# Rotation operator on Pauli parity gates $XX$, $YY$ and $ZZ$

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? $$R(XX)_\theta = e^{-i\frac{\theta}{2}XX} = \cos\left(\frac{\theta}{2}\right)I - i\sin\left(\frac{\theta}{2}\right)XX$$

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.
• Any chance you could give more details on this part ? "hence the eigenvalues of H must be ±1. Let P± be projectors onto the eigenspaces of eigenvalue ±1 respectively. So, $\mathbb{I}=P_++P_− \hspace{4em} H=P_+−P_−$" – Reda Drissi May 10 at 8:40
• $H$ is Hermitian, so we know it has real eigenvalues. Call them $\lambda_i$, with eigenvectors $|\lambda_i\rangle$. Then $H^2$ has the same eigenvectors, with $H^2|\lambda_i\rangle=H\lambda_i|\lambda_i\rangle=\lambda_i^2|\lambda_i\rangle$. So, the eigenvalues of $H^2$ are $\lambda_i^2$. But if $H^2$ is identity, it obviously has eigenvalues 1. So, $\lambda_i=\pm\sqrt{1}=\pm 1$. – DaftWullie May 10 at 9:34
• This means that $H=\sum_{i:\lambda_i=1}|\lambda_i\rangle\langle\lambda_i|-\sum_{i:\lambda_i=-1}|\lambda_i\rangle\langle\lambda_i|$, where I'm assuming the $|\lambda_i\rangle$ form an orthonormal basis, $\sum_i|\lambda_i\rangle\langle\lambda_i|=\mathbb{I}$. So I can choose to define the projector $P_+=\sum_{i:\lambda_i=1}|\lambda_i\rangle\langle\lambda_i|$. – DaftWullie May 10 at 9:37