I am watching a recorded seminar on Youtube given by Prof.Daniel Gottesman (here).

At 38:39 he claims that $M_{1}=Z_1\otimes Z_2$,$M_{2}=Z_2\otimes Z_3$,$M_{3}=Z_4\otimes Z_5$, $M_{4}=Z_5\otimes Z_6$,$M_{5}=Z_7\otimes Z_8$,$M_{6}=Z_8\otimes Z_9$,$M_{7}=X_1\otimes X_2 \otimes X_3 \otimes X_4 \otimes X_5 \otimes X_6$,$M_{8}=X_4\otimes X_5 \otimes X_6 \otimes X_7 \otimes X_8 \otimes X_9$ all have eigenvalue 1. So that they can generate the group, which is the stabilizers of the 9-qubit code.

I think for all $M_j$ for $j=1,...,9$ can have eigenvalues 1 and -1, not only 1.

Do I miss something here?

  • $\begingroup$ In what context do you think they can have value -1 instead of +1? That's like saying you're okay with being given |1> (stabilizer by -Z) after asking for |0> (stabilized by +Z). Sometimes you are okay with that, as long as you're told which, but usually to keep things simple you require the sign to be correct. $\endgroup$ Sep 9, 2021 at 3:46
  • $\begingroup$ @CraigGidney Thank you for your reply. Take M1 as an example, if only 1 qubit between the first and second qubits is bit-flipped, then M1 yields eigenvalue -1. $\endgroup$
    – QubitTy
    Sep 9, 2021 at 4:15
  • $\begingroup$ I don't see the issue. The stabilizers identify a state. You have applied an error and ended up in a different state. The stabilizers have correspondingly changed. Why do you then think the sign doesn't matter? You just showed that it does matter! $\endgroup$ Sep 9, 2021 at 4:29
  • $\begingroup$ My last comment is about the why I think the eigenvalues contains -1 case. And I don't understand why stabilizers would change for different states since they serve for QECC to discover the errors. I think the sign does matter, and that's why I think the argument about why M should be generators of stabilizers for their eigenvalue being only 1 is not right. $\endgroup$
    – QubitTy
    Sep 9, 2021 at 4:44
  • 1
    $\begingroup$ If I ask for the state stabilized by +Z, that state is |0> because Z|0> = |0>. The state |1> is not stabilized by +Z because +Z|1> = -|1> which is not equal to |1>. If I wanted the state |1> I would have asked for the stabilizer -Z instead of +Z. You seem to be thinking of the stabilizer as a prescribed measurement to perform, but the stabilizer also specifies what the result of the measurement must be. A stabilizer without a prescribed result is a mere observable. $\endgroup$ Sep 9, 2021 at 5:01

1 Answer 1


If the operators had only the eigenvalue $1$, then they would be the identity. In the stabilizer formalism you define a state (or space) by a set of operators which all leave the state (or space) invariant. This can be done if there are other eigenvalues as well. In fact you need some other eigenvalues to rule out other states and end up with a unique one which fulfills the requirement.

  • $\begingroup$ Thank you for your reply! It seems that I didn't understand the reason why stabilizer formalism is used in the first place. This is my current understanding: I think the definition of stabilizer here is similar to the stabilizer defined in group theory. It should be confined in a subspace instead of the whole space. And here for a specific M, it works as stabilizer in a subspace where the error doesn't occur (or occur twice). To rule out all possible errors, we must find different stabilizers that have corresponding defined spaces where such error doesn't occur. $\endgroup$
    – QubitTy
    Sep 9, 2021 at 7:10

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