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Could someone please elaborate on the accepted answer to this mathoverflow post?

I'm working on a problem that looks like this \begin{equation}I=\int d^{n} x\, f(\vec x)\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\end{equation} I thought about expanding $f(\vec x)$ as a series and using Wick's theorem for each term, but I can't figure how to do the resulting sum. The post in question seems like it may answer this, but I don't really understand the notation in the first equation of his. In my example, $f(\vec x)=\prod_{i}f_{i}(x_{i})$ if that helps. Any clarification of the aforementioned answer would be awesome.

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In the particular case where $f(\vec x)$ is a sum of expressions, where each expression has a total even power of the $x^i$ like : $f(x) = x_1x_2 + x_1^2 x_2 x_3 + ...$, . we may present general expressions. we have :

$$I(\Sigma)=\int d^{n} x\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\tag{1} = \sqrt{(2 \pi)^ndet \Sigma}$$

We have : $$\langle x_{i_1}\,x_{i_2} x_{i_3}\,x_{i_4}...x_{i_{2n-1}}\,x_{i_{2n}}\rangle= \frac{\int d^{n} x\,(x_{i_1}\,x_{i_2} x_{i_3}\,x_{i_4}...x_{i_{2n-1}}\,x_{i_{2n}})\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}}{\int d^{n} x\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}} \\= \Sigma_{contractions} (\Sigma_{j_1j_2})(\Sigma_{j_3j_4})....(\Sigma_{j_{2n-1}j_{2n}})\tag{2}$$ where a contraction is a repartition 2 by 2 of indices $i_1...i_{2n}$ in pairs $(j_1j_2), (j_3j_4)...(j_{2n-1}j_{2n})$, with no double counting ($(j_1j_2) = (j_2j_1)$)

Now, with $1$ and $2$, you may theorically calculate your integrals, for instance, with $f(x) = x_1x_2x_3x_4$, you have :

$$\begin{equation}I=\int d^{n} x\, f(\vec x)\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\end{equation} = I(\Sigma)(\Sigma_{12}\Sigma_{34} + \Sigma_{13}\Sigma_{24}+\Sigma_{14}\Sigma_{23})\tag{3}$$

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  • $\begingroup$ I get that, but the post seemed to say that the resulting sum of contractions could be represented in terms of the even coefficients of the original function. That's what I'd like to know. $\endgroup$ – TeeJay Oct 5 '13 at 23:39
  • $\begingroup$ @TeeJay : In the math article, you have the very particular choice $\Sigma^{-1} = \mathbb{Id}$, It is obvious, that, with $\Sigma^{-1} = \mathbb{Id}$, the odd terms of the function disappear in the integration. But your question was for a general $\Sigma^{-1}$. If you take my last example, with $f(x) = a^{1234} x_1x_2x_3x_4$, the result would be $\begin{equation}I=\int d^{n} x\, f(\vec x)\, e^{-\frac{1}{2} \vec x \cdot \Sigma^{-1}\cdot \vec x}\end{equation} = I(\Sigma)a^{1234}(\Sigma_{12}\Sigma_{34} + \Sigma_{13}\Sigma_{24}+\Sigma_{14}\Sigma_{23})$,.... $\endgroup$ – Trimok Oct 6 '13 at 7:38
  • $\begingroup$ @TeeJay : ......more generally if you know the decomposition of $f$ into a sum of expressions of total even power in the $x_i$, $f(x) = \large \Sigma_{i_1...i_{2n}} a_{i_1....i_{2n}} x_{i_1}....x_{i_{2n}}$, the result would be, $ I= I(\Sigma)\large \Sigma_{i_1...i_{2n} a_{i_1....i_{2n}}}(\Sigma_{contractions (i_1...i_{2n} \to (j_1j_2)....(j_{2n-1}j_{2n})}\Sigma_{j_1j_2}....\Sigma_{j_{2n-1}j_{2n}})$ $\endgroup$ – Trimok Oct 6 '13 at 7:39

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