-dependent channel propagators; they do not assert unbounded-generator results or a microscopic bath limit. Reduced unitary channels require a product-prepared system and environment, and a fixed finite bath need not yield a semigroup. The entropy inequality is restricted to finite pinching operations; arbitrary channels can lower entropy. These are mathematical results about the stated finite models, not claims that a physical system follows a particular model.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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Finite matrix trace cyclicity and nondegeneracy

Statement

Let $d, e \geq 1$ . For $A \in M_{e \times d} (C)$ and $B \in M_{d \times e} (C)$ , ${Tr}_{e} (A B) = {Tr}_{d} (B A),$ where ${Tr}_{r}$ is the diagonal-sum trace on $M_{r} (C)$ . This includes cyclicity for square matrices. For $C \in M_{d} (C)$ , the bilinear trace pairing is nondegenerate: if ${Tr}_{d} (X C) = 0$ for every $X \in M_{d} (C)$ , then $C = 0$ . No adjoint is inserted in this pairing.

Facts & Assumptions

Given: Integers $d, e \geq 1$ , rectangular matrices $A \in M_{e \times d} (C)$ , $B \in M_{d \times e} (C)$ , a matrix $C \in M_{d} (C)$ , and standard matrix units $E_{i j}$ on $C^{d}$ .

Proof

1.1givenalgebra

By matrix multiplication and the diagonal-sum trace, ${Tr}_{e} (A B) = \sum_{i = 1}^{e} \sum_{j = 1}^{d} A_{i j} B_{j i}$ . These are finite scalar sums, so reordering them gives $\sum_{j = 1}^{d} \sum_{i = 1}^{e} B_{j i} A_{i j} = {Tr}_{d} (B A)$ . Taking $e = d$ gives square trace cyclicity.

2.1givenalgebra∎

If ${Tr}_{d} (X C) = 0$ for all $X$ , choose $X = E_{i j}$ . Direct multiplication gives ${Tr}_{d} (E_{i j} C) = C_{j i}$ , so $C_{j i} = 0$ for every $i, j$ . Thus every matrix entry of $C$ vanishes and $C = 0$ .

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Finite completely positive trace-preserving maps

Definition

Let $d, e \geq 1$ be integers, let $H_{i n} = C^{d}$ and $H_{o u t} = C^{e}$ with their standard inner products, and write $M_{d} (C)$ and $M_{e} (C)$ for the complex linear maps on these spaces. A matrix $X$ is positive semidefinite, written $X \geq 0$ , when it is Hermitian and $⟨ X v, v ⟩ \geq 0$ for every vector $v$ . The finite spectral convention in Finite matrix spectral and state calculus gives positive matrices their orthonormal eigenbases and positive square roots.

A positive map is a complex-linear map $Φ : M_{d} (C) \to M_{e} (C)$ such that $Φ (X) \geq 0$ whenever $X \geq 0$ . For each integer $r \geq 1$ , identify $M_{r} (C) \otimes M_{d} (C)$ with matrices on $C^{r} \otimes C^{d}$ in the ordered product basis. The map $Φ$ is completely positive when ${id}_{r} \otimes Φ$ is positive for every such $r$ , where ${id}_{r}$ is the identity map on $M_{r} (C)$ .

The map $Φ$ is trace-preserving when $Tr (Φ (X)) = Tr (X)$ for every $X \in M_{d} (C)$ . A finite quantum channel is a complex-linear map that is both completely positive and trace-preserving, abbreviated CPTP. Unitality, $Φ (I_{d}) = I_{e}$ , is a separate condition and is not part of this definition.

This is a mathematical definition. A physical preparation, measurement, or open-system model must separately identify when one of these maps represents a physical transformation.

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Finite GKSL generators and units

Definition

Fix a finite dimension $d \geq 1$ , a positive constant $ℏ$ in joule-seconds, a Hermitian matrix $H \in M_{d} (C)$ in joules, and a finite family $L_{1}, \dots, L_{m} \in M_{d} (C)$ with units $s^{- 1 / 2}$ , where $m \geq 0$ . On dimensionless matrices $X \in M_{d} (C)$ , define the finite Gorini–Kossakowski–Sudarshan–Lindblad (GKSL) generator

$L (X) = - \frac{i}{ℏ} [H, X] + \sum_{a = 1}^{m} (L_{a} X L_{a}^{*} - \frac{1}{2} {L_{a}^{*} L_{a}, X}),$

where $[A, B] = A B - B A$ , ${A, B} = A B + B A$ , and $A^{*}$ is the conjugate transpose. Each summand has units $s^{- 1}$ times the units of $X$ . An empty sum when $m = 0$ is zero. The resulting $L$ is a linear map on the matrix space used in the finite-channel definition Finite completely positive trace-preserving maps.

For a compact time interval $I \subset R$ whose coordinate is time in seconds, continuous functions $H : I \to M_{d} (C)$ and $L_{a} : I \to M_{d} (C)$ with $H (t) = H (t)^{*}$ define the corresponding continuous finite GKSL family $L_{t}$ by the same formula at each $t$ . A propagator for this family means a family of linear maps $Λ (t, s)$ , $s \leq t$ in $I$ , satisfying $\partial_{t} Λ (t, s) = L_{t} \circ Λ (t, s)$ and $Λ (s, s) = id$ . This definition alone asserts neither existence nor uniqueness.

