# How to decompose Bloch sphere rotations $e^{\frac{i\theta}{2}(\cos(\phi)\sigma_x + \sin(\phi)\sigma_y)}$ in terms of $R_x,R_y,R_z$?

I learned a formula to represent the rotation around bloch sphere:

$$\theta_{\phi} = e^{\frac{i\theta}{2}(\cos(\phi)\sigma_x + \sin(\phi)\sigma_y)}$$

So that $$\pi_0$$ is the gate $$X$$ and $$\pi_{\frac{\pi}{2}}$$ is the gate $$Y$$.

My question is: how do we get this formula to represent the rotation gate? If we have a gate, say $$\pi_{\frac{\pi}{6}}$$, how do we use the rotation gates like $$R_x$$, $$R_y$$, or $$R_Z$$ to represent the gate?

• Does this help en.wikipedia.org/wiki/… ? Dec 14, 2021 at 18:48
• To clarify: are you asking how of to decompose $\theta_\phi$ in terms of $R_x$, $R_y$ and $R_z$, or are you asking how to write (say) $R_x$ in terms of $\theta_\phi$? Dec 15, 2021 at 8:13
• I am asking the first case. Dec 15, 2021 at 9:03

If we look at $$\cos(\phi)\sigma_x+\sin(\phi)\sigma_y$$ we can write this as $$\cos(\phi)\sigma_x+\sin(\phi)\sigma_y=R_z(2\phi)\sigma_x=R_z(\phi)\sigma_xR_z(-\phi).$$ (I'm taking the convention that $$R_z(\phi)=e^{-i\phi/2}$$). The above equation is easier to verify from right to left, using the anticommutation properties of $$\sigma_x$$ and $$\sigma_z$$, meaning that $$\sigma_xR_z(-\phi)=R_z(\phi)\sigma_x$$, followed by the fact that $$R_z(\phi)R_z(\phi)=R_z(2\phi)$$.
Now, you can write that $$\theta_\phi=e^{-i\theta R_z(\phi)\sigma_xR_z(-\phi)/2}$$ Remember that $$Ue^{-iHt}U^\dagger=e^{-iUHU^\dagger t}$$. Hence, \begin{align*} \theta_\phi&=R_z(\phi)e^{-i\theta\sigma_x/2}R_z(-\phi) \\ &=R_z(\phi)R_x(-\theta)R_z(-\phi) \end{align*}
• In the original equation: $\theta_{\phi} = e^{\frac{i\theta}{2}(\cos(\phi)\sigma_x + \sin(\phi)\sigma_y)}$, there is no -. The final result should be: \begin{align*} \theta_\phi&=R_z(\phi)e^{i\theta\sigma_x/2}R_z(-\phi) \\ &=R_z(\phi)R_x(-\theta)R_z(-\phi) \end{align*} where I add a minus sign to the middle term, right? Jan 4 at 15:40