Phy5645/Cross Section Relation

Consider the scattering of a particle from a real spherically symmetric potential. If ${\frac {\mathrm {d} \sigma (\theta )}{\mathrm {d} \Omega }}$ is the differential cross section and $\sigma$ is the total cross section, show that $\sigma \leq {\frac {4\pi }{k}}{\sqrt {\frac {\mathrm {d} \sigma (0)}{\mathrm {d} \Omega }}}$

for a general central potential using the partial-wave expansion of the scattering amplitude and the cross section.

Solution:

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 (Im f_{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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{l^} (1)= 1} we can write

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

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

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