In quantum mechanics, the state of a single particle in one dimension is a square-integrable function ψ∈L2(R)\psi \in \mathbb{L}^{2}(\mathbb{R}) , assume that a particle is in the following state: \begin{equation} \psi(x) = \sqrt{\frac{e^{-x^{2}}}{\sqrt{\pi}}} \end{equation} Suppose that i change my coordinate labels to y(x)=tan(12erf(x))y(x) = tan(\frac{1}{2}erf(x)) , where: erf(x)=2π∫0xe−t2dterf(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}dt this is a bijective smooth transformation that maps R\mathbb{R} into itself. So it's a perfectly valid new frame of reference. However, the state of the particle in this new system of coordinates is: \begin{equation} \psi(y) = \sqrt{\frac{1}{\pi(y^{2}+1)}} \end{equation} Here's the problem, the position operator is defined by multiplication on a state, for this operator to be well-defined we must have the following quantity to be well-defined: \begin{equation} \langle X\cdot\psi(x)|X\cdot\psi(x)\rangle \end{equation} While this is true in the XX coordinates, it is certainly not true in the YY coordinates, as this integral is not well-defined: \begin{equation} \frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{y^{2}}{y^{2}+1}dy \end{equation} How can the theory be well-defined if the very existence of a physical observable is a coordinate-dependent notion? Given that coordinate transformations carry no physical meaning, it's clear that a quantity that exists in one frame should also exist in another frame, why is this not happening in quantum mechanics and how can the theory make physical sense?