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I Mesons and baryons written in terms of quarks

  1. Jun 18, 2016 #1
    Hello, guys.

    I have not understood what it means when one writes ##\pi^+=u\bar{d}##, for example. I though it simply meant that the ##\pi^+## meson was composed of one up-quark and one anti-down-quark. However, that doesn't explain what writing ##\pi^0=\frac{1}{\sqrt{2}}(d\bar{d}-u\bar{u})## means. I'd say that writing ##\pi^+=u\bar{d}## means that the ##\pi^+## meson's wavefunction is the tensor product of the wavefunctions of ##u## and ##\bar{d}##. Is that it?

    A related question: how can one draw the feynman diagram for something like ##\pi^0\rightarrow A+B##? My problem is that I usually start by writing the quarks which constitute each particle, but in this case the left hand side particle is a composition of two "states". Shall I draw one feynman diagram for ## d\bar{d}## and another one for ##u\bar{u}##?
     
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  3. Jun 18, 2016 #2

    mfb

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    Staff: Mentor

    The ##\pi^0## is something like a superposition of ##d \bar d## and ##u \bar u##. Those are just the valence quarks, however - hadrons are complex objects, and interpreting them as just their valence quark content doesn't give an accurate description.

    For Feynman diagrams involving neutral pions, choose either ##d \bar d## or ##u \bar u##.
     
  4. Jun 20, 2016 #3

    vanhees71

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    The hadrons are ordered according to some (approximate) symmetries, related to QCD. In the case of the light hadrons (the part which consists only of u and d quarks and anti-quarks) this approximate symmetry is the chiral ##\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R## symmetry, which is in the vacuum and at low temperatures and densities spontaneously broken by the formation of the quark condensate, which leads to a split in the masses of otherwise degenerate chiral-partner hadrons. It is also slightly explicitly broken by the finite quark masses.

    The spontaneous breaking of the symmetry leads to the observation that only (approximate) isospin symmetry, i.e., the vector part ##SU(2)_V## is explicit, and that thus there must be 3 psuedoscalar Goldstone modes, and these are the pions. They are isovectors and thus given by the combinations
    $$\vec{\pi}=\bar{\psi} \vec{\tau} \psi,$$
    where ##\psi=(u,d)## is the isospin doublet of quarks (##u## and ##d## being both Dirac-spinor fields). ##\vec{\tau}## are the three Pauli matrices in this isospin space.

    Another important quantum number is the electric charge, and the diagonalization to the charge eigenstates leads to
    $$\pi_{\pm} = \pi_1 \pm \mathrm{i} \pi_2, \quad \pi_0=\pi_3.$$
    From this it's easy to read off the quark content and the basic structure of the wave functions in terms of the naive quark model.
     
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