Exponential function of a Hermitian matrix

Question

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.

Answer 1

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 $\exp(\pi iH/2)=iH$.

Answer 2

The eigenvalues of $H$ are $\lambda =1,1,-1$

According to Cayley-Hamilton Theorem $$ e^{tH}=\alpha I +\beta H + \gamma H^2$$ where $ \alpha,\beta,\gamma$ are functions of t to be found by the equation $$ e^{t\lambda}=\alpha +\beta \lambda + \gamma \lambda^2 $$ and its derivative with repect to $\lambda.$

We find $$ \beta = (e^t+e^{-t})/2, \gamma = 1/2te^t - \beta /2, \alpha = e^t-\beta - \gamma $$

For $t=i\pi /2$ we have $$ \beta = i, \gamma = -\pi /4, \alpha = \pi/4 $$

Thus we get $$e^{(i\pi /2) H}=iH$$

Answer 3

Note that $H^2=I$, thus $$ e^{tH} = I + tH +t^2/{2!}I +t^3/{3!}H +... $$

$$= (1+ t^2/{2!} +t^4/{4!}+...)I + (t +t^3/{3!} + t^5/{5!}+...)H $$

$$= (\sinh t)I + (\cosh t)H$$

For $t=(\pi/2) i$ we get $$ \sinh t=0 $$ and $$ \cosh t=i$$ thus $$e^{(\pi/2) iH} = iH$$