Moving point particles in quantum mechanics

Question

Given a quantum system with a hamiltonian $H$, if initial state of a physical object is $|\psi\rangle $, then state after time $dt$ is $e^{iH\cdot dt}|\psi\rangle$. It is easy to see that $|\langle\psi | e^{iH\cdot dt}|\psi\rangle|^2$ is at least $1-\|H\|\cdot dt$, where $\|H\|$ is the norm of the hamiltonian (or its largest eigenvalue).

Now consider a point particle moving on a line. Its position changes from $x$ to $x+ dx$ in arbitrary small time $dt$. But in quantum mechanics, the states $|x\rangle$ and $|x+dx\rangle$ are mutually orthogonal. Then how can a physically reasonable hamiltonian $H$ describe a transition from $|x\rangle$ to $|x+dx\rangle$, without having infinite norm (that is $\| H\| \rightarrow \infty$, so that the equation $1-\|H\|\cdot dt = 0$ can be satisfied for arbitrarily small $dt$) ? Does this mean that when dealing with point particles, we need hamiltonians of infinite norm and conversely, when dealing with systems with hamiltonians of finite norm, we cannot have point particles?

Answer 1

Your argument doesn't work because the states $|x\rangle$ are very bad.

They are not normalizable and don't belong to the Hilbert space. Rather they should be considered as generalized functions on the Hilbert space, what is called rigged Hilbert space. It can be seen if we discuss wavefunctions in coordinate representation. Proper wavefunction is a square-integrable function that have finite $\int_{-\infty}^{+\infty}|\psi(x)|^2dx$. In contrast $|x\rangle$ is represented by generalized function $\delta(x)$ for which that integral strictly speaking makes no sense and less strictly is "equal" to "infinite number" denoted as $\delta(0)$.

Secondly, you seem to assume that if we take $|x\rangle$ it will remain coordinate eigenstate, just for different $x$. Now, remember Heisenberg inequality $\Delta x\Delta p\geq\frac{\hbar}{2}$. The states $|x\rangle$ being ideally localized in coordinate representation are infinitely delocalized in the momentum representation - the Fourier transform of $\delta$-function is simply plane wave. As result they become instantly delocalized and $\langle x|e^{-\frac{i}{\hbar}\hat{H}t}|\tilde{x}\rangle$ is nonzero for arbitrarily small $\Delta t$.

To realize something similar to the classical point particle you have to consider the wavepacket minimizing $\Delta x\Delta p$. For small $\hbar$ such wavepackets approximately follow classical trajectories however they generally speaking spread with time (one of the few exceptions is harmonic oscillator that have non-spreading coherent states)