I have to solve the following exercise:
Let $U(t)$ be a unitary operator ($t$ is a real parameter) such that $U(0)=\mathbb 1$ (identity). Show that $$U(t) = \exp(itH)$$ to a first order approximation. $H$ is an hermitian operator and $i$ is the imaginary unit.
I found some solutions to similar exercises and they all involve matrix representation or the spectral theorem, but I am supposed to solve this without using them.
I think I succedeed in demonstrating a part of the exercise: let's suppose $$U(t) = \mathbb 1 + itH + \mathcal O(t^2),$$ where $H$ is a generic operator, then $$U^{\dagger}(t) = \mathbb 1 - itH^{\dagger} + \mathcal O(t^2).$$ The first order approximation of the product $U^\dagger(t)U(t)$ is $$ U^\dagger(t)U(t) \approx \mathbb 1 + it(H-H^\dagger),$$ which is equal to the identity if $H = H^\dagger$.
I sitll can't figure out how to find the first order approximation of the operator.
Thanks in advance!
You have a couple of missing assumptions. For starters,you are using regularity of $U(t)$ when you represent it to first order. The other condition you need in this situation is that $U(t)$ is a semigroup, i.e., $U(t+s)=U(t)U(s)$. This in particular implies that $U(t)U(s)=U(s)U(t)$.
With these assumptions:
Knowing that $U'(0)$ exists, you can take $$ H=\tfrac1i\,U'(0). $$ This works because $$ U'(t)=\lim_{h\to0}\frac{U(t+h)-U(t)}h=U(t)\,\lim_{h\to0}\frac{U(h)-U(0)}h=U(t)\,U'(0). $$ The initial value problem $U'(t)=U(t)\,U'(0)$, $U(0)=I$ has unique solution $U(t)=e^{t\,U'(0)}$. So $U(t)=e^{itH}$. The equality $U^*(t)=U^{-1}(t)$ can be written as $$ e^{-itH^*}=e^{-itH}. $$ Calculating the derivative at $t=0$, $$ H^*=H. $$