# Week 7, Wednesday

## 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 $$D = \cup_{j=1}^k n_J C_j$$ consisting of connected components $C_j$ with multiplicity $n_j$. The Feynman rules for $D$ contain a symmetry factor $$\frac{1}{|\mathop{Aut} D|} = \prod_{j=1}^k \frac{1}{n_j! ~ |\mathop{Aut} C_j|}$$ 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 $$Z = \int \mathcal{D}\phi e^{iS} = \sum_k D_k = \prod_j \sum_{n_j} \frac{1}{n_j!} C_j = \prod_j e^{C_j} = e^{\sum_j C_j}.$$ This suggests to define a functional $W[J]$ as $$Z[J] = e^{i W[J]} ,\quad Z_E[J] = e^{W_E[J]}$$ 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 $$\langle \phi(x_1) \cdots \phi(x_n) \rangle_C = \tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \; \cdots \tfrac{1}{i} \frac{\delta}{\delta J(x_n)} \; W[J] \Big|_{J=0},$$ and, in particular, $$\langle \phi(x)\rangle_C = \tfrac{1}{i} \frac{\delta}{\delta J(x_1)} \; W[J] \Big|_{J=0} = \frac{1}{i^2} \frac{1}{Z} \frac{\delta Z}{\delta J(x)} \Big|_{J=0} = \frac{1}{i^2} \langle 0|\phi(x) |0\rangle.$$ 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 $$\label{eq:meanfield} \varphi(J) = \frac{\delta W}{\delta J}$$ It is perhaps suggestive to think of $W$ in $$e^{iW[J]} = \int\mathcal{D}\phi e^{i\int d^4x (\mathcal{L}+J\phi)}$$ 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 $$\Gamma[\varphi] = W[J(\varphi)] – \int d^dx \; \varphi J(\varphi).$$ Functionally differentiating with respect to the mean field yields $$\frac{\delta \Gamma[\varphi]}{\delta \varphi} + J = 0 \quad \Leftrightarrow \quad J = -\frac{\delta \Gamma[\varphi]}{\delta \varphi}$$ 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 $$\frac{\delta \Gamma[\varphi]}{\delta \varphi} \quad\Rightarrow\quad J=0 \quad\Rightarrow\quad \varphi = \frac{\delta W}{\delta J} \; \Big|_{J=0} = \tfrac{1}{i} \langle 0|\phi(x) |0\rangle,$$ 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 $$\Gamma[\varphi] = S[\varphi] + \sum_{1PI} A(G)$$ 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.