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 ?


1 Answer 1


First to be clear about the definition, a quantum code is degenerate if there are two linearly independent (i.e. distinct) Kraus operators that transform the code linearly dependently. The two Kraus operators acting linearly dependently on the code basically means they act the same on the code. Otherwise we say the code is nondegenerate, but this does not mean that two distinct errors will always send the code to two orthogonal subspaces.

Next, the given error correction condition $$\forall (i,j) \in C : \langle \overline{i} | M_{\delta}^{\dagger} M_{\mu} | \overline{j} \rangle = C_{\delta \mu} \delta_{ij}$$ is correct but what follows with the cases is not correct. Nondegenerate does not mean that you always have $C_{\delta\mu} = \delta_{\delta\mu}$ and at this point you cannot say whether the code is degenerate. Like you mention next, it is required to diagonalize the matrix $C_{\delta \mu}$ meaning find Kraus operators $N_{\delta}$ such that $$\forall (i,j) \in C : \langle \overline{i} | N_{\delta}^{\dagger} N_{\mu} | \overline{j} \rangle = D_{\delta \mu} \delta_{ij}$$ where $D_{\delta \mu} = 0$ if $\delta \neq \mu$ (but in general it does not hold that $D_{\mu \mu} = 1$). See Theorem 10.1 of Nielsen and Chuang for the proof of this.

I will not prove this, but it turns out that the code is nondegenerate if and only if $D_{\mu \mu} \neq 0$ for all $\mu$ (this is the correct statement as opposed to $C_\mu = 1$). On the other hand this says that a code is degenerate if and only if there is at least one $\mu$ where $D_{\mu \mu} = 0$. Whether or not the code is degenerate, the constant $D_{\mu\mu}$ is the vector length scaling factor for $N_{\mu}$ and if $\mu \neq \delta$ then $N_{\mu}$ and $N_{\delta}$ send the code to orthogonal subspaces. Here the conceptual difference between nondegenerate and degenerate is that if $D_{\mu \mu} = 0$ then $N_{\mu}$ acts as the zero matrix on the code. In this case we can interpret $N_{\mu}$ as essentially being a non-error for our code and so no orthogonal subspace is taken up for this error. We can also relate this back to the first definition. Suppose again $D_{\mu\mu} = 0$ but $D_{\delta\delta} \neq 0$. We can replace $N_{\mu}$ and $N_{\delta}$ with $\frac{1}{\sqrt{2}}(N_{\delta} + N_{\mu})$ and $\frac{1}{\sqrt{2}}(N_{\delta} - N_{\mu})$ which are linearly independent. Observe that for all $i \in C$, $$\frac{1}{\sqrt{2}}(N_{\delta} + N_{\mu})|\overline{i}\rangle = \frac{1}{\sqrt{2}}(N_{\delta} - N_{\mu})|\overline{i}\rangle$$ so these Kraus operators act identically on the code.


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