A problem related to Wick's theorem from RG analysis of KT transition Recently, I was reading a review paper by John B. Kogut An introduction to lattice gauge theory and spin systems, when he was doing the RG analysis for the X-Y model, on page 702, to go from (7.61a) to (7.61b), seem there is a step where we need to use the following identity:(at least when I use it, I can get the correct result...)
$$
\begin{equation}
\langle[h(x)+h(y)]^{2n}\rangle_0=\frac{1}{n!}C_{2n}^2C_{2n-2}^2\cdots C_{2}^2 \{\langle[h(x)+h(y)]^{2}\rangle_0\}^n
\end{equation}
$$
where the $\langle \cdots \rangle_0$ means average over the free (Gaussian) action of $\textbf{Real bosonic}$ field $h(x)$. (the factor $C_n^k$ is the binomial factors: $C_n^k=\frac{n!}{k!(n-k)!}$)It looks like a Wick's theorem for the "$h(x)+h(y)$". 
I wonder if this relation is true and if it's true, how to get it from Wick's theorem?
 A: To show that the OP is right, one has to show two things :


*

*That Wick's theorem works for the sum of two fields.

*That the combinatorial factors are ok.


First things first. We know that the field is Gaussian (and in fact of zero mean), so the averages of its powers obey Wick's theorem, and that odd powers vanishes. 
One has trivially that $\langle (h_x+h_y)^2\rangle$ obeys Wick's theorem. Assuming that it works for $\langle (h_x+h_y)^{2n}\rangle$ with $n$ an integer, one shows by recurrence that 1 works, for example by using the generating functional and using $\partial_{J_x}+\partial_{J_y}$ to generate the correlation functions, instead of just one derivative.
For the second point, one just uses the standard Wick's theorem argument. One singles out one pair of $(h_x+h_y)^2$. There are $C^2_{2n}$ ways to do that, but $n$ of them are equivalent, so the correct factor is $C^2_{2n}/n$. Then one repeats the argument for the $2n-2$ remaining pairs.
A: I just got an idea of how to prove it:


*

*Write the equation as:


$\langle [h(x)+h(y)]^{2n} \rangle_0=\sum_{ \{\alpha_i=x \ or\  y\} } 
\langle h(\alpha_1)h(\alpha_2) \cdots h(\alpha_{2n})\rangle_0
$
the summation over $\alpha_i's$ contains over all the possible cases, and as for $\langle h(\alpha_1)h(\alpha_2) \cdots h(\alpha_{2n})\rangle_0$, we can use the Wick's theorem


*As for the  $\langle h(\alpha_1)h(\alpha_2) \cdots h(\alpha_{2n})\rangle_0$, we can use the Wick's theorem, and from mathematics, when know that there are $\frac{1}{n!}C_{2n}^2C_{2n-2}^2 \cdots C_{2}^2$ ways to do the contraction, in other words, we will be left with $\frac{1}{n!}C_{2n}^2C_{2n-2}^2 \cdots C_{2}^2$ terms after the contraction of $\langle h(\alpha_1)h(\alpha_2) \cdots h(\alpha_{2n})\rangle_0$. (Remember there is still a sum of $\alpha_i's$ !)

*For each way of the contraction, we will have, for example:
$\begin{align}
&\sum_{ \{\alpha_i \} }  \langle h(\alpha_1)h(\alpha_i)\rangle_0 \langle h(\alpha_j)h(\alpha_k)\rangle_0 \cdots \langle h(\alpha_l)h(\alpha_m)\rangle_0 \\
&= \left( \sum_{\alpha_1,\alpha_i} G(\alpha_1,\alpha_i) \right) \times \left( \sum_{\alpha_j,\alpha_k} G(\alpha_j,\alpha_k) \right) \times \cdots \left( \sum_{\alpha_l,\alpha_m} G(\alpha_l,\alpha_m) \right) \\
&= \left[ \langle [h(x)+h(y)]^{2} \rangle_0 \right]^n
\end{align} $
so we see that for each term given by Wick's theorem, after the $\sum_{ \{\alpha_i \} }$, gives the same result.
Combine with the number of ways of contraction , which is $\frac{1}{n!}C_{2n}^2C_{2n-2}^2 \cdots C_{2}^2$, we can finally arrive at the answer.
