# Feynman diagrams, Feynman rules and corresponding integrals

I would like some basic examples of Feynman diagrams: in particular I would like to understand how a Feynman diagram produces an integral: before I start let me made some remarks in the form of a

Disclaimer 0) I'm a mathematician, not a physicist
1) I'm interested in the theory where there is only one scalar field and the only interaction term in the Lagrangian is a monomial of the form $$\frac{g}{d!}\varphi^d$$-in particular for $$d=3$$ (non physical but nevermind) or $$d=4$$-here $$g$$ is a coupling constant
2) As I understood Feynman diagrams is a device to compute some complicated (functional) integrals-in particular I'm less interested in the physical interpretation of the diagrams: I'm only interested in an analytical information they contain.

Let me also specify concretely Feynman rules (fixing notations and conventions at the same time): 1) Every external line produces a propagator $$\frac{1}{p_i^2+m^2}$$
2) Every internal line produces an expression $$\frac{1}{k_i^2+m^2} \frac{d^Dk_i}{(2 \pi)^D}$$
3) Every vertex (corresponding to the monomial $$\frac{g}{d!}\varphi^d$$ produces delta distribution $$(2\pi)^D g \delta(\sum_{lines entering}k_i-\sum_{lines living}k_j)$$ (here we allow also external lines).
At the end we integrate the product of the all above expressions

For a Feyman diagram $$\Gamma$$ one can define the loop number $$L$$ as $$I-V+1$$ where $$V$$ is a number of vertices, $$I$$ is a number of internal lines. One can also define the so called superficial degree of summability as $$DL-2I$$: if this degree is negative then the integral associated to a graph will be finite. There is some subtelty that the loop number is undefined for graphs which are not connected: however the formula $$I-V+1$$ still makes sense but now this formula does not coincide with the number of loops of geometric realization of the graph.

These notions can be translated into the analytic language: the loop number should be the number of free variables over which (after we get rid of delta ,,functions'') we integrate. The superficial degree of summability comes from power counting: we integrate over $$DL$$ dimensional space and the expression under the integral is a rational function with a $$2I$$ degree polynomial in the denominator. However when I tried to evaluate some examples I met different expressions: since drawing Feynman diagrams is problematic I refer to this picture:

This picture contains several examples in which, by following Feynman rules I get expressions which look strange for me. I will be grateful if could somebody correct me and write the correct form of integrals corresponding to the Feynman graphs given in this image.

Well that's why the superficial degree of divergence is called superficial. If you look at say your first graph it has a bridge (the line in the middle of the dumbbell). In principle you should have a momentum there to be integrated over but conservation of momenta fixes it equal to $$0$$. This results in you being able to take the propagator out of the integral hence the factor $$\frac{1}{m^2}$$. The superficial degree of divergence counts this as a propagator as if it was still in the integral and thus gives $$2D-6$$. Your $$2D-4$$ corresponds to what one could call the "true" degree of divergence.
• That's just linear algebra (or baby homological algebra). Suppose that you have a homogeneous linear system of $V$ equations in $I$ variables, and that these equations are not linearly independent but are constrained by one equation only. Then space of solutions has dimension $I-(V-1)=I-V+1$. For each connected component the momentum conservations at the vertices are not independent relations but are constrained by a single equation. If you have $n$ components then you have to compensate by $+n$ instead of $+1$. Jun 13 '19 at 17:51
• The integral for the before last graph is wrong. It does not disappear. If $p_1,p_2$ are the two momenta coming from the left, i.e., external legs and if $u$ is the momentum in the loop say counterclockwise, then momentum conservation at the only vertex gives $p_1+p_2-u+u=0$ and therefore tells you nothing about $u$ which you still have to integrate. Same mistake in last more complicated graph. Jun 13 '19 at 19:15