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.


1 Answer 1



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?

  • $\begingroup$ 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? $\endgroup$
    – user404720
    Nov 11, 2017 at 18:31
  • $\begingroup$ Yes, but some additional argument is needed for that part (think symmetry). $\endgroup$ Nov 11, 2017 at 18:33
  • $\begingroup$ Okay thank you for your help! $\endgroup$
    – user404720
    Nov 11, 2017 at 18:46
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 11, 2017 at 19:01
  • $\begingroup$ Thank you! I think everything is clear now ^^ $\endgroup$
    – user404720
    Nov 11, 2017 at 19:24

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