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Since the Pauli Z gate equate to a rotation around z axes of the Bloch sphere by $\pi$ radians, the phase of anything that lies on z axes is expected to change by $\pi$ by applying z-gate. As $|0⟩$ and $|1⟩$ lies on z-axis so their phase should change by $\pi$. So, why doesn't z-gate change the phase of $|0⟩$?

Mathematically it can be understood but as per definition of rotation representation of z-gate matrix doesn't seem to be correct as it only changes the phase of $|1⟩$ and not $|0⟩$. Why is it like this?

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    $\begingroup$ Well, it creates a phase difference between the two let’s $|0\rangle$ and $|1\rangle$. If you apply a $Z$ gate to a computational basis state, you don’t do anything, by virtue of the unobservable change in global phase. $\endgroup$ Jul 24, 2022 at 2:48
  • $\begingroup$ Well, $$Z(\alpha|0\rangle+\beta|1\rangle)=\alpha|0\rangle-\beta|1\rangle$$ So $$Z(|0\rangle)=|0\rangle$$, $\endgroup$
    – kludg
    Jul 24, 2022 at 11:58

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You can consider $Z$ gate to be a "negation" in Hadamard basis. If you apply the gate on state $|+\rangle$, you get $|-\rangle$ and vice versa.

In computational basis, state $|0\rangle$ remains unchanged and $|1\rangle$ is changed to $-|1\rangle$. However, you can ignore the minus because it is a global phase. So, there is no action of $Z$ in computational basis. To changes the computational basis states between each other, you can use $X$ gate.

Now you see that both gates rotates the basis states by 180 degrees, or in other words, they negate the states. However, each of the gate does so in different basis.

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A rotation by $\pi\over2$ around the Z-axis is done by applying $iZ$. However, this is equivalent to $Z$ up to unobservable global phase.

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