If $Z[J]$ is the generating functional for the path-integral, could any prove (or more reasonably, refer me to a proof) that $$W[J]\equiv\frac{\hbar}{i}\log\left(Z[J]\right)$$ "generates" only connected diagrams?

So far I've only seen theory-dependent "examples" (basically showing how in $\phi^4$ theory the two-point function from $W$ gives only connected contributions).

I'm looking for a generic systematic proof for a general field theory.


4 Answers 4


The logarithmic relationship is equivalent to $$Z[J]=\exp[iW[J]]$$ where $W$ is the sum of connected diagrams. This formula is trivial to prove via Taylor expansion of the exponential $$\exp(X) = \sum_{n=0}^\infty \frac{X^n}{n!} $$ If we substitute $i$ times the sum of all connected diagrams $iW$ for $X$ in this formula, the term $X^n/n!$ will simply produce the products of $n$ components, i.e. all disconnected (for $n\gt 2$) diagrams with $n$ components.

The combinatorial factor will work, too. Recall that when we evaluate Feynman diagrams, we have to divide by the symmetry factor. The symmetry group of a disconnected, $n$-component diagram includes the permutation group of all the $n$ components if the components are the same, that's why there is $1/n!$ in front of a "fixed single 1-component diagram" to the $n$-th power.

The extra symmetry group from permuting the components is reduced to the product of $n_i!$ over all subgroups of the group of $n$ components that contain the same diagram. But $$ \prod_i \frac{1}{n_i!} $$ is exactly what we get if we calculate $1/n!$ times the coefficient from the expansion of the $n$th-power of the sum of the connected diagrams.


Essentially, the answer to this is nothing but a consequence of the product and chain rule for (functional) derivatives. Think of W as some abstract object where all the Feynman diagrams (up to arbitrary orders) are stored. It does not matter what W looks like exactly. Acting on W[J] with functional derivatives with respect to J, you will start to generate some "unique" diagrams, in the sense that each sequence of applied derivatives with respect to J gives you another diagram, so for example \begin{equation} \frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_2)} \frac{\delta}{\delta J(x_3)} W[J] \end{equation} gives you the 3-point function with end-points $x_{1,2,3}$ etc. Now, think about taking some n functional derivatives from $e^{W[J]}$: \begin{equation} \frac{\delta}{\delta J(x_1)} \ldots \frac{\delta}{\delta J(x_n)} e^{W[J]} \end{equation} The first derivative gives you just $W[J] e^{W[J]}$. Now, you apply the second derivative, where you have to apply the product rule: \begin{equation} \frac{\delta}{\delta J(x_1)} \ldots \frac{\delta}{\delta J(x_{n-2})} \left[ \left( \frac{\delta}{\delta J(x_{n-1})} W[J] \right) + W[J] \frac{\delta}{\delta J(x_{n-1})} \right] e^{i W[z]} \end{equation} Now continue like that until you have taken all $n$ derivatives. You see that more and more terms arise. However, the only term that constantly gives you "new" diagrams is the one, where all derivatives are applied to $W[J]$ and none to the $e^{i W[J]}$ a second time. All the other diagrams are products of terms with fewer derivatives, applied on "different" $W[J]$ and hence correspond to a multiplication of several diagrams that are not full n-point functions.

So or instance, assume that after having applied the functional derivative some $k$-times straight on $W$ and then some $m := n-k$ times on another $e^{i W[J]}$ factor (just by doing product rule) you will get a product of a fully connected $k$-point function multiplied with a fully connected $m$-point function where $k+m = n$.

  • $\begingroup$ I want to find only the connected diagrams for phi 4 theory using tha above formalism. But it seems we need to find functional derivative of logarithm of a differential (inside integral) which acts on an integral containing J's (sources).But isn't this formalism supposed to be more useful compared to other technique where disconnected diagrams also appear. Please suggest or refer to some article so that I can use this to find connected diagrams efficiently. Also I ask is it even possible? Thank you. $\endgroup$
    – quarkonium
    Commented Nov 3, 2016 at 17:30

If you think for a second about permutations:

All permutations are a product of disjoint cycles. So you can write a permutation by multiplying cycles together, then dividing through by the number of ways you can stitch them together because that gives the same permutation.

So to get all permutations, you multiply cycles C by each other, which gives $C^n/n!$ and sum them up. In other words you can generate permutations by taking exponential of cycles, and so the log of permutations are cycles, the connected permutations.

This is made proper using combinatorial species and the symbolic method.

If you think about Z as a generating function for the combinatorial species of the feynman graphs, then taking a logarithm to get connected structures is exactly what you would expect. And if you look at the functional integration in a rough way, basically the same rules you learn about feynman diagrams are the ones given by manipulating the generators for these combinatorial species.


1) A "connected" diagram is a diagram that cannot be separated into two diagrams by cutting a single edge. A single connected diagram is a distinct integral which is a function of a single momentum defined by that edge, so each of these connected diagrams is a simple factor. So any given diagram can be factored into terms that are each represented by these "connected" diagrams.

