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How can we demonstrate hadrons in QCD must be color singlet?

  1. Sep 27, 2012 #1
    Coupling constant of strong interaction is only finitely greater than coupling constant in QED if the quarks insides hadron not very far apart each other,so I do not understand why there are not exist color multiplet states of hadrons,other hand I do not understand why the singlet states have minimum energy(ground states).
     
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  3. Sep 28, 2012 #2
    Now I have heard that we can only believe that there are exist only color singlet states in hadrons.Then what base is for that believing?Why we know that the color singlet states are ''colourless''?
     
  4. Sep 28, 2012 #3

    tom.stoer

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    Using the Gauss law constraint Ga(x) |phys.> = 0 one finds that all physical states must be singulets w.r.t. this "local algebra". Integrating the Ga(x) this becomes Qa (up to surface terms), so one immediately deduces Qa |phys.> = 0.

    Therefore all physical states are color singulets which is synonymous for colorless (a state with vanishing angular momentum is the same as an angular momentum singulet state). And hadrons are of course physical states.

    So it's rather simple to deduce that a single-hadron-state must be a color singulet. What's difficult is to find a reason why a two-hadron-state must consist of two singulets, i.e. why a two-hadron-state with two colored hadrons is forbidden. This is color confinement and is the "really hard problem of QCD".
     
  5. Sep 29, 2012 #4
    How can we ''build'' the Gauss law constraint with which the ''charge'' of physical states are zero?.Then I do not understand: despite that,why in QED (example) the electric charge in many physical states are not equal zero.
     
  6. Sep 29, 2012 #5

    tom.stoer

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    The Gauss law follows as an Euler-Lagrange equation in the A°=0 gauge (A° is unphysical b/c its conjugate momentum vanishes due to the absence of ∂°A° in the Lagrangian)

    The electric field need not be zero; only the total charge must vanish.
     
  7. Sep 29, 2012 #6
    What is ''total charge'' if it is not the ''charge'' that is the origin of electric field(the ''charge'' of U(1) symmetry?)?
     
  8. Sep 29, 2012 #7

    tom.stoer

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    The source of the electric field is not the total charge but the charge density which can be non-zero, of course.
     
  9. Sep 29, 2012 #8
    Then the electric charge of an electron is not the ''total charge''?
     
  10. Sep 29, 2012 #9

    tom.stoer

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    The Gauss law as a field equation reads

    [tex]G(x) = \nabla E(x) + \rho(x) \sim 0[/tex]

    As an equation of constraint acting on physical states |phys> this becomes

    [tex]G(x)\,|\text{phys}\rangle = 0[/tex]

    Integrating the Gauss law (and omitting boundary terms - hand-waving ;-) this becomes

    [tex]\int_{\mathbb{R}^3} d^3x\,G(x) = \int_{\mathbb{R}^3} d^3x\,[\nabla E(x) + \rho(x)] = \int_{\mathbb{R}^3} d^3x\,\rho(x) = Q[/tex]

    Now we apply this charge operator to physical states Q which is

    [tex]Q\,|\text{phys}\rangle = 0[/tex]

    b/c it's nothing else but the integrated Gauss law constraint.

    Therefore physical states have vanishing total charge, i.e. the eigenvalue of Q is zero in the physical Hilbert space. But they allow for non-vanishing electric fields and charge densities.

    And of course this means that a single electron is NOT an physical state!

    (the proof can be made more rigorous when treating boundary terms and operators more carefully)

    EDIT: of course Q is the charge related to the U(1) symmetry; and the Gauss law is related to the generator of local U(1) gauge transformations
     
    Last edited: Sep 29, 2012
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