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In Peskin & Schroeder, "An Introduction to Quantum Field theory", chapter 4, the author derives the 2 point correlation function:

$$\langle \Omega|P{\phi(x)\phi(y)}|\Omega \rangle = \lim_{T\rightarrow \infty(1-i\epsilon)} \frac{\langle 0|P \left [U(T,x^0)\phi_I(x)U(x^0,y^0)\phi_I(y)U(y^0,-T)\right ]|0\rangle}{\langle 0 | U(T,-T)|0\rangle }, \tag{4.30c}$$

where $P$ is the time ordering operator and $U(t,t')$ is the evolution operator $$U(t,t') =\exp\left \{ -i \int^t_{t'}dt H_I(t)\right \}\tag{4.23}$$ $$\langle \Omega|P{\phi(x)\phi(y)}|\Omega \rangle= \lim_{T\rightarrow \infty(1-i\epsilon)} \frac{\langle 0|P \left [\phi_I(x)\phi_I(y)\exp\left \{ -i \int_{-T}^Tdt H_I(t)\right \}\right ]|0\rangle}{\langle 0 | \exp\left \{ -i \int_{-T}^Tdt H_I(t)\right \}|0\rangle }. \tag{4.31}$$

My question is how do the various $U()$ factors like $U(T,x^0)$, $U(x^0,y^0)$ and $U(y^0,T)$ simplify to a final $\exp\left \{ -i \int_{-T}^Tdt H_I(t)\right \}$ in the numerator from Eq. (4.30c) to Eq. (4.31)?

My problem is that $U()$ does not commute with $\phi_I$, so, for example, one cannot really shift $U(T,x^0)$ to the right of $\phi_I(x)$ in Eq. (4.30c). $x^0$ is the $0^{th}$ component of $x$, and hence $U(T,x^0)$ and $\phi_I(x)$ are at the same time $x^0$, and thus the time ordering operator $P$ cannot be used to shift $U(T,x^0)$ to the right of $\phi_I(x)$. To my understanding, time ordering can be used to shiit terms only if they are at different times (according to their time sequence).

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The order you place the operators inside of the time ordering operator doesn't matter. For example $$P[\mathcal{O}(t+1)\mathcal{O}(t)]=P[\mathcal{O}(t)\mathcal{O}(t+1)]=\mathcal{O}(t+1)\mathcal{O}(t).$$ When we write $$P\left [\phi_I(x)\phi_I(y)\exp\left \{ -i \int_T^Tdt H_I(t)\right \}\right ],$$ the time ordering operator rearranges the terms to give $$U(T,x^0)\phi_I(x)U(x^0,y^0)\phi_I(y)U(y^0,-T).$$ Using the former expression is just shorthand.

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  • $\begingroup$ Got it. Thanks. $\endgroup$
    – Angela
    Commented Oct 9, 2020 at 5:04

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