# What Hilbert space have position operator?

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.

• Hilbert spaces that admit a position operator are typically those that have a continuous domain and a well-defined inner product. Examples of such Hilbert spaces include the space of square-integrable functions defined on the real line ($L^2(\mathbb{R})$), the space of square-integrable functions defined on the complex plane ($L^2(\mathbb{C})$), and the space of square-integrable functions defined on the unit circle ($L^2(S^1)$). Commented Dec 23, 2022 at 0:38

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 $$z\in\mathcal E_X:=\{z\in\mathcal E\mid\sum_{x\in\Bbb R}x^2|z_x|^2<\infty\}.$$
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}