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In classical field theory, the on-shell dynamical field variables $\bar{q}$ give a minimum value of the action: $$A=\int dt ~L(\bar{q}(t),\dot{\bar{q}}(t)).$$

In this case, the action is actually a real number, so it makes good sense for it to have some extremal value.

What is the meaning of "extremal" in canonical Quantum Field Theory where the action, $$A=\int d^{4}x ~\mathcal{L}(\bar{\phi}(x),\partial_{\mu}{\bar{\phi}}(x))$$ is instead, an operator?

Qmechanic
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Arnab Barman Ray
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    There's no "action as an operator" in QFT, so your question is pointless. – DanielC Sep 01 '17 at 20:14
  • The fields are operators right? So an integral of some function of these fields should be an operator. – Arnab Barman Ray Sep 01 '17 at 20:16
  • No, the Lagrangian refers to classical fields. In QFT (in the canonical quantization framework), one first takes considers a Lagrangian of classical fields, then transitions to the Hamiltonian framework, and only then quantizes the fields. – J. Murray Sep 01 '17 at 20:30
  • Do you mean to say that the fields used in say the KG lagrangian are not operators? I have just finished reading up some books that seem to say that the scalar field \phi(x) is actually the sum of creation and annihilation operators that create and annihilate particles at x. Do you have any sources? – Arnab Barman Ray Sep 01 '17 at 20:38
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    Related: https://physics.stackexchange.com/q/48030/2451 – Qmechanic Sep 01 '17 at 20:58
  • Ok, so in the path-integral formulation, the fields are (super-)numbers while in the canonical quantization mechanism, the fields are operators. Is that correct? – Arnab Barman Ray Sep 01 '17 at 21:04

1 Answers1

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  1. The action in quantum field theory formally$^1$ enters the operator formalism via the Schwinger action principle. However, not as a genuine variational problem per se. See also this related Phys.SE post.

  2. In contrast to the operator formalism, the action (and its extremals) enter and play a central role in the path integral formulation, in particular in the semi-classical limit. In the path-integral formalism, the fields (and hence the action) are number-valued (rather than operator-valued), cf. this Phys.SE post. So the usual variational calculus applies.

  3. More generally, it is interesting to ponder What should an operator-valued variational problem be? How should we order operators, and what is an extremal operator? These questions are part of the broad mathematical topics of operator theory, functional analysis, convex analysis, and optimization theory.

  4. In physics, the task of optimizing operators usually involves taking a norm or a trace in the end to form a real-valued (rather than an operator-valued) functional. This is e.g. typically the case in matrix models, i.e. we are again back to the case where usual variational calculus applies.

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$^1$ Formally, up to possible operator ordering issues related to turning the action into an operator.

Qmechanic
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  • This is weird, we used the action to derive the conserved Noether charges for some lagrangians while learning the canonical procedure. Was that just in the spirit of introduction, that is do you mean to say that in actual calculations in canonical perturbation theory people don't care about the action? – Arnab Barman Ray Sep 01 '17 at 21:10