# Schwinger-Dyson eqation for a Klein-Gordon-Field

I have a question about the usage of the Schwinger-Dyson-Equation for the Klein-Gordon-Field. $$i <0|T (\delta S / \delta \phi(x) ) \phi(x_1)\ldots |0>+<0|T\delta(x-x_1)\ldots|0>=0 .\tag{22.23}$$ As long as the other insertions aren't near $x$. So $$<0|T((-\partial^2-m^2)\phi(x))\phi(x_1)|0>=i\delta(x-x_1).\tag{22.23'}$$ Now, in Srednicki's QFT book, it is claimed that the Klein-Gordon-Operator should rather be outside the VEV, that "is clear from the path integral formulation". Well it is not clear to me, even having written down the path integral formulation, why it should rather be $$(\partial_x^2+m^2)<0|T\phi(x)\phi(x_1)|0>=(\partial_x^2+m^2)\Delta(x-x_1)=-i\delta (x-x_1). \tag{22.24}$$ That way the derivative also acts on the step functions in the time ordering. For convenience, I'll also write down the path integral representation of the first formula: $$0=\int D\phi \ e^{iS(\phi)}(i \frac{\delta S}{\delta \phi(x)} \phi(x_1) +\delta(x-x_1)).\tag{22.22}$$

• Srednicki is being sloppy. The real justification is that the path-integral time-ordering symbol is the covariant one, not the naïve one, and the former commutes with space-time derivatives. You won't find a proper explanation in Srednicki's book, so you will have to accept his claims and learn to live with it. Nice book nevertheless, if you keep in mind that he is not trying to be precise nor rigorous. – AccidentalFourierTransform Sep 29 '17 at 11:06
• Try replacing the derivative with a finite difference approximation. It should be evident why it commutes with the path integral from this. – Prof. Legolasov Sep 29 '17 at 11:56
• Thank you, I figured it out that way. Path integral and functional derivative commute. – Markus Zetto Sep 29 '17 at 15:25

1. AccidentalFourierTransform's above comment is exact right: The point is that Srednicki's time-ordering $$T$$ should be replaced with covariant time-ordering $$T_{\rm cov}$$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $$T$$.
2. More generally, the formal operator-path integral correspondence reads $$\left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~ Z[J]^{-1}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\tag{A}$$ where the partition function/path integral is $$Z[J]~:=~ \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}.\tag{B}$$ The correspondence (A) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer.
3. Then the Schwinger-Dyson (SD) equations becomes $$\left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(y)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(y)} \right\}\right| \Omega \right>_J ~, \tag{C}$$ cf. e.g. this Phys.SE post.
4. In contrast, if we only use the usual time ordering $$T$$, we do not get the contact term: $$\left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(y)}\right\}\right| \Omega \right>_J~=~0, \tag{D}$$ because the eoms are satisfied in quantum average.