# Are the Yang-Mills equation and its generalization gauge invariant?

I have derived the Yang-Mills equation and its generalization coupled to a current of a scalar field $\phi$ by extremalizing the action describing a $\mathrm{SU}(2)$ scalar field gauge theory:

$$\partial_\mu F^{\mu\nu} +ig\left[ A_{\mu},F^{\mu\nu}\right] = j^\nu$$

where, $\phi$ is a two components scalar field,

$$j^\nu = -ig\left[ (D_\nu \phi)^\star T^a\phi -\phi^\star T^a(D_\nu \phi)\right]T^a$$

where $D_\nu = \partial_\nu + ig A_\nu$. But when I take a gauge transformation:

$$\phi' = e^{-i\omega_a T_a}\phi = U\phi, \quad A'_\mu = UA_\mu U^{-1} -\frac{i}{g}U\partial_\mu U^{-1}$$

I find I can not take same formalism from $j'^{\nu}$ as the original $j^{\nu}$. I think there must be something wrong with my calculation, because the current should be gauge invariant. My question is therefore whether the Yang-Mills equation and its generalization is gauge invariant and how one would show this invariance.

More about my calculation, please comment @ACuriousMind, Thanks you help and analysis.Now I will write down more of my calculation: When I calculate the $j^{\nu}$', I find that the term : $${\phi}^{*\alpha}{\partial _{\upsilon}\phi}^{\alpha }$$ always contains the quality: $${ U }^{ -1 }{ T }^{ a }U$$ as $${\phi}^{*\alpha}{\partial _{\upsilon}\phi}^{\alpha }{ U }^{ -1 }{ T }^{ a }U$$ the ${ U }^{ -1 }{ T }^{ a }U$can not be cancelled by the commutator calculation. Base on your above answer, can I think this phenomena is correct? I'm never work for QFT, and I learning the classical gauge field theory by my-self, please point out my mistake,Please, theanks.

No, the gauge current need not be gauge invariant, since it carries a group index in non-abelian theories. You should recall that both sides of the Yang-Mills equation (and therefore the current itself) are Lie-algebra valued and therefore transform in the adjoint representation. Not even the field strength $F^a_{\mu\nu}$ is gauge invariant, but transforms in the adjoint representation of the gauge group, which is why your action should (hopefully) contain only its trace as $\mathrm{Tr}(F_{\mu\nu}F^{\mu\nu})$.
• @alxandernashzhang: Not really. The equation of motion itself is gauge invariant, since both sides transform in the same representation, so the transformation can be cancelled. Your current should transform as $j_\mu \mapsto Uj_\mu U^{-1}$, if it doesn't, it's the wrong current. Also, please don't post comments as answer - either edit the additional information into the question, or use these comments. – ACuriousMind Oct 15 '14 at 12:26
• ,you mean,if we take a gauge transformation,the new current $j^{\nu}\prime$ must equal to $Uj^{\nu}U^{-1}$? – alxandernashzhang Oct 15 '14 at 12:35
• ,you mean,if we take a gauge transformation,the new current $j^{\nu}\prime$ must equal to $Uj^{\nu}U^{-1}$? And I always have another question about the trace in the energy density gauge field: Can I seem the trace as a standard Matrix trace act on the matrix of $F_{\mu\nu}F^{\mu\nu}$, and how to get the relation: $\frac { 1 }{ 2 } { F }_{ \mu \upsilon }^{ a }{ F }_{ a }^{ \mu \upsilon }=tr({ F }_{\mu \upsilon }{ F }^{ \mu \upsilon })$? I teach myself to learn gauge field theory, my supervisor always not investigate the area, so I have some basic problem like this.thanks – alxandernashzhang Oct 15 '14 at 12:45