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A uniform transpose-fidelity bound for the one-decay channel

Statement

Let $C : L (C^{2}) \to L (C^{3})$ be a channel such that $C (I_{2}) = 2 I_{3} / 3$ and $C (Z) = diag (1 / 3, - 2 / 3, 1 / 3)$ , with $Z = diag (1, - 1)$ . Let $T$ be its transpose recovery for the full logical code $I_{2}$ . Then $F_{e} (T C) \geq 3 / 8$ . More precisely, if $C^{*}$ denotes the adjoint for the Hilbert–Schmidt inner product $⟨ A, B ⟩_{H S} = Tr (A^{†} B)$ and $Q_{0} = I_{2} / \sqrt{2}$ , $Q_{1} = Z / \sqrt{2}$ , $Q_{2} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}) / \sqrt{2}$ , $Q_{3} = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}) / \sqrt{2}$ , then $T C = \frac{3}{2} C^{*} C, F_{e} (T C) = \frac{3}{8} \sum_{j = 0}^{3} ∥ C (Q_{j}) ∥_{H S}^{2} \geq \frac{3}{8} .$ Here $∥ A ∥_{H S}^{2} = Tr (A^{†} A)$ is the sum of squared entry moduli.

Facts & Assumptions

[F1]

For code $I_{2}$ the transpose map has Kraus maps $K_{a}^{†} C (I_{2})^{- 1 / 2}$ , with no output-support ambiguity when $C (I_{2})$ is positive definite. Transpose (Petz) recovery on a code

[F2]

The qubit entanglement fidelity is the maximally entangled overlap and is linear in the recovered channel. Entanglement fidelity on a finite logical space

[F3]

Use finite matrix units, adjoints and ordered tensor coordinates. Finite operator and ket conventions

[F4]

A finite-dimensional channel has a finite Kraus representation $C (X) = \sum K_{a} X K_{a}^{†}$ with normalized adjoint-product sum. Finite channel, Kraus, Choi, and Stinespring equivalence

[F5]

For compatible finite rectangular matrices $A \in M_{e \times d} (C)$ and $B \in M_{d \times e} (C)$ , the diagonal-sum trace satisfies ${Tr}_{e} (A B) = {Tr}_{d} (B A)$ ; in particular, square matrix trace is cyclic. Finite matrix trace cyclicity and nondegeneracy

Proof

Given: The channel, its two specified output matrices and the full logical code.

Goal: Obtain a positive-square expression for its recovered fidelity and a uniform lower bound.

1.1givenF1F3F4F5algebra

Choose a finite Kraus representation by [F4]. Its Hilbert–Schmidt adjoint is $C^{*} (Y) = \sum_{a} K_{a}^{†} Y K_{a}$ , since rectangular trace cyclicity in [F5] verifies the defining adjoint identity entrywise. Because $C (I_{2})^{- 1 / 2} = \sqrt{3 / 2} I_{3}$ , the transpose formula [F1] gives $T (Y) = (3 / 2) C^{*} (Y)$ . Hence $E : = T C = (3 / 2) C^{*} C$ and $⟨ Q, E (Q) ⟩_{H S} = (3 / 2) ∥ C (Q) ∥_{H S}^{2}$ for every matrix $Q$ .

2.1F2F3F5step 1.1algebra

Expanding the qubit maximally entangled projector in matrix units $E_{a b} = ∣ a ⟩ ⟨ b ∣$ in [F2] gives $F_{e} (E) = \frac{1}{4} \sum_{a, b} Tr (E_{a b}^{†} E (E_{a b}))$ . The sum is unchanged under any orthonormal change of matrix basis: substituting its unitary coefficient matrix makes the cross coefficients collapse to Kronecker deltas. The four stated $Q_{j}$ are an orthonormal basis, as multiplication of their entries verifies. Thus $F_{e} (E) = \frac{1}{4} \sum_{j} ⟨ Q_{j}, E (Q_{j}) ⟩_{H S} = \frac{3}{8} \sum_{j} ∥ C (Q_{j}) ∥_{H S}^{2}$ .

3.1givenF3F5step 2.1algebra∎

The first specified output has $∥ C (Q_{0}) ∥_{H S}^{2} = \frac{1}{2} Tr ((2 I_{3} / 3)^{2}) = 2 / 3$ . The second has $∥ C (Q_{1}) ∥_{H S}^{2} = \frac{1}{2} (1 / 9 + 4 / 9 + 1 / 9) = 1 / 3$ . The remaining two squared norms are nonnegative, so their sum with these first two is at least one. Substitution in step 2.1 gives $F_{e} (T C) \geq 3 / 8$ .

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