A. Zee, Quantum Field Theory in a Nutshell · Problem I.3.3

Feynman Diagrams

Physics — QFT

Problem

Show that the advanced propagator defined by

D_{adv}(x - y) = \int \frac{d^4k}{(2\pi)^4} \frac{e^{ik(x-y)}}{k^2 - m^2 - i \text{sgn}(k_0)\epsilon}

(where the sign function is defined by

\text{sgn}(k_0) = +1

k_0 > 0

and

\text{sgn}(k_0) = -1

k_0 < 0

) is nonzero only if

x^0 > y^0

. In other words, it only propagates into the future. [Hint: both poles of the integrand are now in the upper half of the

k_0

-plane.] Incidentally, some authors prefer to write

(k_0 - i\epsilon)^2 - \vec{k}^2 - m^2

instead of

k^2 - m^2 - i \text{sgn}(k_0)\epsilon

in the integrand. Similarly, show that the retarded propagator

D_{ret}(x - y) = \int \frac{d^4k}{(2\pi)^4} \frac{e^{ik(x-y)}}{k^2 - m^2 + i \text{sgn}(k_0)\epsilon}

propagates into the past.