As I know, the $Rz(\frac{\pi}{2})$ gate is equivalent to the Phase gate $S$ up to the global phase. However, I found using qiskit, the $Rz(\frac{-\pi}{2})$ is also equivalent to the Phase gate $S$. I think it should be a $S^\dagger$ gate. Are they really the same up to a global phase?

I further tried the code and found any angles of $Rz$ gate equivalents to the phase gate $S$, which I think definitely is not true.

Here's the code I used: enter image description here


1 Answer 1


The $RZ(\theta)$ gate is defined by: $$RZ(\theta)=\begin{pmatrix}\mathrm{e}^{-\mathrm{i}\frac\theta2}&0\\0&\mathrm{e}^{\mathrm{i}\frac\theta2}\end{pmatrix}=\mathrm{e}^{-\mathrm{i}\frac\theta2}\begin{pmatrix}1&0\\0&\mathrm{e}^{\mathrm{i}\theta}\end{pmatrix}=\mathrm{e}^{-\mathrm{i}\frac\theta2}P(\theta)$$ with $P$ being a phase gate. In particular, $S=P\left(\frac\theta2\right)$. So, as you mentioned, a $RZ\left(\frac\pi2\right)$ gate is equivalent up to a global phase to an $S$ gate. However, it is not equivalent to an $S^\dagger$ gate up to a global phase.

So it means that the unitaries are not equivalent. However, what you're testing is whether the statevectors are equivalent, and it turns out, they are!

The reason for this is that a quantum circuit starts from the state $|0\rangle$. Thus, when applying your first circuit to it, you end up with $\mathrm{e}^{-\mathrm{i}\frac\pi4}|0\rangle$, which is equivalent to $|0\rangle$ up to a global phase. Since the second one leaves $|0\rangle$ unchanged, the final statevector is also $|0\rangle$, which is why your code (correctly) tells you that these two statevectors are equivalent.

Simply put an h gate before the rz and s gates respectively and you will see that the statevectors are no longer equivalent.


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