# Degenerated vs non degenerated code: for both there always exist Kraus bringing to different orthogonal subspaces?

## 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

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 ?