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.