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Finite-dimensional Feynman Diagrams

2. Facts from calculus and their d-dimensional analogues

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

Proposition 1  

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

The identity with  a = 1  is proved by the trick of calculating the 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   v = (v1, ..., vd) and   dv = (dv1 ... dvd), and let   A   be a symmetric   d by d   matrix.

Proposition 2  

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

We use the fact that a symmetric matrix A is diagonalizable: there exists an orthogonal matrix U (so Ut = U-1) such that UAU-1 is the diagonal matrix B whose only nonzero entries are b11, ... , bdd along the diagonal. Then A = U-1BU and vtAv = vt U-1B U v = vtUtB U v = wtB w where w = Uv, using Ut = U-1 and (Uv)t = vtUt. Since U is orthogonal   detU = 1   and the change of variable from v to w does not change the integral:

$\displaystyle \int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf v...
...\int_{{\bf R}^d} d{\bf w} ~~\exp(-{\scriptstyle\frac{1}{2}}{\bf w}^tB~{\bf w})=$

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

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

$\displaystyle (\sqrt{2\pi}/\sqrt{b_{11}})~\cdots~ (\sqrt{2\pi}/\sqrt{b_{dd}})=$

$\displaystyle (2\pi)^{d/2} (\det B)^{-1/2} = (2\pi)^{d/2} (\det A)^{-1/2}.$

Proposition 3  

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

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 b = (b1, ... , bd)

Proposition 4  

$\displaystyle \int_{{\bf R}^d} d{\bf v} ~~\exp(-{\scriptstyle\frac{ 1}{ 2}}{\bf...
...)^{d/2} (\det A)^{-1/2} \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}). $

This is proven exactly like Proposition 2. If we write this integral as Zb then the integral of Proposition 2 is Z0 and this proposition can be rewritten as

$\displaystyle Z_{\bf b} = Z_0 \exp({\scriptstyle\frac{1}{2}}{\bf b}^tA^{-1}{\bf b}) . $

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