# 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?

## 1 Answer

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.