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.