= Solution
With Euclidean source convention $+\int d^dx\,J_a\phi_a$, the <generating functional> is
$$
Z[J]=\int\mathcal D\phi\,
\exp\left[-S[\phi]+\int d^dx\,J_a(x)\phi_a(x)\right].
$$
Here $J_a(x)$ is a nondynamical source. Its <functional derivatives> insert fields:
$$
\frac{1}{Z[0]}\frac{\delta^nZ[J]}
{\delta J_{a_1}(x_1)\cdots\delta J_{a_n}(x_n)}\bigg|_{J=0}
=\langle\phi_{a_1}(x_1)\cdots\phi_{a_n}(x_n)\rangle.
$$
This is why $Z$ “generates” correlation functions.
With the notation of the question, $W[J]=-\log Z[J]$ is the Euclidean <connected generating functional>; it is distinct from a <Wilsonian effective action>. Define the classical field
$$
\Phi_a(x)=\langle\phi_a(x)\rangle_J=-\frac{\delta W}{\delta J_a(x)}.
$$
The <quantum effective action> is the <Legendre transform>
$$
\boxed{\Gamma[\Phi]=W[J]+\int d^dx\,J_a(x)\Phi_a(x)},
$$
where $J$ is eliminated in favor of $\Phi$. Its first derivative is
$$
\frac{\delta\Gamma}{\delta\Phi_a(x)}=J_a(x),
$$
so at zero source the quantum expectation value is a <stationary point> of $\Gamma$. Moreover,
$$
\int d^dz\,
\Gamma^{(2)}_{ac}(x,z)G_{cb}(z,y)=\delta_{ab}\delta^{(d)}(x-y),
$$
where $G_{ab}=\langle\phi_a\phi_b\rangle_{J,\rm conn}$ is the exact <connected correlation function>. Thus the second derivative of $\Gamma$ is the inverse exact propagator.
Perturbatively, $Z[J]$ sums all <Feynman diagrams>, including disconnected ones. The exponential formula for combinatorial structures says that its logarithm selects <connected Feynman diagrams>, so $W$ sums connected diagrams with the sign dictated by the convention above. The Legendre transform removes diagrams that disconnect when one internal line is cut; consequently $\Gamma$ sums <one-particle-irreducible Feynman diagrams>. Equivalently, every connected diagram is a tree assembled from one-particle-irreducible vertices and full propagators, and the Legendre transform inverts that tree construction.
Now let $J'_a=J_bU_{ba}$. Then
$$
Z[J']=\int\mathcal D\phi\,e^{-S[\phi]+\int J_bU_{ba}\phi_a}.
$$
Changing variables to $\phi'_b=U_{ba}\phi_a$ and using invariance of both the action and <functional measure> gives $Z[J']=Z[J]$, hence $W[J']=W[J]$. In the Legendre transform, the pairing obeys
$$
J_a(U_{ab}\Phi_b)=(J_bU_{ba})\Phi_a.
$$
Changing the source variable and using the invariance of $W$ therefore gives
$$
\boxed{\Gamma[U\Phi]=\Gamma[\Phi]}.
$$
The symmetry of the classical action and measure is inherited by the full <quantum effective action> when it has no <quantum anomaly>.
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