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Having worked through almost all calculations in section 4.2 of Peskin & Schroeder's An Introduction to QFT, I still don't get why we can get to Eq. (4.31)

\begin{equation} \langle\Omega|\mathcal{T}\phi(x)\phi(y)|\Omega\rangle=\lim_{T\to\infty(1-i\epsilon)}\frac{\langle 0|\mathcal{T}\left\{\phi_I(x)\phi_I(y)\exp\left[-i\int_{-T}^{T}dtH_I(t)\right]\right\}|0\rangle}{\langle 0|\mathcal{T}\exp\left[-i\int_{-T}^{T}dt H_I(t)\right]|0\rangle} \tag{4.31} \end{equation}

from the previous expression (valid for $x^0>y^0$ but generalizable):

\begin{equation} \langle\Omega|\phi(x)\phi(y)|\Omega\rangle=\lim_{T\to\infty(1-i\epsilon)}\frac{\langle 0|U(T,x^0)\phi_I(x)U(x^0,y^0)\phi_I(y)U(y^0,-T)|0\rangle}{\langle0|\mathcal{T}\exp\left[-i\int_{-T}^{T}dt H_I(t)\right]|0\rangle}. \tag{4.30b}\end{equation}

All $U$s are time ordered exponentials and the fields and interaction Hamiltonian on the right-hand sides are in the interaction picture. I can't wrap my head around the re-ordering of the numerator. Why is it possible to go from the second to the first?

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3 Answers 3

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Obviously the interval from $-T$ to $T$ can be split into three: $(-T, x^0)$, $(x^0, y^0)$, and $(y^0, T)$.

I'd suggest taking the first expression, splitting the exponential into three exponentials one for each sub-range, and applying time-ordering operator by hand. The result is rather self-evident.

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The integral in $(4.31)$ can be written by splitting the interval from −T to T into: $(−T,x^0)$, $(x^0,y^0)$ and $(y^0,T)$ for $x^0<y^0$ while $x^0$ and $y^0$ can be exchanged for $x^0>y^0$. Note that $\phi_I(x)$ and $\phi_I(y)$ also rearrange according to those conditions.

By equation $(4.23)$ we see that the exp in $(4.31)$ is just $U(T,-T)$ which for $x^0<y^0$ becomes $U(T,y^0)U(y^0,x^0)U(x^0,-T)$. The exp part in denominator is thus simply $(UT,-T)$.

Hence, you recover the required equation.

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I will show how your first expression is related to your second for the case $x_0 > y_0$, the other case should be similar.

Note that the time ordering operator is like a sorting algorithm, hence it doesn't matter if we permute something before applying the time ordering operator. In particular we can perform some additional time-ordering by inserting an additional time-ordering operator:
\begin{align} A=\mathcal{T}\left\{\phi_I(x)\phi_I(y)\exp\left[-i\int_{-T}^{T}dtH_I(t)\right]\right\} =& \mathcal{T}\left\{\phi_I(x)\phi_I(y)\mathcal{T}\left\{ \exp\left[-i\int_{-T}^{T}dtH_I(t)\right]\right\}\right\}\\ =&\mathcal{T}\left\{\phi_I(x)\phi_I(y)U(T,-T)\right\} \end{align}

We show the case $x_0 > y_0$. By 4.26 we get $$U(T,-T) = U(T, x_0)U(x_0,y_0)U(y_0,-T),$$ which we substitute.

$A =\mathcal{T}\left\{\phi_I(x)\phi_I(y)U(T, x_0)U(x_0,y_0)U(y_0,-T)\right\}$

Now we want to apply the time ordering. For this we note that the $U(T,x_0)$ contains only operators with the with in the interval $[T,x_0]$, and similar for the terms $U(x_0,y_0)$ and $U(y_0,-T)$. Hence, if we apply the time ordering we get.

$A =U(T, x_0)\phi_I(x)U(x_0,y_0)\phi_I(y)U(y_0,-T),$ which occurs in the numerator of your second expression.

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