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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?

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  • $\begingroup$ Does this help en.wikipedia.org/wiki/… ? $\endgroup$
    – Condo
    Dec 14, 2021 at 18:48
  • $\begingroup$ 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 $\theta_\phi$? $\endgroup$
    – DaftWullie
    Dec 15, 2021 at 8:13
  • $\begingroup$ I am asking the first case. $\endgroup$
    – peachnuts
    Dec 15, 2021 at 9:03

1 Answer 1

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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*}

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  • $\begingroup$ 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? $\endgroup$
    – peachnuts
    Jan 4 at 15:40
  • $\begingroup$ @peachnuts Yes, thank you. corrected. $\endgroup$
    – DaftWullie
    Jan 5 at 8:02

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