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Finite-dimensional Feynman Diagrams




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

The integrals of interest in Physics have the form

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

which we rewrite using the series expansion for the exponential as

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

If U is a polynomial in the coordinate functions v1, ...vd, 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:

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



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

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

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

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:

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

By Wick's Theorem we can rewrite this sum as

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

where the inside sum is taken over all pairings (i1,i2),(i3,i4)(i5i6) 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.


The ``dumbbell'' and the ``theta''are the two 3-valent 2-vertex graphs.

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.


All eight of these labelings correspond to the same product: u123 u456 A-113 A-125 A-146.


All six of these labelings, and their six left-right mirror images, correspond to the same product: u123 u456 A-114 A-125 A-136.

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:

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

where the sum $ \sum_G$ is taken over the set of the topologically distinct trivalent graphs with two vertices (in this case, 2), 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, and |AutG| is the number of automorphisms of the graph G.

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



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