I have a short question about the possible gluon polarization in loop diagrams. For external gluons, we only want the 2 transverse polarizations.

In Peskin-Schroeder it is explained that in Feynman-gauge we need the ghost-loops to cancel the longitudinal polarization states in gluon loops because otherwise unitarity would be violated, which is shown by using the optical theorem on p.512.

But since ghosts do not couple to quarks, how does the argument work when there are loops involving a gluon with a quark, for example the quark self energy? How are the longitudinal polarizations cancelled there? Thanks a lot :).

  • $\begingroup$ The amplitude to produce a single longitudinally polarised gluon is zero (check). $\endgroup$
    – CAF
    Jun 7 '19 at 7:04

The amplitude to produce a pair of longitudinally polarised gluons is a gauge dependent statement - in covariant gauges the unphysical degrees of freedom propagating in loops are indeed removed by ghosts. The generic amplitude for a gluon pair production can be written as $\mathcal M = \epsilon^*_{\mu} \epsilon^*_{\nu} \mathcal M^{\mu \nu}$. One can show that, in such gauges, $$ \begin{cases} \epsilon^*_{+,\mu} \epsilon^*_{+, \nu} \mathcal M^{\mu \nu} \neq 0 \\ \epsilon^*_{-,\mu} \epsilon^*_{-, \nu} \mathcal M^{\mu \nu} \neq 0 \\ \epsilon^*_{+,\mu} \epsilon^*_{-, \nu} \mathcal M^{\mu \nu} = 0 \\ \epsilon^*_{-,\mu} \epsilon^*_{+, \nu} \mathcal M^{\mu \nu} = 0, \end{cases}$$ where $\epsilon^*_{\pm}$ refers to transverse (+) / longitudinal (-) polarisations.

In your question, are you considering the amplitude to produce a ${\it single}$ longitudinally polarised gluon in e.g. $q \rightarrow qg$? In that case, you have $$\mathcal M = \epsilon^*_{\mu} \mathcal M^{\mu} \sim k_{\mu} \mathcal M^{\mu}$$ which vanishes identically (check). That is, there is no amplitude to produce a ${\it single}$ longitudinally polarised gluon.


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