Let $\pi$ be a unitary representation of a Lie group $G$ on a Hilbert space $H_{\pi}$. Thus $\pi:G \to B(H_{\pi})$. One can extend $\pi$ to a representation of $L^1(G)$ via the formula $\pi(f)=\int_{G}f(t)\pi(t)d\mu(t)$ where $\mu$ is the Haar measure. Thus $\pi$ may be viewed as a mapping $\pi: L^1(G) \to B(H_{\pi})$. I will describe two ways for obtaining a representation of the Lie algebra $\mathfrak{g}$ of $G$ from a representation $\pi$.

- One way uses the description of $\mathfrak{g}$ as the Lie algebra of left invariant vector fields. For $X \in \mathfrak{g}$ (viewed as a left invariant vector fields) one defines an operator (in general unbounded) which we denote by $d\pi(X)$ as follows: it acts on vectors of the form $\pi(\varphi)\xi$ where $\varphi \in C^{\infty}_c(G)$ and $\xi \in H_{\pi}$ and the formula is $$d\pi(X)(\pi(\varphi)\xi)=\pi(X(\varphi))\xi$$ where $X(\varphi)$ is understood as the action of vector fields on smooth functions

2.The second way is as follows: for $X \in \mathfrak{g}$ viewed as an element in $T_eG$ we define $d\pi(X)$ as an operator which acts on smooth vectors for $\pi$ (i.e. those vectors $\xi \in H_{\pi}$ such that the map $G \ni t \mapsto \pi(t)\xi$ is smooth) by the formula $$d\pi(X)(\xi)=\frac{d}{dt}\Bigg|_{t=0}\pi(exp(tX))\xi$$

Are these two constructions equivalent?