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: $$ 0=\int D\phi \ e^{iS(\phi)}(i \frac{\delta S}{\delta \phi(x)} \phi(x_1) +\delta(x-x_1)).\tag{22.22} $$

  • 1
    $\begingroup$ 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. $\endgroup$ Sep 29, 2017 at 11:06
  • 1
    $\begingroup$ Try replacing the derivative with a finite difference approximation. It should be evident why it commutes with the path integral from this. $\endgroup$ Sep 29, 2017 at 11:56
  • $\begingroup$ Related: physics.stackexchange.com/q/197313/2451 , physics.stackexchange.com/q/296309/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 15, 2022 at 20:11

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.