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In Dawson's and Nielsen's pedagogical review of the Solovay-Kitaev algorithm, they describe the decomposition of U into $U=VWV^\dagger W^\dagger$, with both $V, W$ being unitary, being rotated by $\phi$ around the x-axis and y-axis, and obeying the following relationship, where $\theta$ is the angle of $U$ about the Bloch sphere:

$$ \sin(\theta / 2) = 2 \sin^2(\phi/2)\sqrt{1 - \sin^4(\phi/2)}$$

Where does this relation come from?

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This simply comes from equation an arbitrary rotation $R_{n}(\theta)$ with the rotation $$ U=R_X(\phi)R_Y(\phi)R_X(-\phi)R_Y(-\phi). $$

The way that I did this calculation, just to verify this claim, was to recognise that, for example $$ R_X(\phi)=\cos\frac{\phi}{2}I+i\sin\frac{\phi}{2}X. $$ Now, if I evaluate $\text{Tr}(U)$, this picks out the identity terms in the product (because I certainly don't want to multiply the whole thing out!), and those terms occur whenever I have a pair of $X$ or a pair of $Y$ (or both or neither). Hence \begin{align} \cos\frac{\theta}{2}&=\cos^4\frac{\phi}{2}+2\cos^2\frac{\phi}{2}\sin^2\frac{\phi}{2}-\sin^4\frac{\phi}{2} \\ &=(\cos^2\frac{\phi}{2}+\sin^2\frac{\phi}{2})^2-2\sin^4\frac{\phi}{2} \\ &=1-2\sin^4\frac{\phi}{2}. \end{align} Now if you use this to express $\sin\frac{\theta}{2}$, you'll find the claimed result.

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  • $\begingroup$ Awesome, thank you, this explains it. $\endgroup$
    – rhundt
    Mar 15, 2021 at 17:15
  • $\begingroup$ Just wanted to clarify, for the last step, one has to factor out $(\cos^2\theta/2 + \sin^2\theta/2)$ $\endgroup$
    – rhundt
    Mar 17, 2021 at 23:15

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