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A degenerate code with a non-diagonal Knill–Laflamme matrix

Example

Take HL=C2, HP=C2⊗C2, with standard basis columns e0=(1,0)T and e1=(0,1)T, and define Vψ=e0⊗ψ and Wψ=e1⊗ψ. Put P0=e0e0∗=(1000) and P1=e1e1∗=(0001). On the first factor use X=(0110) and Z=(100−1). Define a channel with Kraus operators

E1=I43,E2=X⊗I23,E3=Z⊗I23.

The code stores the logical qubit in the second factor. The errors E1 and E3 have the same restriction to the code, while E2 moves the code to the orthogonal first-factor subspace.

Facts & Assumptions

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

[F1]

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

[F2]

Finite-dimensional quantum channel, Finite Kraus representation, and Trace preservation is Kraus normalization specify finite Kraus channels and their trace-preserving normalization.

[F3]

An explicit syndrome recovery corrects a Knill–Laflamme error family: the displayed recovery construction corrects every finite family satisfying the scalar-compression condition.

[F4]

A finite Kraus sum is completely positive: a finite Kraus sum is completely positive.

[F5]

Trace preservation is Kraus normalization: the adjoint-square sum of a trace-preserving Kraus family is the identity.

Verification

1.1givenF2F4F5algebra

Each of I4, X⊗I2, and Z⊗I2 is unitary. Hence ∑a=13Ea∗Ea=13(I4+I4+I4)=I4, so the displayed Kraus map is trace-preserving; it is completely positive because it is a finite Kraus sum.

2.1givenF1step 1.1algebra

With P=VV∗, direct compression gives V∗Ea∗EbV=CabIL,C=13(101010101). Since P=VV∗, this is equivalently PEa∗EbP=CabP for every a,b. In particular, C is non-diagonal and has rank 2.

3.1step 2.1algebra

The orthonormal eigenvectors u+=(1,0,1)/2, u2=(0,1,0), and u0=(1,0,−1)/2 have eigenvalues 2/3,1/3,0, respectively. The corresponding mixed Kraus operators are F+=(I4+Z⊗I2)/6=2/3(P0⊗I2), F2=E2, and F0=(I4−Z⊗I2)/6=2/3(P1⊗I2). The columns u+,u2,u0 form a unitary matrix, so expanding the Kraus sum gives ∑iFiYFi∗=∑aEaYEa∗ for every operator Y.

4.1step 3.1algebra

By the definition of W, one has F+V=2/3V, F2V=1/3W, and F0V=0. The restrictions V,W are isometries with orthogonal ranges, so the positive-eigenvalue error basis has exactly two syndromes; the zero eigenvalue error annihilates the code.

5.1step 4.1algebra

Define recovery Kraus operators R+=V∗ and R2=W∗. Their adjoint products are VV∗=P0⊗I2 and WW∗=P1⊗I2, so R+∗R++R2∗R2=I4 and the recovery is trace-preserving.

6.1step 2.1step 3.1step 5.1F1F3F5algebra∎

For the mixed errors, R+F+V=2/3IL and R2F2V=1/3IL; all other products RiFjV vanish. Thus for every X∈B(HL) the corrected channel returns (2/3+1/3)X=X. The unitary Kraus change preserves the original channel, so this recovery corrects the displayed three-error channel, including its degeneracy; this is the recovery construction in [F3].

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