# Constant term in Lagrangian

Why can we always drop any constant term in a Lagrangian density in quantum field theory?

This issue is somehow related to the constant term being some kind of cosmological constant.

Can you please explain this issue?

• @LoganM In this answer I was mainly thinking of the operator formalism. In the path integral formalism, a constant term in the action factors out right through the integral, $\int \mathrm e^{-S_0-S[\phi]}\mathrm d\phi=\mathrm e^{-S_0}\int\mathrm e^{-S[\phi]}\mathrm d\phi$, which has no effect on correlators (because these are defined as quotients of path integrals). – AccidentalFourierTransform Jun 28 '18 at 16:22
• I misunderstood what you meant by operator formalism. I thought you were talking about defining QFT based on the algebra of OPEs; in this case one does not even generically have a Lagrangian (but if one does then a constant term does not change anything). I now think you're talking about what I would call "canonical quantization", but this is a rather pathological procedure because of the ordering ambiguities you mention. They are only guaranteed to be unimportant as $\hbar\rightarrow 0$, which is why canonical quantization fundamentally can't say much about highly off-shell processes. – Logan M Jun 28 '18 at 16:48