Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chiral symmetry, quark condensate and self-energy

  1. Jul 10, 2009 #1
    Why in QCD chiral symmetry breaking study only the light quarks are taken into account? And why are their masses usually set to zero, Goldstone bosons are found, and then their masses are corrected by letting the quarks have non-zero mass? What happens if one study this symmetry breaking with massive quarks instead?

    And what exactly are quark condensates? I mean, I know they break the chiral symmetry because of their VEV, but what is their meaning?

    What is the relation between a particle's self-energy and its mass? (if any)

    Thanks in advance.
  2. jcsd
  3. Jul 10, 2009 #2
    I just posted a kinda similar question.

    Only the light quarks are taken into account because you need them to be approximately massless for there to be chiral symmetry. Also, when you do chiral perturbation theory, you do an expansion in terms of the meson mass - if this is too big, as it would be for the heavier quarks, then your expansion breaks down.

    The pions are actually pseudo-goldstone bosons, meaning the broken chiral symmetry isnt exact to begin with so the goldstone modes arent massless. Basically this is allowed because their mass is small enough that they can be thought of as "almost" being goldstone bosons in the sense above - the effects due to the meson mass are small.

    When you study the massless theory and find that the pions are the goldstone bosons of the broken chiral symmetry, and then put in realistic up and down masses, you are doing exactly what you said ie studying chiral symmetry with massive quarks. In chiral perturbation theory you do the same thing, that is you introduce a small explicit chiral symmetry breaking mass term.

    The quark condensate as you say is the non-zero VEV of q-bar q. It means that it is possible to form a state with lower energy than the empty vacuum by having a sea of bound states of quarks floating about. The empty vacuum then is unstable and will quickly "decay" into a state filled with a bunch of quarks and gluons. This will be the true ground state of the theory, ie the state of lowest energy.
  4. Jul 10, 2009 #3
    Is this idea of "approximate symmetry" and "pseudo-Goldstone bosons" rigorous? Better put: How rigorous are those ideas?

    Also, is there no chiral symmetry with massive quarks? (I'm kinda not used to the idea of a chiral symmetry)
  5. Jul 11, 2009 #4
    Well, we know of course that chiral symmetry is not exact even for the u and d because of their non-zero masses. But we also have quantiative measures of how wrong we are when we treat an already explicitly broken symmetry as one that gets spontaneously broken. Basically, if the explicity symmetry breaking parameters, ie the quark masses, are small compared to the symmetry breaking scale, then everything is rosy. The errors we introduce are small and we can still get a lot of useful information treating the theory in this way.

    If you want more information, this paper is pretty good:

    There is not even an approximate chiral symmetry of the heavy quarks in the sense above that their mass is too large compared to the symmetry breaking scale for this approximation to be useful.
  6. Jul 11, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook