Suppose we have the Green function G(k)≡∫d4xeikx⟨0|T(∂xμAμ(x)B(0))|0⟩, which in path integral approach is equal to G(k)≡∫d4xeikx∫[∏iDΨi]∂xμAμ(x)B(0)eiS. Since in Eq. (2) all quantities are classical, then it seems that I can rewrite in a form G(k)≡kμΠμ(k), where Πμ(k)≡∫d4xeikx∫[∏iDΨi]Aμ(x)B(0)eiS≡∫d4xeikx⟨0|T(Aμ(x)B(0))|0⟩. But in Eq. (1) T-ordering is present, and quantities A,B are quantum operators. So I can't reduce it to Eq. (3): T(∂μAμ(x)B(0))≡θ(x0)∂μAμ(x)B(0)±θ(−x0)B(0)∂μAμ(x)= =∂μT(Aμ(x)B(0))+δ(x0)[A(0,x),B(0)]±⇒ G(k)≡kμΠμ(k)±∫d3xe−ik⋅x⟨0|[A(0,x),B(0)]±|0⟩ But I can't see why formally Eq.(3) is incorrect. Moreover, I don't understand how nonzero commutator in path integral approach may appear, since all quantities are classical, and (anti)commutators are always zero. I expect that because path integral contains information about all symmetries of a given theory, the commutators will be automatically replaced on Poisson brackets, but I don't see how.
Could you explain please?
Answer
Well, in a nutshell the issue is the following. Recall that the path integral formulation [with a Z2-graded (super)commutative integrand] is derived from the non-commutative operator formalism via a time-slicing procedure. This means that there is an implicit time-order prescription in the definition of the path integral, which manifests itself everytime we pull operators in and out of the path integral. In particular, this is so for a time-derivative, because a time-derivative interferes in a non-trivial way with the time-slicing procedure.
See this, this and this Phys.SE posts for related discussions.
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