Phy5645/Cross Section Relation

The differential cross section is related to the scattering amplitude through

${\frac {\mathrm {d} \sigma (\theta )}{\mathrm {d} \Omega }}=|f_{k}(\theta )|^{2}$

Since $|f|^{2}=(\Re ef)^{2}+(\Im mf)^{2}\geq (\Im mf)^{2},$

we obtain

${\frac {d\sigma (\theta )}{d\Omega }}\geq (\Im m[f_{k}(\theta )])^{2}.$

On the other hand, from the optical theorem we have

$\sigma ={\frac {4\pi }{k}}\Im m[f_{k}(\theta )])\leq {\frac {4\pi }{k}}{\sqrt {\frac {d\sigma (0)}{d\Omega }}}$

For a central potential the scattering amplitude is

$f_{k}(\theta )={\frac {1}{k}}\sum _{l=0}^{\infty }(2l+1)e^{i\delta _{l}}\sin \delta _{l}P_{l}(\cos \theta )$

and, in terms of this, the differential cross section is

${\frac {d\sigma (\theta )}{d\Omega }}={\frac {1}{k^{2}}}\sum _{l=0}^{\infty }\sum _{l^{\prime }=0}^{\infty }(2l+1)(2l^{\prime }+1)e^{i(\delta _{l}-\delta _{l^{\prime }})}\sin \delta _{l}\sin \delta _{l'}P_{l}(\cos \theta )P_{l'}(\cos \theta )$

The total cross section is then

$\sigma ={\frac {4\pi ^{2}}{k^{2}}}\sum _{l=0}^{\infty }(2l+1)\sin ^{2}\delta _{l}.$

Since $P_{l}(1)=1\!$ we can write

${\frac {d\sigma (0)}{d\Omega }}={\frac {1}{k^{2}}}\left[\sum _{l=0}^{\infty }(2l+1)e^{i\delta _{l}}sin\delta _{l}\right]^{2}={\frac {1}{k^{2}}}\left[\sum _{l=0}^{\infty }(2l+1)\sin \delta _{l}\cos \delta _{l}+i\sin ^{2}\delta _{l}\right]^{2}$

$={\frac {1}{k^{2}}}\left[\sum _{l=0}^{\infty }(2l+1)\sin \delta _{l}\cos \delta _{l}\right]^{2}+{\frac {1}{k^{2}}}\left[\sum _{l=0}^{\infty }(2l+1)\sin ^{2}\delta _{l}\right]^{2}\geq {\frac {1}{k^{2}}}\left[\sum _{l=0}^{\infty }(2l+1)sin^{2}\delta _{l}\right]^{2}={\frac {k^{2}\sigma ^{2}}{16\pi ^{2}}}.$

From this, it follows that $\sigma \leq {\frac {4\pi }{k}}{\sqrt {\frac {d\sigma (0)}{d\Omega }}}.$

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