1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.