Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Product-state reduced evolution is a channel

Physical scope

Exact reduced maps of finite-dimensional system and environment models with a fixed product initial assignment and unitary joint evolution. The theorem permits neither an initial-correlation extension nor a continuum bath or Markov limit; its finite example is a conditional model calculation, not an experiment.

Statement

In the finite product-state model, fix any $t \geq 0$ and define the complex-linear map $Φ_{t} : M_{d} (C) \to M_{d} (C)$ by $Φ_{t} (X) = {Tr}_{E} ⁣ (U_{t} (X \otimes η) U_{t}^{*})$ . Then $Φ_{t}$ is a finite quantum channel. The family $t \mapsto Φ_{t}$ need not obey the semigroup law and need not be CP-divisible. The terms “semigroup law” and “CP-divisible” have the definitions of Finite channel families, semigroups, and CP-divisibility.

Statement references

Finite system-environment and Markov model assignments

Finite channel families, semigroups, and CP-divisibility

Partial trace over a finite tensor factor

The partial trace is characterized by the trace pairing and is basis-independent

Finite completely positive trace-preserving maps

Complete finite Choi-Kraus equivalence

Facts & Assumptions

Given: Integers $d, m \geq 1$ , time $t \geq 0$ , system and environment density matrices $ρ \in M_{d} (C)$ and $η \in M_{m} (C)$ , and a unitary $U_{t}$ on $C^{d} \otimes C^{m}$ with the system factor first.

[A1]

Finite system-environment and Markov model assignments: $η$ is fixed independently of every allowed input $ρ$ , the joint initial state is $ρ \otimes η$ , and the reduced map is $Φ_{t} (X) = {Tr}_{E} (U_{t} (X \otimes η) U_{t}^{*})$ ; the parameter $t$ indexing $U_{t}$ is elapsed physical time measured in seconds.

[F1]

Partial trace over a finite tensor factor: in the ordered product basis, $({Tr}_{E} R)_{α β} = \sum_{j} R_{(α, j), (β, j)}$ .

[F2]

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

[F3]

Finite matrix spectral and state calculus: every finite positive matrix has an orthonormal eigenbasis with nonnegative eigenvalues, and its trace is the sum of its diagonal entries in any orthonormal basis.

[F4]

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

[F5]

Complete finite Choi-Kraus equivalence: a finite map of Kraus form $Ψ (X) = \sum_{a} K_{a} X K_{a}^{*}$ is completely positive, and it is trace-preserving exactly when $\sum_{a} K_{a}^{*} K_{a} = I$ .

[F6]

Finite channel families, semigroups, and CP-divisibility: CP-divisibility requires a finite channel $V_{t, s}$ for every $s \leq t$ with $Φ_{t} = V_{t, s} \circ Φ_{s}$ ; the semigroup law requires $Φ_{t + s} = Φ_{t} \circ Φ_{s}$ and $Φ_{0} = id$ .

[F7]

The complex exponential by its power series: $\exp z = \sum_{n = 0}^{\infty} z^{n} / n!$ and $\exp 0 = 1$ .

[F8]

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

[F9]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : for real $θ$ , $\exp (i θ) = \cos θ + i \sin θ$ and $∣ \exp (i θ) ∣ = 1$ .

[F10]

$t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle: there exists $θ \in [0, 2 π)$ with $(\cos θ, \sin θ) = (0, 1)$ .

Proof

technique · finite Kraus construction followed by an explicit two-qubit counterexample

1.1givenA1F1F2F3constructalgebra

By [F3], choose an orthonormal environment eigenbasis $(∣ k ⟩)_{k = 1}^{m}$ with $η = \sum_{k = 1}^{m} p_{k} ∣ k ⟩ ⟨ k ∣$ , $p_{k} \geq 0$ , and $\sum_{k} p_{k} = 1$ . Use the fixed system basis from the model and its ordered product with this environment basis. By [F2], changing the environment basis does not change the partial-trace map in [F1]. For environment indices $j, k$ define the system matrix $K_{j k}$ by $(K_{j k})_{α β} : = \sqrt{p_{k}} (U_{t})_{(α, j), (β, k)}$ . Expanding $X \otimes η$ , conjugating by $U_{t}$ , and applying the coordinate formula [F1] gives, entry by entry, $Φ_{t} (X) = \sum_{j, k = 1}^{m} K_{j k} X K_{j k}^{*}$ for every $X \in M_{d} (C)$ .

