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I am asking for a counterexample or a proof for the following statement: Let $V$ be some (time-independent) potential in $\mathbb{R}^n$, then the following statements are equivalent (under some mild regularity assumptions):

  • For any initial conditions at $t=0$ the classical trajectory exists for all $t>0$.

  • The operator $-\Delta+V$ is essentially self-adjoint on some nice domain, e.g. $C_0^\infty(\mathbb{R}^n)$.

Classic examples of potentials which may not be associated to a self-adjoint Hamiltonians are usually those, whose classical behaviour is already ill-defined, e.g. $-x^4$ in one dimension. From a physical point of view, this does not seem very surprising to me, therefore I am looking for a system, whose "quantum-mechanical ill-definedness" is not directly visible from classical physics. To me, it seems reasonable to expect such a system to exist, since by quantum effects, QM-particles are allowed in regions, where classical particles have no access to. So in particular, there could be a potential which allows a particle to "tunnel to infinity" after finite time.

By elementary ODE theory, under some mild regularity assumptions to $V$ the classical premise is fulfilled if $V$ goes to $-\infty$ not faster than $-r^2$. The corresponding theorem in QM is the Faris-Lavine Theorem, which states that if $-r^2$ is a lower bound to $V$ up to constants, $C_0^\infty(\mathbb{R}^n)$ is a core. This already means, that the counterexample I am looking for is of the form, such that $-r^2$ is not a lower bound. I could imagine it to be something like $-r^3$ plus some positive oscillation whose amplitude grows as $r$ goes to infinity. However, I did not manage to work out the details yet (disproving the existence of a self-adjoint extension is harder than I originally thought), and I also don't think that my idea is so original that nobody came up with it in the last 50 years. Similarly, the fact that the $"\Leftarrow "$-direction is not some well-known theorem makes me think that there should also be a counterexample, however I have no physical intuition how this should look like.

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This issue is discussed here. Regarding the converse: the hydrogen atom's Hamiltonian is self-adjoint and has a perfectly well-defined quantum ground state whose wavefunction is continuous everywhere, but if you send a classical electron directly radially inward, its trajectory will become undefined in finite time when it collides with the nucleus.

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  • $\begingroup$ The quantum potential is still bad for the 1/r potential for non stationary scattering solutions of an electron and positron as r goes to zero, as @tparker said, and in QFT it causes divergences also. So it still less than 'good'. The other way, the SHO is ok in classical theory, but in QM a non zero energy ground state and in QFT a non zero energy vacuum $\endgroup$ – Bob Bee Dec 26 '16 at 3:00
  • $\begingroup$ Thanks for your input! The referenced paper was a pretty interesting read, however it did not answer the question, since all the examples discussed in the paper are in favor of the statement I am seeking to disprove. It explicitly states, that "Whenever the classical equations of motion admit global solutions for arbitrary initial conditions, then the quantization procedure is unambiguous ([...]).", which is pretty much the statement I am seeking to disprove. $\endgroup$ – Daniel Dec 27 '16 at 4:09
  • $\begingroup$ Also, the hydrogen atom does not really make me happy, since the potential is ill-defined at $0$, well-definedness should go under what I called "mild regularity assumptions", maybe I should clarify that. In particular, this behavior does not seem to be reproduceable for any regularized version. $\endgroup$ – Daniel Dec 27 '16 at 4:09
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Counterexamples to my statement are actually featured in the second volume of Reed-Simon, in the Notes to chapter X. The worked-out counterexamples are to be found in

Rauch, Jeffrey; Reed, Michael. Two examples illustrating the differences between classical and quantum mechanics. Comm. Math. Phys. 29 (1973), no. 2, 105--111.

The counterexample to $"\Rightarrow"$ is more or less how I imagined it: The example Rauch and Reed construct is the potential $$ V(x)=\frac{1}{x^2}-x^4+\sum_{i\in\mathbb{N}}\sigma_i(x) $$on $(0,\infty)$, where $\sigma_i$ are bump functions of increasing height, such that $V(k)=k$ for $k\in\mathbb{N}$ and decreasing width. Since these bumps will stop any classical particle from escaping to infinity, it is classically well-behaved. However, a quantum-mechanical particle can tunnel through these barriers and escape to infinity after finite time. It is shown that $\sum_{i\in\mathbb{N}}\sigma_i$ is relatively bounded with respect to the rest of the operator with relative bound $<1$, by the Kato-Rellich theorem this implies, that $-\Delta+V$ is self-adjoint iff $-\Delta+\frac{1}{x^2}-x^4$ is self-adjoint, where the latter is not. The $x^{-2}$ still bothers me a bit, but I think its just an artifact from the calculation, it doesn't really serve any physical purpose.

The counterexample to $"\Leftarrow"$ looks like a smoothed version of $$ V(x)=-\pi^2\sum_{i\in\mathbb{N}}i^4\theta(i-x) $$where $\theta$ is the Heaviside function. Since it goes to infinity like $-x^4$, a classical particle will escape to infinity after finite time, however a wave function will be reflected at each step in such a way, that it does not escape. The proof uses some ODE theory, it shows that the nontrivial solution of $(-\Delta+V)\psi=0$ will not be integrable by infinity, which shows that $V$ is limit-point case near infinity, hence $-\Delta+V$ is e.s.a. on $C_0^\infty(0,\infty)$.

In the paper, there is also a theorem by Wintner, which in the one-dimensional setting assures equivalence if the derivatives of $V$ are not "too big" in a certain sense. I doubt however that this easily generalizes to higher dimensions, since everything in this paper is based on ODE theory.

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