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We all know very illustrative spatial representations of predicted electron orbitals in atoms which are essentially spatial plots of the solutions of wave equations.

In all atoms the electrons occupy the positions around the nucleus, sometimes, spatially symmetric.

But I wonder which form could take orbitals of three particles of nearly-similar mass, such as quarks in a hadron or hadrons in tritium nucleus? Seems there is no place in the center for one of them...

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Interesting question, but it's worth bearing in mind that you are dealing with an entirely different interaction when it comes to quarks and hadrons. So the analysis is immensely more complicated. – David Z Feb 7 '12 at 3:08
Quarks in a hadron don't need orbitals, as they have different quantum numbers (the color number). So they can occupy the same wavefunction/orbital (just like two electrons in a single orbital who have $m_s=\pm\frac{1}{2}$), and the orbital will probably be a very (probabalistically) dense spherical body. – Manishearth Feb 7 '12 at 3:18
There are various quark models and the particle data group has a paragraph on them .They assign S,P,D, etc angular momentum quantum states for various nucleon states, which means an angular momentum, so the supposed center would be the center of mass of the nucleon. But they are just that, models approximating a very complicated ( due to gluon self interaction) dynamics.I agree with @Manishearth that the probability density would be a sort of dense spherically deformed body, not similar to electron orbitals. – anna v Feb 7 '12 at 5:52
@annav Would it really be spherically deformed ? I havent seen that paper yet, but it makes morse sense if it was a complicated, but rotationally symmetric field. Of course, visualising the slew of interactions is pretty hard.. So maybe it may be asymmetric. I'll check out the pdf later. Thanks! – Manishearth Feb 7 '12 at 6:40
Related (nearly a duplicate): Where can I get the most accurate measurements of parton distribution functions?. (Note that the PDFs are generally expressed in momentum space, but that is only a Fourier transform away from position space.) – dmckee Feb 7 '12 at 16:00

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