1.2givenA1F1F7F9constructalgebra

For a two-qubit instance take $Z = diag (1, - 1)$ , $g > 0$ in $s^{- 1}$ , $H_{t o t} = ℏ g Z \otimes Z$ with $ℏ$ in joule-seconds, and $η = ∣ + ⟩ ⟨ + ∣$ where $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ in the chosen environment $Z$ eigenbasis. Put $A = Z \otimes Z$ and $P_{b} = (I + b A) / 2$ for $b = \pm 1$ ; direct multiplication gives $P_{b} P_{b^{'}} = δ_{b b^{'}} P_{b}$ , $P_{+} + P_{-} = I$ , and $A = \sum_{b = \pm 1} b P_{b}$ . Define the matrix exponential for this instance by the power series $U_{t} : = \sum_{n = 0}^{\infty} (- i g t)^{n} A^{n} / n!$ . Since $A^{n} = \sum_{b} b^{n} P_{b}$ , the series converges entrywise by the scalar exponential definition [F7] and equals $\sum_{b} e^{- i b g t} P_{b} = e^{- i t H_{t o t} / ℏ}$ . In the ordered environment $Z$ eigenbasis this is $\sum_{b = \pm 1} V_{b} (t) \otimes ∣ b ⟩ ⟨ b ∣$ , where $V_{b} (t) = diag (e^{- i b g t}, e^{i b g t})$ . By [F9], each scalar phase has modulus one, so $U_{t}$ is unitary and $U_{0} = I$ . The partial trace in [F1] therefore gives $Φ_{t} (X) = \frac{1}{2} (V_{+ 1} (t) X V_{+ 1} (t)^{*} + V_{- 1} (t) X V_{- 1} (t)^{*})$ , so for $X = (\begin{array}{cc} a & z \\ w & c \end{array})$ the diagonal entries are unchanged and the off-diagonal entries are multiplied by $κ (t) = \frac{1}{2} (e^{- 2 i g t} + e^{2 i g t}) = \cos (2 g t)$ .

1.3F7F8F9F10algebra

By [F10], choose $θ \in [0, 2 π)$ with $\cos θ = 0$ and $\sin θ = 1$ ; hence $θ > 0$ . Set $s = θ / (2 g) > 0$ . Euler's identity [F9] gives $e^{i θ} = i$ , and [F7]–[F8] give $e^{- i θ} = (e^{i θ})^{- 1} = - i$ . Applying [F8] again yields $κ (s) = 0$ and $κ (2 s) = \frac{1}{2} (e^{- 2 i θ} + e^{2 i θ}) = - 1$ .

2.1givenF4F5step 1.1algebra

For system basis indices $β, δ$ , unitarity of $U_{t}$ gives $(\sum_{j, k} K_{j k}^{*} K_{j k})_{β δ} = \sum_{k = 1}^{m} p_{k} \sum_{j = 1}^{m} \sum_{α = 1}^{d} \overline{(U_{t})_{(α, j), (β, k)}} (U_{t})_{(α, j), (δ, k)} = \sum_{k} p_{k} δ_{β δ} = δ_{β δ}$ . Thus $\sum_{j, k} K_{j k}^{*} K_{j k} = I_{d}$ , so [F5] makes the Kraus map in step 1.1 completely positive and trace-preserving; by [F4], $Φ_{t}$ is a finite channel.

3.1F3F6step 1.2step 1.3step 2.1algebra∎

Let $X_{⁣ P} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ and $ρ_{\pm} = (I \pm X_{⁣ P}) / 2$ . Since $X_{⁣ P} = X_{⁣ P}^{*}$ and $X_{⁣ P}^{2} = I$ , each $ρ_{\pm}$ is a projection; since $Tr X_{⁣ P} = 0$ , each has trace one. Steps 1.2 and 1.3 give $Φ_{s} (ρ_{+}) = Φ_{s} (ρ_{-}) = I / 2$ , while $Φ_{2 s} (ρ_{+}) = ρ_{-}$ and $Φ_{2 s} (ρ_{-}) = ρ_{+}$ . Consequently $Φ_{2 s} (ρ_{+}) \neq Φ_{s} (Φ_{s} (ρ_{+})) = I / 2$ , so the semigroup law in [F6] fails. If a channel $V_{2 s, s}$ satisfied $Φ_{2 s} = V_{2 s, s} \circ Φ_{s}$ , it would send the equal matrices $Φ_{s} (ρ_{+})$ and $Φ_{s} (ρ_{-})$ to equal outputs, contradicting the two distinct outputs at $2 s$ . By step 2.1, the family consists of channels, so this contradiction proves it is not CP-divisible by [F6].

Remarks

The two-qubit recurrence is a conditional prediction of a finite unitary model. It is not a reported observation and does not establish a physical Markov limit.

Product-state reduced evolution is a channel

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources