# A Feynman rules for a 0-dimensional field theory

1. Oct 6, 2016

### spaghetti3451

Consider the partition function $Z(\lambda)$ of the $0$-dimensional scalar $\phi^{4}$ theory

$Z(\lambda)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}d\phi\ \exp\{-\frac{1}{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4}\}.$

It can be shown that

$Z(\lambda)=\sum\limits_{n=0}^{\infty}c_{n}\lambda^{n},$

where $c_{n}=\frac{(-1)^{n}(2n-1)!!}{(4!)^{n}n!}.$

Can we instead propose a set of Feynman rules to compute the $c_n$'s, or can Feynman rules only be written down for the correlation functions of the theory?

2. Oct 8, 2016

### A. Neumaier

Since you have 0 space-time dimensions, there is nothing to take correlations of.

3. Oct 8, 2016

### vanhees71

It's just a formal Taylor series of an integral. The coefficients are determined by
$$\int_{-\infty}^{\infty} x^{4n} \exp(-x^2/2)=2^{2n+1/2}\Gamma[2n+1/2].$$
The message is that you get a series with 0 radius of convergence.

Since you do something similar with the path integral for $\phi^4$ theory this indicates that also the corresponding Dyson series of perturbation theory is not convergent, but has to be taken as an asymptotic series.

4. Oct 9, 2016

There are interesting 0+1d models where you can write down the partition function, correlation functions from diagrammatics etc., but I'm not really sure what you mean in this case.

5. Oct 11, 2016

### spaghetti3451

I am not talking about a $0+1$-dimensional quantum field theory. This is a $0+0$-dimensional quantum field theory. The field $\phi$ exists at a single point in space-time.

Hold on, if the field $\phi$ exists at a single point in space-time, does it make sense to talk about its dynamics via the partition function, because a field that exists at a single point in space-time should have no dynamics, right?

6. Oct 11, 2016

### spaghetti3451

The series is not convergent, but I presume that the integral could still be convergent. Is that so?

If the series is not convergent, then it is nonsense to write the integral as a Taylor expansion. Why do I still find Taylor expansions and corresponding Feynman diagrams of this integral in QFT textbooks?

7. Oct 12, 2016

### vanhees71

The reason is that perturbation theory is the only way you can evaluate relativistic QFTs in a straight forward way.

8. Oct 12, 2016

### Demystifier

It's an asymptotic series. This means that it is not convergent, but a finite number of terms may give a very good approximation to the "actual" result, provided that you do not take too many terms. For instance, perturbative QED gives a good approximation if you do not go much beyond 137'th order in the expansion in $\alpha=1/137$.

9. Oct 12, 2016

### Demystifier

Yes we can, and you have just written down these "Feynman" rules.

10. Oct 12, 2016

### A. Neumaier

yes, and hence no correlation functions.

11. Oct 12, 2016

### andrien

An example of a zero dimensional qft are quantum gauge theories of matrix models. The basic field involved is a N dimensional matrix M with a gauge symmetry $$M\rightarrow UMU^\dagger$$, where U is a U(N) matrix. The partition function for this gauged matrix model involves a Haar measure dM and the usual path integral will have a gauged out part of the volume factor U(N).

A common example of this model is the Gaussian matrix model which is completely solvable and corelation functions are computed in the usual way of qft, but will involve the introduction of t'hooft "double line formalism" and the notion of "fatgraphs". A simple Gaussian two-point function will be given by the product of two Kronecker delta which in the zero dimensional case will correspond to propagator of the M field. Different topological twisting of the higher corelation function can also be obtained for different surfaces classified by the genus "g" and partition function can ultimately be obtained dependent on only the "Euler characteristic" which will involve the fatgraphs for which g=0 and non-zero g classifying the "Planar" or a "non-planar" structure.

All this construction will only come in handy if your system is expressed as a gauge model and hence will involve a trace of all the M matrices involved (2 for gaussian model). If you impose a gauge symmetry in your model (which is possible) then the construction will work in a similar way with free part removed and coupling constant as \lambda but it won't look like the series you wrote down.

12. Oct 12, 2016

### spaghetti3451

I do not see what the Feynman rules are. Are these the rules used to compute the coefficients of the various terms in the expansion of $Z(\lambda)$?

13. Oct 12, 2016

### vanhees71

Yes, there's not much more to calculation. You can get this also from the generating function,
$$Z(j)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \mathrm{d} \phi \exp(-\phi^2/2+j \phi)=\exp(j^2/2).$$
Then you get the moments
$$M_n=\langle \phi^n \rangle = \frac{2}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \mathrm{d} \phi \phi^n \exp(-\phi^2/2)=\left (\frac{\mathrm{d}^n Z(j)}{\mathrm{d} j^n} \right)_{j=0}.$$
You can find a recursion relation for the derivatives and a graphical description similar to Feynman rules. It's clear that $M_n=0$ if $n$ is odd.

14. Oct 12, 2016

### Demystifier

Yes. More precisely, these are the rules for calculating the zero-point functions, i.e. Feynman diagrams with no external lines, known also as bubbles. In the post above, vanhees71 sketches also how to calculate n-point functions for $n>1$.

15. Oct 12, 2016

### ShayanJ

Can such a model be helpful in any way?

16. Oct 12, 2016

### Demystifier

Higgs physics in the static limit?