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
    Commented 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
    Commented 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. $\endgroup$ Commented Dec 19, 2018 at 16:01
  • $\begingroup$ I think the correct expression for $\sqrt{Z}$ should be $-i \left[\begin{matrix} e^{i\pi/4} & 0\\ 0& e^{-i\pi/4}\end{matrix}\right] = -i e^{i \pi Z/4}$ since its square is $-i e^{i\pi Z/2} = -i(\cos(pi/2) + i\sin(\pi/2) Z) = Z$. $\endgroup$ Commented Dec 14, 2023 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.