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Resolved rank-one transition obstructs exact correction

PDF p. 1: Paragraph following Eq. (1)

Statement

Let $H_{L}$ , $H_{P}$ , and $H_{O}$ be finite-dimensional complex inner-product spaces, let $V : H_{L} \to H_{P}$ be an isometry, and let $P = V V^{†}$ project onto the codespace $C = V (H_{L})$ . Suppose $\dim H_{L} \geq 2$ . Let a quantum channel $N : L (H_{P}) \to L (H_{O})$ have a finite Kraus representation containing the nonzero operator $E = ∣ f ⟩ ⟨ s ∣$ , where $s \in H_{P}$ and $f \in H_{O}$ . If $P s \neq 0$ , the code is not exactly correctable for $N$ . In particular, this applies when source and final states are orthogonal unit vectors in the same physical space.

Here a Kraus representation is an operator sum $N (X) = \sum_{a} E_{a} X E_{a}^{†}$ , and exact correction means that a quantum channel $R : L (H_{O}) \to L (H_{L})$ satisfies $R (N (V X V^{†})) = X$ for every $X \in L (H_{L})$ . The condition $P s \neq 0$ expresses nonzero support of the codespace on the source state.

The dimension restriction matters: an orthogonal rank-one transition can occur in a channel that is exactly correctable on a one-dimensional code with $P s \neq 0$ .

Facts & Assumptions

[F1]

On arbitrary finite-dimensional complex inner-product spaces, the bra-ket conventions give $∣ f ⟩ ⟨ s ∣ x = ⟨ x, s ⟩ f$ , with the root inner product linear in its first argument. Dirac Notation Conversion For the coordinate spaces used in the boundary example, these agree with Finite operator and ket conventions.

[F2]

The code projector, channel and exact decoded recovery have the meanings stated in Code restriction and decoded channel types; finite Kraus representations are operator sums as in Finite Kraus representation.

[F3]

If a finite-dimensional code of dimension at least two is exactly correctable for a channel whose finite Kraus family contains a nonzero operator $x \mapsto ⟨ x, s ⟩ f$ , then its code projector satisfies $P s = 0$ . Dimensional boundary of a rank-one KL obstruction

[F4]

For finite-dimensional complex inner-product spaces $H, K$ , a complex-linear map $Φ : L (H) \to L (K)$ with a finite Kraus family $(K_{a})$ satisfying $\sum_{a} K_{a}^{†} K_{a} = I_{H}$ is a quantum channel. Finite channel, Kraus, Choi, and Stinespring equivalence

Proof

Given: The spaces, isometry, channel, Kraus operator and dimension hypothesis in the statement, with $P s \neq 0$ .

Goal: Prove that exact correction is impossible and exhibit the one-dimensional boundary.

1.1givenF1F2F3

By [F1], the designated Kraus operator has the form required by [F3]. If exact correction existed, [F3] would give $P s = 0$ , contrary to the given support condition. Therefore no exact decoder exists. Orthogonality of $s$ and $f$ is not needed for this implication.

1.2F1F2F4constructalgebra

For the boundary, take $H_{P} = H_{O} = C^{2}$ with orthonormal basis $s, f$ , take $H_{L} = C$ and $V (1) = s$ . Set $E = ∣ f ⟩ ⟨ s ∣$ , $F = ∣ f ⟩ ⟨ f ∣$ and $N (X) = E X E^{†} + F X F^{†}$ . This operator sum is complex-linear and $E^{†} E + F^{†} F = ∣ s ⟩ ⟨ s ∣ + ∣ f ⟩ ⟨ f ∣ = I_{C^{2}}$ . Hence [F4], with $H = K = C^{2}$ , shows that $N$ is a quantum channel. Direct evaluation gives $N (X) = Tr (X) ∣ f ⟩ ⟨ f ∣$ for every $X \in L (C^{2})$ .

2.1F1F2F4step 1.2constructalgebra∎

Define $R_{s} = ⟨ s ∣$ and $R_{f} = ⟨ f ∣$ as maps from $C^{2}$ to $C$ , and set $R (Y) = R_{s} Y R_{s}^{†} + R_{f} Y R_{f}^{†} = Tr (Y)$ . This map is complex-linear and $R_{s}^{†} R_{s} + R_{f}^{†} R_{f} = ∣ s ⟩ ⟨ s ∣ + ∣ f ⟩ ⟨ f ∣ = I_{C^{2}}$ . By [F4], now with $H = C^{2}$ and $K = C$ , $R$ is a quantum channel. For every $X \in L (C)$ , the formula in step 1.2 gives $N (V X V^{†}) = X ∣ f ⟩ ⟨ f ∣$ , identifying endomorphisms of $C$ with scalars, and therefore $R (N (V X V^{†})) = X$ . Yet $P s = s \neq 0$ and $s ⊥ f$ . Thus the asserted one-dimensional example exists.

Remarks

The article states the obstruction for any codespace with support on the source state and cites its Ref. 9. The proposition makes the dimension condition explicit; the example shows why the unqualified wording cannot be extended to arbitrary one-dimensional codes under the stated channel interpretation. The article discusses encoded qubits, for which the code dimension is two.

Depends on

Dependency tree · 2 levels

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 1, paragraph following Eq. (1)