# Question on irreducible versus reducible Feynman graphs

• RedX
In summary, the conversation discusses the functional \mbox{ }e^{iW[J]} and the Legendre transformation to obtain a functional in \phi(x) instead of J(x). It also mentions the calculation of \Gamma[\phi] by diagrammatic methods and how it differs from W[J] by subtracting \int d^4x \mbox{ } J(\phi) \phi and evaluating at J(\phi) = -\frac{\partial \Gamma}{\partial \phi}. This process eliminates 1-particle reducible graphs, resulting in \Gamma[\phi] being given by summing 1-particle irreducible graphs. The proof for this equivalence is not obvious and can be found in various
RedX
Consider the functional:

$$(1) \mbox{ }e^{iW[J]} = \int d \hat{\phi} \mbox{ }e^{i\int d^4x \mbox{ } \mathcal L(\hat{\phi})+J\hat{\phi}}$$

Define a Legendre transformation to get a functional in $$\phi(x)$$ instead of $$J(x)$$:

$$(2) \mbox{ }\Gamma[\phi]=W[J(\phi)]- \int d^4x \mbox{ } J(\phi) \phi$$

where $$J(\phi)$$ is found by solving $$\frac{\partial W[J]}{\partial J}=\phi$$ for J in terms of $$\phi$$ and substituting this expression in for the value $$J(\phi)$$. Also, by differentiating eqn (2) with respect to $$\phi$$, one can show:

$$\frac{\partial \Gamma[\phi]}{\partial \phi}+J(\phi)=0$$

To calculate $$\Gamma[\phi]$$ by diagrammatic methods instead, exponentiate it and substitute the earlier result for $$e^{iW[J]}$$:

$$(3) \mbox{ } e^{i\Gamma[\phi]}= e^{i(W[J(\phi)]- \int d^4x \mbox{ } J(\phi) \phi )} =\int d \hat{\phi} \mbox{ }e^{i\int d^4x \mbox{ } \mathcal L(\hat{\phi})+J(\phi)(\hat{\phi}-\phi)}$$

Now here is what I don't understand. The author of the paper now says:

"A saddle-point evaluation of eqn. (1) gives W[J] as the sum of all
connected graphs that are constructed using vertices and propagators built from
the classical lagrangian, L, and having the currents, J, as external lines. But $$\Gamma[\phi]$$
just differs from W[J] by subtracting $$\int d^4x \mbox{ } J\phi$$, and evaluating the result at the specific configuration $$J(\phi) = -\frac{\partial \Gamma}{\partial \phi}$$. This merely lops off all of the 1-particle
reducible graphs, ensuring that $$\Gamma[\phi]$$ is given by summing 1-particle irreducible
graphs."

How does one see that adding all irreducible graphs is equivalent to evaluating eqn. (3)? In other words, how does doing all that "merely lops off all the 1-particle reducible graphs"?

Well, I wouldn't say that from what the paper says it is obvious ... From my point of view the proof for this has to be constructive. You will probably find one in Zinn Justin book or in Itzykson's. For more pedagogical aspects I would say : Abers and Lee Physics Reports on gauge theories and Iliopoulos, Martin and a 3rd in Rev mod phys about introduction on functional methods

## 1. What is the difference between irreducible and reducible Feynman graphs?

Irreducible Feynman graphs are those that cannot be broken down into smaller subgraphs, while reducible Feynman graphs can be broken down into smaller subgraphs.

## 2. How do irreducible and reducible Feynman graphs affect calculations in quantum field theory?

Irreducible Feynman graphs are important for calculating scattering amplitudes in quantum field theory, while reducible Feynman graphs are often used as a way to simplify calculations.

## 3. Can a reducible Feynman graph have loops?

Yes, a reducible Feynman graph can have loops, as long as it can be broken down into smaller subgraphs.

## 4. Are reducible Feynman graphs always simpler than irreducible ones?

Not necessarily. While reducible Feynman graphs can sometimes simplify calculations, they can also lead to more complicated expressions in certain cases.

## 5. How do we determine if a Feynman graph is irreducible or reducible?

A Feynman graph is irreducible if it cannot be broken down into smaller subgraphs, and it is reducible if it can be broken down into smaller subgraphs. This can be determined by examining the graph and identifying any subgraphs that are connected by a single line.

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