Position operator I'm learning about unbounded operator densely defined on Hilbert spaces from "Reed & Simon, Functional Analysis".
The first example was the position operator, which is defined as follows.
Let $D(T) = \{ \varphi \in L^2(\mathbb R) : \int_\mathbb R x^2|\varphi(x)|^2 {\rm d}x < \infty \}$ and $T\colon D(T)\to L^2(\mathbb R)$ be given by $T\varphi(x) = x\varphi(x)$.
Well, the domain $D(T)$ of $T$ that was chosen is the largest one for which the range of $T$ is in $L^2(\mathbb R)$.
I am wondering: Why such operator is called position operator? What is its meaning for physicists? What is the intuition in defining such operator like this?
Thanks in advance!
 A: The basic answers were already given in the comments but I will try to elaborate: In (non-relativistic single-particle) quantum mechanics, the state of a particle is modeled by a wave-function $\psi$. Typically these functions are modeled by $L^2(\mathbb{R}^d,\mathbb{C}^k)$, that is complex vector valued and square integrable functions over the euclidean $d$-space (for physics it is $d=1,2,3$).
The name 'wave-function' originates from the fact that we will typically model a particles evolution by some wave equation, namely a Schrödinger equation
$$i \partial_t \psi(t) = H\psi(t) \qquad\text{for}\qquad \psi\colon\mathbb{R}\to L^2(\mathbb{R}^d,\mathbb{C}^k),$$
where $H\colon \mathcal{D}(H) \subseteq L^2(\mathbb{R}^d,\mathbb{C}^k)\to L^2(\mathbb{R}^d,\mathbb{C}^k)$ is some self-adjoint but possibly unbounded linear operator.
Now, with the basic setting presented, we state a physical interpretation of a wave function: $|\psi(t,x)|^2$ is proportional to the probability of measuring the particle at position $x$. Consequently, $\int_{\mathbb{R^d}} x |\psi(t,x)|^2 \mathrm{d}{x} = \langle\psi,T\psi\rangle_{L^2}$ is the mean position of the particle at time $t$. For this reason, we refer to $T$ as the position operator.
Further remarks: There are various subtleties around choosing the right model-space and 'giving physical interpretation'. One could also choose wave functions to 'live over momentum space' instead of 'position space' (see here). In this case, the multiplication operator as given by yourself would be the momentum operator. In the end, this boils down to representing a quantum mechanical state in different eigen-basis of different self-adjoint operators, which in our above case was simply the position operator: Given a Hilbert Space $\mathcal{H}$ equipped with a self-adjoint (let's assume cyclic) operator $T$, we find an isometry $U\colon \mathcal{H} \to L^2(\sigma(T))$, such that $U T U^\ast$ acts as a multiplication operator on $L^2(\sigma(T))$. Hence, I could have also answered your question as: It is called the position operator, since we decided to model our wave-functions as $L^2$-functions over the spectrum the position operator :).
