One particle irreducible Feynman diagrams

Question

In quantum field theory Feynman has invented a diagrammatic method to encode various terms in the Taylor decomposition of integrals of the following form below which I will write in a baby version as finite dimensional integral rather than path integral (and using "imaginary time"): $$Z(j_1,\dots,j_n):=\frac{\int_{\mathbb{R}^n} \exp\{-B(x_1,\dots,x_n)+\sum_kP_k(x)+\sum_{i=1}^nj_ix_i\}dx}{ \int_{\mathbb{R}^n} \exp\{-B(x_1,\dots,x_n)+\sum_kP_k(x)\}dx},$$ where $B$ is a positive definite quadratic form on $\mathbb{R}^n$, $P_k$ are homogeneous polynomials. Furthermore one can write $Z(j)=exp\{ W(j)\}$ and it is shown that $W(j)$ is a sum of terms corresponding only to connected diagrams.

In the context of path integrals there is a notion of effective action which in this context is defined as follows. Let $\phi_i:=\frac{\partial W(j)}{\partial j_i}$. Define the Legendre transform $$\Gamma(\phi):=\sum_{i=1}^n \phi_ij_i-W(j).$$

In QFT it is claimed that the Taylor decomposition of $\Gamma(\phi)$ is the sum of terms corresponding to connected one particle irreducible diagrams. I am wondering if a finite dimensional (baby) version of this claim is true, and if this is the case whether there is a reference to a detailed discussion of the finite dimensional case.

Answer 1

Section 5 of Borcherds, Barnard, Lectures on Quantum Field Theory is a discussion of the 0-dimensional spacetime case, which gives finite dimensional integrals.

Answer 2

It is important to distinguish giving a sense to all of this nonperturbatively, i.e., for honest functions of $j$ or $\phi$ versus as formal power series in $j$ or $\phi$. In the nonperturbative context, one should use the Fenchel definition of the transform and has to worry about functions being convex. Since there is talk of 1PI diagrams here, the implicit assumption is this is all about formal power series. In that case the way to understand this is as a very special case of the reversion of power series.

Namely, suppose you have $n$ formal power series $F_1,\ldots,F_n$ in $\mathbb{C}[[x_1,\ldots,x_n]]$. Suppose the the constant terms are zero and the linear term of $F_i$ is $\sum_j a_{i\ell}x_\ell$ where the matrix $A=(a_{i\ell})$ is invertible. Then the system of equations $$ \left\{ \begin{array}{c} y_1=F_1(x_1,\ldots,x_n) \\ \vdots \\ y_n=F_n(x_1,\ldots,x_n) \end{array} \right. $$ has a unique solution, i.e., $x_i$'s in $\mathbb{C}[[y_1,\ldots,y_n]]$ without constant terms. Moreover, one has an explicit formula for this solution as a sum over rooted trees where the leaves correspond to the $y$ variables. See my article "Feynman Diagrams in Algebraic Combinatorics" in SLC 2003 where this explained in detail.

Just apply this to the case where $x=j$, $y=\phi$ and $F$ is the gradient of $W$. One finally, needs to insert the almost tautological expansion of $W$ as a sum over trees, with vertices corresponding to 1PI diagrams, and use some simple cancellations.

Answer 3

The "baby version" is actually the regularization of the path integral on a finite Euclidean lattice, which is what is studied by theoretical physicists working in lattice quantum field theory. The usual results about the effective action do indeed hold in that context.

Answer 4

The $0$-dimensional case that others have alluded to gives exactly the cumulants, and the effective action is the cumulant-generating function. These are treated in many textbooks on probability theory.