Let's say you have a system with which you can perform arbitrary rotations around the X and Z axis. How would you then be able to use these rotations to obtain an arbitrary rotation around the Y axis?

I have seen somewhere that rotation around an arbitrary axis can be achieved by doing three rotations around two fixed axis, that is, $$\hat{R}_\vec{n}(\theta)=R_Z(\gamma)R_X(\beta)R_Z(\alpha)$$ for some angles $\gamma, \alpha, \beta$. But how do you actually use this? What if I want to rotate around the Y axis with an angle of $\theta$ i.e. $\hat{R}_Y(\theta)$? Then how do I figure out what $\gamma,\alpha,\beta$ to use?

Edit: I've found a nice answer on Physics SE.


2 Answers 2


Single-qubit unitaries are just 3D rotations, multiplied by a phase. So in order to find the actual angles, you can resort to the theory of rotation matrices, in particular to Euler's rotation theorem, which states that any rotation is a composition of 3 rotations (the theorem proof is constructive, so you get the actual angles).


Try selecting $\gamma$ and $\alpha$ so that you get the rotation $$ \sqrt{Z}R_X(\beta)\sqrt{Z}^\dagger. $$ There's two little tricks here that make this work. Firstly, the $R_X(\beta)$ has an $\mathbb{I}$ component and an $X$ component (I always get factors of $1/2$ wrong here, so I won't write out the cos and sin functions explicitly unless you define your $R_x$ function). Now, $$ \sqrt{Z}\mathbb{I}\sqrt{Z}^\dagger=\mathbb{I} $$ while there's some funky anti-commutation that goes on with $X$: $$ \sqrt{Z}X\sqrt{Z}^\dagger=ZX=iY, $$ so you've managed to effectively change the $X$ into a $Y$, so you're getting $Y$ rotations.

Of course, this doesn't answer your more general question about how you get a more general rotation....

  • $\begingroup$ Could you explain what $\sqrt{Z}$ means? If $Z=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ then is $\sqrt{Z} = \begin{bmatrix}1&0\\0&i\end{bmatrix}$? $\endgroup$
    – PhysicsMan
    Nov 20, 2018 at 12:18
  • $\begingroup$ $\sqrt{Z}=\left(\begin{array}{cc} e^{i\pi/2} & 0 \\ 0 & e^{-i\pi/2} \end{array}\right)$. $\endgroup$
    – DaftWullie
    Nov 20, 2018 at 13:25
  • $\begingroup$ Well actually, $\sqrt Z$ is indeed $\mathrm{diag} (1,i)$, but it is proportional to $\mathrm{diag} (\exp(-i\pi/2), \exp(i\pi/2)) = \exp(-i \pi Z/2) $, which more closely corresponds to how physicists might realise it or analyst processes involving it. $\endgroup$ Dec 19, 2018 at 16:01

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