The creation, $\hat{a}(k)$, and annihilation, $\hat{a}^\dagger(k)$, operators obey the following commutation (non abelian) algebra
$$[\hat{a}(k), \hat{a}^\dagger(k’)] = \delta(k - k’).$$
I have the following vacuum expectation value (VEV) that I have managed to solve by repeated use of the above commutation relations (until all annihilation operators were shifted to the right, whereupon their operator on the vacuum yields zero)
$$\langle0\vert\prod_{j=1}^N \hat{b}(k_j)\prod_{j=1}^N \hat{b}^\dagger(k'_j)\vert0\rangle = \sum_\sigma\prod_{j=1}^N \delta\left(k_j - k'_{\sigma(j)}\right),$$
where the permutation sum $\sigma$ identifies all the possible combinations of pairing $x_j$ with $x'_k$ for $k = \sigma(j)$. The method of repeated use of the commutation is incredibly laborious and difficult with increasing length of operators. An alternative method of solving this is through the Stirling numbers as reported here. How can I use this method to prove the VEV shown above. I would like to use this new method to find more complicated VEVs.
Any help is appreciated!