# Propagator of a Scalar Field via Path Integrals

• I

## Main Question or Discussion Point

I don't understand a step in the derivation of the propagator of a scalar field as presented in page 291 of Peskin and Schroeder. How do we go from:
$$-\frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_2)} \text{exp}[-\frac{1}{2} \int d^4 x \; d^4 y \; J(x) D_F (x-y) J(y)]|_{J=0}$$
To:
$$-\frac{\delta}{\delta J(x_1)} [ -\frac{1}{2} \int d^4 y \; D_F (x_2-y)J(y) - \frac{1}{2} \int d^4 x \; J(x) D_F (x-x_2)]\frac{Z[J]}{Z_0} |_{J=0} \; \; \;\text{?}$$
Why is the functional derivative with respect to $J(x_2)$ equal to the term within the brackets above?

Related High Energy, Nuclear, Particle Physics News on Phys.org
Well, do you know how to evaluate the following derivatives?
$$\frac{\delta (F[J]G[J])}{\delta J(x)}$$
$$\frac{\delta J(y)}{\delta J(x)}$$
Where $F$, $G$ are functionals and $J$ is a function. Knowing this is almost immediat to obtain Peskin's result.

I see now. I just had to use the product rule and the fact that:
$$\frac{\delta J(y)}{\delta J(x)} = \delta^4(x-y)$$
What really confused me was the $\frac{Z[J]}{Z_0}$ term appearing out of nowhere. But on second look that's just:
$$\frac{Z[J]}{Z_0} = \text{exp}[-\frac{1}{2}\int d^4 x \; d^4 y \; J(x)D_F(x-y) J(y)]$$
Thanks for clearing things out, sorry for the silly question.

Exact, very good