- 42

- 8

ψ→ψ' = e

^{[- i q θ(x)]}ψ.

We then have to introduce a abelian gauge field, A

_{μ}(x) to construct a covariant derivative,

D

_{μ}ψ = ∂

_{μ}ψ + i q A

_{μ}ψ

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

D

_{μ}ψ → D'

_{μ}ψ' = e

^{[- i q θ(x)]}D

_{μ}ψ.

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

A

_{μ}→ A'

_{μ}= A

_{μ}+ ∂

_{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

D

_{μ}ψ = ∂

_{μ}ψ + i g A

_{μ}ψ,

D

_{μ}ψ → D'

_{μ}ψ' = U (D

_{μ}ψ).

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

A

_{μ}→ A'

_{μ}= U A

_{μ}U

^{-1}+ (i/g)(∂

_{μ}U) U

^{-1}.

Infinitesimally,

A

_{μ}→ A'

_{μ}= A

_{μ}- g [A

_{μ}, T

_{a}] θ

^{a}+T

_{a}∂

_{μ}θ

^{a}.

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?