# Infinitesimal transformation of the Yang-Mills field

I am trying to obtain the infinitesimal transformation for the Yang-Mills field $$A_{\mu}$$. I want to show that

$$A^{\prime a}_\mu=A_\mu^a-\partial_\mu \theta^a-g_s f^{a b c} \theta^b A_\mu^c$$

For them I used the finite transformation:

$$A^{\prime}_{\mu} = U A_{\mu} U^{\dagger} + \frac{i}{g} U^{\dagger} \partial_{\mu} U$$

where $$U = e^{i \theta^{a} T^{a}} \approx 1 + i \theta^{a} T^{a}$$. For the first term I obtain its infinitesimal expression

$$U A_{\mu} U^{\dagger} = ( 1 + i \theta^{a} T^{a}) A_{\mu} (1 - i \theta^{a} T^{a})$$ $$= A_{\mu} - i A_{\mu} \theta^{a} T^{a} + i A_{\mu} \theta^{a} T^{a} + \mathcal{O}(\theta^{2}) = A_{\mu}$$

However, for the second term I only get to

$$\frac{i}{g} U^{\dagger} \partial_{\mu} U = - \frac{1}{g} U^{\dagger} \partial_{\mu} \theta^{a} T^{a} U = \frac{-1}{g} \partial_{\mu} \theta^{a} ( 1 - i \theta^{b} T^{b}) T^{a} \left( 1 + \theta^{a} T^{a} \right)$$ $$= \frac{-1}{g} \partial_{\mu} \theta^{a} ( T^{a} + i \theta T^{a} T^{c} - i \theta^{b} T^{b} T^{a} + \mathcal{O}(\theta^{2}) )$$ $$- \frac{1}{g} \partial_{\mu} \theta (T^{a} + i \theta^{c} (T^{a} T^{c} - T^{c} T^{a})$$

using the property $$\left[ T^{a}, T^{c} \right] = i f^{acb} T^{c}$$

$$= - \frac{1}{g} \partial_{\mu} \theta^{a} \left(T^{a} + i \theta^{c} ( i f^{acb} T^{c} ) \right) = - \frac{1}{g} \partial_{\mu} \theta^{a} (T^{a} - \theta^{c} f^{acb} T^{c} )$$

However I don't get the correct infinitesimal transformation. Could anyone give me any suggestions or observations please? Likewise, I am looking for how to deduce the following relationship but I cannot find how $$A_{\mu \nu} = \frac{1}{g} \left[ D_{\mu}, D_{\nu}\right]$$ where $$D_{\mu \nu}$$ is the covariant derivative.

$$A_\mu = A^a_\mu T_a$$ solves your problem ($$A_\mu$$ is a matrix-valued field). Thus, in contrast to your claim, $$U A_\mu U^\dagger \ne A_\mu$$.
• Thanks for the observation, I still can't get what I want. For the first term, I got $$U A_{\mu} U^{\dagger} = A_{\mu} + \theta^{c} f^{b c d} T^{d} A_{\mu}^{b}$$ Commented Feb 26 at 6:06
• @DavidLazaro Simply read off the transformation of the field components by writing $A_\mu = T^d A^d_\mu$. Commented Feb 26 at 6:23