I'm working through Zee's QFT in a Nutshell but there's an integral [I.2 (16)] I couldn't quite derive. The problem is to find $$\langle x_i x_j ... x_k x_l\rangle=\frac{\int ... \int dx_1 ... dx_n x_i x_j ... x_k x_l\exp(-\frac{1}{2}\vec{x}\cdot A\cdot\vec{x})}{\int ... \int dx_1 ... dx_n \exp(-\frac{1}{2}\vec{x}\cdot A\cdot\vec{x})} $$ using $$ \int ... \int dx_1 ... dx_n \exp(-\frac{1}{2}\vec{x}\cdot A\cdot\vec{x}+\vec{J}\cdot\vec{x})=\sqrt{\left(\frac{(2\pi)^n}{\det[A]}\right)}\exp(\frac{1}{2}\vec{J}\cdot A^{-1} \cdot \vec{J}) $$ with $A$ a real symmetric matrix. Zee suggests taking repeated derivatives with respect to $\vec{J}$ and then letting $\vec{J}\rightarrow0$. Here's my attempt at a solution:

The top of the numerator in the first equation can be obtained by differentiating the the above equation by $\frac{\partial}{\partial J_a}$ for $a=i,j,...,k,l.$ Dropping the constant for now, we find that: $$ \int ... \int dx_1 ... dx_n x_i \exp(-\frac{1}{2}\vec{x}\cdot A\cdot\vec{x}+\vec{J}\cdot\vec{x})=\left(\sum_m^n A^{-1}_{im}J_m\right)\exp(\frac{1}{2}\vec{J}\cdot A^{-1} \cdot \vec{J}). $$ This process of taking derivatives can be repeated via the product rule to get messy stuff, but what worries me is that when one takes $\vec{J}\rightarrow0$ the integral will vanish, will it not? Because the sums over $m$ that pop out will all vanish. The answer, Zee says, is not zero, but instead $$ \langle x_i x_j ... x_k x_l\rangle = \sum_{{\rm Wick}}(A^{-1}_{ab})...(A^{-1}_{cd}) $$ "where the set of indices {$a,b,...,c,d$} represent a permutation of the set of indices {$i,j,...,k,l$}. The sum is over all such permutations or Wick contractions." How exactly does one reach this result?

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    $\begingroup$ Just derive your expression a second time w.r.t. $J_m$. Already you should see that there will be a non-zero term. So you shouldn't expect the entire expression to be zero. $\endgroup$ – Raskolnikov May 26 '12 at 10:37
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    $\begingroup$ I agree with Raskolnikov. Your "proof" that it's zero seems to be based on taking an "example" with one $x$ insertion only. You got zero - which is right (and you get zero from the right, Zee's formula as well) - but you apparently extrapolated (by "math induction"?) that the result is zero for many $x$'s, too. But it's not. Why don't you try to insert at least 2 - and then 4 - $x$'s in the product? And don't forget to differentiate the factor $\sum A^{-1} J$ with respect to $J$! $\endgroup$ – Luboš Motl May 26 '12 at 10:44
  • $\begingroup$ Huh, I did so on paper, but I was too lazy to write it out here in LaTeX. I'll try it again haha, sorry! $\endgroup$ – Nilay Kumar May 26 '12 at 11:04
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    $\begingroup$ Ah, got it! I had been doing exactly what you suggested @LubošMotl, it's just that I was making a silly arithmetic mistake. Thanks for the help, guys! (In hindsight, of course 1 (and all odd multiplicities) would vanish - the integrand is odd!) $\endgroup$ – Nilay Kumar May 26 '12 at 11:18

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