# Why can the time-ordered exponentials be brought to the right?

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)

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

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

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

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?

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.

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 $$x^0 while $$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 $$x^0 becomes $$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.

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.