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

In summary: Therefore, the total charge must vanish in physical states due to the Gauss law constraint. In summary, the discussion is about the color singlet states in hadrons and the reason why they are believed to exist. This is due to the Gauss law constraint, which states that all physical states must be singlets with respect to the "local algebra." This means that all physical states must be colorless, which is equivalent to having a vanishing angular momentum. While this is easy to deduce for single-hadron states, it becomes more difficult to explain for two-hadron states with two colored hadrons, leading to the concept of color confinement. The Gauss law constraint can also be applied to the electric charge in QED, leading to
  • #1
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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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  • #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''?
 
  • #3
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".
 
  • #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.
 
  • #5
ndung200790 said:
How can we ''build'' the Gauss law constraint with which the ''charge'' of physical states are zero?.
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)

ndung200790 said:
despite that,why in QED (example) the electric charge in many physical states are not equal zero.
The electric field need not be zero; only the total charge must vanish.
 
  • #6
What is ''total charge'' if it is not the ''charge'' that is the origin of electric field(the ''charge'' of U(1) symmetry?)?
 
  • #7
The source of the electric field is not the total charge but the charge density which can be non-zero, of course.
 
  • #8
Then the electric charge of an electron is not the ''total charge''?
 
  • #9
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
 
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1. What is a hadron in QCD?

A hadron in QCD (Quantum Chromodynamics) is a subatomic particle made up of quarks and gluons, which are the fundamental particles that make up protons, neutrons, and other particles. Hadrons are bound together by the strong force, one of the four fundamental forces of nature.

2. What is a color singlet in QCD?

A color singlet in QCD refers to a hadron that is color-neutral, meaning that the combination of its constituent quarks and gluons result in a total color charge of zero. This is necessary for the stability of the hadron, as individual quarks and gluons cannot exist in isolation due to the strong force.

3. How can we demonstrate that hadrons in QCD must be color singlets?

We can demonstrate this through experimental evidence, such as studying the scattering and decay of hadrons in high-energy particle collisions. The consistency of these observations with the theory of QCD, which predicts that hadrons must be color singlets, provides strong evidence for this principle.

4. Why is it important for hadrons in QCD to be color singlets?

It is important for hadrons in QCD to be color singlets because it is a fundamental property of the strong force and explains the stability of hadrons. This principle also allows us to understand and predict the behavior of hadrons in various high-energy interactions, such as in particle accelerators.

5. Are there any exceptions to the rule that hadrons in QCD must be color singlets?

While the majority of hadrons in QCD are indeed color singlets, there are some exceptions known as exotic hadrons. These are particles that have been observed to have properties that cannot be explained by the conventional quark model and may contain additional quark-antiquark pairs or gluons. However, these particles are still being studied and their exact nature is not fully understood.

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