What Hilbert space have position operator?

Question

I have some questions about the position operator used in physics, to avoid physical context I formulate the questions like this:

Definition 1$^{[1]}$: Let $(\mathcal E, \langle\cdot, \cdot\rangle)$ a Hilbert space over the field of complex numbers. A position operator in $\mathcal E$ is a self-adjoint operator $\hat X:\mathcal E_X\subseteq\mathcal E\rightarrow\mathcal E$ such that for every $x\in\mathbb R$ there are a vector $\varphi_x\in \mathcal E_X$ such that $\hat X(\varphi_x) = x ~ \varphi_x$.

Question 1: What Hilbert spaces admit a position operator?
Can you provide or reference examples of how looks $\hat X$ and $\varphi_x$ in such spaces?

Little context

A physicist have, by axioms, a Hilbert space that admit a position operator and say that such space is [isomorphic to] $L^2(\mathbb R; \mathbb C)$. The position operator is therefore defined as $(\hat X(\Psi))(x) = x~\Psi(x)$, but I read (see) that there are no function $\varphi_y\in L^2(\mathbb R; \mathbb C)$ such that $x~\varphi_y(x) = y~\varphi_y(x)$, then I ask:

Question 2: The space $L^2(\mathbb R; \mathbb C)$ admit some position operator like the above defined?

Notes

[1] - From axioms of quantum mechanics definition 1 is, a priori, the only that one can say about the position operator but a physicist also use some other relations, like the commutator with linear momentum. Therefore I imagine that not all operators satisfying definition 1 serves as position operator for a physicist.

Answer 1

With your own definition of a position operator (Definition 1), we must have: $$\forall x,y\in\Bbb R\quad x\langle\varphi_x,\varphi_y\rangle=\langle\hat X(\varphi_x),\varphi_y\rangle=\langle\varphi_x,\hat X(\varphi_y)\rangle=y\langle\varphi_x,\varphi_y\rangle$$ Hence $(\varphi_x)_{x\in\Bbb R}$ must be a family of pairwise orthogonal vectors. If you implicitely wanted these vectors to be non-zero, such a family (hence such an operator) exists iff the Hilbert dimension of $\cal E$ is at least the cardinality $\mathfrak c$ of the continuum.

The Hilbert dimension of $L^2(\Bbb R;\Bbb C)$ is $\aleph_0$$<\mathfrak c.$

The simplest example of a Hilbert space of Hilbert dimension $\mathfrak c$ is the space $\mathcal E:=\ell^2(\Bbb R;\Bbb C)$ of $\Bbb R$-indexed families $z=(z_x)_{x\in\Bbb R}\in\Bbb C^{\Bbb R}$ of complex numbers such that $\sum_{x\in\Bbb R}|z_x|^2<\infty$ (such families necessarily have a countable or finite support), and a position operator on this space is given by $(\hat X(z))_x=xz_x,$ for all $z\in\mathcal E_X:=\{z\in\mathcal E\mid\sum_{x\in\Bbb R}x^2|z_x|^2<\infty\}.$

Answer 2

The position operators used in physics do not satisfy your definition $\mathbf{1}$.

Your (incomplete) definition of $\ \hat X:\mathscr{D}(\hat X) \subseteq L^2(\mathbb{R};\mathbb{C})\rightarrow \mathbb{C}\ $ provides one example of a position operator that is used in physics, but that operator doesn't satisfy your definition $\mathbf{1}$. Its domain $\ \mathscr{D}(\hat X)\ $ is given by $\ \mathscr{D}(\hat X)=\left\{\Psi\in L^2(\mathbb{R};\mathbb{C})\,\left|\,\int_\limits{\ \ -\infty}^{\ \ \ \infty} \,|x\Psi(x)|^2\,dx<\infty\right.\right\}\ $, the omission of whose specification constitutes the incompleteness of your definition of it. This operator has no eigenfunctions—that is, there doesn't exist any $\ z\in\mathbb{C}\ $ and $\ \Psi_z\in\mathscr{D}(\hat X)\ $ such that $\ \hat X\Psi_z=z\Psi_z\ $—let alone an uncountably infinite number of them. What is true is that every real number $\ x\ $ is an approximate eigenvalue of $\ \hat X\ $—that is, for every $\ \epsilon>0\ $ there exists a function $\ \Psi_{x,\epsilon}\ $, with $\ \left\| \,\Psi_ {x, \epsilon}\,\right\|=1\ $, such that $$ \left\| \,\hat X\Psi_{x, \epsilon}-x\Psi_ {x, \epsilon}\,\right\|\le\epsilon\ , $$ so maybe that's what you should be using in your definition of "position operator".

I have seen some treatments of the subject by physicists where it's claimed that the Dirac delta function shifted by $\ x\ $ is an eigenfunction of $\ \hat X\ $ corresponding to the eigenvalue $\ x\ $. Even when it's properly defined, however, the shifted Dirac delta function $\ \delta_x\ $ doesn't belong to the space $\ L^2(\mathbb{R};\mathbb{C})\ $. For a suitably chosen subspace $\ \cal{S}\ $ of $\ \mathscr{D}(\hat X)\ $, it can be regarded as an element of a dual space $\ \cal{S}^*\ $ of $\ \cal{S}\ $ , defined by $$ \delta_x\Psi=\Psi(x)\ \ \text{ for }\ \ \Psi\in\cal{S}\ . $$ It is then an eigenvector of the algebraic dual $\ \hat{X}^*:\cal{S}^*\rightarrow\cal{S}^*\ $ of $\ \hat{X}\ $ corresponding to the eigenvalue $\ x\ $: $$ \hat{X}^*\delta_x=x\delta_x\ , $$ which, expanded out, simply means \begin{align} \delta_x(\hat{X}\Psi)&=(\hat{X}\Psi)(x)\\ &=x\Psi(x)\\ &=x\delta_x(\Psi)\ \ \text{ for all }\ \ \Psi\in\cal{S}. \end{align}