Is every unitary operator induced by a Hamiltonian? Diving deeper into the mathematical inner workings of quantum mechanics: The set of unitary operators on the Hilbert space $\mathcal{H}$ forms a group. While for finite-dimensional Hilbert spaces, this group is essentially the matrix Lie group $U(n)$ and everything is well-known, in the infinite-dimensional case, this does not seem to be the case. Wikipedia says this group is sometimes referred to as Hilbert group, is this the standard terminology?
Main question: are there unitary operators in this group, that cannot be written as 
$$U=e^{-iHt}$$
for some Hamiltonian $H$?
Secondly, I would like to understand how those Hamiltonians might look. Really, the only thing you get to see in standard QM are Hamiltonians $H$ which are polynomials of the canonical operators $X$ and $P$ or equivalently can be written as polynomials of annihilation and creation operators. Do these Hamiltonians induce a subgroup of the (Hilbert) group mentioned above, are there other known subgroups?
 A: There is a one-to-one correspondence between self-adjoint operators and representations of the abelian group $\mathbb{R}$ as a group of unitary operators $\bigl(U(t)\bigr)_{t\in\mathbb{R}}$ that is continuous, as a function of $t\in\mathbb{R}$, with respect to the strong operator topology. 
The map between self-adjoint operators and strongly continuous unitary groups is given by $H\mapsto e^{-itH}$. This is the content of Stone's theorem mentioned in the comments. However, there are unitary operators that are not generated by a self-adjoint Hamiltonian in the above sense.
For general unitary operators, there is still a way of "taking the logarithm", in some sense. By functional calculus for normal operators (unitary operators are normal), it is possible to represent each unitary operator $U$ as follows:
$$U=\int_{\mathbb{R}} e^{-it} \mathrm{d}P_U(t)\; ,$$
where $P_U(t)$ is a projection-valued spectral family such that $P_U(t)=0$ for all $t<0$, and $P_U(t)=I$ (the identity operator) for all $t\geq 2\pi$. 
However, the formal logarithm that results, i.e.
$$``i \ln U "=\int_{\mathbb{R}} t \, \mathrm{d}P_U(t)$$
is a bounded (albeit self-adjoint) operator, and therefore it does not coincide in general with the self-adjoint generator (take for example the unitary operator $e^{iH}$, with $H$ self-adjoint and unbounded to have a counterexample).
