4
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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"

$\endgroup$
  • $\begingroup$ Thank you for your answer! Do you have references for the "external sources ... as Wilson Loops"? I would really like to read more about that. Regarding the Wong equations: so you're saying that if the current is generated by other degrees of freedom, for instance by classical point particles with color charge, then the action is gauge invariant again? That would make sense, because it would be similar to the QCD case, where the offending term in my post is canceled by a term from the fermions. $\endgroup$ – user9564 Nov 29 '17 at 9:38
  • $\begingroup$ @user9564 The is a Wikipedia article on Wilson lines, that contains some references. Wong''s original paper is: Wong, S.K., Nuovo Cimento 65A, 689–694 (1970), but I think that there are a number of freely available papers on the ArXiv that are more accessible. A Google search will find them. $\endgroup$ – mike stone Nov 29 '17 at 15:14
0
$\begingroup$

Though mike stone have given a excellent answer, I want to collect some basic facts here. For Yang-Mills with external current $$S_{\rm YM}= \int d^4x\Big[-\frac{1}{4}F^a_{\mu\nu}F^{\mu\nu a}-J^{\mu a}A_\mu^a\Big]$$ The gauge transformation is $$A_\mu(x)\,\,\, \longrightarrow \,\,\,\Omega A_\mu(x) \Omega^{-1}+\frac{i}{g} \Omega \partial_\mu\Omega^{-1}$$ and its infinitesimal form is $$A_\mu(x) \,\,\,\longrightarrow \,\,\, A_\mu(x)-\partial_\mu\omega +ig[A_\mu(x),\,\omega(x)]$$ The classical equation of motion is \begin{align} D_\mu F^{\mu\nu }=J^{\nu} \,\,\,\,\,\hbox{or}\,\,\,\,\,\, D_\mu F^{\mu\nu a }=J^{\nu a} \end{align} which implies the gauge transformation of the external current: \begin{align} J^{\mu}\,\,\,\longrightarrow\,\,\,\Omega J^{\mu} \Omega^{-1} \end{align} since \begin{align} F^{\mu\nu}\,\,\,\longrightarrow\,\,\,\Omega F^{\mu\nu} \Omega^{-1}, \,\,\,\,\,\,\,\,\, D_\alpha F^{\mu\nu}\,\,\,\longrightarrow\,\,\,\Omega D_\alpha F^{\mu\nu} \Omega^{-1} \end{align} For the external current, we have \begin{align} D_\nu J^{\nu}=D_\nu D_\mu F^{\mu\nu }=-\frac{1}{2}[D_\mu,\,D_\nu]F^{\mu\nu} =\frac{ig}{2}[F_{\mu\nu},\,F^{\mu\nu}]=0 \end{align}

Gauge invariance of ${\rm tr} (j^\mu A_\mu)$ requires \begin{align} \partial_\mu J^\mu=0 \end{align} so we need independent conditions \begin{align} \partial_\mu J^\mu=0,\,\,\,\,\,\,\,\,\, [A_\mu,\,J^\mu]=0 \end{align} Thus at some extend we make the external currrent term gauge invariance.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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