The Hamiltonian of quantum chromodynamics (like the rest of the Standard Model) is rotationally symmetric. My question is whether these space symmetries are spontaneously broken in the ground state of hadrons.
The cartoon version of a hadron looks like this:
A more realistic (although still simplified) picture looks like this, showing the presence of a huge number of "sea" quarks and antiquarks:
(Image sources: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-structure-of-matter/protons-and-neutrons/)
Q1: Do the ground-state wavefunctions of hadrons preserve or spontaneously break rotational symmetry? (I'm talking about rotations in physical space; of course a hadron is a singlet with respect to "rotations" in SU(3) color space). For a given proton, do the positions of the quarks pick out a special spatial orientation?
For scalar mesons, my guess is that the ground state is unique and the quarks are perfectly rotationally symmetric. But for baryons and vector mesons, the presence of nonzero spin degeneracy makes me wonder whether there's some kind of spin-orbit coupling that connects the hadron's spin angular momentum and the positions of the constituent quarks.
Q2: I know that the position of a particle is a subtle concept in relativistic QFT. But is it possible to come up with some kind of effective position density function $\rho({\bf x}) := \langle 0 | \hat{N}({\bf x}) | 0 \rangle$ within a hadron, where $\hat{N}({\bf x})$ represents a local quark number operator? (Side Q3: is it possible to separately do this for the "primary" quarks and for the antiquarks, or only for the total quark number operator that represents their difference?) If so, what does this position density function look like? Is it rotationally symmetric (this is a special case of Q1)? If so, does the probability density fall off monotonically with distance from the center of the hadron, or does the most likely location lie at a finite distance from the center? (Note that the most likely location is different from the most likely distance from the center, because the latter gets multiplied by a factor of $r^2$ from the Jacobian factor in the change of variables.)
Again, my guess is that for a scalar meson, $\rho({\bf x})$ is rotationally symmetric and monotonically decreases with distance, but I'm not as sure about the hadrons with spin.
