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- Thread starter ndung200790
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tom.stoer

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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".

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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)How can we ''build'' the Gauss law constraint with which the ''charge'' of physical states are zero?.

The electric field need not be zero; only the total charge must vanish.despite that,why in QED (example) the electric charge in many physical states are not equal zero.

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tom.stoer

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Then the electric charge of an electron is not the ''total charge''?

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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

[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

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