Propagator of a Scalar Field via Path Integrals

  • #1
63
1

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?
 

Answers and Replies

  • #2
175
64
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.
 
  • #3
63
1
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.
 
  • #4
175
64
Exact, very good
 

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