Least Action

Nontrivializing triviality..and vice versa.

Gaussian Integrals and Wick Contractions

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Theorem 1: For nonzero real a,

\langle x^{2n}\rangle = \frac{\int_{-\infty}^{\infty}dx\,e^{-\frac{1}{2}ax^2}x^{2n}}{\int_{-\infty}^{\infty}dx\,e^{-\frac{1}{2}ax^2}} = \frac{1}{a^{n}}(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1

Theorem 2: For a real symmetric N\times N matrix A, a real N \times 1 vector x and a real N\times 1 vector J, we have

\int_{-\infty}^{\infty}\dotsi\int_{-\infty}^{\infty}\,dx_1\cdots dx_N\, e^{-\frac{1}{2}x^{T}Ax + J^{T}x} = \left(\frac{(2\pi)^N}{\det(A)}\right)^{1/2}e^{\frac{1}{2}J^{T}A^{-1}J}


Define O to be the real orthogonal matrix which diagonalizes A to its diagonal form D. That is,

A = O^{T} D O

where O^{T}O = OO^{T} = I. Further, let us choose O to be a special orthogonal matrix, so that \det(O) = +1. Also, define Y = OX. Now,

dy_1 \cdots dy_n = |\det(O)| dx_1 \cdots dx_n = dx_1 \cdots dx_n

The argument of the exponential is

-\frac{1}{2}X^{T}AX + J^{T}X = -\frac{1}{2}Y^{T}O(O^{-1}DO)(O^{T}Y) + J^{T}O^{T} Y

which equals

-\frac{1}{2}Y^{T}DY + J^{T}O^{T}Y

The first term is

-\frac{1}{2}Y^{T}DY = -\frac{1}{2}\sum_{i}y_{i}D_{ii}y_{i}

Let C^{T} = J^{T}O^{T} (that is, C = OJ). So the second term is

J^{T}O^{T}Y = C^{T}Y

So the argument of the exponential becomes

-\frac{1}{2}\sum_{i}y_{i}D_{ii}y_{i} + \sum_{i}C_{i}Y_{i}

which can be written as

-\frac{1}{2}\sum_{i}\left[D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2 - \frac{C_i^2}{D_{ii}}\right]

The integrand becomes

e^{-\frac{1}{2}\sum_{i}\left[D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2 - \frac{C_i^2}{D_{ii}}\right]} = \prod_{i}e^{-\frac{1}{2}D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2}e^{\frac{1}{2}C_{i}D_{ii}^{-1}C_{i}}

Hence the quantity to be evaluated is

\left\{\int_{-\infty}^{\infty}\dotsi\int_{-\infty}^{\infty}dy_1 \cdots dy_n \prod_{i}e^{-\frac{1}{2}D_{ii}\left(y_{i}-\frac{C_i}{D_{ii}}\right)^2}\right\}e^{\frac{1}{2}\sum_{i}C_{i}D_{ii}^{-1}C_{i}}

The quantity in curly brackets involves N definite Gaussian integrals, the i^{th} term of which is


using the identity \int_{-\infty}^{\infty} dt\,e^{-\frac{1}{2}at^2} = \sqrt{\frac{2\pi}{a}}.

So the continued product is

\left(\frac{(2\pi)^N}{\det(D)}\right)^{1/2} = \left(\frac{(2\pi)^N}{\det(A)}\right)^{1/2}

whereas the exponential factor sticking outside is

e^{\frac{1}{2}\sum_{i}C_{i}D_{ii}^{-1}C_{i}} = e^{\frac{1}{2}C^{T}D^{-1}C} = e^{\frac{1}{2}(J^{T}O^{T}OA^{-1}O^{-1}OJ)} = e^{\frac{1}{2}J^{T}A^{-1}J}

This completes the proof.

Theorem 3: \langle x_{i} x_{j} \cdots x_{k}x_{l}\rangle = \sum_{\mbox{Wick}}(A^{-1})_{ab}\cdots(A^{-1})_{cd}

Proof: If m is the number of terms in \langle x_{i} x_{j} \cdots x_{k}x_{l}\rangle, differentiate the identity proved in Theorem 2 above m times with respect to each of J_i, J_j, \ldots, J_k, J_l.

As an example, differentiate the RHS of the identity in Theorem 2 with respect to J_{i}. The derivative is

\frac{d}{dJ_{i}}\exp\left\{\frac{1}{2}\sum_{m,n}J_{m}(A^{-1})_{mn}J_{n}\right\} = \frac{1}{2}\left(\sum_{m,n}(\delta_{mi}(A^{-1})_{mn}J_{n} + J_{m}(A^{-1})_{mn}\delta_{ni})\right)\exp\left\{\frac{1}{2}\sum_{m,n}J_{m}(A^{-1})_{mn}J_{n}\right\}

the prefactor of the exponential is

\frac{1}{2}(\sum_{n}(A^{-1})_{in}J_{n} + \sum_{m}J_{m}(A^{-1})_{mi})

which becomes


using the fact that A is symmetric. Repeating this exercise, we see that bringing down x_{i} involves differentiating with respect to J_{i}, and effectively introduces a matrix element of A^{-1}. It is obvious by induction that the sum must run over all possible permutations of matrix indices of A^{-1}. That is, the sum must run over every possible permutation (a, b), \ldots (c, d) of pairing the indices i, j, \ldots k, l. This completes the “proof”.


Written by Vivek

October 9, 2010 at 21:45

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