What's up with Wick's theorem?

2010-11-25T17:58:37Z

Sorry about the dumb title.

I'd like to understand Wick's theorem. More specifically, I have seen it pop up in several different contexts and I am really puzzled by the different statements of it that I have seen. My own background/interest is in moduli of curves, if that helps.

The first version, that is also the only one I have seen more than one time, is in the context of infinite wedge space. Here Wick's theorem is a formula about how to decompose any product of the fermionic operators $\psi_k$ and their adjoints $\psi_k^\ast$ as a sum over normally ordered products. This is for instance how it is explained in Kac-Raina.

A second version is in "Graphs on surfaces" by Lando and Zvonkin as Theorem 3.2.5. Here it is a statement about how to integrate a polynomial against a Gaussian measure on the real line. If $\langle f \rangle$ denotes the integral $\frac{1}{\sqrt{2\pi}}\int_{\mathbb R} f(x) \exp(-x^2/2) dx$, then Wick's theorem states that if f is a product $f = f_1 f_2 \cdots f_{2k}$ of linear polynomials, then $\langle f \rangle$ can be written as an explicit sum of products of pairs $\langle f_if_j\rangle$.

Now I can somehow believe that the two theorems above are talking about the same thing or that the second is a special case of the first. But what really got me scratching my head was the following statement from page 2 of Getzler & Kapranov's paper on modular operads (sorry for the lengthy quote):

[...] As a model for this calculation, take the formula for the enumeration of graphs known in mathematical physics as Wick’s Theorem. Consider the asymptotic expansion of the integral $W(\xi,\hbar) = \log \int \exp \frac 1 \hbar \left( x \xi - \frac{x^2}{2} + \sum_{2g-2+n > 0} \frac{a_{g,n}\hbar^g x^n}{n!}\right) \frac{dx}{\sqrt{2\pi\hbar}}$ considered as a power series in $\xi$ and $\hbar$. (The asymptotic expansion is independent of the domain of integration, provided it contains 0.) Let $\Gamma((g,n))$ be the set of isomorphism classes of connected graphs $G$, with a map $g$ from the vertices Vert(G) of $G$ to $\{0,1,2,...\}$ and having exactly $n$ legs numbered from 1 to $n$, such that $g = b_1 + \sum_{v\in \mathrm{Vert}(G)} g(v)$ where $b_1$ is the first Betti number of the graph. If $v$ is a vertex of $G$, denote by $n(g)$ its valence, and let |Aut(G)| be the cardinality of the automorphism group of $G$. Wick’s Theorem states that $W \sim \frac 1 \hbar \left(\frac{\xi^2}{2} + \sum_{2g-2+n>0} \frac{\hbar^g\xi^n}{n!} \sum_{G\in \Gamma((g,n))} \frac{1}{|\mathrm{Aut}(G)|} \prod_{v\in \mathrm{Vert}(G)} a_{g(v),n(v)}\right)$.

I also heard a version of Wick's theorem at a talk of Rahul Pandharipandhe about two months ago, which I will not be able to state correctly here since I can't really make sense of my notes. In that version of Wick's theorem one studied an $n$-fold product of a variety with itself by interpreting it as a configuration space of $n$ "particles" moving on the variety. The goal was to simplify certain complicated products of cohomology classes given by diagonals (= particles coinciding) and Chern classes of the tangent/cotangent bundle at one of the "particles". This was all done pictorially, and one represented the diagonals as a line connecting the two points, which at least shows some connection with the Wick formalism since I think I have at one point seen these lines between particles also in the context of Feynman diagrams.

Can someone give a hint about how these Wick theorems fit together?

Answer by Theo Johnson-Freyd for What's up with Wick's theorem?

2010-11-25T23:18:27Z

Let's take for granted the Gaussian integration formula, which holds for both bosonic and fermionic integrals, if they are properly interpreted:

Theoreom (Gauss, Wick): Let $X$ be a vector space with a chosen volume form ${\rm d}x$, $f: X \to \mathbb C$ a polynomial, and $a: X^{\vee 2} \to \mathbb C$ a symmetric bilinear form with inverse $a^{-1}\in X^{\vee 2}$ such that the Gaussian measure $\exp(-\frac12 a\cdot x^{\otimes 2}){\rm d}x$ is defined. (So for example $X$ can be bosonic finite-dimensional and $a$ can have positive-definite real part; or $a$ can be invertible pure-imaginary and all integrals can be taken as conditionally convergent; or $X$ can have even-dimensional fermionic parts and the integral can be defined a la Berezin.) Then we have $$ \int_X \sum f^{(n)} \cdot \frac{x^{\otimes n}}{n!} \exp \left(-\frac12 a\cdot x^{\otimes 2}\right){\rm d}x = \sqrt{\det(2\pi a)} \sum f^{(2k)} \cdot \frac{(a^{-1})^{\otimes 2k}}{2^kk!} $$ Or, anyway, when $X$ is finite-dimensional Bosonic this is correct. In the fermionic case, it's off by some $\sqrt{-2\pi}$s, and $\det$ is Berezin's superdeterminant. Here $f^{(n)} : X^{\vee n} \to \mathbb C$ is the $n$th Taylor coefficient of $f$ at $0$; it's a symmetric tensor. In the fermionic case, some care must be taken with words like "symmetric", but I will ignore this subtlety.

