The generating functional is defined as: $$Z[J] = \int \mathcal{D}[\phi] \exp\Big[\frac{i}{\hbar}\int d^4x [\mathcal{L} + J(x)\phi(x)]\Big].$$

I know this object is used as a tool to generate correlation functions by taking functional derivatives, but does it have any interpretation on its own? In this post the answer states that it can be interpreted as the sum of all possible Feynman diagrams. Is this interpretation from Taylor expanding the exponential in $Z[J]$?

  • $\begingroup$ Since it generates all the correlation functions, it is a sum of all possible correlation functions with different numbers of fields for the different terms in the summation. Each correlation function consists of multiple Feynman diagrams. $\endgroup$ Commented Feb 29 at 3:04
  • $\begingroup$ @flippiefanus Wouldn't we need to first act on $Z$ with functional derivatives to do that? How does $Z$ contain any diagrams before differentiating? $\endgroup$
    – CBBAM
    Commented Feb 29 at 3:08

2 Answers 2


The generating functional by definition is an object that encodes the Green's functions in a series as $$Z[J]=\sum_{n=0}^\infty \dfrac{1}{n!}\int dx_1\cdots dx_n J(x_1)\cdots J(x_n) \langle \phi(x_1)\cdots \phi(x_n)\rangle\tag{1}.$$

This is the case because it is defined so that the correlators can be extracted as $$\langle \phi(x_1)\cdots\phi(x_n)\rangle = \dfrac{\delta^n Z[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0}\tag{2}.$$

The fact that $Z[J]$ is given by the expression you wrote can then be seem as a consequence of the Dyson-Schwinger equations, for example. See this answer of mine to a question you asked previously for more into this.

So, answering your current question, since each correlator is a sum of Feynman diagrams it is indeed possible to see $Z[J]$ as a sum over all possible Feynman diagrams in view of its definition, equation (1).

  • $\begingroup$ Wow your answer seems to reveal something a lot deeper than I had originally had in mind. So the functional measure $d\mu$ that is of interested in constructive quantum field theory is the link between your (1) an the $Z[J]$ I wrote? $\endgroup$
    – CBBAM
    Commented Feb 29 at 3:43
  • 2
    $\begingroup$ Generating functionals and measures are related by a kind of Fourier transform (that is the Bochner-Minlos theorem). So yes, there should be a measure underlying $Z$ and people in constructive field theory try to find ways to define it. Informally this measure is $d\mu(\phi) = {\scr D}\phi e^{i S}$, but this is of course not rigorous and not a definition of anything. In fact, defining this measure is extremely hard. $\endgroup$
    – Gold
    Commented Feb 29 at 4:18
  • $\begingroup$ Thank you, this is now clear. Also if you have any suggestions for learning more about constructive field theory that would be great. $\endgroup$
    – CBBAM
    Commented Feb 29 at 4:21
  • 1
    $\begingroup$ Even for free theories one cannot make sense of $d\mu(\phi)=D\phi\ e^{iS}$ as a measure. However, what constructive QFT manages to do in some models like $\phi_{2+1}^{4}$ is to make sense of $d\mu(\phi)=D\phi\ e^{-S_E}$, where $S_E$ is the Euclidean action, as a measure. $\endgroup$ Commented Feb 29 at 14:46

Perturbatively, one can formally argue that the partition function/path integral/functional integral/generating functional can be written as $$\begin{align} Z[J]~=~&\int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S[\phi] +J_k \phi^k \right)\right\} \cr ~\sim~& \underbrace{\exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}}_{\text{bag of vertices}} \underbrace{\exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}}_{\text{bag of propagators}} , \end{align}$$ cf. my Phys.SE answer here. All possible diagrams are generated by performing the functional $J$-differentiations.


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