# S-Matrix Generating Functional (Problem 4.1 in Weinberg)

I'm currently working through Weinberg's QFT book, but I'm somewhat stuck at problem 4.1, which states:

Define generating functionals for the S-matrix and its connected part: $$F[\nu]=1+\sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\int\nu^*(q_1^{\prime})...\nu^*(q_N^{\prime})\nu(q_1)...\nu(q_M)S_{q_1^{\prime}...q^{\prime}_Nq_1...q_M}dq_1^{\prime}...dq^{\prime}_Ndq_1...dq_M$$ and $$F^{C}[\nu]=\sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\int\nu^*(q_1^{\prime})...\nu^*(q_N^{\prime})\nu(q_1)...\nu(q_M)S^C_{q_1^{\prime}...q^{\prime}_Nq_1...q_M}dq_1^{\prime}...dq^{\prime}_Ndq_1...dq_M\, .$$ Derive a formula relating $F[\nu]$ and $F^{C}[\nu]$.

He defined the connected S-matrix in order to satisfy the Cluster-Decomposition principle as: $$S_{\beta\alpha}=\sum_{PART}S^{C}_{\beta_1\alpha_1}S^C_{\beta_2\alpha_2}....\, ,$$ where $\beta$ and $\alpha$ denote the set of initial and final momenta, respectively. For instance, we have: $$S^C_{q^{\prime}q}=S_{q^{\prime}q}=\delta(q^{\prime}-q)$$ or $$S_{q^{\prime}_1q^{\prime}_2q_1q_2}=S^C_{q^{\prime}_1q^{\prime}_2q_1q_2}+\delta(q_1^{\prime}-q_1)\delta(q_2^{\prime}-q_2)+\delta(q_1^{\prime}-q_2)\delta(q_2^{\prime}-q_1)\, .$$ Now, concerning the problem: I assumed the connection should be something like $$F[\nu]=\exp{iF^{C}[\nu]}\, ,$$ but I'm missing the way to generate any $i$'s. $$F[\nu]=1+F^C[\nu]+\sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\int\nu^*(q_1^{\prime})...\nu^*(q_N^{\prime})\nu(q_1)...\nu(q_M)\sum_{K=1}^{R}\delta(q^{\prime}_1-q_1)...\delta(q^{\prime}_K-q_K)S^C_{q_{K+1}^{\prime}...q^{\prime}_Nq_{K+1}...q_M}\binom{N}{K}\binom{M}{K}K!dq_1^{\prime}...dq^{\prime}_Ndq_1...dq_M\,$$ where $R=\min{(M,N)}-1$, since the sum gives zero otherwise. The problems seems to be entirely combinatoric, but somehow, I don't get it right. I would appreciate it, if you could give me a hint?