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Two orthogonal Pauli alternatives bound recovery fidelity

Statement

Let $H = C^{2}$ with its standard ordered basis, and let $R : L (H) \to L (H)$ be a quantum channel: a complex-linear, completely positive, trace-preserving map. Let $U$ be a unitary $2 \times 2$ matrix, so $U^{†} U = U U^{†} = I$ , and put $Z = diag (1, - 1)$ . For any matrix $V$ write ${Ad}_{V} (X) = V X V^{†}$ . Then $F_{e} (R \circ {Ad}_{U}) + F_{e} (R \circ {Ad}_{Z U}) \leq 1$ , where both entanglement fidelities use the maximally mixed input $I / 2$ .

Facts & Assumptions

Given: The space, channel $R$ , and matrices $U, Z$ in the Statement.

Goal: Prove the displayed sum of entanglement fidelities is at most one.

[F1]

The finite operator conventions use the standard basis and conjugate-transpose adjoints. Finite operator and ket conventions

[F2]

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

[F3]

A quantum channel is completely positive and trace-preserving. Every finite-dimensional completely positive map has a finite Kraus representation $R (X) = \sum_{a} K_{a} X K_{a}^{†}$ . Finite-dimensional quantum channel Finite Kraus representation Kraus and Choi characterizations of complete positivity

[F4]

A finite Kraus sum is completely positive, and it is trace-preserving exactly when its Kraus operators satisfy $\sum_{a} K_{a}^{†} K_{a} = I$ . A finite Kraus sum is completely positive Trace preservation is Kraus normalization

Proof

1.1givenF3F4algebra

Since $Z^{†} Z = I$ , both $U$ and $Z U$ are unitary. By [F4], each one-operator sum ${Ad}_{V}$ for $V \in {U, Z U}$ is a channel. By [F3], choose a finite Kraus family $(K_{a})$ for $R$ ; [F4] gives $\sum_{a} K_{a}^{†} K_{a} = I$ . Direct composition gives Kraus families $(K_{a} V)$ for $R \circ {Ad}_{V}$ , and their normalization is $\sum_{a} V^{†} K_{a}^{†} K_{a} V = I$ . Thus the compositions are channels by [F4].

2.1step 1.1F1F2F3algebra

For any matrix $A$ , expansion in the basis in [F1] gives $⟨ Φ ∣ (I \otimes A) ∣ Φ ⟩ = tr (A) / 2$ . Using the Kraus formula and [F2] therefore gives $F_{e} (R \circ {Ad}_{V}) = \frac{1}{4} \sum_{a} ∣ tr (K_{a} V) ∣^{2}$ for $V \in {U, Z U}$ . Put $B_{a} = U K_{a}$ . Finite matrix-index expansion gives $tr (K_{a} U) = tr (B_{a})$ and $tr (K_{a} Z U) = tr (B_{a} Z)$ . If the diagonal entries of $B_{a}$ are $b_{a, 00}, b_{a, 11}$ , then $∣ tr (B_{a}) ∣^{2} + ∣ tr (B_{a} Z) ∣^{2} = ∣ b_{a, 00} + b_{a, 11} ∣^{2} + ∣ b_{a, 00} - b_{a, 11} ∣^{2} = 2 (∣ b_{a, 00} ∣^{2} + ∣ b_{a, 11} ∣^{2}) \leq 2 tr (B_{a}^{†} B_{a})$ .

3.1givenstep 1.1step 2.1F1algebra∎

Since $U^{†} U = I$ , the normalization in step 1.1 gives $\sum_{a} tr (B_{a}^{†} B_{a}) = tr (\sum_{a} K_{a}^{†} K_{a}) = tr (I) = 2$ . Summing the inequality in step 2.1 and multiplying by $1 / 4$ yields $F_{e} (R \circ {Ad}_{U}) + F_{e} (R \circ {Ad}_{Z U}) \leq 1$ , as required.

Depends on

Used by · 1

Pauli-complementary qubit recovery fidelity inequality

Dependency tree · 2 levels

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