Optical theorem in QFT

I've been working with the Optical theorem in the case in which final and initial states are equals and I have the following doubt. Let's write the scattering matrix $$S$$ as:

$$S = 1 + i·T \tag1$$

where $$T$$ is the transition matrix. Therefore, the Optical theorem is:

$$2·Im(T) = T^\dagger T \implies 2·Im(T_{ii}) = \sum_a|T_{ai}|^2 \tag2$$

In Eq. (2), we have imposed the particular case commented above with $$a$$ any state in between. So my doubt arises from here: if, for Eq. (1), $$ = 1 + i \implies S_{ii} = 1 + iT_{ii},\tag3$$ then for that particular case, when I computed the $$T_{ii}$$ and took $$|S_{ii}|^2 = probability\ of\ the\ process\ i \rightarrow i,\tag4$$ I will get a probability greater than 1. But that isn't possible.

What am I misunderstanding?

• What happened to the crossterm $-2{\rm Im} T_{ii}$ in eq. (4)? Oct 22, 2018 at 20:00
You might be assuming the matrix element $$T_{ii}$$ to be real. If so, then
$$\lvert S_{ii} \rvert^2 = 1 + \lvert T_{ii} \rvert^2 > 1$$
Without such an assumption, \begin{align*} \lvert S_{ii} \rvert^2 &= 1 + \lvert T_{ii} \rvert^2 - 2\mathrm{Im}(T_{ii})\\ &= 1 + \lvert T_{ii} \rvert^2 - \sum_a \lvert T_{ai} \rvert^2 \\ &= 1 - \sum_{a \neq i} \lvert T_{ai} \rvert^2 \end{align*}
which is smaller than $$1$$.