Why does S-matrix unitarity imply the cross section $\sigma$ $\propto$ $\frac {1}{s}$? I'm currently learning for an oral exam in theoretical physics and as a learning aid protocols of older exams exist. In one protocol the question was asked: 

Why is the scattering cross section $\sigma$ at least proportional to $\frac {1}{s}$. The correct answer was that this is the case because the S-Matrix is unitary. Furthermore in the answer it was mentioned, that $M$, defined as usual by $S= 1 + i (2 \pi)^4 \delta(...) M$ is unrestricted because $S$ is unitary. 

Does anyone have an idea where I can read about this? I searched in quite a few books, but wasn't able to find anything related. (I do understand that S needs to be unitary and know the relationship between the cross section and the S-matrix) 
Any idea or explanation would be awesome!
 A: I wonder if this may be of some help.  
The optical theorem  relates the total cross-section of a scattering with the forward scattering amplitude. For example for a $2\to2$ scattering you get that:
$\left\langle n\left|S\right|m\right\rangle \equiv\delta_{mn}+i\left(2\pi\right)^{4}\delta^{4}\left(p_{m}^{\mu}-p_{n}^{\mu}\right)\left\langle n\left|\mathcal{T}\right|m\right\rangle $
$\sigma_{t}^{12}=\frac{1}{2\vert p_{1}\vert\sqrt{s}}\mbox{Im}\left\langle n\left|\mathcal{T}\right|n\right\rangle$
What is called amplitude is $\mathcal{A}(s,t)=\left\langle n\left|\mathcal{T}\right|m\right\rangle $. In the Mandelstan variables the optical theorem reads:
$\sigma_{t}^{12}=\frac{1}{2\vert p_{1}\vert\sqrt{s}}\mbox{Im}\mathcal{A}\left(s,t=0\right)$
This is a consequence of unitarity and is not difficult to show. One can say then that $p_1\sim\sqrt{s}$, at least in the Regge limit, thus for a constant amplitude $\sigma_t\sim \frac{1}{s}$. I think one may later argue that is unphysical to obtain amplitudes that grows with negative exponents of $s$. 
