Past exam of the quantum field theory course of the University of Oxford / 2020 / 3 / a / Solution

Keep $\lambda$ for the cubic coupling and $j$ for the external source. Expanding the interaction part of the scalar field path integral yieldsA functional derivative of the source exponential inserts $i\varphi (y)$. In particular, ${\delta }_j(y{)}^3{e}^{i\int j\varphi }=-i\varphi (y{)}^3{e}^{i\int j\varphi }$, so $-i\lambda \varphi (y{)}^3$ can be replaced by $\lambda {\delta }_j(y{)}^3$. The functional derivatives commute and can be taken outside the free path integral. Resumming the exponential series givesThis is a formal perturbation series; a real cubic scalar potential is unbounded and the expression does not assert a stable nonperturbative vacuum state.

To evaluate the free Gaussian path integral, use integration by parts to writeThe inverse kernel isCompleting the square with $\varphi =\chi -K^{-1}j$ changes the exponent to $i\int \chi K\chi /2-i\int j{K}^{-1}j/2$. The translation-invariant functional measure makes the remaining Gaussian path integral precisely $Z[0;0]$, henceFor $ϵ>0$, the exponent includes $-ϵ\int {\varphi }^2/2$, which damps the regulated finite-dimensional integrals. The Feynman i-epsilon prescription displaces the positive-energy pole below the real $p^0$ axis and the negative-energy pole above it, selecting the time-ordered vacuum expectation value. Indeed the Feynman propagator is ${\Delta }_F=iD$. Thus $iϵ$ specifies both convergence and the vacuum state boundary conditions, with $ϵ\to 0^{+}$ taken after integration.