# Bounded Operators, Unitary group

It's clear to me that if H is a self-adjoint bounded operator on a Hilbert space, then the bounded operators

$$U_t :=\sum_{n=0}^\infty (iHt)^n / n!$$

are unitary for all $t\in \mathbb{R}$. How do I show the converse, that if the $U_t$ are unitary for all $t$, then $H$ is self-adjoint? Is there some simple obvious way to show this, without referring to Stone's theorem for unbounded operators? It's also pretty clear that $$U_t^* = U_t^{-1} = U_{-t}$$ but what do I do next? The complex exponential is not an injective map. (Problem 5.1.6 of Davies Linear Operators and their Spectra).

For all $t$ and for all $x,y$, $$0 = \frac{d}{dt}(x,y)=\frac{d}{dt}(U_t x,U_t y) = (U_t iHx,U_t y)+(U_t x,U_t iHy)$$ Set $t=0$ in order to obtain the following for all $x,y$: $$(Hx,y)=(x,Hy).$$