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 $|f|^{2}=(Ref)^{2}+(Imf)^{2}\geq (Imf)^{2}$

therefore, ${\frac {\mathrm {d} \sigma (\theta )}{\mathrm {d} \Omega }}\geq (Imf_{k}(\theta ))^{2}$

On the other hand, from the optical theorem we have

$\sigma ={\frac {4\pi }{k}}Imf_{k}(\theta ))\leq {\frac {4\pi }{k}}{\sqrt {\frac {\mathrm {d} \sigma (0)}{\mathrm {d} \Omega }}}$

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

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

The total cross section is $\sigma ={\frac {4\pi ^{2}}{k^{2}}}\sum _{l=0}^{\infty }(2l+1)sin^{2}\delta _{l}$

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

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

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

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

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