# Zee's Nutshell: Feynman diagrams "baby problem": Connected vs. Disconnected

On page 47 of A. Zee's QFT in a Nutshell, he explains how disconnected Feynman diagrams can be built from lower-order connected diagrams:

I don't know how to understand formula $(6)$.

I understand that there must be a way to "decompose" $Z(J,\lambda)$ into connected Feynman diagrams. After all, if we know the coefficients for the "straight line" $|$ and for the vacuum bubble $8$, we should be able to calculate the coefficient for the combined diagram $|8$ from the "parts".

But how exactly does that work? Maybe I can work out the details if I have one concrete example of a term $W(J,\lambda)$, but all my efforts so far were fruitless.

To show my efforts, here's one example: I tried to use $(6)$ on the concrete example $Z(2,1)$:

$$\frac{5}{16m^6}(-\lambda)J^2 = Z(J=2, \lambda=1) \overset{(6)}{=} Z(J=0, \lambda=1) e^{W(J, \lambda)} = \frac{1}{8m^4}(-\lambda)e^{W(J, \lambda)}$$

$$e^{W(J, \lambda)} = \frac{5}{2m^2}J^2$$

But how is that useful? How can I calculate $W(J,\lambda)$ without having $Z$?

EDIT: If you don't know the book, the "baby problem" is the integral

$$Z(J) = \int_{-\infty}^{+\infty}dq e^{-\frac 12m^2q^2-\frac{\lambda}{4!}q^4+Jq}$$

This is then expanded in $J$ terms, which yields integrals that can be solved using Wick's theorem. The result is a double series in $\lambda$ and $J$ with coefficients

$$Z(J^{2a}, \lambda^b) = \frac{(2a+4b-1)!!}{b!(2a)!(4!)^b} \frac{1}{m^{2a+4b}}(-\lambda)^b J^{2a}$$

• I'm not exactly sure what your question is. The equation $(6)$ is just the statement that the sum $Z(J)$ over all diagrams decomposes into the product of the disconnected and the partially connected diagrams. Commented Oct 30, 2015 at 17:16
• My question is, what is $W(J, \lambda)$ exactly? I know it is something like $Z(J, \lambda)$, but "contains" only connected diagrams. What is it quantitatively? Is there a formula to calculate it? As I said, maybe I can understand it if I have some concrete example, with the exact numbers calculated. An example like $Z(J=0, \lambda=1)=\frac{1}{8m^2}(-\lambda)$, but for $W$.
– Bass
Commented Oct 30, 2015 at 17:57
• Related: physics.stackexchange.com/q/107049/2451 , physics.stackexchange.com/q/129080/2451 and links therein. Commented May 6, 2017 at 13:32

First, $(6)$ is just the definition of $W$. Since $e^W = Z(J)/Z(0)$ (suppresing the $\lambda$ dependence), it is given by the sum of all diagrams divided by the sum of all diagrams with no external legs. Now, when you have a disconnected diagram, it factors into the product of its connected subdiagrams. In your notation, the amplitude for $|8$ is just the amplitude for $|$ times the amplitude for $8$.
Now a magical thing happens. When you sum all diagrams (connected and disconnected) to get $Z$, you can sort of divide it into factors, where each factor is the sum of many copies of a single connected diagram. If you have $n$ copies you need to raise the diagram to the power $n$ and divide by $n!$ to account for normalization, so what ends up happening is that $Z$ is proportional to the exponential of the sum of all connected diagrams:
$$Z(J) \propto \exp(\sum_I C_I)$$
Where each $C_I$ is a connected diagram. Since $Z$ was also proportional to $e^W$, we get that $W$ is the sum of connected diagrams; this is the quantity we calculate when actually doing Feynman diagrams. This is because calculating observables amounts to taking functional derivatives of $Z$ with respect to $J$ and then setting $J=0$, which brings down factors of $W$.
By the way, Zee doesn't care about the proportionality constant but it makes sense to have $Z(0)=\langle 0 | 0\rangle = 1$; therefore, if we say $Z(J) = \exp(W(J))$ and omit vaccum diagrams from $W$, we get $Z(0)=1$.