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So the Feynman path integral tells us to sum over all the possible paths a particle can take and to weight each path by the exponential of i times the action along the path.

What does a loop-integral look like in this picture? That is, as the loop momentum goes to infinitiy, does the action vary or stay the same? If it stays the same then it seems any path with a loop would diverge. But if the action varies with loop momentum, maybe the oscillation of the exponential can cancel, or at least sum to a finite value. This might be something like the argument that the classical path appears naturally in the FPI formalism.

So does the action change when a loop momentum increases?

If I'm going to find my answer in the linked question I would be glad to delete this one. And I am planning to go through PSI 2014/2015 - Quantum Field Theory II in the near future. So, if the answer is "go learn that" I am happy with that answer also.

Qmechanic
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Dan
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1 Answers1

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It turns out that "the path integral" is a very bad name for what is really a large set of different techniques that involve path integrals. I can think of at least 4 different quantities that can all be called a path integral (the generating functional for disconnected diagrams, the generating functional of connected diagrams, the generating functional for the one-particle irreducible effective action, and the Wilson path integral on a lattice with some cutoff energy scale $\Lambda$), and for each there are different techniques and approximations for evaluating them. Which is to say, there is a lot to learn, and it's probably best to start learning from the beginning without too many preconceived notions of how things work.

Having said that, you are asking good questions.

It's a little hard to directly map what you are computing with a loop integrals, to the value of the action. The result of one Feynman diagram, is not really a path integral, but more like a piece of a path integral. Having just complained about different meainings of the phrase "the path integral," I'm now going to choose one -- where the path integral is used to an effective action $\Gamma[\Phi]$ that can be used to generate correlation functions (which in turn tell us about observables in the theory). If $Z$ is the path integral and $\Gamma$ is the one-particle irreducible effective action (the name isn't so important), then \begin{eqnarray} e^{i \Gamma} &=& Z \\ \Gamma &=& \Gamma_{\rm tree} + \hbar \Gamma_{1-loop} + \cdots \end{eqnarray} Then we can construct a power series for, say, $\Gamma_{1-loop}$ \begin{equation} \Gamma_{1-loop}[\phi] = \Gamma_{1-loop, 2} \phi^2 + \Gamma_{1-loop, 3} \phi^3 + \Gamma_{1-loop, 4} \phi^4 + \cdots \end{equation} where the operators $\Gamma_{1-loop, n} \phi^n$ are complicated non-local operators acting on the fields. For example taking $n=2$, we might have $\Gamma_{1-loop, 2} \phi^2 = \int d^4 x_1 d^4 x_2 k(x_1, x_2) \phi(x_1) \phi(x_2)$ for some kernel function $k(x_1, x_2)$. Then to compute $\Gamma_{1-loop, 4} $, we compute a sum of Feynman diagrams with one loop \begin{equation} \Gamma_{1-loop, 4} = F^{(1)}_{1-loop, 4} + F^{(2)}_{1-loop, 4} + \cdots \end{equation} For simplicity, I've suppressed the fact that both the left and right side depend on the 4 positions of the four field operators $\phi$, and am ignoring the fact that this calculation is almost always actually done in momentum space. In other words, the sum of all the loop diagrams with a given structure will give you one piece of the result of doing the path integral, to one-loop order. How many Feynman diagrams you have to sum to get $\Gamma_{1-loop, 4}$ will depend on what kind of interactions you have in the theory. Focusing now thinking on just a single Feynman diagram, such as $F^{(1)}_{1-loop, 4}$, you see you aren't getting the value of a path integral per se, but only one term in one way of computing the value of a path integral (which in turn may just be a step to the full answer you are looking for). So, the direct link you are asking for in your question between the value of the action in the path integral, and the loop momentum in a Feynman diagram, is difficult to make rigorously.

However, taking a broader view of your question, I think it's fair to say that whether or not a given Feynman diagram diverges does not directly depend on the oscillating exponential factor. It has more to do with the structure of the diagram.

Nevertheless, I think you are grasping toward an important piece of intuition, which is that the (ultra-violet) divergences arise in quantum field theory because of situations where we look at fluctuations in the quantum field in an arbitrarily small region of spacetime. In the path integral, these fluctuations arise when looking at correlations of fields that are very closed together. In Feynman diagram happens, the divergences happen when the momentum (~1/distance) goes to infinity. So, there is a "moral" connection of the kind you are suggesting, even if not a direct, mathematical one between the two quantities you had in mind.

The oscillating factor is built into the loop expansion, although not in the way you were anticipating. Individual diagrams will diverge or not, but this has nothing to do with the oscillating factor. If you compute the version of the path integral that is the generating functional for the one particle irreducible effective action, you find that the WKB / saddle point approximation of the path integral is precisely the loop expansion. The WKB approximation effectively takes into account these cancellations due to the oscillating factor, and orders the results by how strong the cancellation is. The tree diagrams dominate the answer because they are computed using the part of the integrand that does not oscillate rapidly and cancel; the 1-loop diagrams are the first correction to that; the 2-loop diagrams are the next correction; and so on. So by computing tree diagrams, and then one loop diagrams, and so on, and summing the results to a given order, we are taking advantage of that cancellation you mentioned.

Andrew
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