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.

  • $\begingroup$ 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, $\mathbb{I}=P_++P_− \hspace{4em} H=P_+−P_−$" $\endgroup$ – Reda Drissi May 10 at 8:40
  • $\begingroup$ $H$ 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, $\lambda_i=\pm\sqrt{1}=\pm 1$. $\endgroup$ – DaftWullie May 10 at 9:34
  • $\begingroup$ 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 $P_+=\sum_{i:\lambda_i=1}|\lambda_i\rangle\langle\lambda_i|$. $\endgroup$ – DaftWullie May 10 at 9:37

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