In quantum mechanics, given a Hermitian operator $A$, it gives rise to a symmetry/unitary operator by exponentiation $e^{i\lambda A}$, which can be properly defining using the eigenvector expansion, i.e. we define this exponentiation in the basis of $A$ as
$$e^{i\lambda A } = \sum_{n} e^{i\lambda a_n} |a_n\rangle \langle a_n|.$$
On the other hand, by Stone's theorem, given a strongly continuous one-parameter family of unitary operators $U(\lambda)$ we get a Hermitian operator $A$ such that
$$U(\lambda)=e^{i\lambda A},$$
where $A = U'(0)$.
Given, then, a Lie group $G$ and a smooth curve $\gamma : I\subset \mathbb{R}\to G$, we get an element of the Lie algebra as $\gamma'(0)$.
It turns out though that if $G$ acts unitarily on the state space of a system by $U : G\to \mathcal{L}(\mathcal{H})$, we can consider a curve $\gamma$ on $G$ and get a one-parameter family of unitary operators $U(\gamma(\lambda))$ on $\mathcal{H}$.
In the examples I looked at, it turns out that the Hermitian operators derived from Stone's theorem from these one-parameter families "$(U\circ\gamma)'(0)$" correspond to the Lie algebra elements $\gamma'(0)$.
The examples I saw of this are basically the Poincare group $G = P(1,3)$ acting by $U(a,\Lambda)$ so that if we consider the one-parameter families obtained by the coordinate lines of the usual coordinate system on $G$ we get that the unitary one-parameter families are for example translations in each of the $4$ directions. The associated observables by Stone's theorem are the $4$-momentum components. They seem related to the Lie algebra elements of the group, I just don't know how to make this connection really precise.
My question is: Given Lie groups, Lie algebras and one-parameter families of unitary operators in a quantum mechanical state space, is there really a connection between the observables and the Lie algebra? How does one make this connection precise?
