Gaussian Integrals: Wick's Theorem

We saw in the last update that the generating function $Z[J]$ can be expressed as
$$Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J}$$
(at least as long as we've normalize things so that $Z[0] = 1$. Now the wonderful thing is that this is something we can compute explicitly:
$$Z[J] = \sum_{n = 0}^{\infty} \frac{(\frac{1}{2} A^{-1}_{ij} J^i J^j)^n}{n!} = \sum_{n=0}^\infty \frac{(A^{-1}_{ij} J^i J^j)^n}{2^n n!}$$

For example, in the one-dimensional case (taking $A = 1$) we get
$$Z[J] = \sum_{n=0}^\infty \frac{J^{2n}}{2^n n!}$$
On the other hand, by the definition of the generating function we have
$$Z[J] = \sum_{n=0}^\infty \frac{\langle x^n \rangle}{n!} J^n$$
Comparing coefficients, we find
$$\frac{\langle x^{2n} \rangle}{(2n)!} = \frac{1}{2^n n!}$$
so that
$$\langle x^{2n} \rangle = \frac{(2n)!}{2^n n!}.$$
Let's give a combinatorial description. Given $2n$ objects, in how many ways can we divide them into pairs? If we care about the order in which we pick the pairs, then we have
$${2n \choose 2}{2n - 2 \choose 2} \cdots {2n-(2n-2) \choose 2} = \frac{(2n)!}{2^n}$$
Of course, there are $n!$ ways of ordering the $n$ pairs, so after dividing by this (to account for the overcounting) we get exactly the expression for $\langle x^{2n} \rangle$. This is the first case of Wick's theorem.

Now consider the general multidimensional case. Given $I = (i_1, \cdots, i_{2n})$, we define a contraction to be
$$\langle x^{j_1} x^{k_1} \rangle \cdots \langle x^{j_n} x^{k_n} \rangle$$
where $j_1, k_1, \cdots, j_n, k_n$ is a choice of parition of $I$ into pairs.

Theorem (Wick's theorem, Isserlis' theorem) The expectation value
$$\langle x^{i_1} \cdots x^{i_{2n}} \rangle$$
is the sum over all full contractions. There are $(2n)!/ 2^n n!$ terms in the sum.

Proof This follows from our formula for the power series of the generating function. The reason is that the coefficient of  $J^I$ in $(\frac{1}{2} A^{-1}_{ij} J^i J^k)^n$ is exactly given by summing products of $A^{-1}_{ij}$ over partitions of $I$ into pairs, and the $n!$ in the denominator takes care of the overcounting.

Next up: perturbation theory and Feynman diagrams.

Introduction to Gaussian Integrals

As a warm-up for more serious stuff, I'd like to discuss Gaussian integrals over $\mathbb{R}^d$. Gaussian integrals are the main tool for perturbative quantum field theory, and I find that understanding Gaussian integrals in finite dimensions is an immense aid to understanding how perturbative QFT works. So let's get started.


The Basics

Let $A$ be some $d \times d$ symmetric positive definite matrix. We are interested in the integral
$$\int_{-\infty}^\infty \exp(-\frac{x \cdot Ax}{2}) dx.$$
Out of laziness, I will suppress the limits of integration and just write this as
$$\int e^{-S(x)} dx.$$
where $S(x) = x \cdot Ax / 2$. Now for a function $f(x)$, we define the expectation value $\langle f(x) \rangle$ to be
$$\langle f(x) \rangle_0 = \int f(x) e^{-S(x)} dx$$
Occasionally, we might care about the normalized expectation value
$$langle f(x) \rangle = \frac{\langle f(x) \rangle_0}{\langle 1 \rangle_0} = \frac{1}{\langle 1 \rangle_0} \int f(x) e^{-S(x)} dx.$$
We mostly care about asymptotics, so we will typically think of a function $f(x)$ as being a polynomial (or Taylor series). So what we're really interested in is
$$\langle x^I \rangle = c\int x^I e^{-S(x)} dx,$$
where $I$ is a multi-index.

The Partition Function

Let us define $Z[J]$ by
$$Z[J] = \int e^{-S(x) + J \cdot x} dx.$$

Now the great thing is that
$$\langle x^I \rangle = \left. \frac{d^I}{dJ^I} \right|_{J = 0} Z[J],$$
so that once we know $Z[J]$, we can calculate anything. So let's try to compute it. We have

$$\begin{align} (Ax - J) \cdot A^{-1} (Ax - J) &= (Ax - J) \cdot (x - A^{-1} J) \\\ &= x \cdot Ax - x \cdot J - J \cdot x + J \cdot A^{-1} J \\\ &= x \cdot Ax - 2 x \cdot J + J \cdot A^{-1} J. \end{align}$$
So we see that
$$-\frac{1}{2} x \cdot A x + J \cdot x = \frac{1}{2} J \cdot A^{-1} J -\frac{1}{2} (x-A^{-1}J) \cdot A(x - A^{-1} J).$$
So, after a change of variales $x \mapsto x - A^{-1} J$ we find
$$Z[J] = e^{\frac{1}{2} J \cdot A^{-1} J} Z[0].$$
Now the argument in the exponential is
$$\frac{1}{2} A^{-1}_{ij} J^i J^j$$
So we find that
$$\langle x^i x^j \rangle = \frac{d^2}{dx^i dx^j} Z[J]|_{J = 0} = A^{-1}_{ij}.$$

Now we are ready to prove Wick's theorem and discuss Feynman diagrams, which we'll do in the next post.