# Least Action

Nontrivializing triviality..and vice versa.

## Gaussian Integrals and Wick Contractions

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”.