# Feynman diagrams: sum of square of diagrams or square of sum of diagrams?

I’m asking this question in the context of the discussion of the optical theorem in section 7.3 of Peskin and Schroeder. The optical theorem states that the imaginary part of a scattering amplitude with the same initial and final state is given by the square of the amplitude for the initial state to scatter into an arbitrary final state, summed over final states and integrated over the final state phase space. In symbols, this can be written (equation 7.49): $$2\, \mathrm{Im}\, \mathcal{M}(a \to a) = \sum_f \int \mathrm{d} \Pi_f |\mathcal{M}(a \to f)|^2$$ In the book, there is a discussion of how this identity arises in the Feynman diagram expansion. In particular, there is the following figure (7.6) describing Bhabha (electron-positron) scattering: Here time runs from left to right. My problem is that there is a second diagram at order $\alpha$ which contributes to electron-positron scattering, the $s$-channel diagram. There are thus two diagrams which contribute to $\mathcal{M}(a \to f)$ on the right hand side of the (order $\alpha^2$ part of the) optical theorem, and these should be summed before the squared modulus is taken. There are also more diagrams at order $\alpha^2$ which contribute to the left hand side. Considering only the terms where the 'final state' $f$ is an electron-positron pair, I believe that the optical theorem in this case should read: Where the circular loop here is supposed to be an electron-positron loop. Expanding the square on the right hand side will give cross-terms. If we can apply the optical theorem to each individual diagram, as appears to have been done in figure 7.6, then we would conclude that these cross-terms must be zero, which is not the case.

What's going on here?