2) Consider all of the possible connected diagrams $C_i$ indexed by $i$. For any specific diagram, the vectors of counts $n_i$ of $C_i$'s comprising the diagram can be taken as the index over diagrams $D_{[n_i]}$. So, the sum over diagrams $\sum_{[n_i]} D_{[n_i]} = \sum_{[n_i]} \prod_i \frac{C_i^{n_i}}{n_i!}$ where the $\frac{1}{n_i!}$ counts the $C_i^{n_i}$ only once. Partitioning these into clusters of size $N$, where $\sum_i n_i = N$ yields $\sum_{[n_i]} D_{[n_i]} = \sum_N \sum_{[n_i], \sum_i n_i = N}\prod_i \frac{C_i^{n_i}}{n_i!} = \sum_N \frac{1}{N!}\left(\sum_i C_i\right)^N =\exp\left(\sum_i C_i \right)$

There are $n_i!$ combinations of $\delta J_1 \delta J_2 ...$ that generate $C_i$. The number of combinations for the $\prod_i C_i^{n_i}$ terms is $\frac{N!}{\prod_i n_i!}$ counting all the diagrams generated by the $\frac{\delta}{\delta J}$'s from $Z[J]$ with $N$ clusters.

3) The argument can be extended to include both connected diagrams $\sum_i C_i$ attached by 1 edge and vacuum bubbles $\sum_j B_j$ attached by no edges. Then for these diagrams that include both clusters and bubbles, $\sum_{[n_i]} D_{[n_i]} = \exp\left(\sum_i C_i + \sum_j B_j \right) = \exp\left(\sum_i C_i \right) \exp\left( \sum_j B_j \right)$ Identifying of $Z[J] = \sum_{[n_i]} D_{[n_i]}$, $Z[0] = \exp\left(\sum_j B_j \right)$, and $W[J] = \sum_i C_i$, this reduces to $$Z[J] = Z[0]\exp(W[J])$$

4) The above gives a sense of how the $\ln Z$ comes about, but this doesn't take into account the $J$'s. Ultimately, the details come from counting $J$'s, and ensuring that the number of derivatives pulled down by the $\frac{\delta}{\delta J(*)}$'s are all accounted for, since the $Z[J]$ terms come from $$Z[J] = \int {\cal{D}} \psi \exp\left(-\frac{1}{2}\psi(*)K(*,*)\psi -V(\psi) +J(*)\psi(*)\right) = C \exp\left(-V\left( \frac{\delta}{\delta J(*)}\right)\right)\exp\left(-\frac{1}{2}J(*)K^{-1}(*,*)J(*)\right)$$

Consider some set of diagrams $C_k(x_1, x_2, ...)$ and $$W[J] = \sum_{j=0}^\infty \int dx_1 dx_2 \cdots dx_j \frac{1}{j!} C_k(x_1, x_2,... x_j)J(x_1)J(x_2)\cdots J(x_j),$$ shortened by writing the integrals in terms of '*'s: $$W[J] = \frac{1}{n!}\sum_{j=0}^\infty\frac{1}{j!} C_j(*)J^j(*),$$ and related to the correlation/Green functions such that $$Z[J] = \exp\left(W[J]\right).$$ With $J=0$, this reduces to $Z[J=0] = \exp\left(C_0\right),$ so that $$Z[J] = Z[0]\exp\left(\sum_{k=1}^\infty \frac{1}{k!} C_k(*)J^k(*)\right) = Z[0]\sum_{n=2}^\infty \frac{1}{n!} G_n(*)J^n(*).$$ The $Z[0]$ contain loops and unconnected "vacuum bubbles" that tend to diverge. This implies $$\sum_{n=2}^\infty \frac{1}{n!} G_n(*)J^n(*) = \exp\left(\sum_{k=1}^\infty \frac{1}{k!} C_k(*)J^k\right) = \sum_{l=0}^\infty \frac{1}{l!}\left(\sum_{k=1}^\infty \frac{1}{k!} C_k(*)J^k\right)^l.$$ Then $C_1 = 0$, $G_2 = C_2$. For even $V$, $C_3 = 0$, $G_4 = C_4 + 3C_2^2$, ... The $C$'s are seen to correspond to factorable terms contributing to the diagrams. Such factors emerge in $k$ space as components that attach through a single edge - that is, cutting a single edge isolates the graphical component. Such subgraphs represent distinct factors, and can be identified from $C \exp\left(-V\left( \frac{\delta}{\delta J(*)}\right)\right)\exp\left(-\frac{1}{2}J(*)K^{-1}(*,*)J(*)\right)$ to each order in $J$. These subgraphs are called "one-particle irreducible" or 1PI diagrams.

These may be written out in terms of "exponential Bell polynomials" which may be defined in terms of the generating function $$\exp\left(u\sum_{j=0}^\infty \frac{x_j t^j}{j!}\right) = \sum_{n,k \ge 0} \frac{t^n u^k}{n!} B_{n,k}(x_1, x_2,...,x_{n-k+1}).$$ By this: $$G_n =\frac{1}{n!} \sum_{k=1}^n B_{n,k} (C_1, C_2,...,C_{n-k+1}).$$ The $B_{n,k}$ have the form

$$ B_{n,k}(C_1, C_2,...,C_{n-k+1}) = \sum_{\{j_l\}} \frac{n!}{j_1! j_2! \cdots j_{n-k+1}!} \left( \frac{C_1}{1!} \right)^{j_1} \left(\frac{C_2}{2!}\right)^{j_2} \cdots \left(\frac{C_{n-k+1}}{(n-k+1)!}\right)^{j_{n-k+1}}$$


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