4

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
  • 220,844
Arnab Barman Ray
  • 688
  • 7
  • 26

1 Answers1

2

  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.

--

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

Qmechanic
  • 220,844