<p>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 $\lvert \psi \rangle$.</p>

CC BY-SA 3.0
2016-05-11T20:28:36.363Z
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!.
Harper
CC BY-SA 3.0
2016-05-11T06:52:52.747Z
2016-05-11T06:52:52.747Z
Joe Bebel
<p>In Nielsen and Chuang's "<a href="https://books.google.ca/books?id=-s4DEy7o-a0C" rel="nofollow">Quantum Computation and Quantum Information</a>", <a href="https://books.google.ca/books?id=-s4DEy7o-a0C&amp;pg=PA463" rel="nofollow">Section 10.5.3</a>, the authors claim the following:</p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>However, it seems to me that there is a simple counterexample: </p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>Am I missing anything or the quoted claim is indeed false?</p>

CC BY-SA 3.0
2016-05-10T13:41:07.550Z
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 ?
Frédéric Grosshans
CC BY-SA 3.0
2016-05-10T13:45:22.680Z
<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 $g_1,…,g_n$.
Frédéric Grosshans
CC BY-SA 3.0
2016-05-11T20:25:52.010Z
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
Harper
<quantum-computing><quantum-information>
Can a measurement commuting with the stabilizer of a state disturb the state?
Harper
# Can a measurement commuting with the stabilizer of a state disturb the state?

(Asked by: username_0 on 2016-05-09)

In Nielsen and Chuang's "[Quantum Computation and Quantum Information](https://books.google.ca/books?id=-s4DEy7o-a0C)", [Section 10.5.3](https://books.google.ca/books?id=-s4DEy7o-a0C&pg=PA463), 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?

**Tags:** quantum-computing, quantum-information

> **Comments:**
> **username_1**: 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 ?  
> **username_1**: <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 $g_1,…,g_n$.  
> **username_0**: 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  

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## 1 answers

### Answer by username_2 on 2016-05-11. Score: 1

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 $\lvert \psi \rangle$.

> **Comments:**
> **username_0**: 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!.  

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