koolmodee said:
Well, I thought what I write was implied in the equations in the Srednicki book.
What you wrote does not make sense. The functional derivative on your right-hand is not acting on anything.
koolmodee said:
But then I don't see how we get from the first line two the second in 7.16. with your equation.
Let
Z(f)=\langle 0|0\rangle_f
From 7.11,
Z(f)=\exp K(f)
where
K(f)={i\over 2}\int dt\,dt'\,f(t)G(t-t')f(t')
By the chain rule,
{\delta\over\delta f(t_2)}Z(f)={dZ\over dK}\;{\delta\over\delta f(t_2)}K(f)
and since Z=\exp K, dZ/dK = \exp K = Z. Now we use
{1\over i}\,{\delta K(f)\over\delta f(t_2)}=<br />
{1\over i}\,{i\over 2}\int dt\,dt'\left[\left({\delta f(t)\over\delta f(t_2)}\right)G(t-t')f(t')+f(t)G(t-t')\left({\delta f(t')\over\delta f(t_2)}\right)\right]
{}\qquad\qquad={1\over 2}\int dt\,dt'\Bigl[\delta(t-t_2)G(t-t')f(t')+f(t)G(t-t')\delta(t'-t_2)\Bigr]
{}={1\over 2}\int dt'\,G(t_2-t')f(t')+{1\over 2}\int dt\,f(t)G(t-t_2)
{}=\int dt'\,G(t_2-t')f(t')
where, to get the last line, we use G(t-t_2)=G(t_2-t), and change the dummy integration variable in the 2nd term from t to t', so that it is then the same as the first term.
