Is nucleus a quantum system? I was going through this question, now I wonder what are necessary criteria for describing a system as quantum system.
Is it so that we have evidence of nuclear shell structure that imply that nucleus is a quantum system? As shell numbers turns out to discrete which leads that nucleon can have only certain fixed energy levels and not continuous energy levels?
Is it so that we have so many identical protons and neutrons confined to very small volume?
Is it so that nuclear dimensions are very small and at such dimensions classical models does't work work well?
How to describe a system as quantum or classical?
I am a beginner simple answers will be appreciated.
 A: One criterion for whether a system is “strongly quantum-mechanical” is whether the typical interaction length scales are large or small compared to a typical wavelength of an interacting particle.  If wavelengths are short, interference effects tend to average away, and a particle-like model is useful.  But if wavelengths are long, interference effects play a major role in the system’s dynamics.
There’s a hint of this at the beginning of every textbook on quantum mechanics, in an exercise where the student is invited to compute the de Broglie wavelength
$$ \lambda = h / p$$
of a macroscopic object, like a sports ball.  For instance, the de Broglie wavelength of $\rm10\,kg$ bowling ball crawling boringly down the lane at $\rm1\,m/s$ is comparable to the “Planck length,” which is substantially smaller than the bowling ball’s diameter. This suggests, consistent with my personal experience, that wavelike interference is not a major factor in ten-pin bowling.
But there’s a flaw in this pedagogical argument: what about a bowling ball at rest, in the limit where the momentum $p\to 0$? Those objects don’t typically behave quantum-mechanically either, which suggests that we might separately consider the interaction wavelengths of the entire object and also its constituent parts.  Something interesting happens if we do this with the various components of an atom in a solid.
It’s usually easier to think about energies than about momenta.  For objects trapped in some potential, we can handwavingly use the “virial theorem” to state that binding energies $U$ and kinetic energies $T$ have roughly the same magnitude, and that non-relativistic objects have momentum $p = \sqrt{2mT} \sim \sqrt{2m|U|}$.  So the de Broglie relation becomes
$$ \lambda \sim \frac{hc}{\sqrt{ (mc^2) |U|}}, $$
where $m$ is the mass of the particle in question, $U$ is the binding energy, and I’ve one-squiggled away a factor of two in order to discourage credulous readers from taking this too quantitatively.  The unit-conversion constant has value $hc = \rm 1240\,eV\,nm = 1240\,MeV\,fm$.
A nucleon in a nucleus has mass $mc^2 = \rm 1000\,MeV$ and typical single-particle binding energy $U\sim\rm10\,MeV$, so its wavelength inside a nucleus is $\lambda\sim\rm10\,fm$.  That’s substantially larger than the size of even the largest nuclei, so every nucleus is a strongly quantum-mechanical system.
An atom’s valence electrons have mass $mc^2 = \rm 500\,keV$ and typical binding energies of a few eV (which is why ions capturing electrons emit visible-ish light).  That gives a typical de Broglie wavelength around a nanometer, which is substantially larger than the typical angstrom-ish size of an atom, so the atom’s electron cloud is a strongly quantum-mechanical system.  For that matter, this suggests that the valence electrons in a solid with angstrom-ish atomic separations have wavelengths which smear them out over several atoms. This is a handwaving explanation of why we talk about the “sea” of conduction electrons in a metal, and why the band structure of electron excitations in semiconductors depends on the structure of the entire crystal.
However, suppose we look at the atoms in a solid.  In that case the masses are typically tens of GeV, but the room-temperature excitation energies are around $kT = \frac{1}{40}\rm\,eV$.  That gives a de Broglie wavelength of somewhere under one angstrom, substantially smaller than the typical distance between two atoms in a solid.  This is why we can productively use the semi-classical picture of a crystal as a set of pointlike atoms vibrating around their well-defined locations on a lattice: the atoms are well-localized.
A fun application of this approach is to try and (again, hand-waving-ly) predict the temperature at which helium transitions from a classical liquid to a strongly quantum-mechanical superfluid.
A: It is predominantly because of the exceedingly small size of an atomic nucleus. On those scale lengths, quantum effects become dominant, and classical dynamics can no longer be used to describe the behavior of a nucleus.
