# Functional derivative of generating function QFT

I am reading Srednicki book for QFT. In page 69 they realize a functional derivative of the following function: $$Z[J]= \exp\left(\frac{i}{2}\int d^4xd^4x'J(x)\Delta(x-x')J(x') \right)$$ with $$\Delta(x-x')=\lim_{\epsilon\rightarrow 0}\int \frac{d^4k}{(2\pi)^4}\frac{\exp\{ik(x-x')\}}{k^2+m^2-i\epsilon}$$ the usual Feynman propagator.

At some point they say that $$\frac{\delta}{\delta(J(x_1))}Z[J]=iZ[J]\int d^4x'\Delta(x_1-x')J(x')$$

I've tried to derive that result but I am not use to functionals having two integrates and I am not very used to functional derivatives.

Any help for the derivation of this result is appreciated.

HINT

In these situations, it's usually helpful to try to translate it into a discrete context and see if you can make sense of it there. For instance, let $$Z(J) = \exp\left(\frac{i}{2} \sum_{i,j}J_i \Delta_{ij}J_j\right).$$

Can you see that $$\frac{\partial}{\partial J_k}Z(J) = i Z(J) \sum_j\Delta_{kj}J_j$$ using the fact that $\Delta_{ij}$ is symmetric?

• I'd like to say that $\Delta_{ij}$ is symmetric, but in the general case is it true that $\Delta(x-x')=\Delta(x'-x)$? $e^{iA}\neq e^{-iA}$ in general, right? Nov 11, 2017 at 18:31
• Yes, but some additional argument is needed for that part (think symmetry). Nov 11, 2017 at 18:33
• Okay thank you for your help! Nov 11, 2017 at 18:46
• in retrospect 'think symmetry' is kind of silly advice for how to see a symmetry relation, but hopefully it led you to the right place of substituting $k\to -k$ in the integral. Nov 11, 2017 at 19:01
• Thank you! I think everything is clear now ^^ Nov 11, 2017 at 19:24