Suppose we have the Green function$$G(k) \equiv \tag 1\int d^4x e^{ikx}\langle 0| T\left(\partial^{x}_{\mu}A^{\mu}(x)B(0)\right)|0\rangle ,$$which in path integral approach is equal to $$\tag 2 G(k) \equiv \int d^4x e^{ikx}\int \left[\prod_{i}D\Psi_{i}\right] \partial_{\mu}^{x}A^{\mu}(x)B(0)e^{iS}.$$Since in Eq. $(2)$ all quantities are classical, then it seems that I can rewrite in a form$$\tag 3 G(k) \equiv k_{\mu}\Pi^{\mu}(k), $$where$$\Pi^{\mu}(k) \equiv \int d^4x e^{ikx} \int \left[\prod_{i}D\Psi_{i}\right]A^{\mu}(x)B(0)e^{iS} \equiv \int d^4x e^{ikx}\langle 0| T(A_{\mu}(x)B(0))|0\rangle. $$But in Eq. $(1)$ $T$-ordering is present, and quantities $A, B$ are quantum operators. So I can't reduce it to Eq. $(3)$:$$T(\partial_{\mu}A^{\mu}(x)B(0)) \equiv \theta (x_{0})\partial_{\mu}A^{\mu}(x)B(0) \pm \theta (-x_{0})B(0)\partial_{\mu}A^{\mu}(x) = $$$$=\partial_{\mu}T(A^{\mu}(x)B(0)) + \delta (x_{0})[A(0,\mathbf x), B(0)]_{\pm} \Rightarrow$$$$G(k) \equiv k_{\mu}\Pi^{\mu}(k) \pm \int d^{3}\mathbf x e^{-i\mathbf k \cdot \mathbf x}\langle 0|[A(0, \mathbf x), B(0)]_{\pm}|0\rangle $$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?
