Optical theorem

Write the S-matrix as $S=1+iT$. Unitarity gives $-i(T-T^{†})=T^{†}T$, so the diagonal forward matrix element obeys $2\mathrm{Im}{T}_{ii}={\sum }_f\int d{\Phi }_f|T_{fi}{|}^2$. Thus the imaginary part of a forward scattering amplitude measures the total inclusive transition rate. For an electromagnetic source, the hadronic electromagnetic-current spectral density supplies the inclusive spectral sum.

For a real elastic scattering potential the first Born forward amplitude is real, while the leading total scattering cross-section is quadratic in the potential. The second Born amplitude supplies an imaginary part from its outgoing Green function. Symmetrizing the two interaction positions makes its imaginary part equal to $k/(4\pi )$ times the integrated squared first Born amplitude. The identity therefore compares quantities at the same perturbative order; a real first Born amplitude alone cannot test it against a nonzero scattering cross-section.

For an isotropic elastic amplitude normalized by an incident unit plane wave and outgoing $f{e}^{ikr}/r$, total cross section is $4\pi |f{|}^2$. The optical theorem gives the displayed identity. At a resonance where the real part of $1/f$ vanishes, $f=i/k$ and the cross section is $4\pi /k^2$.