Gaussian integrals and Wick's theorem

This post has been migrated from my old blog, the math-physics learning seminar.

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.