2025-05-30
Solutions to Principles of Quantum Mechanics
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目录

Chapter 19 Scattering Theory
19.1 Introduction
19.2 Recapitulation of One-Dimensional Scattering and Overview
19.3 The Born Approximation (Time-Dependent Description)
19.4 Born Again (The Time-Independent Approximation)
19.5 The Partial Wave Expansion
19.6 Two-Particle Scattering

Chapter 19 Scattering Theory

19.1 Introduction

19.2 Recapitulation of One-Dimensional Scattering and Overview

19.3 The Born Approximation (Time-Dependent Description)

Exercise 19.3.1 Show that

σYukawa=16πr02(gμr02)211+4k2r02\sigma_{\text{Yukawa}}=16 \pi r_0^2\left(\frac{g \mu r_0}{\hbar^2}\right)^2 \frac{1}{1+4 k^2 r_0^2}

where r0=1μ0r_0=1 \mu_0 is the range. Compare σ\sigma to the geometrical cross section associated with this range.

Exercise 19.3.2 (1) Show that if V(r)=V0θ(r0r)V(r)=-V_0 \theta\left(r_0-r\right),

dσdΩ=4r02(μV0r022)2(sinqr0qr0cosqr0)2(qr0)6\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=4 r_0^2\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \frac{\left(\sin q r_0-q r_0 \cos q r_0\right)^2}{\left(q r_0\right)^6}

(2) Show that as kr00k r_0 \rightarrow 0, the scattering becomes isotropic and

σ16πr029(μV0r022)2\sigma \cong \frac{16 \pi r_0^2}{9}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2

Exercise 19.3.3 Show that for the Gaussian potential, V(r)=V0er2/r03V(r)=V_0 e^{-r^2 / r_0^3},

dσdΩ=πr024(μV0r022)2eq2r02/2σ=π22k2(μV0r022)2(1e2k2r02)\begin{aligned} \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} & =\frac{\pi r_0^2}{4}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \mathrm{e}^{-q^2 r_0^2 / 2} \\ \sigma & =\frac{\pi^2}{2 k^2}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2\left(1-\mathrm{e}^{-2 k^2 r_0^2}\right) \end{aligned}

[Hint: Since q2=2k2(1cosθ)q^2=2 k^2(1-\cos \theta), d(cosθ)=d(q2)/2k2\mathrm{d}(\cos \theta)=-\mathrm{d}(q^2) / 2 k^2.]

Exercise 19.3.4 Verify the above claim for the Gaussian potential.

19.4 Born Again (The Time-Independent Approximation)

Exercise 19.4.1 Derive the inequality (19.4.44).

19.5 The Partial Wave Expansion

Exercise 19.5.1 Show that for a 100MeV100-\mathrm{MeV} (kinetic energy) neutron incident on a fixed nucleus, lmax2l_{\max} \cong 2. [Hint: The range of the nuclear force is roughly a Fermi=105A˚=10^{-5} \text{\AA}. Also c200MeV F\hbar c \cong 200 \mathrm{MeV}~\mathrm{F} is a more useful mnemonic for nuclear physics.]

Exercise 19.5.2 Derive Eq. (19.5.18) and provide the missing steps leading to the optical theorem, Eq. (19.5.21).

Exercise 19.5.3 (1) Show that σ04πr02\sigma_0 \rightarrow 4 \pi r_0^2 for a hard sphere as k0k \rightarrow 0.

(2) Consider the other extreme of kr0k r_0 very large. From Eq. (19.5.27) and the asymptotic forms of jlj_l and nln_l show that

sin2δlkr0sin2(kr0lπ/2)\sin ^2 \delta_l \xrightarrow[k r_0 \rightarrow \infty]{ } \sin ^2\left(k r_0-l \pi / 2\right)

so that

σ=l=0lmax=kr0σl4πk20kr0(2l)sin2δldl2πr02\begin{aligned} \sigma=\sum_{l=0}^{l_{\max }=k r_0} \sigma_l & \cong \frac{4 \pi}{k^2} \int_0^{k r_0}(2 l) \sin ^2 \delta_l \mathrm{d} l \\ & \cong 2 \pi r_0^2 \end{aligned}

if we approximate the sum over ll by an integral, 2l+12 l+1 by 2l2 l, and the oscillating function sin2δ\sin ^2 \delta by its mean value of 1/21 / 2.

Exercise 19.5.4 Show that the ss-wave phase shift for a square well of depth V0V_0 and range r0r_0 is

δ0=kr0+tan1(kktankr0)\delta_0=-k r_0+\tan ^{-1}\left(\frac{k}{k^{\prime}} \tan k^{\prime} r_0\right)

where kk^{\prime} and kk are the wave numbers inside and outside the well. For kk small, kr0k r_0 is some small number and we ignore it. Let us see what happens to δ0\delta_0 as we vary the depth of the well, i.e., change kk^{\prime}. Show that whenever kkn=(2n+1)π/2r0k^{\prime} \simeq k_n^{\prime}=(2 n+1) \pi / 2 r_0, δ0\delta_0 takes on the resonant form Eq. (19.5.30) with Γ/2=2kn/μr0\Gamma / 2=\hbar^2 k_n / \mu r_0, where knk_n is the value of kk when k=knk^{\prime}=k_n^{\prime}. Starting with a well that is too shallow to have any bound state, show k1k_1^{\prime} corresponds to the well developing its first bound state, at zero energy. (See Exercise 12.6.9.) (Note: A zero-energy bound state corresponds to k=0k=0.) As the well is deepened further, this level moves down, and soon, at k2k_2^{\prime}, another zero-energy bound state is formed, and so on.

