I am trying to find out how the $X\otimes X^{1/2}$ operator propagates through a $CZ$ one.

Below a circuit example.


1 Answer 1


The X gate is trivial to get through. The hard part is the $\sqrt{X}$.

A useful trick for propagating complicated gates through a controlled gate is to break it down into cases by controlling the complex gate. Split it into "if ON then gate" and "if OFF then gate" then individually get the controlled operations through, then merge them back together on the other side.

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The reason this is handy is because oppositely controlled operations commute and same-controlled operations can be dealt with as if they weren't controlled at all. The main caveat is you have to turn any changes in the global phase of the uncontrolled case into changes in the phase kickback once you reintroduce the control.

As you can see, moving a Clifford operation in an anticommuting basis across a controlled operation has a tendency to create more controlled operations.

Another way to see what's going on here is to realize moving the $\sqrt{X}$ across the controlled $Z$ turns the $Z$ into a $Y$. Then, because $Y = iXZ$, you can split the $CY$ into $S \cdot CX \cdot CZ$ to get the $CZ$ back.

  • $\begingroup$ Wow that is indeed a really handy trick $\endgroup$
    – chrysaor4
    Oct 24, 2022 at 16:30

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