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.


1 Answer 1


$A_\mu = A^a_\mu T_a$ solves your problem ($A_\mu$ is a matrix-valued field). Thus, in contrast to your claim, $U A_\mu U^\dagger \ne A_\mu$.

  • $\begingroup$ Thanks for the observation, I still can't get what I want. For the first term, I got $$ U A_{\mu} U^{\dagger} = A_{\mu} + \theta^{c} f^{b c d} T^{d} A_{\mu}^{b} $$ $\endgroup$ Commented Feb 26 at 6:06
  • 1
    $\begingroup$ @DavidLazaro Simply read off the transformation of the field components by writing $A_\mu = T^d A^d_\mu$. $\endgroup$
    – Hyperon
    Commented Feb 26 at 6:23

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