In both statistical field theory and quantum field theory one computes average values / time ordered expectation values of functionals of fields with the path integral. I have two related questions:

  1. Is it the path integral alone that make quantum fields non-commutative or are other constraints also required?

  2. Are statistical fields non-commutative?


1) It's exactly the opposite of what you said - in the Hamiltonian operator formalism, the fields are non-commutative; the Lagrangian path integral is an equivalent reformulation of the problem in which the fields are commutative. Field ordering doesn't matter in the path integral; the original "non-commuting-ness" effectively gets absorbed into the integration measure.

2) I guess it depends on exactly what you mean by "statistical fields," but nope, anything that goes into a path integral is commutative.

  • $\begingroup$ 1)Ordering does matter inside the path integral in the case of equal times: If we evaluate the equal time commutator of the momentum conjugate with the field inside the path integral, the answer is nonzero. 2)By statistical field I mean a field which has no time dependence. $\endgroup$ – Luke Feb 22 '17 at 8:26
  • $\begingroup$ @Luke, no all the fields inside the path integral all commuting c-number fields. There aren't any commutators inside of path integrals. The fields in the path integral correspond to normal-ordered particle creation/annihilation operators, so there is no ordering ambiguity. Also, statistical partition functions for quantum many-body systems have imaginary time dependence, although classical many-body systems do not. $\endgroup$ – tparker Feb 22 '17 at 9:06
  • $\begingroup$ +1, with the caveat that Grassmann numbers are anticommutative. $\endgroup$ – knzhou Mar 13 at 14:07
  • $\begingroup$ @knzhou Good point. I guess that means this answer's wrong, but I'm too lazy to fix it. $\endgroup$ – tparker Mar 14 at 2:38

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