The operator $-i \frac{d}{dx} : H^1(\mathbb{R}) \subset L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ is self adjoint hence it is unitary equivalent to a multiplication operator
i.e. $-i \frac{d}{dx}= U^* M_f U$ where $U:L^2(\mathbb{R}) \rightarrow L^2(\Omega,\mathcal{A},\mu)$, $UH^1(\mathbb{R}^n)\subseteq D(M_f)$ and $L^2(\Omega,\mathcal{A},\mu)$ is a $\sigma$-finite measure space. $M_f$ is multiplications operator where $f:\Omega \rightarrow \mathbb{R}$ is $\mathcal{A}$-$\mathcal{B}(\mathbb{R})$-measurable.
What is $U$ and $f$ here?
Using the advice given by "peek-a-boo":
I get for $U f:=\frac{1}{\sqrt{2\pi}} \int\limits_{\mathbb{R}} e^{-i x \overline{y}}f(y)dy$
$U(-i f^\prime(x))=\frac{1}{\sqrt{2\pi}} \int\limits_{\mathbb{R}} e^{-i x \overline{y}}(-if^{\prime}(y))dy=\frac{1}{\sqrt{2\pi}} \int\limits_{\mathbb{R}}ixe^{-i x \overline{y}}(-if(y))dy=U(f(x))\cdot x$
Thus
$$-if^\prime(x)=U^{-1}M_x U f$$
for all $f\in H^1(\mathbb{R})=H_0^1(\mathbb{R})$.
Yet I just assumed that $g(-\infty)=g(\infty)=0$ for $g\in H_0^1(\mathbb{R})$ such that $e^{-i x\cdot\infty}f(\infty)=e^{i x\cdot\infty}f(-\infty)=0$. Is this actually true? And why?
As I know Elements from $H_0^1(\Omega)$ where $\Omega=[a,b]$ ($a<b$) vanishes on the boundary. This can be justified by the characterization:
But how is it in generall if $\Omega$ equal the whole space i.e. $\Omega=\mathbb{R}^n$???