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

Question

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?

Answer 1

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.

Answer 2

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".

Answer 3

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.