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\begin{document}
\title[Feynman Diagrams]{Finite-dimensional Feynman Diagrams}

\maketitle

Feynman diagrams are a fundamental tool for the investigation and explanation
of phenomena in quantum field theory. Their origin, however, 
is purely mathematical: they
give a convenient way of organizing and encoding certain important calculations.

In this column we will look at a finite-dimensional calculation that shares many
of the formal properties of the calculation of interest. 

The mathematics involved is more technical than usual in this series of columns,
but it is quite concrete. With the restriction to
finite-dimensionality we will see exactly where the ``diagrams'' come into 
the picture, using no more than fairly elementary procedures
from calculus and linear algebra. So any third or fourth-year undergraduate
should be able to follow the details, while a broader audience should
be able to gain from the examples  an accurate picture of the whole
procedure.

\section{Facts from calculus and their $d$-dimensional analogues}

The basic fact from calculus that powers the whole discussion is:

\begin{proposition}
$$\int_{-\infty}^{\infty} dx~~ e^{ -\frac{\scriptstyle a}{2}x^2} = \sqrt{\frac{2\pi}{a}}.$$
\end{proposition}

The identity with $a=1$ is proved by the familiar trick of calculating the {\em square} of the
integral in polar coordinates. The general identity follows by change of variable
from $x$ to $x{\sqrt a}$ .

This fact generalizes to higher-dimensional integrals. Set 
${\bf v} = (v^1, \dots, v^d)$
and $d{\bf v} = dv^1\cdots dv^d$, and let $A$ be a symmetric $d$ by $d$ matrix.

\begin{proposition}
$$\int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf v}^tA~{\bf v}) = (2\pi)^{d/2} (\det A)^{-1/2}.$$ 
\end{proposition}


We work the calculation for the case $A$ diagonalizable: in that case there 
exists an orthogonal matrix
$U$ (so $U^t = U^{-1}$) such that $UAU^{-1}$ is the diagonal matrix $B$ whose only
nonzero entries are  $b_{11}, \dots, 
b_{dd}$ along the diagonal.  Then $A = U^{-1}B~U$ and ${\bf v}^tA~{\bf v} = 
{\bf v}^tU^{-1}B~U{\bf v} = {\bf v}^tU^tB~U{\bf v} = {\bf w}^tB~{\bf w}$ where
${\bf w} = U{\bf v}$, using $U^t = U^{-1}$ and $(U{\bf v})^t = {\bf v}^tU^t$. Since $U$
is orthogonal $\det U =1$ and the change of variable from ${\bf v}$ to ${\bf w}$ does not
change the integral:

$$\int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf v}^tA~{\bf v}) =\int_{{\bf R}^d} d{\bf w} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf w}^tB~{\bf w})=$$
$$\int_{{\bf R}^d} d{\bf w} ~~\exp(b_{11}(w^1)^2 +\cdots + b_{dd}(w^d)^2) =$$

$$\int_{{\bf R}^d} d{\bf w} ~~\exp(b_{11}(w^1)^2)~ \cdots ~\exp( b_{dd}(w^d)^2) =$$

$$(\int_{-\infty}^{\infty} dw^1~\exp(b_{11}(w^1)^2))~\cdots~(\int_{-\infty}^{\infty} dw^d~\exp(b_{dd}(w^d)^2))=$$

$$(\sqrt{2\pi}/\sqrt{b_{11}})~\cdots~ (\sqrt{2\pi}/\sqrt{b_{dd}})=$$
$$(2\pi)^{d/2} (\det B)^{-1/2} = (2\pi)^{d/2} (\det A)^{-1/2}.$$

The argument in general uses two additional observations: both sides of
the equation vary continuously as functions of the entries in $A$, and any
matrix $A$ with complex coefficients can be approximated to arbitrary
accuracy by diagonalizable matrices.

\begin{proposition}
$$\int_{-\infty}^{\infty} dx \, e^{-\frac{a}{2}x^2 + bx} = 
\sqrt{\frac{2\pi}{a} } e^{b^2/2a}\, .$$
\end{proposition}

This follows from Proposition 1 by completion of the square in the exponent and a change of variables.

