A Gauge transformation of gauge fields in the adjoint representation

1. Mar 23, 2017

spaghetti3451

In some Yang-Mills theory with gauge group $G$, the gauge fields $A_{\mu}^{a}$ transform as
$$A_{\mu}^{a} \to A_{\mu}^{a} \pm \partial_{\mu}\theta^{a} \pm f^{abc}A_{\mu}^{b}\theta^{c}$$
$$A_{\mu}^{a} \to A_{\mu}^{a} \pm \left(\partial_{\mu}\theta^{a}-A_{\mu}^{b}f^{bac}\theta^{c}\right)$$
$$A_{\mu}^{a} \to A_{\mu}^{a} \pm \left(\partial_{\mu}\theta^{a}-iA_{\mu}^{b}(T^{b}_{\text{adj}})^{ac}\theta^{c}\right),$$

where $T^{a}_{\text{adj}}$ is the adjoint representation of the gauge group $G$ and the gauge parameters $\theta^{a}$ are seen to transform in the adjoint representation of the gauge group $G$.

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Why does this mean that the gauge fields $A_{\mu}^{a}$ transform in the adjoint representation?

Should the transformation of the gauge fields $A_{\mu}^{a}$ in the adjoint representation not be given by

$$A_{\mu}^{a} \to A_{\mu}^{a} \pm i\theta^{b}(T^{b}_{\text{adj}})^{ac}A_{\mu}^{c}?$$

2. Mar 28, 2017

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Mar 29, 2017

dextercioby

No, not again the $\pm$. :(. What book are you reading?

4. Mar 30, 2017

vanhees71

This sign makes me already crazy, because of course there are the two choices to define the covariant derivative, and of course physicists use both in the literature, but to write always both is just confuses the issue without any additional value. So just use one sign,
$$\mathrm{D}_{\mu}=\partial_{\mu} + \mathrm{i} g \mathcal{A}_{\mu},$$
where $\mathrm{i} \mathcal{A}_{\mu}$ is in the Lie algebra of the gauge group (the imaginary $\mathrm{i}$ in most of the literature comes from the fact that physicists prefer hermitean over antihermitean matrices, one example for an exception is the textbook by Itzykson and Zuber, who use antihermitean gauge fields).

In QED you usually have $g=-\mathrm{e}$, because electrons are negatively charged. There the $-$ makes sense to me :-), but it's of course completely arbitrary, which sign convention you choose, and no physics is changed by flipping this sign. In (naive) perturbation theory you get anyway only results with $\alpha=g^2/4 \pi$ (for QED it's $\alpha_{\text{em}}=e^2/4 \pi \simeq 1/137$).