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In chapter 21 (p.127-129) of Srednicki's book the quantum action $\Gamma(\phi)$ is defined in formula (21.1) I won't repeat here (it's quite long). Then he considers the following path integral:

$$Z_\Gamma(J) = \int \mathcal{ D} \phi \exp\left[ i\Gamma(\phi) + i \int d^dx J\phi\right] = \exp[iW_\Gamma(J)].\tag{21.4}$$

adding:

$W_\Gamma(J)$ is given by the sum of connected diagrams (with sources) in which each line represents the exact propagator, and each $n$-point vertex represents the exact 1PI vertex $\bf{V}_n$.$W_\Gamma(J)$ would be equal $W(J)$ if we included only tree diagrams in $W_\Gamma(J)$.

Isolating the tree-level contribution to the path integral by means of introduction of the dimensionless parameter called $\hbar$ we have the following path integral:

$$Z_{\Gamma,\hbar}(J) = \int \mathcal{ D} \phi \exp\left[ \frac{i}{\hbar}\left(\Gamma(\phi) + i \int d^dx J\phi\right)\right] = \exp[iW_{\Gamma,\hbar}(J)]\tag{21.6}$$

It is followed by the following text:

In a given connected diagram with sources, every propagator (including those connected to sources) is multiplied by $\hbar$, every source by $1/\hbar$, and every vertex by $1/\hbar$.

  1. I don't get this assertion. There is no hint neither why it should be like this.

Later on $W_{\Gamma,\hbar}(J)$ is developed in a series of loop occurrence in diagrams, or in different words a series in orders of $\hbar$:

$$W_{\Gamma,\hbar}(J)= \sum_{L=0}^{\infty} \hbar^{L-1} W_{\Gamma,L}(J).\tag{21.8}$$

I can understand the formula under the assumption that I take the above cited assertion for correct/granted. Srednicki's keeps on saying:

If we take the formal limit of $\hbar\rightarrow 0$, the dominant term is the one with $L=0$, with is given by the sum of tree diagrams only. This is just what we want. We conclude that:

$$W(J) = W_{\Gamma, L=0}(J).\tag{21.9}$$

  1. This assertion is also curious (I don't understand it ). I thought that $W(J)$ would be the sum of all connected diagrams whereas here it seems to be only a subset, i.e. all connected diagrams without loops. I assume that it is related with what was said above (the reason I included the part above).

Actually, it seems that Srednicki's explanation of the quantum action is elegant, so I would really appreciate if somebody could explain the mentioned assertions to me what they actually mean.

Qmechanic
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Frederic Thomas
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1 Answers1

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  1. The $\hbar$-weights are a direct consequence of eq. (B9) in my Phys.SE answer here:

  • A vertex has $\hbar$-weight$=-1$.

  • An internal propagator comes with $\hbar$-weight$=+1$ (rather than $-1$) because an internal propagator is accompanied by 2 source differentiations (which each carry $\hbar$-weight$=+1$).

  • By the same token, an external leg (=source+propagator) has $\hbar$-weight$=0$ because it is accompanied by 1 source differentiation. Hence a source has $\hbar$-weight$=-1$.

  1. $W(J)$ is by definition a sum of all possible connected diagrams made out of bare/free propagators and (amputated) bare vertices. The statement is that it is also the sum $W_{\Gamma, L=0}(J)$ of all possible trees made out of full propagators and (amputated) 1PI vertices. This is also explained in my Phys.SE answer here.
Qmechanic
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