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The title explains it all. Suppose one needs to find z-y decomposition of unitary matrix H or T. What is the step by step process to find it?

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For a 2x2 unitary matrix given by four complex numbers $\begin{pmatrix} a & b\\ c & d \end{pmatrix}$, the aspects of the $z$-$y$ decomposition $e^{i\alpha}R_z(\beta)R_y(\gamma)R_z(\delta)$ can be worked out quickly as follows:

  1. $\gamma$ is related to the magnitude between two elements on the same row. You can interpret it as $2 \arctan(\frac{|b|}{|a|})$ giving a value between 0 to $\pi$, with a value of $\pi$ if $a$ is 0.
  2. $\beta$ and $\delta$ together are related to the complex phases of the four matrix elements. Normally, $\beta$ is the difference in phase between $c$ and $a$ and $\delta$ is the difference in phase between $-b$ and $a$. If $\gamma$ is 0, then instead $\beta + \delta$ is equal to the difference in phase of $d$ and $a$, and if $\gamma = \pi$ then $-\beta + \delta$ is the difference in phase of $-b$ and $c$. In either of those cases, their individual values do not matter. The $\gamma = 0$ case corresponds to entirely $z$ rotations.
  3. $\alpha$ is the global phase after all this, not related to the relations between elements but a scalar multiple of the entire matrix. To get $\alpha$, take the phase of $a$ in the matrix and add $\frac{\beta}{2} + \frac{\delta}{2}$. If $a$ is 0 thanks to $\gamma$ being $\pi$, you can add $-\frac{\beta}{2} + \frac{\delta}{2}$ to $c$'s phase instead.

So a decompositions of $H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$ in terms of $\{\alpha, \beta, \gamma, \delta\}$ is $\{\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\}$: $a$ and $c$ have the same phase and the difference in phase of $-b = -1$ and $a = 1$ is $\pi$, all elements are equal in magnitude so $\gamma$ is $\frac{\pi}{2}$. The global phase $\alpha$ is then just $\frac{\delta}{2}$ since $a$ already has a phase of 0.

$T = \begin{pmatrix} 1 & 0\\ 0 & e^{i \pi/4} \end{pmatrix}$ can be decomposed into any decomposition where $\gamma = 0$, $\alpha = \frac{\pi}{8}$ and $\beta + \delta = \frac{\pi}{4}$. Note that $T$ is a $\frac{\pi}{4}$ rotation around the $Z$ axis, so, with $R_y(0)$ being the identity, the two $R_z$ transformations are just additive, with the $\alpha = \frac{\pi}{8}$ being the global phase used to make the top-left element 1.

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