# Srednicki QFT Chapter 29: Feynman diagrams for calculating the effective action

I am trying to work my way through Srednicki Chapter 29 on Wilson's approach to renormalisation. However I am unsure why the Feynman diagrams Srednicki considers and calculates in this chapter are the correct one.

In the chapter, we consider a $\phi^4$ theory in Euclidean space with path integral $$Z(J) = \int D\phi \ e^{-S_{E} + \int J \phi} \tag{29.4}$$ where the Euclidean Action $$S_E = \int d^4x \left( \frac{1}{2}Z_{\phi} \partial_{\mu} \phi \partial_{\mu}\phi + \frac{1}{2}Z_{m} m_{ph}\phi^2 + \frac{1}{4!}Z_{\lambda}\lambda_{ph}\phi^4\right).\tag{29.5}$$

As far as I understand, we then impose some momentum cut-off $\Lambda$ and split the field $$\phi (x) = \varphi (x) + \chi (x) ,$$ where $\varphi (x)$ has support in momentum space only for $|k| < \Lambda$ while $\chi$ has support only for $|k| > \Lambda$. This should split $$D \phi = D\varphi D\chi,$$ and the action becomes

$$S_E = \int d^4x \left( \frac{1}{2}Z_{\phi} \partial_{\mu} \varphi \partial_{\mu}\varphi + \frac{1}{2}Z_{m} m_{ph}\varphi^2 + \frac{1}{4!}Z_{\lambda}\lambda_{ph}\varphi^4\right) + \int d^4x \left( \frac{1}{2}Z_{\phi} \partial_{\mu} \chi \partial_{\mu}\chi + \frac{1}{2}Z_{m} m_{ph}\chi^2 + \frac{1}{4!}Z_{\lambda}\lambda_{ph} \left( \chi^4 + 4 \chi^3 \varphi + 6 \chi^2 \varphi^2 + 4 \chi \varphi^3 \right) \right).$$

Now we want to integrate out the high momentum modes to get an effective action

$$Z(J) = \int D\varphi e^{-S_{eff}(\varphi) + \int J \varphi}, \tag{29.9}$$

where

$$S_{eff}(\varphi) = - \log \left( \int D\chi e^{-S_E(\varphi , \chi)} \right).\tag{29.10}$$

Srednicki then says that to calculate the parameters multiplying the operators appearing in the effective lagrangian

$$L_{eff}(\varphi) = \frac{1}{2}Z(\Lambda) \partial_{\mu} \varphi \partial_{\mu}\varphi + \frac{1}{2} m(\Lambda)^2 \varphi^2 + \frac{1}{4!}\lambda (\Lambda)\varphi^4 + \sum_{d \geq 6} \sum_{i} c_{d,i}(\Lambda) \mathcal{O}_{d,i}\tag{29.11}$$

we need to sum over the 1PI diagrams with the correct number of external $\varphi$ lines and internal $\chi$ propagators.

Now, what I do not understand is why we only need to sum over the 1PI diagrams. To me, the formula for the effective action would suggest that we should sum over all connected* diagrams with only internal $\chi$ propagators and not just the 1PI diagrams. So for example, for calculating the coefficient of $\varphi^6$, why do I not consider a diagram joining two vertices with 3 external $\varphi$ lines with a single $\chi$ line?

I don't have the book in front of me but I think one should not take this explanation of Wilson's RG too literally. If you insist on an exact identity $$Z[J]=\int D\phi\ e^{-S_E+J\phi}=\int D\varphi\ e^{-S_{eff}(\varphi)+J\varphi}$$ then, in principle, the effective action will not be given by a local effective lagrangian. Namely the higher operator terms will not be like $$\int dx \ \varphi(x)^n$$ but more like $$\int\cdots\int dx_1\cdots dx_n\ K(x_1,\ldots,x_n)\varphi(x_1)\cdots\varphi(x_n)$$ for some nonlocal kernels $K$ which are made of the connected diagrams with $\chi$ propagators. One could write a local approximation which amounts to replacing the last quantity by say $$\int\cdots\int dx_1\cdots dx_n\ K(x_1,\ldots,x_n)\varphi(x_1)^n$$ so the contribution of the diagrams involves the effective couplings $$\int\cdots\int dx_2\cdots dx_n\ K(x_1,\ldots,x_n)$$ $$=\int\cdots\int dx_2\cdots dx_n\ K(0,x_2,\ldots,x_n)$$ by translation invariance. If you now write this in momentum space you see that the $\chi$ connected graphs are being evaluated at zero external momentum. If the graph is not 1PI, there is a bridge or separating internal line which should have momentum zero flowing through it. But this a $\chi$ propagator and by construction it vanishes for momenta $<\Lambda$ and in particular zero. In conclusion, in principle one should include all connected graphs but the only survivors of the zero momentum evaluation are the 1PI ones.

