LemmaStatement: AI-generatedProof: AI-generated

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Qubit coherence channels and fidelity functionals

Statement

Let $H = C^{2}$ have the ordered orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ . For $λ \in C$ with $∣ λ ∣ \leq 1$ , let the complex-linear map $D_{λ}$ send a matrix $X = (x_{i j})$ to the matrix with diagonal entries $x_{00}, x_{11}$ and off-diagonal entries $λ x_{01}, \overline{λ} x_{10}$ . Then $D_{λ}$ is a quantum channel and its entanglement fidelity for the maximally mixed input is $F_{e} (D_{λ}) = (1 + Re λ) / 2$ . For every unit vector $∣ ψ ⟩ = a ∣ 0 ⟩ + b ∣ 1 ⟩$ , its pure-state infidelity $1 - ⟨ ψ ∣ D_{λ} (∣ ψ ⟩ ⟨ ψ ∣) ∣ ψ ⟩$ is $2 ∣ a ∣^{2} ∣ b ∣^{2} (1 - Re λ)$ .

Composition satisfies $D_{μ} \circ D_{λ} = D_{μ λ}$ . For a finite instrument $(I_{j})_{j \in J}$ from $H$ to $H$ , the flag here means its outcome label $j$ , and discarding it means taking the nonselective map $I_{n s} (X) : = \sum_{j \in J} I_{j} (X)$ . If its branches are $I_{j} = p_{j} D_{λ_{j}}$ , with $p_{j} \geq 0$ , $\sum_{j} p_{j} = 1$ and $∣ λ_{j} ∣ \leq 1$ , then this map is the channel $\sum_{j} p_{j} D_{λ_{j}}$ ; its entanglement fidelity and pure-state infidelity at any fixed $∣ ψ ⟩$ are the respective $p_{j}$ -weighted averages of the preceding expressions. For $p_{j} > 0$ the conditional output is $D_{λ_{j}} (ρ)$ ; zero-weight branches contribute zero and have no assigned conditional state.

Statement references

Finite operator and ket conventions

Entanglement fidelity on a finite logical space

Finite-dimensional quantum channel

Finite instruments and conditional states

Facts & Assumptions

[F1]

The specified coordinates, bras, outer products and ordered tensor basis use the finite operator conventions. Finite operator and ket conventions

[F2]

For a qubit channel, $F_{e} (E) = ⟨ Φ ∣ (i d \otimes E) (∣ Φ ⟩ ⟨ Φ ∣) ∣ Φ ⟩$ , where $∣ Φ ⟩ = (∣ 00 ⟩ + ∣ 11 ⟩) / \sqrt{2}$ . Entanglement fidelity on a finite logical space

[F3]

A quantum channel is a complex-linear, completely positive, trace-preserving map. Complete positivity means positivity after tensoring with the identity on every finite ancillary matrix space. Finite-dimensional quantum channel

[F4]

An instrument's outcome weight is the trace of its branch output, and its conditional state is that output divided by the weight when positive; no conditional state is assigned at zero weight. Finite instruments and conditional states

Proof

Given: The parameters, maps, vector and flagged branches in the statement.

Goal: Establish channel validity, the fidelity formulas, composition and branch averaging.

1.1F1F3givenalgebra

Put $r = ∣ λ ∣$ and choose $z = λ / r$ if $r > 0$ , and $z = 1$ if $r = 0$ . Let $U = diag (1, \overline{z})$ and $Z = diag (1, - 1)$ . Direct multiplication gives $D_{λ} (X) = ((1 + r) / 2) U X U^{†} + ((1 - r) / 2) U Z X Z U^{†}$ . The coefficients are nonnegative and sum to one. After tensoring with any finite identity map, each summand sends a positive matrix to a positive matrix by conjugation; hence their sum is positive. Both conjugating matrices are unitary, so the trace is preserved. Thus $D_{λ}$ is a quantum channel.

1.2F1givenalgebra

The input projector has diagonal entries $∣ a ∣^{2}, ∣ b ∣^{2}$ and $01$ entry $a \overline{b}$ . Its output overlap with $ψ$ is $∣ a ∣^{4} + ∣ b ∣^{4} + 2 ∣ a ∣^{2} ∣ b ∣^{2} Re λ$ . Since $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , subtracting this from one gives the stated infidelity. Entrywise multiplication also gives $D_{μ} \circ D_{λ} = D_{μ λ}$ .

2.1F1F2step 1.1algebra

Expand $∣ Φ ⟩ ⟨ Φ ∣$ as one half of the sum of $∣ j j ⟩ ⟨ k k ∣$ over $j, k \in {0, 1}$ . The two diagonal terms contribute $1 / 4$ each to the overlap in [F2], while the two off-diagonal terms contribute $λ / 4$ and $\overline{λ} / 4$ . Therefore $F_{e} (D_{λ}) = (1 + Re λ) / 2$ .

3.1F2F3F4step 1.1step 2.1step 1.2givenalgebra∎

For each density operator $ρ$ , trace preservation gives $tr I_{j} (ρ) = p_{j}$ . The conditional-state rule therefore gives the stated branch output when $p_{j} > 0$ , and assigns none when $p_{j} = 0$ . By the definition in the Statement, discarding the outcome label gives $I_{n s} = \sum_{j} I_{j} = \sum_{j} p_{j} D_{λ_{j}}$ . This sum is complex-linear and completely positive: after every finite amplification it is a sum of positive maps with nonnegative coefficients, by step 1.1. For every matrix $X$ , its trace is $\sum_{j} p_{j} tr X = tr X$ , so it is a channel by [F3]. Both the overlap defining entanglement fidelity and the pure-state output overlap are linear in that sum; because $\sum_{j} p_{j} = 1$ , the corresponding infidelity is also the weighted sum of branch infidelities. Substituting steps 2.1 and 1.2 proves the claimed averages.

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