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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.

Independent-preparation reduced unitary channel

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources