Quantum Computing Asked on December 8, 2020
My question is very related to Intuition about Knill-Laflamme QEC conditions but in the particular case of Stabilizer code.
In Nielsen&Chuang, the theorem 10.8 on page 466 gives error-correction conditions for stabilizer codes. It is stated as the following:
Let S be the stabilizer for a stabilizer code $C(S)$. Suppose ${E_j}$ is a set of operators in $G_n$ such that $forall (j,k), E_j^{dagger} E_k notin N(S)-S$. Then ${E_j}$ is a set of correctable error for the code $C(S)$
Notations:
$G_n$: n-Pauli group
$S$: Stabilizer group associated to $C(S)$. I assume $S=<g_1,…,g_l>$ where the $g_i$ are independant generators.
$N(S)$: normalizer, which is actually equal to the centralizer of $S$, $Z(S)$ (the set of all element in $G_n$ which commute with $S$)
I want to have an intuitive feeling about what it means. For me the conditions here are much more restrictive than what it is enough to have. Which is not necesseraly absurd as it is only a sufficient condition that is provided but I want to check.
Let’s assume that a potential error $E_j$ occured. What happens is that we will measure all the generators. Different scenario can occur:
$I$: $E_j$ commute with all generators.
$II$: $E_j$ anticommute with at least one of them.
In $I$: two subcases:
$I.a$: $forall |psirangle in C(S)$ $E_j |psi rangle = |psi rangle$: no error actually occured.
$I.b$: $exists|psirangle in C(S)$ $E_j |psi rangle neq |psi rangle$: an error occured.
$I.a$ is not a problem, $I.b$ would lead to an uncorrectable error.
Now, for $II$:
Let’s assume that for us, $forall i in A, E_i$ and $E_j$ would give the same measurement for the generators (but those error are different). This is how the ensemble $A$ is defined here. What is important is that if I apply the "wrong" correction: $E_i^{dagger}$, I will fix the state.
Thus the condition should be:
$forall i in A, E_i^{dagger} E_j$ should commute with all the generators (such that after error+correction we are in $C(S)$ and verify: $forall |psi rangle in C(S): E_i^{dagger} E_j |psirangle = |psi rangle$
So the questions:
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