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Chapter 14: Problem 14

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

Short Answer

Expert verified
The S-wave scattering phase shift \(\delta_{0}\) can be found by breaking down the region into three parts and finding the solution to the wave equation for each region. Next, by applying boundary conditions to ensure the wavefunction and its derivative are continuous at \(r=a\) and \(r=b\), we get the equations that can be used to calculate \(\delta_{0}\). The exact expression will depend on the parameters given in a specific scenario; solving those equations will give the precise solution.

Step by step solution

01

Consider different regions

Given the area where the particle exists, one can break down the region into three parts based on the potential energy \(V(r)\). Region I: Inside the potential (i.e., 0=b). Using such a structural simplification helps in solving the wave equations accurately for each region.
02

Solving wave equation for Region I

Inside the potential, the potential energy \(V(r) = 0\), thus the wave equation becomes a simple radial part of the free particle Schrödinger equation. Solving this equation yields plane wave solutions. As we are considering s-wave scattering (l=0), the radial solution will have the form \(R = A \cdot r + B \cdot r^{-1}\). But because the solution must be regular at the origin i.e., r=0, the \(B \cdot r^{-1}\) term must be dropped. Hence, the solution for region I will be \(R_I = A \cdot r\).
03

Solving wave equation for Region II

In region II, the potential energy \(V(r) = V_0\), and the energy of the particle is given as \(E>V_0\). Here, we can write energy as \(E=V_0+E'\), where \(E'>0\). The wave equation in this region can be solved to give solutions that look like linear combinations of Bessel functions of modulus k of the first and second kinds, \(J_{0}(kr)\) and \(Y_{0}(kr)\). However, because we must have a finite solution at \(r=b\), we take the solution for region II as \(R_{II} = C \cdot J_{0}(kr)\).
04

Solving wave equation for Region III

In region III, again \(V(r) = 0\). But here, we find the solution beyond the potential barrier. Hence, solving the wave equation, we get the radial solution in the form \(R = F \cdot r\cos((kr-\delta_0)\) where F is the scattering amplitude and \(kr\) is the wave number. So, the solution for region III is \(R_{III} = F \cdot r \cdot \cos((kr-\delta_0)\). Here, \(\delta_0\) is the scattering phase shift we need to calculate.
05

Applying boundary conditions

At \(r=a\) and \(r=b\), we need to ensure that both the wavefunction and its derivative are continuous. This means that \(R_I = R_{II}\) and \(dR_I/dr = dR_{II}/dr\) at \(r=a\), and \(R_{II} = R_{III}\) and \(dR_{II}/dr = dR_{III}/dr\) at \(r=b\). Solving these equations simultaneously will give us the expression for the S-wave scattering phase shift \(\delta_{0}\).

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Most popular questions from this chapter

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 \(K_{l}=\tan \delta_{l}\)

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.

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.

Equation 14.3 gives the form of the S matrix for a one-dimensional system in which a particle is scattered from an abrupt blip in the potential energy. Write down the analogous expression for scattering from a comparable dip in the potential energy. Proceed to compute the transmission probability for positive energies given that the particle is incident from the left.
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