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--- gaussian [2025/12/02 14:41] – spencer
+++ gaussian [2025/12/02 15:04] (current) – [The Wick formula] spencer
@@ Line 40: / Line 40: @@
     \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.
+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*}
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 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.