Calculate the angular components of the flux density, \(J_{\theta}\) and \(J_{\varphi},\) for the scattered wave $$\psi=f_{k}(\theta, \varphi) \frac{\mathrm{e}^{\mathrm{i} k r}}{r}$$ and confirm that in the limit \(r \rightarrow \infty,\) only the radial component \(J_{r}\) given in Justification 14.3 need be retained.

Short Answer

Expert verified
The angular components of the flux density (\(J_{\theta}\) and \(J_{\varphi}\)) can be calculated from the given wave function using vector calculus. As \(r \rightarrow \infty\), all angular components vanish, leaving only the radial component \(J_{r}\) as expected.

Step by step solution

01

Wave Function Analysis

The wave function given is \(\psi=f_{k}(\theta, \varphi)\frac{\mathrm{e}^{\mathrm{i} k r}}{r}\). To find the angular components of the flux density (current density), we'll first need to determine the gradient of the wave function using vector calculus.
02

Calculate the Angular Components

The general formula for current density J is given by \( J = i[\psi∗∇ψ - ψ∇ψ∗] \). Applying this formula, we'll find both \(J_{\theta}\) and \(J_{\varphi}\) by calculating the gradient of the wave function and using their respective unit vectors.
03

Limit to Infinity

Finally, we'll examine the limit of the expression as \(r \rightarrow \infty\). We expect that all angular components of the flux density go to zero, leaving only the radial component \(J_{r}\), as stated in the problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

Show that in the limit of low energies, the scattering phase shift for P-wave scattering by a hard sphere is proportional to \((k a)^{3}\) and therefore is negligible compared to the S-wave scattering phase shift. Hint. Use the asymptotic forms given in eqn \(14.32 \mathrm{c}\)

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.

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 \(r=a\)

Use the Born approximation to calculate the differential cross-section for scattering from the spherical square-well potential (Section 14.5 ). Hint. Use integration by parts to determine the scattering amplitude.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free