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So I've started going down the QFT rabbit hole aided by Schwartz's book "Quantum Field Theory and the Standard Model". On chapter 7, the first method used to find the position-space Feynman Rules, is assuming that Quantum Fields respect the E-L equations (supported by the fact that it was shown in Chapter 2 that scalar fields in free theories do so), and using this to find the Schwinger-Dyson equations.

My question is, conceptually, why should Quantum Fields obey the Euler Lagrange Equations. Doing so suggests that some kind of action must be extremized, but in Quantum Mechanics, the path integral formulation shows us that the action isn't extremized at the quantum level, at least not in QM. Maybe this will be resolved once I get to chapter 14 where the book apparently explores path integrals in QFT, but the assumption still bothers me either way.

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2 Answers 2

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Remember non relativistic QM and the Heisenberg picture:

$$\frac{dX}{dt}=-\frac{i}{\hbar} [X, H]=\frac{P }{m}$$

$$\frac{dP}{dt}=-\frac{i}{\hbar} [P, H]=-V'(X) $$

These turn out looking like the classical Hamilton's equations because of the correspondence between the Poisson bracket and the commutator.

Now you can use the first equation to eliminate $P$ for $\frac{dX}{dt}$ to get something like Newton's second law:

$$m\frac{d^2}{dt^2}X=-V'(X) $$

which corresponds to an Euler Lagrange equation.

In general, the operator equations equations will look like Hamilton's equations because of the Poisson bracket-commutator correspondence. Hamilton's equations are related to the EL equation through a Legendre transform. But I wouldn't say that any action is being minimized in the Quantum Theory, because action written using operators isn't a scalar.

Also, the path integral formulation is for the evolution of wavefunctions, not operators.

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    $\begingroup$ Thanks for the answer, this makes sense. In retrospect this isn't as puzzling, I guess I'm not yet used to the fact that the main objects are field operators. But I still wonder if something like Schwinger's Quantum Action principle can be formulated here... $\endgroup$
    – FranDahab
    Dec 20, 2022 at 4:03
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    $\begingroup$ @FranciscoNemeCamposDahab I haven't seen a book talk about an operator stationary action principle. But it's fun to imagine, yes. $\endgroup$
    – Ryder Rude
    Dec 20, 2022 at 4:06
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  1. On one hand, in the operator formalism (section 7.2), the Heisenberg EOMs are imposed as exact operator identities.

  2. On the other hand, in the path integral formalism (section 7.1), the Euler-Lagrange (EL) equations only hold in a quantum-average sense up to quantum corrections, cf. the Schwinger-Dyson (SD) equations.

  3. Concerning the consistent co-existence of both formalisms, see e.g. this & this Phys.SE posts.

References:

  1. M.D. Schwartz, QFT & the standard model, 2014; section 7.1 + 7.2.
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  • $\begingroup$ This was pretty enlightening, thanks a bunch. $\endgroup$
    – FranDahab
    Dec 20, 2022 at 13:39

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