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As a part of textbook exercise, Y gate is to be constructed using H,Z and X-gates, just like we have $X = HZH$. is there some way/process/intuition to find such combinations or it is just like we need to use hit and trial method to find such combination?

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First, you got a mistake there. The correct decomposition is $$X = HZH.$$ The way you find the decomposition is by understanding that $HZH$ performs the change of basis. The right-most $H$ brings a vector from the $Z$ basis into the $X$ basis. Then $Z$ applies the $NOT$ operation in the $X$ basis. Finally, the left-most $H$ brings a vector into the $Z$ basis again.

Another way of looking at it is realizing that $$X = HZH= H Z H^{\dagger}$$ is an eigendecomposition of $X$ where $H$ is a matrix of eigenvectors of $X$ and $Z$ is a diagonal matrix of eigenvalues of $X$.

Following this logic, we can quickly decompose $Y$ by just knowing its eigenvectors and eigenvalues. So we get $$Y = (SH) Z (SH)^{\dagger}.$$

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  • $\begingroup$ made the edit for the typing mistake...thanks !!! $\endgroup$ Aug 18 at 5:47

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