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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.

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    $\begingroup$ The Shor code is a stabilizer code; in particular, its stabilizer group is abelian ... btw if you would take a non-abelian subgroup, the stabilized subspace would be $\{0\}$. $\endgroup$ Jan 27 at 14:01
  • $\begingroup$ My question is about the commuting property. The shor code has anti-commuting elements $\endgroup$ Jan 27 at 14:18
  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/4541/55 $\endgroup$
    – glS
    Jan 27 at 21:41
  • $\begingroup$ There are "gauge codes" such as the Bacon-Shor code where the stabilizers that are being measured do not all commute, but the stabilizer group describing the state of the system at any one time is always generated by a set of commuting stabilizers (assuming all logical qubits are in stabilizers states, of course). $\endgroup$ Jan 27 at 22:50

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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.

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