0
$\begingroup$

While I have a decent knowledge of general relativity (and, of course, classical mechanics), I am quite a novice when it comes to quantum mechanics, so I apologize if this is a rather basic question.

In a proton, neutron, or other hadron, the constituent quarks are bound together, thanks to color confinement. The strong nuclear force gets stronger as quarks are moved further apart, so it is thought to be impossible to "see" an isolated quark. My question is this:

A quark's wavefunction extends throughout all of space (prior to any de-coherence), so how can it be considered "bound" to the other quark(s) with which it constitutes a hadron?

My logic is that the quark could be anywhere, not simply tightly bound to the others; this flies in the face of quark confinement. Where am I wrong?

Improve this question
$\endgroup$
5
  • $\begingroup$ A quark's wavefunction extends throughout all of space (prior to any de-coherence)... Why would this be? $\endgroup$
    – user4552
    Commented Aug 17, 2014 at 18:55
  • $\begingroup$ I had thought a position-space wavefunction wasn't bounded inside a certain area. $\endgroup$
    – HDE 226868
    Commented Aug 17, 2014 at 18:58
  • 2
    $\begingroup$ Think of a harmonic oscillator. All eigenstates are bound states but extend to infinity. However, their amplitude rapidly becomes quite small. $\endgroup$
    – Urgje
    Commented Aug 17, 2014 at 20:05
  • $\begingroup$ @HDE226868: I had thought a position-space wavefunction wasn't bounded inside a certain area. Yes, but that's also true for a hydrogen atom. $\endgroup$
    – user4552
    Commented Aug 17, 2014 at 21:14
  • $\begingroup$ So, is it "energy levels" that bind the quarks, or simple probability? $\endgroup$
    – HDE 226868
    Commented Aug 17, 2014 at 23:05

1 Answer 1

Reset to default
1
$\begingroup$

A free quark wavefunction does not exist. Instead, inside nucleons quarks are relativistic and asymptotically free, which means that they only behave like individual particles for sufficiently large momentum/energy transfer.

Imagine a classical solid state analog: if you apply a very small, slowly acting force to a single atom of a crystal, the force will apply to all of the crystal, as long as the atom stays in place. What matters for the dynamics, in this case, is the mass of the whole crystal, not the mass of the atom. You can basically move a whole crystal my pushing on a single atom... but only very slowly.

Now make that force a very fast, strong acting one: the atom gets now dislocated from the crystal. The effective dynamics will be very different: the force will move the atom, but the crystal, as a whole, will not experience the full force and it will not move with the atom.

The quantum mechanical situation in a nucleon is similar: a high energy, high momentum transfer interaction can "kick" a part of the nucleon that behaves like an "individual" quark, and this can be approximated with a single particle interaction. In contrast, a low energy interaction will not just move the quark, but it will also change the other constituents of the nucleon, so one can't chose a simple single particle wavefunction approximation.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.