These are finite-dimensional mathematical definitions. They do not assert that a physical system follows this generator or that a microscopic bath has a Markovian limit.

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Complete finite Choi-Kraus equivalence

Statement

Let $d, e \geq 1$ and let $Φ : M_{d} (C) \to M_{e} (C)$ be complex-linear. Choose the standard input basis $(∣ i ⟩)_{i = 1}^{d}$ , let $E_{i j} = ∣ i ⟩ ⟨ j ∣$ , and use the ordered tensor convention input-reference $\otimes$ output. Define the (unnormalized) Choi matrix

$J (Φ) : = \sum_{i, j = 1}^{d} E_{i j} \otimes Φ (E_{i j}) \in M_{d} (C) \otimes M_{e} (C) .$ Then $Φ$ is completely positive if and only if $J (Φ) \geq 0$ , and this holds if and only if there are $q \leq d e$ matrices $K_{1}, \dots, K_{q} : C^{d} \to C^{e}$ such that $Φ (X) = \sum_{a = 1}^{q} K_{a} X K_{a}^{*} (X \in M_{d} (C)) .$

The empty sum is allowed when $Φ = 0$ . For any such Kraus representation, $Φ$ is trace-preserving if and only if $\sum_{a = 1}^{q} K_{a}^{*} K_{a} = I_{d}$ .

Facts & Assumptions

Given: Integers $d, e \geq 1$ , a complex-linear map $Φ : M_{d} (C) \to M_{e} (C)$ , the standard ordered bases, and the input-reference $\otimes$ output Choi convention in the Statement.

[F1]

Finite completely positive trace-preserving maps: complete positivity means that ${id}_{r} \otimes Φ$ preserves positive semidefinite matrices for every finite $r \geq 1$ ; trace preservation means $Tr Φ (X) = Tr X$ for every $X$ .

[F2]

Finite matrix spectral and state calculus: every finite Hermitian matrix has an orthonormal eigenbasis with real eigenvalues, and positive square roots are defined by its finite spectral projections.

[F3]

Finite matrix trace cyclicity and nondegeneracy: for rectangular factors, ${Tr}_{e} (A B) = {Tr}_{d} (B A)$ ; the square bilinear trace pairing is nondegenerate.

Proof

1.1givenF1algebra

Let $Ω = \sum_{i = 1}^{d} ∣ i ⟩ \otimes ∣ i ⟩$ in the input-reference and input tensor factors. Then $Ω Ω^{*} = \sum_{i, j} E_{i j} \otimes E_{i j} \geq 0$ , and $({id}_{d} \otimes Φ) (Ω Ω^{*}) = J (Φ)$ . If $Φ$ is completely positive, [F1] applied with ancilla dimension $d$ gives $J (Φ) \geq 0$ .

1.2givenF2construct

Suppose $J (Φ) \geq 0$ . By [F2], choose an orthonormal eigenbasis of $J (Φ)$ , absorb each positive eigenvalue's square root into its eigenvector, and write $J (Φ) = \sum_{a = 1}^{q} ∣ w_{a} ⟩ ⟨ w_{a} ∣$ , where $q = rank J (Φ) \leq d e$ ; if $J (Φ) = 0$ , take $q = 0$ . For each $a$ , uniquely define $K_{a} : C^{d} \to C^{e}$ by $w_{a} = \sum_{i} ∣ i ⟩ \otimes K_{a} ∣ i ⟩$ .

2.1step 1.2algebra

Expanding each rank-one term gives $∣ w_{a} ⟩ ⟨ w_{a} ∣ = \sum_{i, j} E_{i j} \otimes K_{a} E_{i j} K_{a}^{*}$ . Comparing the coefficients of the linearly independent first-factor matrix units in $J (Φ)$ yields $Φ (E_{i j}) = \sum_{a} K_{a} E_{i j} K_{a}^{*}$ . Since $(E_{i j})$ is a basis and both sides are linear in the input, $Φ (X) = \sum_{a} K_{a} X K_{a}^{*}$ for every $X$ .

3.1givenF1F3step 2.1algebra∎

For any $r \geq 1$ and any $P \geq 0$ on $C^{r} \otimes C^{d}$ , the Kraus formula gives $({id}_{r} \otimes Φ) (P) = \sum_{a} (I_{r} \otimes K_{a}) P (I_{r} \otimes K_{a})^{*} \geq 0$ ; hence $Φ$ is completely positive. For a Kraus representation, cyclicity [F3] gives $Tr Φ (X) = Tr (X \sum_{a} K_{a}^{*} K_{a})$ . Thus $\sum_{a} K_{a}^{*} K_{a} = I_{d}$ implies trace preservation, and if trace preservation holds then testing $Tr (X (\sum_{a} K_{a}^{*} K_{a} - I_{d})) = 0$ on every matrix unit $X = E_{i j}$ forces every entry of the difference to vanish.

