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In the check matrix representation of stabilizers, one does not care about the global phase. Now why is that? As far as I understand if I have a quantum computation, it can be computationally more efficient to keep track of how the stabilizers transform to find the unique quantum state they stabilize in the end (result of our quantum computation), rather than doing the calculation and keeping track of how the state transforms in each step. Now when neglceting the global phase would that still be possible? For example, suppose we have $X_1 X_2$ and $Z_1 Z_2$, they stabilize the unique state $\phi^+$. On the other hand, if we have $-X_1 X_2$ and $Z_1 Z_2$, they stabilize $\phi^-$ (I don't know whether this is the unique state). So as we can see, the global phase of our operators does have an impact on which state is stabilized. Or maybe I haven't quite understood the usage of the stabilizer formalism yet.

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The sign on the stabilizers is not what people mean by "global phase". For example, the state stabilized by $+Z$ is $|0\rangle$ and the state stabilized by $-Z$ is $|1\rangle$. These states differ by more than just global phase.

The global phase issue is that $+Z$ stabilizes $|0\rangle$ and also $-|0\rangle$ and also $i|0\rangle$ and more generally any state of the form $e^{i \theta} |0\rangle$. Given only the stabilizers, you can't tell what the value of $\theta$ is supposed to be. In most contexts this isn't an issue, but it's a problem if (for example) you are approximating a non-Clifford state as a linear combination of stabilizer states. The different terms of that linear combination need to have the correct phases in order for the sum to come out right.

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