Time ordering and Schwinger-Dyson equation for a Klein-Gordon field

Question

I have a question about the usage of the Schwinger-Dyson Equation for the Klein-Gordon Field. $$\begin{align} i <0|T (\delta S / \delta \phi(x) ) \phi(x_1)\ldots |0> &\cr +<0|T\delta(x-x_1)\ldots|0>&=0 .\end{align}\tag{22.23}$$ 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 $$\begin{align} (\partial_x^2+m^2)<0|T\phi(x)\phi(x_1)|0>=&(\partial_x^2+m^2)\Delta(x-x_1)\cr =&-i\delta (x-x_1). \end{align}\tag{22.24} $$ 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: $$ 0=\int D\phi \ e^{iS(\phi)}(i \frac{\delta S}{\delta \phi(x)} \phi(x_1) +\delta(x-x_1)).\tag{22.22} $$

Answer 1

1. AccidentalFourierTransform's above comment is exactly 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 $T$.

   This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

2. More generally, the formal correspondence/dictionary between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$\begin{align} \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~& \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\cr ~=~& \frac{1}{Z[J]}F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J],\end{align}\tag{B} $$ where $F$ is an arbitrary functional and where the partition function/path integral is $$\begin{align} Z[J]~:=~& \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\cr S[\phi;J]~:=~&S[\phi]+J_k\phi^k, \end{align}\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer.

3. Now to the main point: Notice how the dictionary (B) naturally talks to $T_{\rm cov}$ rather than $T$: If the functional $F$ doesn't contain time derivatives, it doesn't matter whether we use $T$ or $T_{\rm cov}$. However, if $F$ does contain time derivatives, they get applied outside the correlator, i.e. the time-order is $T_{\rm cov}$.

4. Example. If $$ F[\phi]~=~\prod_{i=1}^n \left(\frac{\partial}{\partial t_i} \right)^{m_i} \phi(t_i),\tag{D}$$ then $$\begin{align}\frac{1}{Z[J]}&F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J]\cr ~\stackrel{(D)}{=}~&\frac{1}{Z[J]}\left[\prod_{i=1}^n \left(\frac{\partial}{\partial t_i} \right)^{m_i}\right]\int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\prod_{j=1}^n\phi(t_j)\cr ~\stackrel{(B)}{=}~&\left[\prod_{i=1}^n \left(\frac{\partial}{\partial t_i} \right)^{m_i}\right]\left< \Omega \left| T\left\{ \prod_{j=1}^n\phi(t_j)\right\} \right| \Omega \right>_J\cr ~\stackrel{(D)}{=}~&\left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J. \end{align}\tag{E}$$

5. Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~, \tag{F}$$ cf. e.g. this Phys.SE post.

6. In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~0, \tag{G}$$ because the EOMs are satisfied in quantum average, cf. e.g. this Phys.SE post.