The generalization to $d$ dimensions replaces $a$ with $A$ as before and $b$ with
the vector ${\bf b} = (b^1,\dots,b^d)$.

\begin{proposition}
$$\int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} + {\bf b}^t{\bf v}) = (2\pi)^{d/2} (\det A)^{-1/2} \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}).  $$
\end{proposition}

This is proven exactly like Proposition 2.
If we write this integral as $Z_{\bf b}$, then the integral of Proposition 2 is $Z_0$
and this proposition can be rewritten as
$$ Z_{\bf b} = Z_0 \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}) . $$

\section{$m$-point Functions}
For any choice of $m$ (not necessarily different) indices $i_1, \dots, i_m$ between
1 and $d$, define the {\em $m$-point function} $\<v^{i_1}, \dots ,  v^{i_m}\>$ as follows:
$$\<v^{i_1} ,\dots , v^{i_m}\> = \frac{1}{Z_0}\int_{{\bf R}^d} d{\bf v} ~~\exp({\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v})v^{i_1},\dots , v^{i_m}.$$

The $m$-point functions are a step towards the ultimate aim of our calculation.
They enter at this moment because {\em they can be calculated
by repeated differentiation of} $Z_{\bf b}$. 

For example, note that
$$\frac{\partial  Z_{\bf b}}{\partial b^i} = \frac{\partial}{\partial b^i}\int_{{\bf R}^d} d{\bf v} ~~\exp({\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} + {\bf b}^t{\bf v}) =$$
$$ \int_{{\bf R}^d} d{\bf v} ~~ \frac{\partial}{\partial b^i}\exp({\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} + {\bf b}^t{\bf v}) = $$
$$ \int_{{\bf R}^d} d{\bf v} ~~ \exp({\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} + {\bf b}^t{\bf v}) v^i.$$

So the 1-point function $\<v^i\>$  is given by

$$ \<v^i\> = \frac{1}{Z_0} \frac{\partial Z_{\bf b}}{\partial b^i}|_{{\bf b} =0}.$$ 

Similarly the $m$-point function $\<v^{i_1} ,\dots , v^{i_m}\>$ is given by

$$\<v^{i_1} ,\dots , v^{i_m}\> =\frac{1}{Z_0} (\frac{\partial}{\partial b^{i_1}}\cdots \frac{\partial}{\partial b^{i_m}}Z_{\bf b})_{\textstyle |_{{\bf b} =0}} = $$
$$ \frac{\partial}{\partial b^{i_1}}\cdots \frac{\partial}{\partial b^{i_m}} \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})_{\textstyle |_{{\bf b} =0}}.$$

\section{Wick's Theorem}

Calculating high-order derivatives of a function like $\exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})$ can be very messy. A useful theorem reduces the calculation
to combinatorics.

\begin{theorem} {\rm ({\bf Wick's theorem})}

$$ \frac{\partial}{\partial b^{i_1}}\cdots \frac{\partial}{\partial b^{i_m}} \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})_{\textstyle |_{{\bf b} =0}} = \sum A^{-1}_{\textstyle  i_{p_1},i_{p_2}} \cdots
A^{-1}_{\textstyle   i_{p_{m-1}},i_{p_m}},$$
where the sum is taken over  all  pairings $ (i_{p_1},i_{p_2}), \dots,  (i_{p_{m-1}},i_{p_m})$
of  $i_1, \dots, i_m$.
\end{theorem}

Let us calculate a couple of examples. 

To begin, it is useful to write $\exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})$
with ${\bf b}^tA^{-1}{\bf b} = \sum A^{-1}_{i,j}b^ib^j$ (the sum running from 1 to $d$),
using the series expansion $\exp x = 1 + x + x^2/2! + x^3/3! + \cdots.$ The typical term
will be $(1/n!)(1/2^n)(\sum A^{-1}_{i,j}b^ib^j)^n$. This term is a homogeneous polynomial
in the $b^i$ of degree $2n$.  

