0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.