Take the tour ×  In the early days of quantum electrodynamics,
the most popular gauge chosen was the Gupta-Bleuler gauge
stating that for physical states, $$\langle \chi | \partial^\mu
A_\mu | \psi \rangle = 0.$$  However, this gauge is practically
never used now. Why? Is there anything wrong or inappropriate
with the Gupta-Bleuler gauge?  How is the Gupta-Bleuler gauge
related to the $R_\xi$ gauge?  share|improve this question I
think that you've got to emphasize that the gauge is not the
same as $\partial^\mu A_\mu=0$. Link to wikipedia will do. – 
Kostya Feb 10 '11 at 9:27 add comment  1 Answer  See:
http://en.wikipedia.org/wiki/BRST_formalism#Pre-
BRST_approaches_to_gauge_fixing  The short summary version:  The
Gupta-Bleuler gauge (Lorentz), like many others (Coulomb), works
fine for QED, but is technically challenging/impossible to push
to the non-Abelian case (of Yang-Mills theory). The modern view
point is via something called the BRST transform, which is
capable of dealing with the full general case.