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Chapter 14: Problem 20

The reactance matrix, \(K\), defined in relation to the scattering matrix through \(K=\mathrm{i}(1-S)(1+S)^{-1},\) also appears in scattering theory. Show for elastic scattering by a central potential with partial wave \(l\) that \(K\) is a \(1 \times 1\) matrix with element \(K_{l}=\tan \delta_{l}\)

Short Answer

Expert verified
The element \( K_l \) in the reactance matrix is equal to \( \tan(\delta_l) \), which is the tangent of the phase shift in the elastic scattering by a central potential with partial wave \( l \).

Step by step solution

01

Define the Reactance Matrix

The reactance matrix \( K \) is related to the scattering matrix \( S \) by the formula \( K = \mathrm{i}(1 - S)(1 + S)^{-1} \).
02

Elastic Scattering in Central Potential

For elastic scattering by a central potential with partial wave \( l \), the scattering matrix \( S \) has an element \( S_l = e^{2i\delta_l} \), where \( \delta_l \) is the phase shift.
03

Plug Scattering Matrix Into Reactance Matrix

Substitute the expression for \( S_l \) in to the expression for \( K \). The matrix \( K \) then becomes \( K = \mathrm{i}(1 - e^{2i\delta_l})(1 + e^{2i\delta_l})^{-1} \). Simplify the equation to obtain \( K_l \).
04

Simplify the Equation

Simplify the equation to get \( K_l = \mathrm{i}(e^{2i\delta_l}-1)(e^{-2i\delta_l} +1 )^{-1} = i \cdot i \cdot \tan(\delta_l) = \tan(\delta_l) \)

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Most popular questions from this chapter

Consider the scattering of an electron by an atom of atomic number \(Z .\) The interaction potential energy can be approximated by the screened Coulomb potential energy \(V(r)=-\left(Z e^{2} / 4 \pi \varepsilon_{0} r\right) \mathrm{e}^{-r / a},\) where \(a\) is the screening length. (a) Use the Born approximation to calculate the differential cross-section for scattering from the screened Coulomb potential. (b) Proceed to evaluate the integral scattering cross-section. (c) In the limit \(a \rightarrow \infty, V(r)\) becomes exactly the Coulomb potential energy. Evaluate the differential and integral cross-sections obtained in parts (a) and (b) in this limit.

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.

By considering flux densities, explain the appearance of the factor \(k_{\alpha} / k_{\alpha_{\rho}}\) in eqn 14.93 for the differential crosssection for scattering from an initial state \(\alpha_{0}\) to a final state \(\alpha\)

A particle of mass \(m\) is scattered off a central notential \(V(r)\) of the form $$V(r)=\left\\{\begin{array}{lll} \infty & \text { if } & r=0 \\ V_{0} & \text { if } & 0V_{0},\) find an expression for the S-wave scattering phase shift \(\delta_{0} .\) Hint. Require that the wavefunction and its first derivative be continuous at \(r=a\)

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.
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