If a potential function in 3D doesn't have bound states in quantum mechanics, does that mean it doesn't have non-circular orbits in classical physics?

Question

As I understand it quantum mechanics can be used to describe objects that appear to behave classically, meaning that any classical orbit can be treated as a bound state in quantum mechanics.

I know a statement is logically equivalent to its contrapositive. This means that IF having no bound states in quantum mechanics implies no stable orbits in classical physics, then having stable non-circular orbits in classical physics would imply having bound states in quantum mechanics, and as I understand it can lead to the wrong conclusion if one tries to derive something in quantum mechanics using classical physics.

So if a potential has no bound states in quantum physics does this mean it has no stable non-circular orbits in classical physics?

Answer 1

No. An example is the finite spherical "square well": $$ V(r) = \begin{cases} -V_0 & r<r_0 \\ 0 & r > r_0\end{cases} $$ A sufficiently "shallow" potential of this type has no bound states, but classically has an infinite number of stable non-circular orbits (any trajectory with a negative energy is bound.) In fact, it has no circular orbits at all.

If the "sharp corners" of the potential are distasteful, one could envision a similar potential with the corners "rounded off", which should have much the same properties. Such a potential could allow circular classical orbits as well, so long as the corners weren't too rounded.

The rough reason why your intuition doesn't work, by the way, is that the classical-to-quantum correspondence only generally works for sufficiently high energies.