Exercise 19.5.5 Show that even if a potential absorbs particles, we can describe it by

Sl(k)=ηl(k)e2iδlS_l(k)=\eta_l(k) \mathrm{e}^{2\mathrm{i} \delta_l}

where η(<1)\eta(<1), is called the inelasticity factor.

(1) By considering probability currents, show that

σinel=πk2l=0(2l+1)[1ηl2]σel=πk2l=0(2l+1)(1+ηl22ηlcos2δl)\begin{aligned} \sigma_{\text{inel}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left[1-\eta_l^2\right] \\ \sigma_{\mathrm{el}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left(1+\eta_l^2-2 \eta_l \cos 2 \delta_l\right) \end{aligned}

and that once again

σtot=4πkImf(0)\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \operatorname{Im} f(0)

(2) Consider a “black disk” which absorbs everything for rr0r \leqslant r_0 and is ineffective beyond. Idealize it by η=0\eta=0 for lkr0l \leqslant k r_0; η=1\eta=1, δ=0\delta=0 for l>kr0l>k r_0. Show that σel=σinelπr02\sigma_{\text{el}}=\sigma_{\text{inel}} \simeq \pi r_0^2. Replace the sum by an integral and assume kr01k r_0 \gg 1. (See Exercise 19.5.3.) Why is σinel\sigma_{\text{inel}} always accompanied by σel\sigma_{\mathrm{el}}?

Exercise 19.5.6 (The Optical Theorem) (1) Show that the radial component of the current density due to interference between the incident and scattered waves is

jrintr(kμ)1rIm[ieikr(cosθ1)f(θ)cosθ+ieikr(1cosθ)f(θ)]j_r^{\mathrm{int}} \underset{r \rightarrow \infty}{\sim}\left(\frac{\hbar k}{\mu}\right) \frac{1}{r} \operatorname{Im}\left[\mathrm{i} \mathrm{e}^{\mathrm{i} k r(\cos \theta-1)} f^*(\theta) \cos \theta+\mathrm{i} \mathrm{e}^{\mathrm{i} k r(1-\cos \theta)} f(\theta)\right]

(2) Argue that as long as θ0\theta \neq 0, the average of jrint j_r^{\text {int }} over any small solid angle is zero because rr \rightarrow \infty. [Assume f(θ)f(\theta) is a smooth function.]

(3) Integrate jrint j_r^{\text {int }} over a tiny cone in the forward direction and show that (see hint)

forward conejrintr2dΩ=(kμ)4πkImf(0)\int_{\text{forward cone}} j_r^{\text{int}} r^2 \mathrm{d}\Omega=-\left(\frac{\hbar k}{\mu}\right) \frac{4 \pi}{k} \operatorname{Im} f(0)

Thus, if we integrate the total current in the region behind the target, we find that the interference term (important only in the near-forward direction, behind the target) produces a depletion of particles, casting a “shadow”. The total number of particles (per second) missing in the shadow region is given by the above expression for the integrated flux. Equating this loss to the product of the incident flux k/μ\hbar k / \mu and the cross section σ\sigma, we regain the optical theorem. [Hint: Since θ\theta is small, set sinθθ\sin \theta \approx \theta, cosθ=1\cos \theta=1 or 1θ2/21-\theta^2 / 2 using the judgment. In evaluating the upper limit in the θ\theta integration, use the idea introduced in Chapter 1, namely, that the limit of a function that oscillates as its argument approaches infinity is equal to its average value.]

19.6 Two-Particle Scattering

Exercise 19.6.1 (1) Starting with Eqs. (19.6.16) and (19.6.17), show that the relation between dσ/dΩ\mathrm{d}\sigma/\mathrm{d}\Omega and dσ/dΩL\mathrm{d}\sigma/\mathrm{d}\Omega_L is

dσdΩLθ0=dσdΩ2θ04cosθ0\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega_L}\right|_{\theta_0}=\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}\right|_{2 \theta_0} 4 \cos \theta_0

(2) Show that θLπ/2\theta_L \leqslant \pi / 2 by using just energy and momentum conservation.

(3) For unequal mass scattering, show that

tanθL=sinθcosθ+(m1/m2)\tan \theta_L=\frac{\sin \theta}{\cos \theta+\left(m_1 / m_2\right)}

where m2m_2 is the target mass.

Exercise 19.6.2 Derive Eq. (19.6.21) using Eq. (19.3.16) for fc(θ)f_c(\theta).

Exercise 19.6.3 Assuming f=f(θ)f=f(\theta) show that (dσ/dΩ)π/2=0(\mathrm{d}\sigma/\mathrm{d}\Omega)_{\pi/2}=0 for fermions in the triplet state.