Is the existence of a sole particle in an hypothetical infinite empty space explicitly forbidden by QM?

Question

Suppose the universe is completely empty with one sole particle trapped in it. To simplify, I will only be looking at the one dimensional case. However, all arguments are applicable for three dimensions. The solution of the Schrödinger equation $\hat{H} \psi = E \psi $ with $\hat{H} = \frac{\hat{p}^2}{2m}$ and $\quad \hat{p} = -i\hbar \frac{\mathrm d}{\mathrm d x}$ for a free particle ($V=0$) is then given by $\psi(x,t)=A\exp{i(kx-wt)}$ with constants $w=\frac{E}{\hbar}$ and $E=\frac{\hbar^2k^2}{2m}$ which can easily be shown. In order to meet the criteria of QM, $\langle \psi|\psi\rangle$ needs to be normalized (=1). If that is not possible, there is no way such a quantity can be interpreted as a probability density. However, if you try to normalize the density you will find:

\begin{align*} \langle\psi|\psi\rangle &=\int_{-\infty}^{\infty} \psi \psi^* dx \\ &= |A|^2 \int_{-\infty}^{\infty}\exp{i(kx-wt)} \exp{-i(kx-wt)}\mathrm dx \\ &=|A|^2 \int_{-\infty}^{\infty}1\cdot\mathrm dx\\ &=\infty\,. \end{align*}

Thus, it is not possible to find said normalization. Asking around, I found that such cases are normally treated as an approximation of a very wide (yet finite) potential well, or in the 3-dimensional case, a box. This allows a fairly simple solution which can be normalized. However it only represents an approximation. In a rigorous point of view there exist no such solutions which satisfy the postulates of Quantum Mechanics. Does that mean that the laws of QM fundamentally prohibit a infinite empty universe with only one sole particle trapped in it?

Answer 1

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be normalizable/non-normalizable? and Rigged Hilbert space and QM for some further reading)

But this doesn't mean no free particles exist, it just means that there are no momentum eigenstates for them. E.g. for any smooth compactly supported function $f$, we have that $\psi(x) = \int f(p)\mathrm{e}^{\mathrm{i}px}\mathrm{d}p$ is normalizable (because in Fourier space, its square integral is just $\int \lvert f(p)\rvert^2\mathrm{d}p$, which is certainly finite), and this is a perfectly fine wavefunction for a perfectly fine state - it just isn't an eigenstate of the time evolution, it's a "smooth superposition" of them. We could just have taken $f$ as a wavefunction directly, which would also be fine, but this way $\psi$ is what one calls a "wavepacket", at least for suitably narrow $f$.