Remarks

The input-reference $\otimes$ output order is fixed here. Watrous writes the output factor first; the two Choi matrices are related by the tensor-factor swap unitary, which preserves positivity. This finite result asserts neither an infinite-dimensional classification nor a physical bath limit.

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Channel duals and norm closure

Statement

Let $Φ : M_{d} (C) \to M_{e} (C)$ be a finite channel as defined in Finite completely positive trace-preserving maps. There is a unique linear map $Φ^{†} : M_{e} (C) \to M_{d} (C)$ satisfying the trace-pairing identity

$Tr (Φ (X) A) = Tr (X Φ^{†} (A)) (X \in M_{d} (C), A \in M_{e} (C)) .$

If $Φ (X) = \sum_{a} K_{a} X K_{a}^{*}$ , then $Φ^{†} (A) = \sum_{a} K_{a}^{*} A K_{a}$ ; in particular, $Φ^{†}$ is completely positive and unital. Composition and finite convex combinations of channels are channels. A limit in the induced operator norm $∥ Ψ ∥_{\infty \to \infty} : = \sup_{∥ X ∥_{\infty} \leq 1} ∥ Ψ (X) ∥_{\infty}$ , where $∥ X ∥_{\infty} : = \sup_{∥ v ∥ = 1} ∥ X v ∥$ , of channels $M_{d} (C) \to M_{e} (C)$ is also a channel. Moreover, a norm limit of completely positive maps of these fixed finite dimensions is completely positive.

Facts & Assumptions

Given: Integers $d, e \geq 1$ , a finite channel $Φ : M_{d} (C) \to M_{e} (C)$ , and finite matrix traces.

[F1]

Finite completely positive trace-preserving maps: a channel is complex-linear, completely positive under every finite ancilla extension, and trace-preserving.

[F2]

Complete finite Choi-Kraus equivalence: every finite completely positive map has a finite Kraus form $Φ (X) = \sum_{a} K_{a} X K_{a}^{*}$ ; trace preservation is equivalent to $\sum_{a} K_{a}^{*} K_{a} = I_{d}$ .

[F3]

Finite matrix trace cyclicity and nondegeneracy: the finite trace cycles rectangular products, and the matrix-unit pairing shows that ${Tr}_{d} (X D) = 0$ for every $X$ forces $D = 0$ .

Proof

1.1givenF2F3algebra

Using the Kraus form in [F2] and finite trace cyclicity [F3], $Tr (Φ (X) A) = \sum_{a} Tr (K_{a} X K_{a}^{*} A) = Tr (X \sum_{a} K_{a}^{*} A K_{a})$ . Thus $Φ^{†} (A) = \sum_{a} K_{a}^{*} A K_{a}$ satisfies the trace-pairing identity.

1.2givenF2algebra

If $Ψ (Y) = \sum_{b} L_{b} Y L_{b}^{*}$ is another channel, then $(Ψ \circ Φ) (X) = \sum_{a, b} (L_{b} K_{a}) X (L_{b} K_{a})^{*}$ , and $\sum_{a, b} (L_{b} K_{a})^{*} (L_{b} K_{a}) = \sum_{a} K_{a}^{*} (\sum_{b} L_{b}^{*} L_{b}) K_{a} = \sum_{a} K_{a}^{*} K_{a} = I_{d} .$ If $Φ_{r} (X) = \sum_{a} K_{r, a} X K_{r, a}^{*}$ and $p_{r} \geq 0$ , $\sum_{r} p_{r} = 1$ , then $\sum_{r} p_{r} Φ_{r}$ has Kraus matrices $\sqrt{p_{r}} K_{r, a}$ , whose adjoint-square sum is $\sum_{r} p_{r} I_{d} = I_{d}$ . In both cases [F2] gives a channel.

2.1givenF1F2F3step 1.1algebra

If $B, C \in M_{d} (C)$ both satisfy the pairing identity for a fixed $A$ , then $Tr (X (B - C)) = 0$ for every $X$ . By [F3], taking $X = E_{i j}$ gives every entry of $B - C$ equal to zero, so uniqueness follows. The Kraus form in step 1.1 is completely positive by [F2], and $Φ^{†} (I_{e}) = \sum_{a} K_{a}^{*} K_{a} = I_{d}$ by [F1] and [F2].

3.1givenF2algebra∎

Let $Φ_{n} \to Φ$ in the stated induced norm, with each $Φ_{n}$ completely positive. Since $∥ E_{i j} ∥_{\infty} = 1$ , each block $Φ_{n} (E_{i j})$ converges to $Φ (E_{i j})$ ; the finite Choi sums therefore converge in matrix norm. Every $J (Φ_{n}) \geq 0$ by [F2], and for every vector $v$ , $⟨ J (Φ) v, v ⟩ = \lim_{n} ⟨ J (Φ_{n}) v, v ⟩ \geq 0$ , so $J (Φ) \geq 0$ . The equivalence in [F2] gives complete positivity. If each $Φ_{n}$ is trace-preserving, then for every $X$ , continuity of finite matrix trace gives $Tr Φ (X) = \lim_{n} Tr Φ_{n} (X) = Tr X$ ; thus $Φ$ is trace-preserving as well.

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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.

