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I read about the stabilizer code as a set characterized by elements stabilized by an abelian group, i.e one whose elements commute.

I can imagine that requiring commutation helps to define new codes starting from classical linear codes.

However I don't see why I often read about that as a necessary condition, when common codes (e.g. the Shor code) doesn't satisfy it.

Adam Zalcman
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Daniele Cuomo
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1 Answers1

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We do not require stabilizers to commute. We require them to jointly stabilize a non-trivial subspace. As a consequence, they commute.

Suppose $P$ and $Q$ are anti-commuting $n$-qubit Pauli operators which jointly stabilize a subspace $C$ of the $n$-qubit Hilbert space. Let $|\psi\rangle\in C$. Then $P|\psi\rangle=|\psi\rangle$ and $Q|\psi\rangle=|\psi\rangle$, so

$$ PQ|\psi\rangle=|\psi\rangle.\tag1 $$

However, $PQ=-QP$, so

$$ PQ|\psi\rangle=-QP|\psi\rangle=-|\psi\rangle.\tag2 $$

Combining $(1)$ and $(2)$ we have $|\psi\rangle=-|\psi\rangle$ so $|\psi\rangle = 0$. Therefore, $C$ is trivial.

Thus, we see that non-abelian subgroups of the Pauli group stabilize the trivial subspace. Consequently, the stabilizer group of any quantum error correcting code is abelian.

In particular, stabilizer generators of the Shor's 9-qubit code, e.g.

$$ \begin{align} g_1&=ZZIIIIIII\\ g_2&=IZZIIIIII\\ g_3&=IIIZZIIII\\ g_4&=IIIIZZIII\\ g_5&=IIIIIIZZI\\ g_6&=IIIIIIIZZ\\ g_7&=XXXXXXIII\\ g_8&=IIIXXXXXX \end{align}\tag3 $$

clearly commute pairwise and thus generate an abelian group.

Adam Zalcman
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