What's up with Wick's theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:11:59Z http://mathoverflow.net/feeds/question/47350 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47350/whats-up-with-wicks-theorem What's up with Wick's theorem? Dan Petersen 2010-11-25T17:58:37Z 2010-11-25T23:18:27Z <p>Sorry about the dumb title.</p> <p>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.</p> <p>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.</p> <p>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 <em>f</em> 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 $\langle f_if_j\rangle$. </p> <p>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 &amp; Kapranov's paper on modular operads (sorry for the lengthy quote): </p> <blockquote>[...] 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 $W \sim \frac 1 \hbar \left(\frac{\xi^2}{2} + \sum_{2g-2+n>0} \frac{\hbar^g\xi^n}{n!} \sum_{G\in \Gamma((g,n))} \frac{1}{|\mathrm{Aut}(G)|} \prod_{v\in \mathrm{Vert}(G)} a_{g(v),n(v)}\right)$. </blockquote> <p>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. </p> <p>Can someone give a hint about how these Wick theorems fit together?</p> http://mathoverflow.net/questions/47350/whats-up-with-wicks-theorem/47385#47385 Answer by Theo Johnson-Freyd for What's up with Wick's theorem? Theo Johnson-Freyd 2010-11-25T23:18:27Z 2010-11-25T23:18:27Z <p>Let's take for granted the Gaussian integration formula, which holds for both bosonic and fermionic integrals, if they are properly interpreted:</p> <p><strong>Theoreom (Gauss, Wick):</strong> 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 <code>$$\int_X \sum f^{(n)} \cdot \frac{x^{\otimes n}}{n!} \exp \left(-\frac12 a\cdot x^{\otimes 2}\right){\rm d}x = \sqrt{\det(2\pi a)} \sum f^{(2k)} \cdot \frac{(a^{-1})^{\otimes 2k}}{2^kk!}$$</code> 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.</p> <p><strong>Proof:</strong> Integrate by parts. $\Box$</p> <p>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.</p> <p>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.</p> <p>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 <code>$$\int_X \exp \frac1\hbar\left( - a\cdot \frac{x^{\otimes 2}}2 + \sum_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right) {\rm d}x$$</code> 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: <code>$$= \int_X {\rm d}x \exp \left(-\frac12 a\cdot x^{\otimes 2}\right) \times \sum_{m\geq 0} \frac1{m!} \left( \sum_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right)^m$$</code> 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)}.</p> <p>Then we integrate by connecting them up. All together, we get: <code>$$= \sum_{\text{graphs }\Gamma} \frac{\hbar^\# \operatorname{ev}(\Gamma)}{\lvert\operatorname{Aut}\Gamma\rvert}$$</code> 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.</p> <p>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.</p> <p><strong>Exercise:</strong> Redo the above calculations with$O(\hbar)$corrections to the exponent of the initial integral, to end up with the precise Getzler-Kapranov result.</p> <hr> <p>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.</p> <p>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.</p> <p>One good reference is the first chapter or two of Costello's recent book on QFT.</p>