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

Question

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?

Answer 1

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*}