# Can a measurement commuting with the stabilizer of a state disturb the state?

In Nielsen and Chuang's "Quantum Computation and Quantum Information", Section 10.5.3, the authors claim the following:

With a system in state $|\psi\rangle$ with stabilizer $g_1,...,g_n$, if a measurement $g$ that is commute with all $g_i$ is applied to the state, then either $g$ or $-g$ is a stabilizer of the system and the measurement doesn't disturb the state.

However, it seems to me that there is a simple counterexample:

Let $g$ be the $Z_1$ operator in the system, thus it commutes with all stabilizers as $g|\psi\rangle$ is still a valid state in the system. However neither $g$ nor $-g$ is a stabilizer of the code.

Am I missing anything or the quoted claim is indeed false?

• What is $Z₁$ ? A logical $Z$ operator on some error-corrected logical qubit ? Or $Z$ applied to the first physical qubit, with a specific set of stabilizers ? May 10 '16 at 13:41
• <rereads Nielsen&Chuang> Note that the first possibility is excluded by the fact that this applies when we have $n$ stabilizers, i.e. the state is fully defined by $g_1,…,g_n$. May 10 '16 at 13:45
• Thank you @FrédéricGrosshans, you are correct and I missed this. The stabilizers mentioned here are not the code's stabilizers, but the state's stabilizers May 11 '16 at 20:25

The quoted claim from the book is true. It is not true that $Z_1$ must always commute with all stabilizers of $\lvert \psi \rangle$. For example, if $X_1X_2$ stabilizes $\lvert \psi \rangle$ (e.g., if $\lvert \psi \rangle$ was an EPR pair) then $Z_1$ does not commute with $X_1X_2$ and $Z_1$ is not a stabilizer of $\lvert \psi \rangle$.
• Here $Z_1$ is not the $Z$ gate applied to the first physical qubit, but the first logical qubit. I think @FrédéricGrosshans 's answer above is the correct result, but thank you for the help!. May 11 '16 at 20:28