## 1PI Effective Action

Let us start by taking a step back and reformulate some of the
combinatorics of Feynman diagrams in terms of algebra of the
generating function. First, consider a Feynman diagram
\begin{equation}
D = \cup_{j=1}^k n_J C_j
\end{equation}
consisting of connected components $C_j$ with multiplicity $n_j$. The
Feynman rules for $D$ contain a symmetry factor
\begin{equation}
\frac{1}{|\mathop{Aut} D|} =
\prod_{j=1}^k \frac{1}{n_j! ~ |\mathop{Aut} C_j|}
\end{equation}
To evaluate the path integral we have to sum over all diagrams,
connected or not. Writing $C_j$ for both the Feynman diagram and the
expression obtained by applying the Feynman rules, we find
\begin{equation}
Z = \int \mathcal{D}\phi e^{iS} =
\sum_k D_k =
\prod_j \sum_{n_j} \frac{1}{n_j!} C_j =
\prod_j e^{C_j} = e^{\sum_j C_j}.
\end{equation}
This suggests to define a functional $W[J]$ as
\begin{equation}
Z[J] = e^{i W[J]}
,\quad
Z_E[J] = e^{W_E[J]}
\end{equation}
in Minkowski and Euclidean signature, respectively. The $W[J]$ then
has an expansion into a sum over connected Feynman diagrams. We can
define “connected” correlation functions
\begin{equation}
\langle \phi(x_1) \cdots \phi(x_n) \rangle_C =
\tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \;
\cdots
\tfrac{1}{i} \frac{\delta}{\delta J(x_n)} \;
W[J] \Big|_{J=0},
\end{equation}
and, in particular,
\begin{equation}
\langle \phi(x)\rangle_C =
\tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \;
W[J] \Big|_{J=0} =
\frac{1}{i^2} \frac{1}{Z}
\frac{\delta Z}{\delta J(x)} \Big|_{J=0} =
\frac{1}{i^2} \langle 0|\phi(x) |0\rangle.
\end{equation}
The vev of $\phi$ is, of course, $0$ in $\phi^4$-theory because of the
$\phi\mapsto -\phi$ symmetry. To get something non-trivial, we should
*not* set $J=0$, this amounts to measuring the response of the
field to an external source. Hence, we define the *mean field* as
\begin{equation}
\label{eq:meanfield}
\varphi(J) = \frac{\delta W}{\delta J}
\end{equation}
It is perhaps suggestive to think of $W$ in
\begin{equation}
e^{iW[J]} = \int\mathcal{D}\phi e^{i\int d^4x (\mathcal{L}+J\phi)}
\end{equation}
as an “effective action” which incorporates all the quantum effects
from the path integral. However, it depends on the source $J$ and not
on the field. Hence, to define something that deserves the name, we
have to replace the $J$-dependence with a dependence on the mean field
$\varphi$. The right way to do this is to use the Legendre transform,
just like how we change between $(q, \dot q)$ and $(q,p)$-dependence
in classical mechanics. Hence, we invert $\varphi=\varphi(J)$ to
obtain $J=J(\varphi)$ and define the *1PI* effective action} as
\begin{equation}
\Gamma[\varphi] =
W[J(\varphi)] – \int d^dx \; \varphi J(\varphi).
\end{equation}
Functionally differentiating with respect to the mean field yields
\begin{equation}
\frac{\delta \Gamma[\varphi]}{\delta \varphi} + J = 0
\quad \Leftrightarrow \quad
J = -\frac{\delta \Gamma[\varphi]}{\delta \varphi}
\end{equation}
We interpret this in the following way: The stationary point of the
effective action $\Gamma[\varphi]$ is where the mean field satisfies
$J(\varphi)=0$. Hence
\begin{equation}
\frac{\delta \Gamma[\varphi]}{\delta \varphi}
\quad\Rightarrow\quad
J=0
\quad\Rightarrow\quad
\varphi =
\frac{\delta W}{\delta J} \;
\Big|_{J=0} =
\tfrac{1}{i} \langle 0|\phi(x) |0\rangle,
\end{equation}
that is, the stationary points of the effective action determine the
quantum one-point function. A similar calculation for higher
derivatives shows that all correlation functions are determined by the
higher derivatives of $\Gamma[\phi]$.

By this argument, the classical solutions of the effective action
$\Gamma$ are equivalent to full path integral with the action $S=\int
d^4x \mathcal{L}$. In terms of Feynman diagrams, this means that the
sum over all tree level $\Gamma$-diagrams equals the sum over all
connected $S$-diagrams. It remains to explain why $\Gamma$ is called
the *one-particle irreducible* (1PI) effective action. This is
because it can be expanded
\begin{equation}
\Gamma[\varphi] = S[\varphi] + \sum_{1PI} A(G)
\end{equation}
where the sum runs over all 1PI graphs $G$ and $A(G)$ is the
corresponding amputated diagram. A mathematical proof is quite
involved, but we can understand it by the following heuristic
argument. According to this equation, the $\Gamma$-Feynman rules have
a different type of vertex for each 1PI $S$-Feynman graph. The
correspondence between all $\Gamma$ and $S$-Feynman graphs is obtained
by shrinking each 1PI $S$-subgraph to the corresponding
$\Gamma$-vertex.