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If we suppose that $XX$ is the tensor product of $X$ with $X$ such as $XX = X \otimes X$

How would we calculate the rotation operator of this $XX$ gate.

Does this work? If so why? $$ R(XX)_\theta = e^{-i\frac{\theta}{2}XX} = \cos\left(\frac{\theta}{2}\right)I - i\sin\left(\frac{\theta}{2}\right)XX $$

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This does indeed work because this construction works for any matrix $H$ where $H^2=\mathbb{I}$. $$ R_\theta(H)=e^{-i\theta H/2}=\cos\frac{\theta}{2}\mathbb{I}-i\sin\frac{\theta}{2}H $$

There are several ways that you could prove this. I think the easiest way is to realise that because $H^2=\mathbb{I}$, then the eigenvalues of $H^2$ are 1, and hence the eigenvalues of $H$ must be $\pm 1$. Let $P_{\pm}$ be projectors onto the eigenspaces of eigenvalue $\pm 1$ respectively. So, $$ \mathbb{I}=P_++P_-\qquad H=P_+-P_- $$ Furthermore, by definition of how a function can be applied to a matrix (you apply the function to each of the eigenvalues), $$ R_\theta(H)=e^{-i\theta/2}P_++e^{i\theta/2}P_-. $$ Now we can substitute for $P_{\pm}$: $$ R_\theta(H)=e^{-i\theta/2}(\mathbb{I}+H)/2+e^{i\theta/2}(\mathbb{I}-H)/2 $$ Group the terms of $\mathbb{I}$ and for $H$ and simplify.

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