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.