While quantizing a non-Abelian gauge theory covariantly, we first demand that the BRST charge acting on the physical states of the Hilbert space must be zero. However such physical states still have an unequal number of ghost and antighost particles and as the book is claiming, longitudinal bosons. To get rid of them he then applies the ghost number operator and picks out those states from the physical Hilbert space which are invariant under it.

So here's my question. I've often come across the statement that BRST symmetry is somehow related to gauge invariance. Is this true? If it is, why do you impose the further requirement that the gauge bosons must only be transversely polarized? I mean shouldn't the BRST invariance (which implies gauge invariance which again implies transverse polarizations) be enough to guarantee that?

  • 1
    $\begingroup$ Is this from some reference? $\endgroup$
    – Qmechanic
    Commented Dec 12, 2019 at 18:33
  • $\begingroup$ Yup, I found it in Ashok Das's book on QFT. Page number 559, right under equation 13.49. $\endgroup$ Commented Dec 12, 2019 at 18:35

1 Answer 1

  1. Ref. 1 only shows [via a non-abelian generalization of the abelian eq. (13.31)] that the physical Hilbert space ${\cal H}_{\rm phys}$ [defined to be BRST-closed and have zero ghost number] can not have longitudinal gluons by using the Lorenz gauge condition and the EL equation for the Lautrup-Nakanishi auxiliary field $F^a$.

    Longitudinal gluons are possible with other gauge-fixing conditions.

  2. For the connection between gauge and BRST symmetry, see e.g. this Phys.SE post.


  1. A. Das, Lectures on QFT, (2008); chapter 13.

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