Differences

This shows you the differences between two versions of the page.

--- gaussian [2025/12/02 14:04] – created spencer
+++ gaussian [2025/12/02 15:04] (current) – [The Wick formula] spencer
@@ Line 14: / Line 14: @@
 \int_{\mathbb{R}^n} e^{-\pi |y|^2} \mathrm{d}y = 1.
 \]
-More generally, we might change coordinates by an invertible linear transformation $x = Sy$, whence $|y|^2 = |Sx|^2$ and
+More generally, we might change coordinates by an invertible linear transformation $Sx = y$, whence $|y|^2 = |Sx|^2$ and
 \[
-\int_{\mathbb{R}^n} \frac{e^{-\frac{1}{2} |Sx|^2}}{(2\pi)^{n/2}\det(S)} \mathrm{d}y = 1;
+\int_{\mathbb{R}^n} \frac{e^{-\frac{1}{2} |S^{-1} x|^2}}{(2\pi)^{n/2}}\det(S) \mathrm{d}y = 1;
 \]
 that is,
 \[
-\int_{\mathbb{R}^n} \frac{e^{-\frac{1}{2} |Sx|^2}}{\det(\sqrt{2\pi} S)}  \mathrm{d}y = 1;
+\int_{\mathbb{R}^n} \frac{e^{-\frac{1}{2} |Sx|^2}}{\det(\sqrt{2\pi} S^{-1})}  \mathrm{d}y = 1;
 \]
+Note that $|Sx|^2 = x^T A x$ with $A = S^T S$, and $\det(A) = \det(S)^2$; thus, for any symmetric $A$
+\[
+\int_{\mathbb{R}^n} \frac{e^{-\frac{1}{2} x^T A x}}{\sqrt{\det(2\pi A^{-1})}}  \mathrm{d}x = 1;
+\]
+taking $A = \lambda I$ recovers our earlier formulae.
+Each of these integrals is a measure on $\mathbb{R}^n$.
+Let $\mu_A$ denote the measure in this last formulation; a //Gaussian integral// is an integral $\int f(x) \mu_A(x)$ with respect to such a Gaussian measure.
+Let $\langle f \rangle_A$ denote this integral, or just $\langle f \rangle$ if $A$ is understood.
+Let $\mu_{\lambda}$ denote $\mu_{\lambda I}$, and let $\mu$ denote $\mu_1$.
+===== The Wick formula =====
+Gaussian integrals are particularly nice to evaluate.
+There's the usual Calculus 1 problem
+\[
+    \langle x \rangle = 0
+\]
+since $x e^{-\frac{x^2}{2}} = -\frac{\mathrm{d}}{\mathrm{d}x} e^{-\frac{x^2}{2}}$ is a total derivative.
+We likewise compute
+\begin{align*}
+\langle x^2 \rangle &= \int_{-\infty}^\infty x \frac{\mathrm{d}}{\mathrm{d}x} e^{-\frac{x^2}{2}} \mathrm{d} x\\
+&= \langle 1 \rangle = 1.
+\end{align*}
+Continuing this computation, clearly for all $n \ge 1$, $\langle x^{2n-1} \rangle = 0$ and
+\[
+    \langle x^{2n} \rangle = (2n-1) \langle x^{2n-2} \rangle = (2n-1)!!
+\]
+is a double factorial.
+The goal is to extend this formula into the multivariate setting, and therefore find a rule to evaluate for any monomial the quantity $\langle x_{i_1} \cdots x_{i_k}\rangle$.
+Certainly if the degree of the monomial is odd, its Gaussian integral is zero; some variable must occur in odd degree, and then the single variable proof suffices.
+Moreover, integrating by parts gives $\langle x_i x_j \rangle_A = (A^{-1})_{ij}$.
+**Theorem ** (Wick). Let $f_1, \ldots, f_{2n}$ be linear polynomials. Then
+\[
+  \langle f_1 \cdots f_{2n} \rangle_A = \frac{1}{2^n n!} \sum \langle f_{p_1} f_{q_1} \rangle_A \cdots \langle f_{p_n} f_{q_n} \rangle_A,
+\]
+where the sum runs over permutations of $(1, \ldots, 2n)$.
+Note that $\frac{(2n)!}{2^n n!} = (2n-1)!!$, as we would expect.
+The formula can be made more computationally efficient by summing just over those permutations for which the $p_i$ are increasing and $p_j < q_j$ for each $j$; there are $(2n-1)!!$ of these, so no normalization is needed.
+This is probably only really useful when $n = 4$ or $6$, after which there is already an explosion in the number of terms needed.
+A consequence of Wick's formula is that the Gaussian integral of a monomial is a sum of products of matrix coefficients, and thus Gaussian integration against a monomial is a kind of a permanent-type operation applied to $A^{-1}$; thus Cramer's rule implies that Gaussian integrals of polynomials are always rational functions in the entries of $A$.
+The proof of the Wick formula is simple; assume that the input polynomials are all monomial, and after an orthogonal change of coordinates assume that $A$ is diagonal. Then integrate by parts until you're done.