1

Consider a $1D$ field theory correlation from the path integral perspective as follows $$\langle\phi(t_1)\phi(t_2) \rangle = Z^{-1}\int [\mathcal{D}\phi]~\phi(t_1)\phi(t_2) e^{iS[\phi]/\hbar}.\tag{1}$$ Now, under a time translation $t_i \to t_i+\epsilon$ we have $$\delta_{\epsilon}\phi = \epsilon\dot\phi.\tag{2}$$

Now, let me look at $$\delta_{\epsilon}\langle\phi(t_1)\phi(t_2) \rangle = \epsilon Z^{-1} \int [\mathcal{D}\phi]~(\dot{\phi}(t_1)\phi(t_2)+\phi(t_1)\dot{\phi}(t_2)) e^{iS[\phi]/\hbar}.\tag{3}$$ I am assuming $\delta_{\epsilon}S[\phi] = 0$. So, no terms from the exponential. Now the question is how is the above expression supposed to vanish? Should there be some Jacobian terms from the path integral measure as well?

Qmechanic
  • 220,844
Dr. user44690
  • 1,650

1 Answers1

1

Hint: Use the Schwinger-Dyson (SD) equation$^1$ $$\langle \underbrace{\delta_{\epsilon}F[\phi]}_{=\epsilon_0\{\dot{\phi}(t_1)\phi(t_2)+\phi(t_1)\dot{\phi}(t_2)\}}\rangle ~+~ \frac{i}{\hbar} \langle F[\phi]\underbrace{\delta_{\epsilon}S[\phi]\rangle}_{=\epsilon_0\{L(t_f)-L(t_i)\}=0}~=~0,\tag{A}$$ with functional $$ F[\phi]~=~\phi(t_1)\phi(t_2)\tag{B}$$ and infinitesimal variation $$ \delta_{\epsilon} ~=~\int\! dt ~\underbrace{\epsilon(t)}_{=\epsilon_0}\dot{\phi}(t)\frac{\delta}{\delta\phi(t)}. \tag{C}$$

$^1$ The infinitesimal Jacobian in the path integral is formally proportional to $\delta^{\prime}(0)$, which vanishes in e.g. dimensional regularization (DR), cf. e.g. this Phys.SE post. Concerning boundary terms in SD equations, see e.g. this Phys.SE post.

Qmechanic
  • 220,844