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Chapter 14: Problem 9

Show for the elastic scattering of a particle by a central potential \(V(r)\) that approaches zero more rapidly than \(1 / r\) as \(r \rightarrow \infty\) that the integral cross-section can be written as $$\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \mathrm{im} f_{k}(0)$$ where im \(f_{k}(0)\) is the imaginary part of the forward scattering amplitude \((\theta=0) .\) This is the so-called optical theorem. Hint. The Legendre polynomials are required to satisfy \(P_{l}(1)=1\) for all values of \(l\)

Short Answer

Expert verified
The optical theorem for particle scattering in a central potential, \(\sigma_{\text{tot}} = \frac{4\pi}{k} \cdot \text{im} f_k(0)\), is proven using the characteristics of the scattering amplitude, the Legendre polynomials, and Parseval's theorem.

Step by step solution

01

Define the Scattering Amplitude

We first define the scattering amplitude \(f(\theta)\) at a scattering angle \(\theta\). It is known that \(f(\theta)=\sum^{\infty}_{l=0}(2l+1)f_{l}P_{l}(\cos \theta)\). Here \(f_{l}\) corresponds to the partial wave amplitude, \(P_{l}(x)\) are Legendre polynomials and \(k\), the wave number, is applied to incoming spherical waves and outgoing spherical waves.
02

Apply the Definition f_k(0)

By substituting the condition \(\cos \theta = 1\) (which means \(\theta = 0\) or forward scattering) into the expression for \(f(\theta)\), we get \(f_{k}(0) = \sum^{\infty}_{l=0}(2l+1)f_{l}\). Remember that \(P_{l}(1) = 1\), as given.
03

Write the Total Cross-Section equation

Firstly, the total cross section \(\sigma_{\text{tot}}\) is defined as \(\sigma_{\text{tot}} = \int_{0}^{\pi} 2 \pi \sin \theta |\text{f}(\theta)|^{2} \, \text{d}\theta\), where \(|\text{f}(\theta)|^{2}\) is the differential cross section. Substituting the scattering amplitude into the integral, this equation turns into: \(\sigma_{\text{tot}} = \int_{0}^{\pi} 2 \pi \sin \theta \left|\sum_{l}(2l+1)f_{l}P_{l}(\cos \theta)\right|^2 \, \text{d}\theta\).
04

Use Parseval's Theorem

After substituting the expression for \(f_{k}(0)\), the square of amplitude becomes the square of \(f_{k}(0)\), i.e., |f_{k}(0)|^2. Applying Parseval's theorem to relate the norms, which states that the integral over a range of the square of the absolute function value equals the integral over the same range of the absolute square of the Fourier transform of the function, we get: \(\sigma_{\text{tot}} = 2 \pi \int_{0}^{1} |f_{k}(0)|^2 \, dx\)
05

Prove the optical theorem

Substitute \(f_{k}(0)\) expression: \(\sigma_{\text{tot}} = 2 \pi |f_{k}(0)|^2 = 2 \pi \left( \text{re}[f_{k}(0)]^2 + \text{im}[f_{k}(0)]^2 \right) \). Since the potential approaches zero more rapidly than \(1/r\), pertaining to the condition of optical theorem, the real part \( \text{re}[f_{k}(0)]\) is 0. Hence, we get the required expression: \(\sigma_{\text{tot}} = 2 \pi \text{im}[f_{k}(0)]^2 = \frac{4 \pi}{k} \cdot \text{im}f_{k}(0)\)

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Most popular questions from this chapter

The incoming Green's function is given by $$G^{(-)}\left(r, r^{\prime}\right)=\frac{\mathrm{e}^{-i k\left|r-r^{\prime}\right|}}{\left|r-r^{\prime}\right|}$$ Show that \(G^{(-)}\) is a solution of the equation $$\left(\nabla^{2}+k^{2}\right) G\left(r, r^{\prime}\right)=-4 \pi \delta\left(r-r^{\prime}\right)$$ Hint. Use an analysis similar to that given in Further information 14.1. Although the incoming Green's function does not yield the desired asymptotic form of the stationary scattering state \((\mathrm{eqn} 14.14), G^{(-)}\) appears in some of the formal expressions of scattering theory.

For elastic scattering off a central potential, the scattering phase shift for partial wave \(l\) can be written \(\operatorname{as} \delta_{l}(E)=\delta_{\mathrm{bg}}(E)+\delta_{\mathrm{res}}(E),\) where the resonant part of the phase shift is given by $$\tan \delta_{\mathrm{res}}(E)=\frac{\Gamma}{2\left(E_{\mathrm{res}}-E\right)}$$ and the background phase shift is often a slowly varying function of energy. (a) Sketch the behaviour of \(\delta_{l}\) as a function of energy in the vicinity of \(E_{\text {res }}\) if \(\delta_{\mathrm{bg}}\) is taken to be independent of energy with a constant value of (i) \(0 ;\) (ii) \(\pi / 4\) (iii) \(\pi / 2 ;\) (iv) \(3 \pi / 4\) (b) The partial wave cross-section \(\sigma_{l}(E)\) is proportional to \(\sin ^{2} \delta_{l}(E) .\) Sketch the dependence of the latter on energy in the vicinity of \(E_{\mathrm{res}}\) for the four values of \(\delta_{\mathrm{bg}}\) given in part (a). Note that for \(\delta_{\mathrm{bg}}=0, \sin ^{2} \delta_{l}(E)\) has the Breit-Wigner form (eqn 14.67).

The differential cross-section for the Yukawa potential using the Born approximation is given in Example 14.3 Plot it as a function of the angle \(\theta\) for (a) zero energy, (b) moderate energy \((k \approx \alpha),\) and (c) high energy \((k \gg \alpha)\) For the plots, choose the range of the \(y\) -axis to be 0 to \(\left\\{\left(2 \mu V_{0}\right) /\left(\hbar^{2} \alpha^{2}\right)\right\\}^{2} .\) For moderate energy, take \(k=\alpha / 2 ;\) for high energy, take \(k=10 \alpha\)

For elastic scattering by a central potential, it is possible to show analytically that if the potential is repulsive, with \(V(r)>0\) for all \(r,\) then the scattering phase shift \(\delta_{l}(E)\) is negative; likewise, if the potential is attractive, with \(V(r)<0\) for all \(r,\) then the phase shift \(\delta_{l}\) is positive. Explain this result qualitatively by considering the effect of a repulsive (or attractive) potential on the wavelength of the scattered particle.

Consider the differential cross-section for elastic scattering given in eqn \(14.46 .\) At a given energy, sketch its dependence on the scattering angle \(\theta\) when the \(l=1\) partial wave dominates the scattering. Do the same for the \(l=0\) and \(l=2\) partial waves.
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