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It is known that in QFT the Euler-Lagrange equations are used to obtain the equations of the quantum fields. Nevertheless, from the path integral's point of view (where you integrate over all $\it{paths}$/possible field configurations, even if they don't satisfy Euler-Lagrange) the use of this equations makes only sense if you are looking for the classical field, since this one satisfies the least action principle.

So, why do we use these equations to obtain the ones for the quantum field?

Vicky
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    "It is known that in QFT the Euler-Lagrange equations are used to obtain the equations of the quantum fields." - No such thing is known to me, so I do not understand what you are trying to ask. What do you mean by "equations of the quantum fields"? The quantum analogue of the classical E-L equations are the Schwinger-Dyson equations – ACuriousMind Jan 02 '19 at 21:22
  • The Euler-Lagrange equations are classical equations for classical fields. QFT evolves the fields in the same way QM evolves operators: they either do evolve (Heisenberg), don't (Schroedinger) or do but weirdly (interaction). – jacob1729 Jan 02 '19 at 21:23
  • @ACuriousMind For instance, by E-L eqs. applied to the Lagrangian of a scalar field, you can obtain the Klein-Gordon eq., the eq. of the field. – Vicky Jan 02 '19 at 21:38
  • ...and the problem with that is? You seem to want to say that we cannot use the KG equation because the fields the path integral integrates over do not obey it in general. That is correct - you cannot claim that the integration variable of the path integral obeys the KG equation. But I very much doubt that anyone else claims that. Please give an explicit example of a case where the KG equation is used and where you think it is not justified. – ACuriousMind Jan 02 '19 at 21:44
  • That's not my complain. What I'm asking is why do we use E-L eqs. in QFT if they are classical eqs. that we can only justify in classical mechanics by Hamilton principle. It doesn't have to hold in quantum physics, so how do we know E-L eqs. are valid to obtain the field eqs.? – Vicky Jan 02 '19 at 21:47
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    Possible duplicate: https://physics.stackexchange.com/q/234774/50583, see also https://physics.stackexchange.com/q/83113/50583 for caveats. – ACuriousMind Jan 02 '19 at 23:44

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