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Independent-preparation reduced unitary channel

Statement

Let $H_{S} = C^{d}$ and $H_{E} = C^{m}$ , with $d, m \geq 1$ , in fixed ordered orthonormal bases. For a bounded operator $A$ on a complex Banach space, define $e^{A} : = \sum_{n = 0}^{\infty} A^{n} / n!$ by its operator-norm convergent series (Every finite-dimensional normed space is Banach, The exponential series of a bounded operator). Let $H_{t o t} = H_{t o t}^{*}$ be a matrix on $H_{S} \otimes H_{E}$ with energy units joules, let $ℏ > 0$ have units joule-seconds, and set $U_{t} = e^{- i t H_{t o t} / ℏ}$ for $t \in R$ measured in seconds. Let $σ_{E} \geq 0$ with $Tr σ_{E} = 1$ , and define

$Φ_{t} (ρ) : = {Tr}_{E} ⁣ (U_{t} (ρ \otimes σ_{E}) U_{t}^{*}) (ρ \in M_{d} (C)) .$

For each $t$ , $Φ_{t} : M_{d} (C) \to M_{d} (C)$ is a finite quantum channel. The theorem concerns the product preparation $ρ \otimes σ_{E}$ ; it makes no claim that a correlated initial joint state defines a map of the system marginal alone.

Facts & Assumptions

Given: Finite dimensions $d, m \geq 1$ , a Hermitian $H_{t o t}$ , $ℏ > 0$ , a positive trace-one matrix $σ_{E}$ , and the ordered product basis $H_{S} \otimes H_{E}$ .

[F1]

Finite completely positive trace-preserving maps: a channel is a complex-linear map that is completely positive on every finite ancilla extension and preserves the trace of every input matrix.

[F2]

Finite matrix spectral and state calculus: finite positive matrices have orthonormal eigenbases with nonnegative eigenvalues, and finite traces are cyclic.

[F3]

Finite unitary dynamics and stationary Gibbs states: for a finite Hermitian Hamiltonian $H$ , the matrix $e^{- i t H / ℏ}$ is unitary for every real $t$ . Its current carrier is a draft; this use is conditional on that exact finite-dimensional statement.

[F4]

Partial trace over a finite tensor factor: in ordered orthonormal bases $(e_{a})$ , $(f_{j})$ , the finite partial trace is defined by $({Tr}_{E} R)_{a b} = \sum_{j} R_{(a, j), (b, j)}$ .

[F5]

The partial trace is characterized by the trace pairing and is basis-independent: the preceding coordinate map is the unique map characterized by the finite trace pairing and is independent of the chosen environment basis.

[F6]

Complete finite Choi-Kraus equivalence: a finite completely positive map has a Kraus representation, and it is trace-preserving when the Kraus adjoint-square sum is the identity.

[F7]

The exponential series of a bounded operator: the operator-norm series defines $e^{t A}$ , differentiates to $A e^{t A}$ , and satisfies $e^{t A} e^{s A} = e^{(t + s) A}$ on a complex Banach space.

[F8]

Every finite-dimensional normed space is Banach: a finite-dimensional complex normed space is Banach.

Proof

1.1givenF2F3F7F8choose

Let $A_{t} = - i t H_{t o t} / ℏ$ , a bounded operator on the finite-dimensional complex space $H_{S} \otimes H_{E}$ ; boundedness follows from the finite matrix-coordinate estimate. By [F8], that space is Banach; [F7] defines $U_{t} = e^{A_{t}}$ by an operator-norm convergent series. Since $A_{t}^{*} = - A_{t}$ , termwise adjunction gives $U_{t}^{*} = e^{- A_{t}}$ , and [F7] gives $U_{t}^{*} U_{t} = e^{- A_{t}} e^{A_{t}} = e^{0} = I$ . By [F2], choose an orthonormal eigenbasis $(∣ k ⟩)_{k = 1}^{m}$ of $σ_{E}$ , with $σ_{E} = \sum_{k} p_{k} ∣ k ⟩ ⟨ k ∣$ , $p_{k} \geq 0$ , and $\sum_{k} p_{k} = 1$ .

2.1givenF4F5step 1.1algebra

For system basis indices $a, b$ , environment indices $j, k$ , and the matrix entries of $U_{t}$ , define $(K_{j k})_{a b} : = \sqrt{p_{k}} (U_{t})_{(a, j), (b, k)}$ . Applying the coordinate formula [F4] to $R = U_{t} (ρ \otimes σ_{E}) U_{t}^{*}$ gives $(Φ_{t} (ρ))_{a b} = \sum_{j, k, c, d} (K_{j k})_{a c} ρ_{c d} \overline{(K_{j k})_{b d}} = {(\sum_{j, k} K_{j k} ρ K_{j k}^{*})}_{a b} .$ Thus $Φ_{t} (ρ) = \sum_{j, k} K_{j k} ρ K_{j k}^{*}$ for every matrix $ρ$ . The basis independence in [F5] ensures that this is the same partial-trace map if a different environment basis is used.

