# Functional derivative of the generating functional with respect to the source term

To be specific, let us use the notations in W. Metzner et al., Rev. Mod. Phys. 84,299 (2012). The generating functional $$G[\eta, \bar\eta]$$ is given by [Eq. 4] \begin{align} G[\eta, \bar\eta] = - \ln \int D\psi D\bar\psi e^{-S[\psi,\bar\psi]} e^{(\bar\eta,\psi)+(\bar\psi,\eta)}. \end{align}

I want to know what is the functional derivative $$\partial G /\partial \eta$$? I have two guesses.

First, Below Eq. 9, the authors showed that \begin{align} \frac{\partial G }{ \partial \eta} = \bar\psi \end{align} and in Eq. 19 the authors have used a trick where $$V[\psi,\bar\psi]$$ is replaced by $$V[\partial_{\bar\eta}, \partial_{\eta}]$$. This clearly shows that the functional derivative of the generating functional $$G[\eta,\bar\eta]$$ with respect to the source field $$\eta$$ is the conjugate field $$\bar\psi$$.

My second guess is that
\begin{align} \frac{\partial{G}}{\partial \eta} = -\frac{\int D\psi D\bar\psi [-\bar\psi] e^{-S[\psi,\bar\psi]}e^{(\bar\eta,\psi)+(\bar\psi, \eta)}}{\int D\psi D\bar\psi e^{-S[\psi,\bar\psi]}e^{(\bar\eta,\psi)+(\bar\psi, \eta)}}. \end{align} This is the expectation value of the field operator with finit source. In the zero souce case, this gives the expectation value \begin{align} \frac{\partial{G}}{\partial \eta}\Big{|}_{\eta = \bar\eta = 0} = -\frac{\int D\psi D\bar\psi [-\bar\psi] e^{-S[\psi,\bar\psi]}}{\int D\psi D\bar\psi e^{-S[\psi,\bar\psi]}} = \langle \bar\psi \rangle \end{align} By taking the second-order derivative, we have \begin{align} \frac{\partial^2{G}}{\partial\bar\eta\partial \eta}\Big{|}_{\eta = \bar\eta = 0} = -\langle \bar\psi \rangle \langle \psi \rangle + \langle \psi \bar\psi \rangle = \langle \psi \bar\psi \rangle_{c} \end{align} This coincides with the Eq. 7 with $$m = 1$$.

I do not know how to reconcile these two cases.

So what is the functional derivative of the generating functional $$G[\eta, \bar\eta]$$ with respect to the source field $$\eta$$? Is it the conjugate field $$\bar\psi$$ or its expectation value $$\langle \bar\psi\rangle$$ with finit source?

• It's always an expectation value. Commented Jul 18, 2023 at 8:15
• Thank you very much. Commented Jul 18, 2023 at 12:47

The functional derivative $$\frac{\delta W_c[J]}{\delta J_k}$$ of the generating functional of connected diagrams $$W_c[J]$$ wrt. a sources $$J_k$$ is both

1. the classical field $$\phi_{\rm cl}^k$$ and

2. the expectation value $$\langle \phi^k \rangle_J$$

(up to sign conventions in the case of a Grassmann-odd field), cf. e.g. eq. (4) in my Phys.SE answer here.

Confusingly, Ref. 1 uses the same notation for a field and its classical field, which likely caused OP's question.

References:

1. W. Metzner, M. Salmhofer, C. Honerkamp, V. Meden & K. Schönhammer, Rev. Mod. Phys. 84 (2012) 299.
• Thank you very much. Commented Jul 18, 2023 at 12:47