A classical localization of a wave function?

Question

We know that the state of a quantum particle defined on the real line is represented by its wave function $\psi(x)$ that is the position probability amplitude. We also know that the momentum probability amplitude $\phi(p)$ is a Fourier transformation of $\psi(x)$, where $p=\hbar k$ and $k$ is the Fourier conjugate variable of $x$. Moreover, the Heisenberg uncertainty principle dictates how well the functions $\psi$ and $\phi$ can be localized simultaneously: $\Delta_{x_0} (\psi)\Delta_{p_0} (\phi)\geq \hbar / 2$, where $x_0$ and $p_0$ are the corresponding expectation values.

How to mathematically show that the both functions $\psi$ and $\phi$ can be sharply localized if the constant $\hbar$ becomes arbitrary small positive number?

When this becomes clear, it is easy to obtain an arbitrary well localized "classical" spacetime curve from the Ehrenfest theorem. Perhaps this approach is much more convenient than the path integral solution to classical limit.

Answer 1

In general, it is not true that the uncertainties both become small as $\hbar \to 0$ - for instance, $\Delta x$ might not be function of $\hbar$ at all and $\Delta p$ a linear function of $\hbar$, so that $\lim_{\hbar \to 0} \Delta x = \Delta x\neq 0$. This is the case for example when you start with a Gaußian $\psi(x) = \mathrm{e}^{(x-x_0)^2/(2\Delta x^2)}$ with fixed $\Delta x$. Nothing in the formalism of quantum mechanics says that wavefunctions must functionally depend on $\hbar$ at all.

This is a basic reason why the classical limit of quantum mechanics is not merely taking $\hbar \to 0$ and doing nothing else. See this question and its linked questions for more discussion of what this limit actually means and when it provides a meaningful transition to classical mechanics. Note also that Ehrenfest's theorem does not in general produce classical results because $\langle V(x)\rangle \neq V(\langle x\rangle)$ and if you think that $\langle x\rangle$ should be the classical position then it is $V(\langle x\rangle)$ that would have to appear in the "classical" equation of motion, but the theorem only gives you $\langle V(x)\rangle$.

More specifically to the question, the usual examples for states that "become classical" as $\hbar \to 0$ in the sense that both uncertainties diminish are coherent states, whose simplest versions have $\Delta x = \Delta p = \sqrt{\frac{\hbar}{2}}$ so that both uncertainties go to 0 as $\hbar \to 0$.