3.1step 1.1step 2.1F1F6algebra∎

For any system indices $b, d$ , unitarity in step 1.1 gives ${(\sum_{j, k} K_{j k}^{*} K_{j k})}_{b d} = \sum_{k} p_{k} \sum_{a, j} \overline{(U_{t})_{(a, j), (b, k)}} (U_{t})_{(a, j), (d, k)} = \sum_{k} p_{k} δ_{b d} = δ_{b d} .$ Hence $\sum_{j, k} K_{j k}^{*} K_{j k} = I_{d}$ ; by [F6], the Kraus formula in step 2.1 is completely positive and trace-preserving, which is the channel property in [F1].

Remarks

The finite construction is shown schematically below. ${Ad}_{U_{t}}$ means conjugation $R \mapsto U_{t} R U_{t}^{*}$ ; the arrows depict the defined composition and are not used as a proof.

diagram failed to render:
\rho\otimes\sigma_E \arrow[r,"\operatorname{Ad}_{U_t}"] \arrow[dr,"\Phi_t"'] & U_t(\rho\otimes\sigma_E)U_t^* \arrow[d,"\operatorname{Tr}_E"] \\
& \Phi_t(\rho)

This finite calculation does not imply a semigroup law for $t \mapsto Φ_{t}$ or a weak-coupling/Markov limit.

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Schwarz and trace-norm channel contraction

Statement

Let $Φ : M_{d} (C) \to M_{e} (C)$ be a finite channel, and define $∥ X ∥_{\infty} = \sup_{∥ v ∥ = 1} ∥ X v ∥$ and $∥ X ∥_{1} = Tr \sqrt{X^{*} X}$ . Its trace dual $Φ^{†} : M_{e} (C) \to M_{d} (C)$ from Channel duals and norm closure satisfies

$∥ Φ^{†} (A) ∥_{\infty} \leq ∥ A ∥_{\infty} (A \in M_{e} (C)) .$ For every $X \in M_{d} (C)$ , the Schrödinger map contracts the trace norm: $∥ Φ (X) ∥_{1} \leq ∥ X ∥_{1} .$

The second inequality applies to arbitrary complex matrices, not only density matrices.

Facts & Assumptions

Given: Integers $d, e \geq 1$ , a finite channel $Φ : M_{d} (C) \to M_{e} (C)$ , and finite matrix operator and trace norms as defined in the Statement.

[F1]

Finite completely positive trace-preserving maps: complete positivity includes positivity of the ancilla extension ${id}_{2} \otimes Φ$ , and trace preservation is exact on all input matrices.

[F2]

Channel duals and norm closure: the trace dual is unital and completely positive and satisfies $Tr (Φ (X) A) = Tr (X Φ^{†} (A))$ .

[F3]

Finite matrix spectral and state calculus: the standard finite-dimensional inner product is taken linear in its first argument; finite Hermitian matrices have orthonormal spectral decompositions and positive matrices have positive square roots defined by spectral calculus; the finite trace is cyclic.

Proof

1.1givenF2F3algebra

By [F2], $Φ^{†}$ is positive. For any Hermitian $H \in M_{e} (C)$ , its finite spectral decomposition [F3] is $H = \sum_{λ} λ P_{λ}$ , where each $λ$ is real and each spectral projection $P_{λ}$ is positive semidefinite. Hence every $Φ^{†} (P_{λ})$ is Hermitian, so $Φ^{†} (H)$ is Hermitian. Write an arbitrary $A$ as $A = H + i K$ , where $H = (A + A^{*}) / 2$ and $K = (A - A^{*}) / (2 i)$ are Hermitian. By complex linearity, $Φ^{†} (A^{*}) = Φ^{†} (H) - i Φ^{†} (K) = Φ^{†} (A)^{*}$ .

1.2givenF3constructalgebra

For any $n \geq 1$ and $Z \in M_{n} (C)$ , use the standard inner product $⟨ u, v ⟩ = \sum_{k} u_{k} \overline{v_{k}}$ , linear in its first argument, and define $∣ u ⟩ ⟨ v ∣$ as the map $x \mapsto ⟨ x, v ⟩ u$ . Since $Z^{*} Z$ is positive, with $⟨ Z^{*} Z v, v ⟩ = ∥ Z v ∥^{2}$ , [F3] diagonalizes it in an orthonormal basis $(v_{j})$ with eigenvalues $s_{j}^{2} \geq 0$ . For each $s_{j} > 0$ , put $u_{j} = Z v_{j} / s_{j}$ . Then $⟨ u_{j}, u_{k} ⟩ = δ_{j k}$ , and $Z = \sum_{s_{j} > 0} s_{j} ∣ u_{j} ⟩ ⟨ v_{j} ∣$ . The positive square root of $Z^{*} Z$ has eigenvalues $s_{j}$ , so $∥ Z ∥_{1} = \sum_{j} s_{j}$ .

2.1givenF1F2step 1.1algebra

For $A \in M_{e} (C)$ , the block matrix $(\begin{array}{cc} A^{*} A & A^{*} \\ A & I_{e} \end{array})$ is positive semidefinite because its quadratic form at $(x, y)$ is $∥ A x + y ∥^{2}$ . Apply ${id}_{2} \otimes Φ^{†}$ : by [F2] the trace dual is completely positive, and the definition in [F1] makes its extension by the two-dimensional ancilla positive. By unitality in [F2] and adjoint preservation from step 1.1, the resulting block is $(\begin{array}{cc} Φ^{†} (A^{*} A) & Φ^{†} (A)^{*} \\ Φ^{†} (A) & I_{d} \end{array}) \geq 0$ .