Differentiating $m$ times a homogeneous polynomial
of degree $2n$ and evaluating at zero will give zero unless $k=2m$. So the job is to analyze
the result of $2n$ differentiations on $(1/n!)(1/2^n)(\sum A^{-1}_{i,j}b^ib^j)^n$. 

The differentiation carried out most frequently in these calculations is
$$\frac{\partial}{\partial b^{k}}(\frac{1}{2}\sum_{i,j=1}^d A^{-1}_{i,j}b^ib^j) = \sum_{i=1}^d A^{-1}_{i,k}b^i,$$
where we use the symmetry of the matrix $A^{-1}$, a direct consequence of the symmetry of $A$. 

In what follows $\frac{\partial}{\partial b^i}$ will be abbreviated as $\partial_i$. 

%\sum A^{-1}_{i,k}b^i

\begin{itemize}
\item $n=1$.
$$\partial_2 \partial_1(\frac{1}{2}\sum A^{-1}_{i,j}b^ib^j) = $$
$$\partial_2 ( \sum_j A^{-1}_{1,j}b^j  ) = $$
$$ A^{-1}_{1,2}~,$$

using the symmetry of the matrix $A^{-1}$. The same calculation shows
that $\partial_1 \partial_1(\frac{1}{2}\sum A^{-1}_{i,j}b^ib^j) = A^{-1}_{1,1}~.$ 

\end{itemize}
Note that $(1,2)$ and $(2,1)$ count as the same pairing. 

\begin{itemize}
\item $n=2$.
$$\partial_4\partial_3\partial_2\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = $$
$$\partial_4\partial_3\partial_2
(1/2)(\sum A^{-1}_{i,j}b^ib^j)( \sum A^{-1}_{1,j}b^j) =$$
$$\partial_4\partial_3
[(\sum A^{-1}_{2,j}b^j)( \sum A^{-1}_{1,j}b^j) +(1/2)(\sum A^{-1}_{i,j}b^ib^j)A^{-1}_{1,2}] = $$
$$\partial_4[A^{-1}_{2,3}( \sum A^{-1}_{1,j}b^j)+(\sum A^{-1}_{2,j}b^j)A^{-1}_{1,3}
+(\sum A^{-1}_{3,j}b^j)A^{-1}_{1,2}] = $$
$$A^{-1}_{2,3}A^{-1}_{1,4} + A^{-1}_{2,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,2}~.
$$ 

Similarly:

$$\partial_4\partial_3\partial_1\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = 2A^{-1}_{1,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,1}~.$$ 



$$\partial_4\partial_1\partial_1\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = 3A^{-1}_{1,4}A^{-1}_{1,1}~.$$



$$\partial_4\partial_4\partial_1\partial_1(1/2!)(1/2^2)(\sum A^{-1}_{i,j}b^ib^j)^2 = 2 A^{-1}_{1,4}A^{-1}_{1,4} + A^{-1}_{4,4}A^{-1}_{1,1}~.$$

And

$$\partial_1\partial_1\partial_1\partial_1(\sum A^{-1}_{i,j}b^ib^j)^2 = 3 A^{-1}_{1,1}A^{-1}_{1,1}~.$$
\end{itemize} 

\section{The first appearance of graphs} 

In the last section we calculated some 2 and 4-point functions:

$$\<v^1,v^2\> =  A^{-1}_{1,2}$$
$$\<v^1,v^1\> =  A^{-1}_{1,1}$$
$$\<v^1,v^2,v^3,v^4\> = A^{-1}_{2,3}A^{-1}_{1,4} + A^{-1}_{2,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,2}$$
$$\<v^1,v^1,v^3,v^4\> = 2A^{-1}_{1,4}A^{-1}_{1,3} + A^{-1}_{3,4}A^{-1}_{1,1}$$
$$\<v^1,v^1,v^1,v^4\> = 3A^{-1}_{1,4}A^{-1}_{1,1}$$
$$\<v^1,v^1,v^4,v^4\> = 2A^{-1}_{1,4}A^{-1}_{1,4} + A^{-1}_{4,4}A^{-1}_{1,1}$$
$$\<v^1,v^1,v^1,v^1\> = 3A^{-1}_{1,1}A^{-1}_{1,1}~.$$ 

It is convenient to represent each of products appearing on the right as a {\em graph},
where the vertices represent the indices of the coordinates $v^i$ appearing in the $m$-point
function, and each $A^{-1}_{1,j}$ becomes an edge from vertex $i$ to vertex $j$.
 
