Non-Abelian gauge theory action with external currents: gauge invariance?

Question

I am currently trying to work out the gauge invariance of the Yang-Mills action coupled to an external source and I noticed something weird.

The Yang-Mills action in presence of an external source $J^a_\mu$ reads

$S = \intop_x - \frac{1}{4} F^{a\mu\nu}F^a_{\mu\nu} + J^{a\mu} A^a_\mu = S_{YM} + S_{J}$.

The pure Yang-Mills part $S_{YM}$ is invariant under gauge transformations $V(x) \in \text{SU(N)}$ via

$\mathcal{A}_\mu = A^a_\mu t^a \rightarrow V \mathcal{A}_\mu V^\dagger - i V \partial_\mu V^\dagger$.

However, $S_J$, the part where we coupling to the external source happens, is not gauge-invariant. I know that the source should transform gauge covariantly via

$\mathcal{J_\mu} = J^a_\mu t^a \rightarrow V \mathcal{J}_\mu V^\dagger$,

but this does not render $S_J$ gauge-invariant. In fact I obtain

$S_J \rightarrow S_J - 2i\intop_x tr(V \mathcal{J_\mu} \partial^\mu V^\dagger)$,

which does not vanish. What's exactly happening here?

Note: if we look at the Abelian case and repeat the same steps we could actually make the action invariant by requiring charge conservation $\partial_\mu J^\mu=0$.

I know that the QCD action is gauge-invariant and in that case I obtain an extra term from the matter fields which cancels against terms from $S_J$. Does this mean that Yang-Mills with external sources is meaningless? Does Yang-Mills only make sense if the current $J^a_\mu$ comes from "dynamical" fields like quarks?

Answer 1

It is not possible to couple a conventional $c$-number source to a quantized non-abelian gauge field and maintain gauge invariance. For a current $J^\mu=\lambda^a J^\mu_a$, gauge invariance requires covariant current conservation $$ 0= \nabla_\mu J^\mu\equiv \partial_\mu J^\mu + [A_\mu,J^\mu]=0 $$ and this requires a $c$-number(the first term) to equal a term containing an the operator $A_\mu$. It is because of this problem that external sources are introduced into a gauge theory as Wilson Loops $$ W= P\exp\{ i\int A_\mu dx^\mu\} $$ where $P$ denoters a path-ordered integral of the matrix valued ($A_\mu = A_\mu^a \lambda^a$) gauge field.

It is possible to add a kind of classical source to a classical gauge action by using the "Wong equations" in which the internal gauge degrees of freedom in the source are replaced by classical dynamics in a "co-adjoint orbit"

Answer 2

The precise statement that you wanted to prove in your question is not correct, to my understanding.

‘Gauge invariance’ of the action with source: $$ \quad S(A,J)= \int d^4x [-\frac{1}{4}F^{a\mu\nu}F_{a\mu\nu} + J^{a\mu}A_{a\mu} ] = \int d^4x [-\frac{1}{4} Tr[F^{\mu\nu}F_{\mu\nu}] + Tr[J^{\mu}A_{\mu}] ] $$ does not mean $S(A, J) = S(A', J')$ where both $A_\mu$ and the source $J_\mu$ perform a transformation. It actually means that: $$ S(A,J) = S(A^U, J) $$ where only $A_\mu \to A^U_\mu$ performs a transformation, and the source $J_\mu$ remains unchanged!!!

You can check in the Faddeev-Popov method that this is what we really need.

Now I’m gonna try to prove the above. The first term in the action is of course gauge invariant and well known.

For the invariance of the second term, we can compute the infinitesimal transformation (Srednicki 71.7) $$ A_\mu(x) \to A_\mu(x) - D^{adjoint}_\mu \theta(x) $$ where $D^{adjoint}_\mu$ is the covariant derivative on the adjoint bundle. And use the similar argument as Stokes's theorem: $$ \delta(Tr(J^\mu A_\mu)) = - Tr(J^\mu D^{ad}_\mu \theta(x)) = - D^{ad}_\mu (Tr(J^\mu \theta(x))) + Tr((D^{ad}_\mu J^\mu) \theta(x)) $$ But I'm not sure whether my derivation is correct....