It is well known that for every unitary operator $\hat U$ an exponential of the form
$$
\hat U = e^{i\hat H}
$$
exists ($\hat H$ is hermitian). But I can only prove it the other way round:
$$
(e^{i\hat A})^\dagger = \sum_{n=0}^\infty \frac{(-i)^n(A^n)^\dagger}{n!}=e^{-i\hat A^\dagger}=e^{-i\hat A}
$$
with $A$ hermitian.
Now suppose
$$
\hat U\hat U^\dagger = e^{i\hat A} e^{-i\hat A}=1
$$
so $\hat U$ has to be unitary. So now I have proven the statement that for every hermitian operator there exists a unitary operator, right?
Now how do I know that I can always find an exponential form of a unitary operator? Is this statement eventually true "in both directions"?
Best Answer
A unitary operator is a diagonalizable operator whose eigenvalues all have unit norm. If we switch into the eigenvector basis of U, we get a matrix like: \begin{bmatrix}e^{ia}&0&0\\0&e^{ib}&0\\0&0&e^{ic}\\\end{bmatrix} which is obviously the exponential of a diagonal hermitian matrix.