Calculations with a potential function, ``Feynman Rules''

The integrals of interest in Physics have the form

$$Z_U = \int d{\bf v} \exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} + \hbar U({\bf v})),$$

which we rewrite using the series expansion for the exponential as

$$Z_U = \int d{\bf v} \exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v}) \sum_n \frac{\textstyle 1} {\textstyle n!} (\hbar U({\bf v}))^n.$$

If $U$ is a monomial in the coordinate functions $v^1, \dots, v^d$, then each term in the sum of integrals is a sum of $m$-point functions, and can be evaluated by our method, which can be written symbolically as:

$$Z_U = Z_0\exp (\hbar U(\frac{\partial}{\partial{\bf b}})) \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}) _{\textstyle \vert _{{\bf b} =0}}.$$

Example: This example is formally like the ``$\varphi^3$ theory.'' We take $U({\bf v}) = \frac{1}{3!}\sum_{i,j,k} u_{ijk}v^iv^jv^k$ and analyze

$$Z_U = \int d{\bf v} \exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} +\frac{1}{3!} \hbar\sum_{i,j,k} u_{ijk}v^iv^jv^k) =$$

$$Z_U = Z_0\exp (\frac{1}{3!}\hbar \sum_{i,j,k} u_{ijk}\partial_i\p... ...riptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}) _{\textstyle \vert _{{\bf b} =0}},$$

using the abbreviation $\partial_i = \frac{\partial}{\partial b^i}$ as before.

Let us compute the terms of degree 2 in $\hbar$.

These terms will involve 6 derivatives; their sum is:

$$\sum_{i,j,k}\sum_{i',j',k'}u_{ijk}u_{i'j'k'}~\partial_i~\partial_... ...riptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}) _{\textstyle \vert _{{\bf b} =0}}.$$

By Wick's Theorem we can rewrite this sum as

$$\sum_{i,j,k}\sum_{i',j',k'} \sum A^{-1}_{i_1,i_2}A^{-1}_{i_3,i_4}A^{-1}_{i_5,i_6}u_{ijk}u_{i'j'k'}$$

where the inside sum is taken over all pairings $(i_1,i_2),(i_3,i_4),(i_5,i_6)$ of $i,j,k,i',j',k'$.

These pairings can also be represented by graphs, very much in the same way that we used for $m$-point functions: there will be one trivalent vertex for each $u$ factor, and one edge for each $A^{-1}$. In this case there will be exactly two distinct graphs, according as the number of (unprimed, primed) index pairs is 1 or 3.

Summing over all possible labellings of these graphs will give some duplication, since each graph has symmetries that make different labellings correspond to the same pairing. The ``dumbbell'' graph has an automorphism (symmetry) group of order eight, whereas the ``theta'' graph has an automorphism group of order twelve.

Keeping this in mind, we may rewrite the coefficient of $\hbar^2$ as:

$$\sum_G \frac{\textstyle 1}{\textstyle \vert{\rm Aut~}G\vert} \sum... ... edge~labellings} \prod_v u_{\rm vertex~label} \prod_e A^{-1}_{\rm edge~label},$$

where the sum $\sum_G$ is taken over the set of the topologically distinct trivalent graphs with two vertices (in this case, 2), and the products are taken over the set of all vertices $v$ (here there are 2) and the set of all edges $e$ (here there are 3) respectively.

In general, the ``Feynman rules'' for computing the coefficient of $\hbar^{2n}$ in the expansion of $Z_U$ are stated in exactly this way, except that the sum $\sum_G$ is over trivalent graphs with $2n$ vertices (and $3n$ edges).