{title2=$Z[J]$}
{wiki}

A generating functional is a <path integral> with a source $J$ coupled to a field. Its <functional derivatives> with respect to $J$ generate time-ordered correlation functions.

Generating functional

Inside the <generating functional>, multiplication by a field can be replaced by a <functional derivative> of the source factor:
$$
\phi(x)e^{i\int J\phi}=\frac1i\frac{\delta}{\delta J(x)}e^{i\int J\phi}
=-i\frac{\delta}{\delta J(x)}e^{i\int J\phi}.
$$
Expanding the interaction exponential, making this replacement in every term, and resumming gives
$$
\boxed{
Z_n[J]=\exp\!\left[-ig\frac{(-i)^n}{n!}
\int d^dx\,\frac{\delta^n}{\delta J(x)^n}\right]Z_0[J]}.
$$
This formal identity assumes a common regulator, a source-independent normalization, and permission to interchange the <path integral>, power series, and functional derivatives. A normalized functional with $Z_n[0]=1$ requires division by the same expression evaluated at $J=0$, which removes connected <Vacuum Feynman diagrams>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 1 / c / Solution

With left <functional derivative>[functional derivatives] for the Grassmann sources, replace fields in the interaction by $\phi\mapsto i^{-1}\delta/\delta J$, $\psi\mapsto i^{-1}\delta/\delta\bar\eta$, and $\bar\psi\mapsto-i^{-1}\delta/\delta\eta$. Thus
$$
Z[\bar\eta,\eta,J]=
\exp\left[-ig\int d^4x
\left(-\frac1i\frac\delta{\delta\eta(x)}\right)
\left(\frac1i\frac\delta{\delta J(x)}\right)
\left(\frac1i\frac\delta{\delta\bar\eta(x)}\right)
\right]Z_{0,f}[\bar\eta,\eta]Z_{0,b}[J].
$$
The ordering displayed fixes the Grassmann signs and makes this an explicit source functional with no dynamical fields.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 2 / c / Solution

{title2=$\delta F/\delta\phi$}
{wiki}

The functional derivative is defined by the first variation
$$
F[\phi+\epsilon h]-F[\phi]
=\epsilon\int\frac{\delta F}{\delta\phi(x)}h(x),dx+o(\epsilon).
$$


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