# 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!

• Are you talking about what cross-section? Is it the total cross-section $\sigma^{tot}_{ab}$ for $a+b\rightarrow anything$? If this is the case it is well known that $\sigma^{tot}_{ab}\propto\log^2 s$ is compatible with unitarity (Froissart-Martin bound) and in fact observed experimentally in various processes Commented Oct 6, 2014 at 13:25
• FYI: In the title, if you wanted to use an arrow for 'implies', it's $\implies$ rather than $\rightarrow$. Commented Oct 6, 2014 at 13:26

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 $s$.

• Thank you! I will read about the optical theorem in this context and report back
– jak
Commented Oct 9, 2014 at 11:21