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}

Proof:

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

\sqrt{\frac{2\pi}{D_{ii}}}

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

\sum_{j}(A^{-1})_{ij}J_{j}

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”.

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Written by Vivek

October 9, 2010 at 21:45

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