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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?

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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

1 Answer 1


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.

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Just a few comments: the Gupta-Bleuler gauge is easy to impose in electrodynamics: negative frequencies are treated one way, positive ones in the other way. The field is linear and has no self-interactions. The latter condition is not satisfied for SU(2) and SU(3) gauge fields that are equally important in the Standard Model. So there's no simple variation of the Gupta-Bleuler method there, and one must use a more powerful method, e.g. BRST quantization, and then there's no reason not to treat the photons in the same logical way. Gupta-Bleuler is just obsolete. – Luboš Motl Feb 10 '11 at 13:28

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