An X gate where there is a phase shift $\phi$ to the applied sinusoidal wave $U = e^{-i\frac{\theta}{2}(cos(\phi)\sigma_x+sin(\phi)\sigma_y)}$ is equivalent to a series of gates $Z_{-\phi}X_{\theta}Z_{\phi} = e^{i\phi\sigma_z/2}e^{-i\theta\sigma_x/2}e^{-i\phi\sigma_z/2}$. What is the math used to get from $U$ to the series of gates?
1 Answer
One of the key steps is a relation that says $$ Ue^{iH}U^\dagger=e^{iUHU^\dagger} $$ for any unitary $U$. So, if I consider $$ Z_{-\phi}X_{\theta}Z_{\phi} $$ then really all I need to work out to see the equivalence is what goes on in the exponent: $$ Z_{-\phi}XZ_{\phi}=XZ_{2\phi}=X(I\cos\phi-i\sin\phi Z)=X\cos\phi-Y\sin\phi $$ (I seem to have a different sign to what you were expecting. I could easily have messed up e.g. whichever convention you're taking for $X_{\theta}$.)
To prove my main claim, you could think about the Taylor expansion of the exponential: $$ e^{-iUHU^\dagger}=\sum_{n=0}^{\infty}\frac{1}{n!}(-iUHU^\dagger)^n. $$ When you write out $$ (UHU^\dagger)^n=UHU^\dagger UHU^\dagger\ldots $$ then all the $U^\dagger U$ pairs are identity, and you just get $UH^nU^\dagger$.