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Context of my question

I call: $\mathcal{M}(\rho)=\sum_a M_a \rho M_a^{\dagger}$ an error map, $C$ the code space.

A CPTP recovery operation exists if and only if, the Kraus operator of the error map verify the Knill-Laflamme condition:

$$\forall (i,j) \in C : \langle \overline{i} | M_{\delta}^{\dagger} M_{\mu} | \overline{j} \rangle = C_{\delta \mu} \delta_{ij}$$

With $C_{\delta \mu}$ an Hermitian matrix and $\langle \overline{i} | \overline{j} \rangle=\delta_{ij}$ (the family $| \overline{i} \rangle$ forms an orthonormal basis of $C$).

From this condition we can distinguish two cases:

  • Non degenerated quantum code: $C_{\delta \mu}=\delta_{\delta \mu}$
  • Degenerated quantum code otherwise

Additional remarks

There always exist a set of Kraus operator such that $C_{\delta \mu} = C_{\delta} \delta_{\delta \mu}$: using the freedom to choose the Kraus we can show that diagonalizing $C$ on the rhs, is equivalent to take another set of equivalent Kraus operator on the lhs. I can edit with the derivation if necessary but it is really just a matter of writing the thing. Thus for any map $\mathcal{M}$, we can in principle have a set of Kraus satisfying:

$$\forall (i,j) \in C : \langle \overline{i} | M_{\delta}^{\dagger} M_{\mu} | \overline{j} \rangle = C_{\mu} \delta_{\delta \mu} \delta_{ij}$$

The distinction between degenerated and non degenerated can then be rephrased as:

  • Non degenerated quantum code: $C_{\mu}=1$
  • Degenerated quantum code otherwise

My question

Because of this last remark, it shows that there always exist a set of Kraus operator such that an error map satisfying Knill-Laflamme condition would bring two different codeword to two orthogonal subspace (we can generalize by linearity to any two vector living in $C$).

Thus for me what remains is that the "conceptual" difference between degenerated and non degenerated code is that:

Whatever the Kraus we use to represent the map, each Kraus of a non degenerated code will always put two different error into two orthogonal subspace.

For degenerated code it is not always the case but there exists a set of Kraus such that it is the case. And for those the difference is that the length of the vector will be modified in a different manner for each different kind of Kraus.

Would you agree ?

Can we give a physical meaning to this change in length when we work with "the good Kraus" for degenerated code ?

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