Edit as per AFT's doubts: An excellent account of the use of the position space operation of moving points $x_1,\ldots,x_n$ to all sitting at say $x_1$, in order to do renormalization, is in Section II.2 of the book "From Perturbative to Constructive Renormalization" by Vincent Rivasseau. For the even more mathematically inclined, see also the recent article by Martin Hairer "An analyst's take on the BPHZ theorem".

• I'm not sure I buy this explanation (IMHO the replacement $\phi_1\cdots\phi_n\to\phi^n$ does not seem to be a valid approximation in any sense whatsoever). For now, note that there is a copy of Srednicki's book for free on his webpage. – AccidentalFourierTransform Nov 30 '17 at 21:16
• @AccidentalFourierTransform: have you heard of the local potential approximation and the derivative expansion? The replacement $\varphi_1\cdots \varphi_n\rightarrow \varphi^n$ is the very heart of renormalization in QFT. The intuition behind it is that $K$ is made of hard $\chi$ propagators which decay fast in position space if the points get further away from each other. On the other hand this nonlocal kernel is coupled with external soft or slowly varying fields $\varphi(x_i)$. – Abdelmalek Abdesselam Nov 30 '17 at 22:11
• Perhaps you are right. I just don't know. – AccidentalFourierTransform Dec 1 '17 at 11:25
• @AccidentalFourierTransform: I'm sure you know but probably in a different language: momentum space. In the BPHZ framework you renormalize a subgraph with superficial degree of divergence $\delta\ge 0$ by subtracting the order $\delta$ Taylor expansion around zero momentum. This is the basic operation recursively iterated in Zimmermann's forest formula. If you change coordinates to position space then this subtraction, in the easiest $\delta=0$ case, is the difference between $\varphi_1\cdots\varphi_n$ and $\varphi_1^n$. I will edit my answer to add a detailed reference. – Abdelmalek Abdesselam Dec 1 '17 at 12:31

The reason is the linked-cluster theorem. It states that given an action $S(\chi)$, then the Feynman diagrams generated by

$Z = \int \mathcal{D}[\chi] e^{iS(\chi)}$

are sometimes disconnected, e.g. you may obtain upon evaluation of an order in the perturbative series expansion not only one single feynman diagram which is connected, you may obtain multiple diagrams that are independent of each other. Let $W$ be the generating functional for Feynman diagrams that are all connected. Then, linked-cluster theorem states that

$W = \log Z$.

Because in the effective action computation you will have exactly that logarithm, you have only add up all connected diagrams and therefore you can ignore disconnected ones. Reducible Feynman diagrams are made to irreducible ones.

Moreover, if $Z[J]$ depends on the source field $J$, from which you can derive all kind of correlation functions by taking derivatives, then you can show that by taking $J$-derivatives from the functional $W[J]$, you will obtain all possible cumulants. The standard deviation

$\sigma_{XY} = <0|T(XY)|0> - <0|X|0><0|Y|0>$

for two observables $X,Y$ and time ordering operator $T$ is a simple form of cumulant, because it measures a new statistical information, the deviations and ignores contributions from simple averages $<0|X|0>$ by subtracting these.

• I appreciate your comment about the connected diagrams, that was my mistake for not including this. However, even with this, I am still not sure why Srednicki restricts to only the 1-particle irreducible diagrams as opposed to all connected diagrams. – Rhinosaur Nov 30 '17 at 20:14
• Note that $\text{irreducible diagrams}\subset\text{connected diagrams}\subset \text{all diagrams}$ as given by $\Gamma,W,Z$. Connected is not the same as irreducible. – AccidentalFourierTransform Nov 30 '17 at 20:15
• You compute Gaussian moments in variable $\chi$ in this case and as you compute these you will get a graph, where a vertex will be connected with at least two propagators (except the $\chi \phi^3$-term that is simply an external leg). The $\chi^2 \phi^2$ has two factors $\chi$ that can link to other vertices of the diagram after computation of the Gauss integrals. So if you cut one link, it will link to at least one vertex and connectivity is preserved. – kryomaxim Nov 30 '17 at 20:39

Srednicki is not implying that the Wilsonian effective action is the 1PI effective action, if that's what OP is asking. See also this Phys.SE post.

Rather Srednicki is merely pointing out that the partition function (29.9) for the effective Wilsonian action (like any partition function) can most conveniently be analyzed via the Legendre transformation to 1PI diagrams.