The differential cross section is related to the scattering amplitude through
d σ ( θ ) d Ω = | f k ( θ ) | 2 . {\displaystyle {\frac {d\sigma (\theta )}{d\Omega }}=|f_{k}(\theta )|^{2}.}
Since | f | 2 = ( ℜ e f ) 2 + ( ℑ m f ) 2 ≥ ( ℑ m f ) 2 , {\displaystyle |f|^{2}=(\Re e\,f)^{2}+(\Im m\,f)^{2}\geq (\Im mf)^{2},}
we obtain
d σ ( θ ) d Ω ≥ ( ℑ m [ f k ( θ ) ] ) 2 . {\displaystyle {\frac {d\sigma (\theta )}{d\Omega }}\geq (\Im m[f_{k}(\theta )])^{2}.}
On the other hand, the optical theorem states that
σ = 4 π k ℑ m [ f k ( 0 ) ] , {\displaystyle \sigma ={\frac {4\pi }{k}}\Im m[f_{k}(0)],}
so that
d σ ( 0 ) d Ω ≥ k 2 σ 2 16 π 2 . {\displaystyle {\frac {d\sigma (0)}{d\Omega }}\geq {\frac {k^{2}\sigma ^{2}}{16\pi ^{2}}}.}
From this, it follows that σ ≤ 4 π k d σ ( 0 ) d Ω . {\displaystyle \sigma \leq {\frac {4\pi }{k}}{\sqrt {\frac {d\sigma (0)}{d\Omega }}}.}
Back to Central Potential Scattering and Phase Shifts
![{\displaystyle {\frac {d\sigma (\theta )}{d\Omega }}\geq (\Im m[f_{k}(\theta )])^{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb3e932d4b3d2da6c7aec3ef3d2c6954e723b0f4)
![{\displaystyle \sigma ={\frac {4\pi }{k}}\Im m[f_{k}(0)],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65449244b1134bd3afbf6d5b0e7732fd91a045cb)
