I am reading P&S (Peskin's and Schroeder's book on QFT), Chapter 16.3 entitled Ghosts and Unitarity. The authors employ the optical theorem to calculate the imaginary part of a $f\bar{f}\rightarrow f\bar{f}$ amplitude that also contains a gauge boson loop. Diagrams of this sort are given by taking the diagrams in Figure 16.2 and gluing them together with diagrams that correspond to the reverse processes (i.e. 2 incoming bosons $\rightarrow$ 1 fermion and 1 antifermion) in all the possible ways.
According to the authors, applying the optical theorem results in the imaginary part of this combined amplitude being reduced to a multiplication of sub-amplitudes corresponding to processes of the form $f\bar{f}\rightarrow bb$ (i.e. sum wrt all the possible ways in which the process $f\bar{f}\rightarrow bb$ can occur) with sub-amplitudes corresponding to the processes of the form $bb\rightarrow f\bar{f}$ (i.e. sum wrt all the possible ways in which the process $bb\rightarrow f\bar{f}$ can occur), integrated over the phase space corresponding to the momenta of the gauge bosons (and also divided by a symmetry factor related to the exchange of the two abovementioned momenta).
My first question is related in my observation that in the optical theorem [formula (7.49)], there is a complex conjugate of the amplitude, corresponding to an interaction going from the final state to an intermediate state. On the other hand, when the authors apply the optical theorem, they have an amplitude (not it's complex conjugate) that corresponds to an intermediate state going to the final state. In other words, according to Eq. (7.49), we should have $$\mathcal{M}^*(f\bar{f}\rightarrow bb)$$ multiplied with $\mathcal{M}(f\bar{f}\rightarrow bb)$ and integrated over the momenta $k_1$ and $k_2$ (gauge boson momenta) in Eq.(16.37), but instead we have $$\mathcal{M}(bb\rightarrow f\bar{f})$$ Is there a relation between the two? How do I see this mathematically?
My second question has to do with the $i\mathcal{M}_{\text{ghost}}$ (representing the $g\bar{g}\rightarrow f\bar{f}$ procedure) amplitude in Eq. (16.42). The latter is given by $$i\mathcal{M}_{\text{ghost}}= ig\bar{\upsilon}(p_+)\gamma_{\mu}t^cu(p) \frac{(-i)}{(k_1+k_2)^2}(gf^{abc}k_1^{\mu})$$ whereas the corresponding $i\mathcal{M}_{\text{ghost}}'$ (where the prime denotes the amplitude representing the procedure $f\bar{f}\rightarrow g\bar{g}$) is given by $$i\mathcal{M}_{\text{ghost}}'= ig\bar{u}(p)\gamma_{\mu}t^c\upsilon(p_+) \frac{(-i)}{(k_1+k_2)^2}(-gf^{abc}k_2^{\mu})$$ The authors claim that the relative minus sign is somehow related to the ghost loop, but I do not see that that way. I think that when we have a vertex with a gauge boson, a ghost and an anti-ghost and when the latter two "particles" are outgoing, then we have a factor $(gf^{abc}k^{\mu})$, where $k$ is the momentum of the anti-ghost. If on the other hand, the two latter particles are ingoing, then there is a minus sign in the Feynman rule corresponding to the vertex at hand and this is my explanation for the minus sign. Are the two explanations related? Or is mine wrong?
Any help will be appreciated.
P.S.#1: $f$ stands for a fermion, $\bar{f}$ stands for an anti-fermion and $b$ stands for a gauge boson, whereas $g$ stands for a ghost and $\bar{g}$ stands for an anti-ghost.
P.S.#2: I have seen the possible related post Form of the optical theorem in non-Abelian theory, but this is focused on other aspects of the calculation...