Proof: Integrate by parts. $\Box$

Now the trick is to interpret the RHS combinatorially: you should draw each summand as a graph with one vertex, labeled $f^{(2k)}$, and $k$ self-loops, labeled $a^{-1}$; then if you count automorphisms correctly, the denominator of the summand is the number of automorphisms of the graph, and the numerator is the "evaluation" of the graph as a picture of tensor contractions. You can also draw the left hand summands combinatorially: a vertex with $n$ incoming strands corresponds to $f^{(n)}$, and the $n!$ counts the symmetries.

What would have happened if you had not just a single polynomial but a product? You can draw $\frac1{m!} f^{(m)} \otimes \frac1{n!} g^{(n)}$ as two vertices, one labeled $f$ with $m$ incoming strands and the other labeled $g$ with $n$ incoming strands, and the symmetries are correct, and using $\frac1{m!} f^{(m)} \otimes \frac1{n!} g^{(n)} = \binom{m+n}{m,n} (f^{(n)}\otimes g^{(m)})$ and the above theorem, the RHS now can be taken as a sum over all graphs (possibly disconnected) with two vertices, one labeled $f$ and the other labeled $g$, where the edges are still valued $a^{-1}$ and a labeled graph is weighted by its automorphism group.

Ok, so now let's try to calculate asymptotics of integrals, and I will henceforth ignore the "determinant" prefactors. My domain of integration will always be a "small" neighborhood of $0$. I'm interested in the $\hbar \to 0$ asymptotics of $$ \int_X \exp \frac1\hbar\left( - a\cdot \frac{x^{\otimes 2}}2 + \sum_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right) {\rm d}x $$ First rescale $x \mapsto \sqrt\hbar x$; this just rescales the overall integral by $\hbar^{\dim X/2}$, and I'm dropping those terms. Then the integral is $\exp\bigl( - a\cdot \frac{x^{\otimes 2}}2 + O(\hbar) \bigr)$, so keep the $O(1)$ part in the exponent but expand the $O(\hbar)$ part out: $$ = \int_X {\rm d}x \exp \left(-\frac12 a\cdot x^{\otimes 2}\right) \times \sum_{m\geq 0} \frac1{m!} \left( \sum_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right)^m$$ Expanding the sum further, the $b$s still look like vertices with $n\geq 3$ incoming strands (and $n!$ symmetries), but now I get to have $m$ many of them, weighted by the $m!$ symmetries from permuting the vertices. So the sum is over all collections of trivalent-and-higher vertices, counted with symmetry, and an $n$-valent vertex is valued $\hbar^{\frac n 2 - 1}b^{(n)}.

Then we integrate by connecting them up. All together, we get: $$ = \sum_{\text{graphs }\Gamma} \frac{\hbar^\# \operatorname{ev}(\Gamma)}{\lvert\operatorname{Aut}\Gamma\rvert} $$ Graphs can be disconnected, empty, have parallel edges and self loops, etc. We compute $\operatorname{ev}(\Gamma)$ by assigning $a^{-1}$ to each edge, $b^{(n)}$ to an $n$-valent vertex, and contracting tensors. The power on $\hbar$ is $-1$ for each vertex, and $\frac12$ for each half-edge (each part of an edge that arrives at a vertex), i.e. it's $-1$ for each vertex, $+1$ for each edge, i.e. it's the negative of the Euler characteristic of the graph.

Finally, let me get to the version from Getzler-Kapranov. Above, I used the fact that if $\star$ is a sum of connected things (counted with symmetry), then $\exp(\star)$ is a sum (counted with symmetry) over all possible collections of disjoint copies of $\star$. We've ended up with a sum-with-symmetry over disjoint things. Taking $\log$ gives the sum of connected things. For a connect graph, the negative Euler characteristic is precisely one less than the first Betti number.

Exercise: Redo the above calculations with $O(\hbar)$ corrections to the exponent of the initial integral, to end up with the precise Getzler-Kapranov result.

Finally, I should say one more thing about Feynman diagrams. Feynman and Dyson disagreed about the meaning of Feynman's diagrams: Feynman thought of them as pictures of particles interacting, and Dyson thought of them they way I do above: as graphs computing the asymptotics of integrals.

The point is the following. The most important integrals like those above that physicists care about come from particularly nice quantum field theories, where $X$ is a space of sections of some vector bundle (with some extra structure) on a Riemannian manifold, and $a$ is the Laplacian, and the $b^{(n)}$ are all "local". In this situation, $a^{-1}$ computes the "heat flow" for the Riemannian manifold, which is nothing more nor less than an integral over all paths connecting two endpoints of some "amplitude for a free particle to travel along this path". Then $\operatorname{ev}(\Gamma)$ can be interpreted as an integral over all embeddings of $\Gamma$ into your manifold of: the amplitude for your particle to travel along the edges, times an amplitude for an "interaction" at the vertices.

One good reference is the first chapter or two of Costello's recent book on QFT.

What's up with Wick's theorem? - MathOverflow

What's up with Wick's theorem?

Answer by Theo Johnson-Freyd for What's up with Wick's theorem?