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.