2.2step 1.2F3choosealgebra

For any $n \geq 1$ , $Z \in M_{n} (C)$ , and $B \in M_{n} (C)$ with $∥ B ∥_{\infty} \leq 1$ , write $Z_{a b} = \sum_{s_{j} > 0} s_{j} (u_{j})_{a} \overline{(v_{j})_{b}}$ using step 1.2. Expanding both diagonal-sum traces gives $Tr (B Z) = \sum_{a, b} B_{a b} Z_{b a} = \sum_{s_{j} > 0} s_{j} \sum_{a, b} B_{a b} (u_{j})_{b} \overline{(v_{j})_{a}}$ and $Tr (Z B) = \sum_{a, b} Z_{a b} B_{b a} = \sum_{s_{j} > 0} s_{j} \sum_{a, b} B_{a b} (u_{j})_{b} \overline{(v_{j})_{a}}$ . The inner sum is $⟨ B u_{j}, v_{j} ⟩$ for the inner product fixed in step 1.2. For each unit $v_{j}$ , $0 \leq ∥ B u_{j} - ⟨ B u_{j}, v_{j} ⟩ v_{j} ∥^{2} = ∥ B u_{j} ∥^{2} - ∣ ⟨ B u_{j}, v_{j} ⟩ ∣^{2}$ , so both trace magnitudes are at most $\sum_{j} s_{j} = ∥ Z ∥_{1}$ . Extend the two finite orthonormal lists $(u_{j})$ , $(v_{j})$ to orthonormal bases and choose a unitary $B$ carrying each $u_{j}$ to $v_{j}$ ; then both trace magnitudes equal $\sum_{j} s_{j}$ . Thus $∥ Z ∥_{1} = \sup_{∥ B ∥_{\infty} \leq 1} ∣ Tr (B Z) ∣ = \sup_{∥ B ∥_{\infty} \leq 1} ∣ Tr (Z B) ∣$ , including $Z = 0$ .

3.1step 2.1F2F3algebra

In the positive block from step 2.1, evaluate the quadratic form at $(x, - Φ^{†} (A) x)$ . This gives $⟨ (Φ^{†} (A^{*} A) - Φ^{†} (A)^{*} Φ^{†} (A)) x, x ⟩ \geq 0$ for every $x$ , so $Φ^{†} (A)^{*} Φ^{†} (A) \leq Φ^{†} (A^{*} A)$ . Also $A^{*} A \leq ∥ A ∥_{\infty}^{2} I_{e}$ ; positivity and unitality in [F2] then give $Φ^{†} (A)^{*} Φ^{†} (A) \leq ∥ A ∥_{\infty}^{2} I_{d}$ , proving $∥ Φ^{†} (A) ∥_{\infty} \leq ∥ A ∥_{\infty}$ .

4.1step 3.1step 2.2F2algebra∎

Apply the second dual norm formula from step 2.2 in output dimension $e$ to $Φ (X)$ . For each $A \in M_{e} (C)$ with $∥ A ∥_{\infty} \leq 1$ , step 3.1 gives $∥ Φ^{†} (A) ∥_{\infty} \leq 1$ , so the same formula in input dimension $d$ bounds $∣ Tr (X Φ^{†} (A)) ∣ \leq ∥ X ∥_{1}$ . The trace-pairing identity [F2] therefore gives $∥ Φ (X) ∥_{1} = \sup_{∥ A ∥_{\infty} \leq 1} ∣ Tr (Φ (X) A) ∣ = \sup_{∥ A ∥_{\infty} \leq 1} ∣ Tr (X Φ^{†} (A)) ∣ \leq ∥ X ∥_{1}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

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.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Pinching raises finite entropy

Statement

Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ . Fix an ordered orthonormal basis $(e_{a})_{a = 1}^{d}$ of $H$ and the unitary coordinate map $U : H \to C^{d}$ given by $U (\sum_{a} z_{a} e_{a}) = (z_{1}, \dots, z_{d})$ . Represent each endomorphism $A : H \to H$ by $U A U^{*} \in M_{d} (C)$ , using the same symbol for the operator and its matrix in these fixed coordinates. Let $P_{0}, \dots, P_{r - 1}$ be pairwise orthogonal projections, meaning $P_{j} = P_{j}^{*} = P_{j}^{2}$ and $P_{j} P_{k} = 0$ for $j \neq k$ , with $r \geq 1$ and $\sum_{j = 0}^{r - 1} P_{j} = I$ . Define the pinching map $Δ : M_{d} (C) \to M_{d} (C)$ by $Δ (X) = \sum_{j = 0}^{r - 1} P_{j} X P_{j}$ . For every positive semidefinite trace-one matrix $ρ \in M_{d} (C)$ , $Δ$ is a finite channel and

$S (Δ (ρ)) \geq S (ρ),$

where $S (ρ) = - k_{B} \sum_{λ} λ \log λ$ , the sum is over eigenvalues with multiplicity, $0 \log 0 = 0$ , the logarithm is natural, and $k_{B} > 0$ has units joule per kelvin. The theorem concerns only this finite pinching operation; it does not say that every channel increases entropy.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space with the fixed orthonormal coordinates in the Statement, the stated orthogonal projections, $k_{B} > 0$ , and a positive semidefinite trace-one matrix $ρ \in M_{d} (C)$ .

