= Solution
Since $A$ is a <bounded operator>, its exponential <power series> converges in <operator norm>. The identity $A^2=I$ gives $A^{2k}=I$ and $A^{2k+1}=A$. Separating the even and odd powers therefore gives
$$
e^{-i\theta A}=\sum_{k=0}^\infty\frac{(-1)^k\theta^{2k}}{(2k)!}I-i\sum_{k=0}^\infty\frac{(-1)^k\theta^{2k+1}}{(2k+1)!}A=\boxed{\cos\theta\,I-i\sin\theta\,A.}
$$
The algebraic identity does not require $A$ to be Hermitian. Hermiticity is additionally needed for this exponential to be a <unitary operator> for real $\theta$.
Back to article page
