# A Different gauge coupling constants

1. Dec 5, 2018

### Roy_1981

Hi all, I am sure I am missing something really elementary, but I would really appreciate someone pointing it out to me. So, if we consider the situation in abelian gauge symmetry, say for fermion matter ψ, of charge q. The transformation law for ψ is,

ψ→ψ' = e[- i q θ(x)] ψ.

We then have to introduce a abelian gauge field, Aμ(x) to construct a covariant derivative,

Dμ ψ = ∂μ ψ + i q Aμ ψ

in order to make the kinetic term in the matter action gauge invariant:

Dμ ψ → D'μ ψ' = e[- i q θ(x)] Dμ ψ.

From this equation, the gauge transformation of the gauge field follows,

Aμ → A'μ = Aμ + ∂μ θ.

Thus the gauge transformation for the gauge field is independent of the charge of the matter, q. In principle I could have started from a fermion or scalar with a different charge, say q' and the gauge field transformation equation won't care about that. This is all nice!

However, things change the moment I replicate the steps for a non-abelian gauge symmetry. The transformation law for the matter is now,

ψ→ψ' = U ψ, U = e[- i g Ta θa(x)]

with g playing the role of charge q. Then we introduce the gauge-covariant derivative

Dμ ψ = ∂μ ψ + i g Aμ ψ,

Dμ ψ → D'μ ψ' = U (Dμ ψ).

This results in the gauge transformation law for the gauge field itself,

Aμ → A'μ = U Aμ U-1 + (i/g)(∂μ U) U-1 .

Infinitesimally,

Aμ → A'μ = Aμ - g [Aμ, Ta] θa +Taμ θa .

Now this looks not so nice because the color charge of the matter, g appears explicitly in the gauge field transformation law. Had we started from a matter field with a different color charge, g', then one would be led to a different gauge field transformation law!

Does it mean unlike in the abelian case there can be only one common color charge g for all color-charged particles in the same representation? In electrodynamics electron and proton can in principle have different charge, at least classically. What is the problem here?

2. Dec 7, 2018

### Orodruin

Staff Emeritus
Yes.

The gauge coupling constant must also be the same for all representations, although the exact coupling relations will depend on the representation (the term appearing in the covariant derivative is proportional to $g A_\mu^a \tau^a_r$, where $\tau^a_r$ is the representation of the gauge group generator.

The point is that the gauge self-interactions set the size of the coupling whereas they are absent for an abelian theory.

Technically, particles with different charge in an Abelian theory are in different representations.

3. Dec 7, 2018

### Roy_1981

Thanks, this is what I was guessing as well (that the gauge self-coupling fixes the coupling with everything else and in the abelian theory there is no self-coupling to constrain other couplings) but the following sentence confuses me:

Technically, particles with different charge in an Abelian theory are in different representations.

In the U(1) theory, the generator is identity and the representation is one-dimensional e.g. the proton and electron, and hence they are in the same representation (the one-dimensional e-i q θ(x)). In what sense different values of q's are different representations, while being the same one-dimensional matrix rep.? Usually when we talk abt different representations, we mean different matrix ranks. Here in the case of the two fermions both are in the one-dimensional rep. Do you mean the generator is not identity but the charge times the identity q*1? I appreciate your response.

4. Dec 7, 2018

### Orodruin

Staff Emeritus
Since the Lie algebra of an Abelian theory is commutative (every Lie bracket is zero), you can multiply the representation of the generator with an arbitrary constant and it will still satisfy the same commutation relations and vector space structure. Since this will mean that the corresponding fields will transform differently under the gauge transformation, they will be in different representations. There is nothing* stopping you from assigning different one-dimensional representations to different fields, but once you have assigned it to one the relative charges can be determined by experiment.

There are also cases of non-Abelian groups where you have different representations of the same dimension, the most striking example being SU(3) where $\bf 3$ and $\bar{\bf 3}$ are different three-dimensional representations.

* Unless you start looking at the cancellation of quantum anomalies.

5. Dec 7, 2018

### Roy_1981

Since the Lie algebra of an Abelian theory is commutative (every Lie bracket is zero), you can multiply the representation of the generator with an arbitrary constant and it will still satisfy the same commutation relations and vector space structure.

Actually this is a bit more subtle. Even for non-vanishing commutators, you can change basis and have different structure constants/ commutation relations. There is no canonical version for Lie algebra structure constants and it depends on how you choose a basis. If you want, you can rescale all the generators and get different structure constants/ different commutation relations but the representation remains the same. In particular for the abelian case, scaling up the generator (unity) would mean you remain in the same vector space (representation). Do you agree?

6. Dec 7, 2018

### Orodruin

Staff Emeritus
No, I do not agree. I agree that it does not matter what generators you choose, but given a particular set of generators the structure constants are fixed and so must be the same for each representation. Each fixed generator must map to a particular element of the Lie algebra (up to unitary equivalence) in order to achieve this. This is not the case when all the structure constants are zero.

7. Dec 7, 2018

### Roy_1981

I agree that it does not matter what generators you choose, but given a particular set of generators the structure constants are fixed and so must be the same for each representation. Each fixed generator must map to a particular element of the Lie algebra (up to unitary equivalence) in order to achieve this. This is not the case when all the structure constants are zero.

But, different vector spaces/ representations are supposed to represent different physical particles. Then how does scaling a generator by a constant (as you are suggesting for the U(1) case by multiplying it with a constant q) represent a change in the vector space itself? A generator Ta and αTa are not equivalent in the sense you cannot find a similarity transformation STaS-1 = α Ta, correct? Yet, all I have done is scale the generator (a linear transformation of the same vector space). I am having difficulty in reconciling scaling of (a set of) generator(s) to lead to/give a new representation (a new vector space).

8. Dec 7, 2018

### Orodruin

Staff Emeritus
You are not scaling the generator of the actual Lie algebra, you are changing how it is represented as an operator on the vector space.

No, this is not what you have done. Your generator is still the same, it is the representation of that generator on the vector space in which the fermion lives that has changed. For example, in QED you can (in principle, let us forget quantum anomalies for the moment) have particles of different charge. Those particles are going to transform differently under U(1) transformations, meaning that they belong to vector spaces with different representations of the gauge group.

Also, please use the quote function instead of just copying text. Otherwise it appears as if you are saying something that I actually said. This is typically seen misrepresentation and is anyway a bad practice.

9. Dec 7, 2018

### Roy_1981

"in QED you can (in principle, let us forget quantum anomalies for the moment) have particles of different charge. Those particles are going to transform differently under U(1) transformations, meaning that they belong to vector spaces with different representations of the gauge group."

I see how I am misreading the situation. Indeed, they are different one-dimensional vector spaces with different representation of generators, none of them are unity, they are the respective q's. Nothing to do with scaling the generators or commutators. Thanks for your inputs.