[A1]

The finite entropy supplier used below carries the full Axiom of Choice through its current trace-cross-log dependency closure (The Axiom of Choice). The phase-average and spectral calculations in this item use only finite sums; no stronger choice principle is introduced.

[F1]

Finite completely positive trace-preserving maps: a finite channel is completely positive and trace-preserving.

[F2]

Complete finite Choi-Kraus equivalence: a map with Kraus operators $P_{j}$ is completely positive, and it is trace-preserving when $\sum_{j} P_{j}^{*} P_{j} = I$ .

[F3]

Quantum spectral entropy and relative-entropy domains: for a positive trace-one matrix with eigenvalues $λ$ , $S (ρ) = - k_{B} \sum_{λ} λ \log λ$ , with zero terms interpreted as $0$ .

[F4]

Finite mixing, product entropy, and bipartite subadditivity: entropy is concave under finite mixtures of finite-dimensional density matrices, including rank-deficient inputs. The current draft has a hash-current Step 1 ready record after owner resolution, but has not received independent proof acceptance; this deduction is conditional on that exact finite-mixing clause and its stated Axiom of Choice.

[F5]

Finite matrix spectral and state calculus: finite Hermitian matrices have orthonormal spectral decompositions, and unitary conjugation preserves their eigenvalues.

[F6]

The complex exponential by its power series: for $z \in C$ , $\exp z$ is defined by its convergent complex power series and $\exp 0 = 1$ .

[F7]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential: $\exp (z + w) = \exp z \exp w$ for all $z, w \in C$ .

[F8]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ : $\exp z = 1$ exactly when $z \in 2 π i Z$ .

Proof

1.1givenF1F2algebra

The map $Δ$ has Kraus matrices $P_{j}$ , and $\sum_{j} P_{j}^{*} P_{j} = \sum_{j} P_{j} = I$ . Therefore [F2] gives complete positivity and trace preservation, which is the channel property in [F1].

1.2givenF6F7F8algebra

Let $ω = \exp (2 π i / r)$ and $V = \sum_{j = 0}^{r - 1} ω^{j} P_{j}$ . By [F7], $ω^{n} = \exp (2 π i n / r)$ for positive integers $n$ ; for negative $n$ , addition and $\exp 0 = 1$ give the corresponding reciprocal identity. The kernel fact [F8] then gives $ω^{r} = 1$ and $ω^{n} \neq 1$ for every integer $n$ with $0 < ∣ n ∣ < r$ . Orthogonality gives $V V^{*} = V^{*} V = \sum_{j} P_{j} = I$ , so $V$ is unitary. For $X \in M_{d} (C)$ , $\frac{1}{r} \sum_{k = 0}^{r - 1} V^{k} X V^{- k} = \sum_{j, ℓ = 0}^{r - 1} (\frac{1}{r} \sum_{k = 0}^{r - 1} ω^{k (j - ℓ)}) P_{j} X P_{ℓ} = \sum_{j} P_{j} X P_{j} = Δ (X),$ Indeed the inner sum is $1$ when $j = ℓ$ ; otherwise $q = ω^{j - ℓ}$ satisfies $q^{r} = 1$ and $q \neq 1$ , so $\sum_{k = 0}^{r - 1} q^{k} = (q^{r} - 1) / (q - 1) = 0$ . When $r = 1$ , there are no off-diagonal terms and the same identity holds.

2.1givenA1F3F4F5step 1.2algebra∎

Each $V^{k} ρ V^{- k}$ has the same eigenvalues as $ρ$ , by [F5], hence the same entropy by [F3]. Applying the finite-mixture concavity [F4] with equal weights $1 / r$ and using step 1.2 gives $S (Δ (ρ)) \geq r^{- 1} \sum_{k} S (V^{k} ρ V^{- k}) = S (ρ)$ , conditional on [A1] and the stated draft supplier interface.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Continuous finite GKSL channel propagators

Statement

Let $I = [s_{0}, s_{1}] \subset R$ be a compact time interval in seconds, $d \geq 1$ , $H : I \to M_{d} (C)$ continuous with $H (t) = H (t)^{*}$ in joules, and $L_{a} : I \to M_{d} (C)$ continuous with finite $a = 1, \dots, m$ in $s^{- 1 / 2}$ . At each $t \in I$ , form $L_{t}$ by Finite GKSL generators and units. For every $s \leq t$ in $I$ , the propagator $Λ (t, s)$ solving

$\partial_{t} Λ (t, s) = L_{t} \circ Λ (t, s), Λ (s, s) = id,$

exists uniquely and is a finite channel. No semigroup law is asserted when $t \mapsto L_{t}$ varies.