\section{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:
$$Z_U = Z_0\exp (\hbar U(\frac{\partial}{\partial{\bf b}})) \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})
_{\textstyle |_{{\bf b} =0}}.$$ 

\noindent
{\em 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
$$Z_U = \int d{\bf v} \exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf v}^tA~{\bf v} +\frac{1}{3!} \hbar\sum_{i,j,k} u_{ijk}v^iv^jv^k) =$$
$$Z_U = Z_0\exp (\frac{1}{3!}\hbar \sum_{i,j,k} u_{ijk}\partial_i\partial_j\partial_k) \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})
_{\textstyle |_{{\bf b} =0}},$$
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:

$$\sum_{i,j,k}\sum_{i',j',k'}u_{ijk}u_{i'j'k'}~\partial_i~\partial_j~\partial_k~\partial_{i'}~\partial_{j'}~\partial_{k'}~\exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b})
_{\textstyle |_{{\bf b} =0}}.$$

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.

%figure showing dumbbell and theta

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 {\em automorphism} (symmetry) group of order eight, 
whereas the ``theta'' graph
has an automorphism group of order twelve. 

%figure showing 8 and 6 labellings

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

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

\section{Correlation functions}

The way path integrals are used in quantum field theory is, very roughly
speaking, that the probability amplitude of a process going from point $v_1$ to
point $v_2$ is an integral over all possible ways of getting from
$v_1$ to $v_2$. In our finite-dimensional model, each of these
``ways'' is represented by a point ${\bf v}$ in ${\bf R}^d$, and
the probability measure assigned to that way is $\frac{1}{Z_U}\exp(-{\scriptstyle\frac{1}{2}}{\bf v}^tA~{\bf v} +\hbar U({\bf v}))v^1v^2~d{\bf v}$. The integral is then
what we called before a 2-point function
$$\<v^1,v^2\> = \frac{1}{Z_U}\int_{{\bf R}^d}d{\bf v}~ \exp(-{\scriptstyle\frac{1}{2}}{\bf v}^tA~{\bf v} +\hbar U({\bf v}))v^1v^2,$$
and what we will now call a {\em corellation function}.

We continue with the example of the cubic potential $$U({\bf v}) = \frac{1}{3!}\sum_{i,j,k} u_{ijk}v^iv^jv^k$. By our previous calculations,
$$\<v^1,v^2\> = \frac{1}{Z_U}\partial_1\partial_2 \exp(\hbar \sum_{i,j,k}
u_{ijk}\partial_i\partial_j\partial_k)(\exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}))
_{\textstyle |_{{\bf b} =0}}.$$

In terms of Wick's Theorem and our graph interpretation of pairings, this
becomes:

$$\sum_G
\frac{\textstyle \hbar^n}{\textstyle |{\rm Aut~}G|}
\sum_{\rm edge~labellings} \prod_v u_{\rm vertex~label} \prod_e A^{-1}_{\rm edge~label},$$
where now the sum is over all graphs $G$ with two single-valent vertices (the ends)
labeled 1,2 and $n$ 3-valent vertices.

%insert figure with correlation function graph (graph4)

Here is a typical graph occurring in the calculation of the coefficient
of $\hbar^6$ in $\<v^1,v^2\>$.

$k$-point corellation functions are similarly defined and calculated. Here
is where we begin to see the usual ``Feynman diagrams.''


Here is a graph occurring in the calculation of $\hbar^2$ in
$\<v^1,v^2,v^3,v^4\>$.







\end{document}