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I am learning the stabilizer code through Nielsen & Chuang which give the following hypothesis to have correctable error for stabilizer codes:

Let S be the stabilizer for a stabilizer code C(S). Suppose ${E_j}$ is a set of operators in $G_n$ such that $\forall (j,k) E^{\dagger}_j E_k \notin N(S) - S$. Then, $\{E_i\}$ is a correctable set of errors for $C(S)$.

Where $N(S)$ is the normalizer of $S$ which is also equal to the centralizer $Z(S)$.

I agree with this statement after having gone through the proof but the initial condition is a little abstract for me. I would like to have intuition of what it means.

This is why I looked at https://en.wikipedia.org/wiki/Stabilizer_code#Stabilizer_error-correction_conditions which seems to give a nice way to interpret it.

In summary, they say that the condition as given in Nielsen & Chuang, are equivalent to say:

An error $E \in \{E_j\} \subset G_n$ (the n-Pauli group) is :

  • correctable if it anti-commutes with at least one of the generators
  • correctable and commutes with all the generator $\Leftrightarrow$ $E \in S$

Which means that an error is correctable if it is NOT in $Z(S)-S$.

However I do not understand why the two proposition are equivalent.

In conclusion, my questions:

What is the physical intuition behind the abstract condition $\forall (j,k), E^{\dagger}_j E_k \notin N(S) - S$ ?

If it is the one given by the wikipedia page, then why is the wikipedia definition equivalent to the one in N&Chuang ?

Marco Fellous-Asiani
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  • "correctable if it commutes with all the generator ⇔ ∈" -- This (your $\Leftrightarrow$) is incorrect. What you want to say is "It is an element of the stabilizer". The issue are exactly errors which commute with all stabilizers but which are not contained in it. This is exactly N(S)-S. – Norbert Schuch Aug 31 '19 at 15:53
  • @NorbertSchuch thank you for your answer. You are right, I edited. However as I understand that the $E$ must not be in $Z(S) - S$ to have something correctable, I am not sure it is equivalent to $E_j^{\dagger} E_k \notin Z(S) - S$ which is the statement of the theorem. – Marco Fellous-Asiani Aug 31 '19 at 16:15
  • @NorbertSchuch Indeed if $E_j$ and $E_k$ are both NOT in $Z(S)-S$ such that they do not commute with exactly the same generators, $E_j^{\dagger} E_k$ will be in $Z(S)-S$. So to say that errors must not be in $Z(S)-S$ is not equivalent for me to say that $\forall (j,k), E_j^{\dagger}E_k \notin Z(S)-S$ – Marco Fellous-Asiani Aug 31 '19 at 16:20
  • "correctable and commutes with all the generator ⇔ ∈" -- Do you mean "correctable if ..."? But then it is not correct either, $E\in S$ is stronger than "commutes with all generators". – Norbert Schuch Aug 31 '19 at 16:30
  • The error $E$ verifies the property "it is correctable and it commutes with all the generators" iff $E \in S$. I agree that $E \in S$ is stronger than commutes with all the generators but it is not stronger than commute with all the generators and is correctable. – Marco Fellous-Asiani Aug 31 '19 at 16:35
  • @NorbertSchuch But this specific thing is more a detail about my general problem that is: if the condition is to say that to be correctable an error must either be in $S$ either be outside of $Z(S)-S$ which I can understand intuitively, I don't get why it is equivalent to the $E_j^{\dagger} E_k \notin Z(S)-S$ (as we have a product of errors in this statement) – Marco Fellous-Asiani Aug 31 '19 at 16:37
  • The latter condition has (roughly) the interpretation: "The difference between two errors should either anticommute with the stabilizer (in which case they can be distinguished), or be part of the stabilizer (in which case it would not matter which error you correct)." -- – Norbert Schuch Aug 31 '19 at 17:18
  • @NorbertSchuch For me there is also a last possible case that: $E_j^{\dagger} E_i \in S$ (so they commute with the stabilizer) but $E_i$ and $E_j$ are both not in the stabilizer. Taking $E_i$ and $E_j$ that anticommute with the same element of the stabilizer and commute with all the others would be an example. And in this case we couldn't know in principle if the error was $E_i$ or $E_j$ – Marco Fellous-Asiani Aug 31 '19 at 18:49
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    That is correct, but in that case, it does not matter whether you correct for E_i or E_j, since the difference in the correction is an element of the stabilizer, so it does not change your encoded state! – Norbert Schuch Aug 31 '19 at 18:57
  • @NorbertSchuch ooh great thanks I think I understood. Excepted if you planned to do so I will write an answer summarizing all this maybe – Marco Fellous-Asiani Aug 31 '19 at 19:06
  • Please go ahead! – Norbert Schuch Aug 31 '19 at 19:17

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