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15
votes
2answers
2k views

How does non-Abelian gauge symmetry imply the quantization of the corresponding charges?

I read an unjustified treatment in a book, saying that in QED charge an not quantized by the gauge symmetry principle (which totally clear for me: Q the generator of $U(1)$ can be anything in $\mathbb{...
8
votes
1answer
703 views

Why is the gauge potential $A_{\mu}$ in the Lie algebra of the gauge group $G$?

If we have a general gauge group whose action is $$ \Phi(x) \rightarrow g(x)\Phi(x), $$ with $g\in G$. Then introducing the gauge covariant derivative $$ D_{\mu}\Phi(x) = (\partial_{\mu}+A_{\mu})\...
20
votes
2answers
3k views

Gauge fields — why are they traceless hermitian?

A gauge field is introduced in the theory to preserve local gauge invariance. And this field (matrix) is expanded in terms of the generators, which is possible because the gauge field is traceless ...
6
votes
2answers
673 views

Interpretation of the field strength tensor in Yang-Mills Theory

In Yang-Mills theory the field strength tensor $F_{\mu \nu}$ can be calculated as \begin{equation} F_{\mu\nu} \equiv \frac{i}{g} [D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu]...
7
votes
2answers
618 views

Why are the “coupling constants” constant?

The coupling constants (in the gauge theory) fix an inner product on the lie algebra of the gauge group and we use it to define strength of the fields. we are using ad-invariant inner products which ...
0
votes
0answers
27 views

Legal values of spin-1 field can take: $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, ..?

For the spin-1/ boson field $A_\mu$, we may choose it to be a vector which needs to be real $\mathbb{R}$ usually for photon field. The field strength $F= dA$ is also real. Same for the nonabelian ...