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Generating functional in field theory in 0 dimensions

+ 3 like - 0 dislike
164 views

I am following these set of notes http://www.damtp.cam.ac.uk/user/dbs26/AQFT/chaps1+2.pdf on field theory on zero dimensions and I have some concerns about the generating functional. In equation (2.1) we are told that
$$
Z=\int_{\mathbb{R}}d\phi\,e^{-S(\phi)}
$$
My first question is, why the Euclidean signature? Why can't we consider theories with
$$
Z=\int_{\mathbb{R}}d\phi\,e^{iS(\phi)}
$$
with a Minkowski version of the exponent? Also, in field theories the generating functional carries a source $J$. I would guess that in the zero dimensional version we should have also a source. $Z$ would be a function instead of a functional but we could still have $J$ right?
$$
Z(J)=\int_{\mathbb{R}}d\phi\,e^{-S(\phi)-J\phi}
$$
So why is there no mention of this?

asked Jun 12 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ no revision ]

$Z$ is a function of $S$, and when you change $S(\phi)$ to $S(\phi)+J\phi$ you get the version with a source. But as long as no perturbation theory is done, there is no need for a source term. 

The Euclidean signature is what is needed in statistical mechanics. This is indeed their context since Skinner is primarily interested in the partition function.

Why is there an integral over $\mathbb{R}$? The integration is over the space of all field configurations, why is it identified with the real line?

A zero dimensional real scalar field is simply a map from a connected zero dimensional manifold (one point) to $\mathbb{R}$ and thus the space of field configurations is simply $\mathbb{R}$ itself.

As for the generating functional, sure your reference might not mention it but others do. See chapter 9 of Mirror Symmetry by Hori et al and eqn (3.15) in Tom Banks' Modern QFT book.
 

@ArnoldNeumaier You say that there is no need for the source term if you don't work perturbatively. I don't get that. To my understanding, you put the source term to have an object $Z[J]$ that encodes all green's functions,which you get via functional differentiation. This perspective has nothing to do with perturbation theory. What do you mean exactly?

In statistical mechanics (which is the context) one doesn't need all Green's functions. The latter are needed only for doing perturbation theory. This is all I had meant. In any case, after having discussed $Z(S(\phi))$ one can apply all results of this discussion to $Z(S(\phi)+J\phi)$ and gets the case with source.

1 Answer

+ 3 like - 0 dislike

I actually took this class at Cambridge very recently, so I feel very qualified to answer.

As for your first question, the entire course is taught in the context of the Euclidean signature. This is primarily because it is taught within the mathematics department, and a vast majority of work done in mathematical QFT is done in a Euclidean signature. This way, we don't have to worry about pseudo-Remannian manifolds and the partition function is a bit more well-defined as the action $S$ is bounded from below.

As for your second question: you are correct. Typically, a generating functional contains a linear "source" term. And you are free to add one if you want. However, in the context of the course, this chapter simply serves as an introduction to some advanced topics (symmetries and Ward identities, the graph theoretic interpretation of Feynman diagrams, effective field theory, and supersymmetry) in a context where the topics themselves aren't obscured in messy integrals.

I hope this helped!

answered Jun 18 by Bob [ no revision ]

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