Given
$$H = \begin{pmatrix}\sin \theta & 0 & \cos \theta \\ 0 & 1 & 0 \\ \cos \theta & 0 & -\sin \theta \end{pmatrix}$$
where $\theta=\pi/6$, then what is $\exp{ \left( i \frac{\pi}{2} H \right)}$?
I tried to calculate in the following way
$e^{(i\pi H)/2}=[e^{(i\pi/2)}]^H=i^H$. I do not know how to proceed.
Best Answer
The eigenvalues of this matrix are $1$ (double) and $-1$ (simple). As it is real symmetric, it is diagonalisable. If $v$ is an eigenvector then $Hv=\pm v$. Therefore $$\exp(\pi i H/2)v=\exp(\pm\pi i/2)v =\pm iv=iHv.$$ Hence $\exp(\pi iH/2)=iH$.