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Knill–Laflamme Recovery and Subsystem Codes — Examples

1 · Prerequisites

2 · Summary

The worked example writes the code and two error ranges as explicit isometries, computes the diagonal Knill–Laflamme matrix, and verifies that the two inverse-decoding branches form a trace-preserving recovery map.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Orthogonal error spaces and a recovery branch

Example

Let $H_{L} = C^{2}$ , $H_{P} = C^{2} \otimes C^{2}$ , with standard basis columns $e_{0} = (1, 0)^{T}$ and $e_{1} = (0, 1)^{T}$ , define $P_{0} = e_{0} e_{0}^{*} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $P_{1} = e_{1} e_{1}^{*} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ , as well as $V ψ = e_{0} \otimes ψ$ and $W ψ = e_{1} \otimes ψ$ . For $X = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , define $E_{0} = I_{4} / \sqrt{2}$ and $E_{1} = (X \otimes I_{2}) / \sqrt{2}$ . The channel has two orthogonal error restrictions and a diagonal Knill–Laflamme matrix.

Facts & Assumptions

Given: The finite-dimensional encoding and the two displayed error operators in the Example.

[F1]

Quantum code as an isometric encoding and its code projector supplies the isometric encoding and projector convention.

[F2]

Finite-dimensional quantum channel, Finite Kraus representation, A finite Kraus sum is completely positive, and Trace preservation is Kraus normalization supply the finite channel, Kraus, complete-positivity, and normalization rules.

[F3]

An explicit syndrome recovery corrects a Knill–Laflamme error family: orthogonal normalized error restrictions admit a CPTP recovery by projection and inverse decoding.

Verification

1.1givenF2algebra

Both $I_{4}$ and $X \otimes I_{2}$ are unitary, so $E_{0}^{*} E_{0} + E_{1}^{*} E_{1} = I_{4}$ . The map $N (Y) = E_{0} Y E_{0}^{*} + E_{1} Y E_{1}^{*}$ is therefore a channel.

2.1givenF1step 1.1algebra

By the definitions of $V$ and $W$ , $W^{*} V = 0$ , $V^{*} V = W^{*} W = I_{L}$ , and $(V^{*} E_{a}^{*} E_{b} V)_{a, b} = (\begin{array}{cc} I_{L} / 2 & 0 \\ 0 & I_{L} / 2 \end{array}) .$ The normalized error restrictions are $V$ and $W$ , so their ranges are orthogonal and each syndrome has weight $1 / 2$ .

3.1step 2.1algebra

Define recovery Kraus operators $R_{0} = V^{*}$ and $R_{1} = W^{*}$ . Then $R_{0}^{*} R_{0} + R_{1}^{*} R_{1} = V V^{*} + W W^{*} = I_{4}$ , so the recovery is trace-preserving.

4.1step 2.1step 3.1F1F3algebra∎

The only nonzero corrected error products are $R_{0} E_{0} V = I_{L} / \sqrt{2}$ and $R_{1} E_{1} V = I_{L} / \sqrt{2}$ . Hence for every $Y \in B (H_{L})$ the corrected channel returns $\frac{1}{2} Y + \frac{1}{2} Y = Y$ . This is the orthogonal-syndrome branch of [F3].