Facts & Assumptions

Given: A compact interval $I$ , continuous finite matrices $H (t), L_{a} (t)$ with the stated units, and the induced finite-dimensional norm on linear maps.

[F1]

Finite GKSL generators and units: each fixed $L_{t}$ has finite GKSL form.

[F2]

Finite GKSL generates CPTP semigroups: for finite dimension $d \geq 1$ and each fixed $u$ , $e^{τ L_{u}}$ , defined by the operator-norm series, is a channel for every $τ \geq 0$ .

[F3]

Channel duals and norm closure: compositions and norm limits of finite channels are channels.

[F4]

Finite completely positive trace-preserving maps: a finite channel is completely positive and trace-preserving.

[F5]

Every finite-dimensional normed space is Banach: every finite-dimensional normed space over $R$ or $C$ is complete.

[F6]

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 with the usual norm bound, additive intervals, and an indefinite integral differentiable to the original curve.

Proof

1.1givenF1F5F6constructalgebra

Let $E = {End}_{C} (M_{d} (C))$ with the induced map norm. It has finite complex dimension $d^{4}$ , so it is complete as a real normed space by [F5]. Set $M : = \sup_{u \in I} ∥ L_{u} ∥ < \infty$ . For fixed $s$ , define $D_{0} (t) = I$ and recursively $D_{n} (t) = \int_{s}^{t} L_{u} D_{n - 1} (u) d u$ for $n \geq 1$ , using [F6]. Induction with the integral norm bound gives $∥ D_{n} (t) ∥ \leq M^{n} (t - s)^{n} / n!$ ; hence $\sum_{n = 0}^{\infty} D_{n} (t)$ converges uniformly on $[s, s_{1}]$ . Calling its sum $Λ (t, s)$ , uniform convergence and boundedness of $L_{u}$ allow passage through the integral, giving $Λ (t, s) = I + \int_{s}^{t} L_{u} Λ (u, s) d u$ . Also $∥ Λ (u, s) ∥ \leq e^{M (u - s)}$ .

1.2givenF2F3algebra

Let $t_{j} = s + j (t - s) / n$ be a uniform partition of $[s, t]$ and freeze $L_{u}$ to $L_{t_{j}}$ on each interval $[t_{j}, t_{j + 1})$ . The resulting propagator is the time-ordered composition $e^{(t - t_{n - 1}) L_{t_{n - 1}}} \circ \dots \circ e^{(t_{1} - t_{0}) L_{t_{0}}}$ . Each factor is a channel by [F2], and the composition is a channel by [F3].

2.1step 1.1F6algebra

By the fundamental theorem [F6], the integral equation in step 1.1 is differentiable and satisfies $\partial_{t} Λ (t, s) = L_{t} \circ Λ (t, s)$ , with $Λ (s, s) = I$ . If another solution differs by $D$ , then $D (t) = \int_{s}^{t} L_{u} D (u) d u$ . With $C = \sup_{u \in [s, s_{1}]} ∥ D (u) ∥ < \infty$ , repeated application of the integral norm bound gives $∥ D (t) ∥ \leq C M^{n} (t - s)^{n} / n!$ for every $n \geq 0$ . This tends to zero, proving uniqueness.

3.1step 1.1step 1.2F3F4F6algebra∎

Let $δ_{n} = \sup_{u \in [s, t]} ∥ L_{u}^{(n)} - L_{u} ∥$ ; uniform continuity gives $δ_{n} \to 0$ . Subtract the integral equations for the frozen propagator $Λ_{n}$ and $Λ$ . For $s \leq u \leq t$ , the norm difference is bounded by $M \int_{s}^{u} ∥ Λ_{n} (v, s) - Λ (v, s) ∥ d v + δ_{n} (t - s) e^{M (t - s)}$ , using the bound in step 1.1. Iterating this scalar integral inequality yields $∥ Λ_{n} (t, s) - Λ (t, s) ∥ \leq δ_{n} (t - s) e^{2 M (t - s)} \to 0$ . The channel set is closed under this norm limit by [F3], so $Λ (t, s)$ is a channel. For $s = t$ it is $I$ , also a channel.

5 · Examples, experiments and counterexamples

None yet.

Finite completely positive trace-preserving maps

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

Finite matrix trace cyclicity and nondegeneracy

Statement

Facts & Assumptions

Proof

Finite completely positive trace-preserving maps

Definition

Finite GKSL generators and units

Definition

Complete finite Choi-Kraus equivalence

Statement

Facts & Assumptions

Proof

Remarks

Channel duals and norm closure

Statement

Facts & Assumptions

Proof

Finite GKSL generates CPTP semigroups

Statement

Facts & Assumptions

Proof

Independent-preparation reduced unitary channel

Statement

Facts & Assumptions

Proof

Remarks

Schwarz and trace-norm channel contraction

Statement

Facts & Assumptions

Proof

Norm-continuous finite channel semigroups are GKSL

Statement

Facts & Assumptions

Proof

Pinching raises finite entropy

Statement

Facts & Assumptions

Proof

Continuous finite GKSL channel propagators

Statement

Facts & Assumptions

Proof

5 · Examples, experiments and counterexamples

Sources