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?


1 Answer 1


You can drop a constant term in the Lagrangian because it doesn't affect the Euler-Lagrange equations, and therefore the equations of motion are the same.

A constant term does have an effect in the actual value of, say, the Hamiltonian (and the rest of Noether charges). But you can only measure differences in energies, and therefore a constant term in the Hamiltonian is again irrelevant (the same can be said about all the scalar Noether charges; if a charge has a Lorentz index then the constant term vanishes by symmetry). In other words, a constant term doesn't affect predictions of conserved operators.

If you include gravity, then a constant term becomes relevant, and you cannot drop it. But the quantisation program doesn't specify a canonical way to choose a constant term: we must resort to experiments to measure it (in other words, the cosmological constant, like most parameters in a theory, cannot be predicted but is an input to the model instead). Or put it another way: if you consider gravity, then the vacuum energy is no longer irrelevant, but it is still arbitrary: no theory can predict its value (as far as we now today).

  • $\begingroup$ It is true that in QFT a constant term is unimportant, but the reason isn't that the EoM are unchanged. In QFT the EoM are not imposed; rather they are stationary points of the action and so in the semiclassical limit the EoM are satisfied to a good approximation. But in general we need to path-integrate over all field configurations, not only the ones satisfying the EoM. Different Lagrangians which lead to the same EoM can nonetheless correspond to different quantum theories, and some examples of this are well-known, e.g. the Nambu-Goto action vs Polyakov action in bosonic string theory. $\endgroup$
    – Logan M
    Jun 28, 2018 at 16:20
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    $\begingroup$ @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). $\endgroup$ Jun 28, 2018 at 16:22
  • $\begingroup$ Yes, it is trivial for exactly the reason you state, but I don't see what the operator formalism vs path integral formalism has to do with it. The operator formalism still needs off-shell information, even in finite-order perturbation theory (as it must, of course, since the operator formalism can be derived from the path integral formalism when a path integral exists). $\endgroup$
    – Logan M
    Jun 28, 2018 at 16:38
  • $\begingroup$ @LoganM In the operator formalism, the operators satisfy the EL equations, up to ordering ambiguities. $\endgroup$ Jun 28, 2018 at 16:41
  • $\begingroup$ 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. $\endgroup$
    – Logan M
    Jun 28, 2018 at 16:48

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