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I'm going over Weinberg Quantum effective action section in his second book, I understood his derivation and reasoning up to the equation:

$$iW[J]=\int_{\substack{Connected\\trees}}\left[\mathcal{D}\phi\right]exp\left\{i\Gamma[\phi]+i\sum_r\int d^4x\phi(x)J(x)\right\}.\tag{16.1.16}$$

But then he argues that any connected graph for $iW[J]$ can be regarded as a tree, as long as, the vertices are 1PI subgraphs. Then he continues to reason that the effective action can be written as:

$$i\Gamma[\phi_0]=\int_{\substack{1\text{PI}\\connected}}[\mathcal{D}\phi]exp\left\{iS[\phi+\phi_0]\right\}.\tag{16.1.17}$$

Or as:

$$exp\left\{i\Gamma[\phi_0]\right\}=\int_{1\text{PI}}[\mathcal{D}\phi]exp\left\{iS[\phi+\phi_0]\right\}.\tag{16.1.18}$$

  1. Can someone clarify how he came to this conclusion?

  2. In his derivation of the effective potential he starts from the action:

$$S[\phi]=-\int d^4x\left[\lambda+\frac{1}{2}\partial_\rho\phi\partial^\rho\phi+\frac{1}{2}m^2\phi^2+\frac{1}{24}g\phi^4\right]\tag{16.2.1}$$ and wants to calculate the effective action to one loop order so he shifts the action by $\phi_0$ and since he only wants up to one loop order the only relevant term after the shift is:

$$\frac{1}{2}\partial_\rho\phi\partial^\rho\phi+\frac{1}{2}\mu^2\phi^2\,\,\,\,,\,\,\,\,\text{Where:}\,\,\mu^2=m^2+\frac{1}{2}g^2\phi_0^2.$$

I agree with this since this is the only term that can create one loops when expanded in $g$. But then he writes the desired quantity he wants to calculate:

$$exp\left(i\Gamma^{\text{one loop}}[\phi_0]\right)=\int[\mathcal{D}\phi]exp\left\{-\frac{i}{2}\int d^4x\left[\partial_\rho\phi\partial^\rho\phi+\mu^2\phi^2\right]\right\}.$$

And proceeds with the calculation. But what about the other uninteresting terms that don't contribute to the one loop order? how is he justified to omit them out? shouldn't it be this:

$$exp\left(i\Gamma^{\text{one loop}}[\phi_0]\right)=\int[\mathcal{D}\phi]exp\left\{-\frac{i}{2}\int d^4x\left[\partial_\rho\phi\partial^\rho\phi+\mu^2\phi^2\right]+\int d^4x\left[\text{irrelevant terms}\right]\right\}$$

and thus his integration technique is not valid?

Raeed Mundow
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1 Answers1

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Regarding the second question if anyone is interested in the justification for his calculation, it was actually simple, the effective action is equal to the 1PI diagrams created by the classical action (with a shift), the quadratic term is the only one that can create diagrams of one loop, so when we do a perturbative calculation we do an expansion and we get a bunch of terms but we are only interested in terms that can create one loop and one loop only, so we throw out all the extra garbage and we are left with the full expansion of the relevant term I wrote in the question. Another way to look it is to say that the one loop diagrams are created by the quadratic term so we can say:

\begin{align} exp\left\{i\Gamma^{1-\text{loop}}[\phi_0]\right\} = \sum_{\substack{\text{one loop diagrams}}}=\int[d\phi]e^{-\frac{i}{2}\int d^4x\left[\partial_\rho\phi\partial^\rho\phi+\mu^2\phi^2\right]} \end{align}

I'm still a newbie to think of diagrams instead of math, someone whos experienced would have seen immediately.

Edit - It is the whole partition function since this partition function can only generate 1 loop "bubble", easy to see when you remember that $\mu^2=m^2+\frac{g^2\phi_0^2}{2}$, so when you do an expansion you and try to contract a connected diagram, you only end up with a loop.

Raeed Mundow
  • 97