Charge Quantization: Exploring e, Q, g & Weak Charges

[JustinLevy]

I'd like to focus in on some info from a previous thread that seemed too good to pass up https://www.physicsforums.com/showthread.php?p=3156965

My guess was that JustinLevy didn't see the distinction between "charge e" and "charge Q".

One must distinguish between the coupling constant e, g, ... in QED, QCD, ... which could have any value, and the charge Q, Q^a, ... as qm generators of U(1), SU(n), ... The latter one is quantized in the sense of the first Casimir Q^aQ^a. But how is this related to the coupling constant? The charge operator in QCD is something like

$$Q^a = \int d^3x\, g\,\bar{\psi}_i (T^a)_{ik}\psi_k$$

The Casimir operator has a discrete spectrum, but still g is an arbitrary multiplicative constant which is not "quantized"

Yes I was making that confusion, and I'd like to understand this a bit better. I have three questions to follow up if you don't mind.

Is there a technical name for these "charge e" and "charge Q", to help distinguish them? I'm realizing now looking back that I've made this mistake before, and if there are technical terms to help distinguish them it would be great.

For the electromagnetic force, is the "g" coupling the same for all particles, and that is why the "Q" quantization leads to "charge e" quantization? That makes it seems much less mysterious, but then I don't understand why people hope to find a magnetic monopole to 'help explain' charge quantization. Maybe I am missing or mixing up things again.

I never see "weak charges" listed for particles, so this is probably a naive question: Is the "weak charge" (the one equivalent to the "charge e") quantized for the weak force -- Or does the symmetry breaking ruin this?

 

[tom.stoer]

Hi Justin,

I'll try to answer the questions you raised.

'e' or in general 'g' are called 'coupling constants'. Q, or in non-Abelian gauge theories the operators Q^a are called 'charges'. For Q^a on can derive a quantization rule due to the commutator algebra

$$[Q^a, Q^b] = if^{abc}Q^c$$

From this relation, which reflects directly the classical algebra, one can derive (again algebraically) that there is "charge quantization" just like angular momentum quantization. But for abelian gauge theories (QED with U(1)) there is no such relation, therefore charge quantization in U(1) cannot be derived from this fact.

For the eletromagnetic coupling constant e there is an argument based on magnetic monopoles that e must be quantized. But this is totally different from the above mentioned considerations.

 

[JustinLevy]

To see if I understand: So for the strong force, the Q^a are quantized. But the "color coupling" need not be quantized.

But is the color coupling actually quantized anyway (like the electric charge or particles)? Or does each quark have a different color coupling (or would that violate the symmetry)?

But for abelian gauge theories (QED with U(1)) there is no such relation, therefore charge quantization in U(1) cannot be derived from this fact.

Because the electroweak model starts from SU(2)xU(1), and after breaking the generator of the electric U(1) is actually a combination of the original SU(2) and U(1) generators ... does that original "heritage" containing part of SU(2) allow it to get charge quantization for the electric interaction?

Also, since the weak force is a broken symmetry, does this mean its charges aren't properly quantized?

 

[tom.stoer]

To see if I understand: So for the strong force, the Q^a are quantized. But the "color coupling" need not be quantized.

But is the color coupling actually quantized anyway (like the electric charge or particles)? Or does each quark have a different color coupling (or would that violate the symmetry)?

The couplings are not quantized. The color charges in QCD are quantized; the electric charges of quarks are quantized. The color couplings of all quarks are identical. The coupling in the Dirac term looks like

$$\bar{\psi}^i_f\gamma^\mu(D_\mu)^{ik}\psi^k_f$$

with

$$(D_\mu)^{ik} = \partial_\mu\delta^{ik} - igA_\mu^{ik}$$

and

$$A_\mu^{ik} = A_\mu^a (T^a)^{ik}$$

You see that there's a summation over the colors i, k and over the flavors f. There is one single coupling g.

This structure is replicated in QED (where one simply drops i,k) and to the electro-weak theory.

Also, since the weak force is a broken symmetry, does this mean its charges aren't properly quantized?

Roughly speaking all arguments apply to the electro-weak case as well as the algebraic structure is not affected by the symmetry breaking which happens at the level of the state vectors (which we haven't discussed so far), not at the level of the operators.

 

[JustinLevy]

...

$$\bar{\psi}^i_f\gamma^\mu(D_\mu)^{ik}\psi^k_f$$

with

$$(D_\mu)^{ik} = \partial_\mu\delta^{ik} - igA_\mu^{ik}$$

...
You see that there's a summation over the colors i, k and over the flavors f. There is one single coupling g.

Based on how the terms are written there, I don't see why the coupling is necessarily the same for each flavor. Is there something in the math that requires this, or is this just an experimental fact?

 

[tom.stoer]

If you re-sort the terms you get

$$-ig \bar{\psi}^i_f \gamma^\mu A^(ik)_\mu \psi^k_f$$

with a sum over the colors i,k and the flavors f. You see that the gauge field A does not carry flavor (which is natural as flavors are quark-attributes) and that g doesn't, either. If you would now allow for flavor dependent coupling g_f you can no longer compensate the local gauge transformations acting on the fermions with a gauge transformation of a flavor-neutral gauge field. So using a flavor-dependent g_f would automatically mix gauge and flavor group. Doing this would mean "gauging flavor" and would result in a different theory - which we do not observe in nature.

So "gauge invariance under color-SU(3) and non-gauged flavor-SU(6)" together require that the total symmetry "factorizes" and that color and flavor must not mix. These facts are experimentally verified and mathematically well-understood. So it's the combination of a mathematical principle with the application of this principle to observations = nature which requires this.