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 100−MeV (kinetic energy) neutron incident on a fixed nucleus, lmax≅2. [Hint: The range of the nuclear force is roughly a Fermi=10−5A˚. Also ℏc≅200MeVF 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 σ0→4πr02 for a hard sphere as k→0.
(2) Consider the other extreme of kr0 very large. From Eq. (19.5.27) and the asymptotic forms of jl and nl show that
if we approximate the sum over l by an integral, 2l+1 by 2l, and the oscillating function sin2δ by its mean value of 1/2.
Exercise 19.5.4 Show that the s-wave phase shift for a square well of depth V0 and range r0 is
δ0=−kr0+tan−1(k′ktank′r0)
where k′ and k are the wave numbers inside and outside the well. For k small, kr0 is some small number and we ignore it. Let us see what happens to δ0 as we vary the depth of the well, i.e., change k′. Show that whenever k′≃kn′=(2n+1)π/2r0, δ0 takes on the resonant form Eq. (19.5.30) with Γ/2=ℏ2kn/μr0, where kn is the value of k when k′=kn′. Starting with a well that is too shallow to have any bound state, show k1′ 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=0.) As the well is deepened further, this level moves down, and soon, at k2′, 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)e2iδl
where η(<1), is called the inelasticity factor.
(1) By considering probability currents, show that
(2) Consider a “black disk” which absorbs everything for r⩽r0 and is ineffective beyond. Idealize it by η=0 for l⩽kr0; η=1, δ=0 for l>kr0. Show that σel=σinel≃πr02. Replace the sum by an integral and assume kr0≫1. (See Exercise 19.5.3.) Why is σinel always accompanied by σel?
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
(2) Argue that as long as θ=0, the average of jrint over any small solid angle is zero because r→∞. [Assume f(θ) is a smooth function.]
(3) Integrate jrint over a tiny cone in the forward direction and show that (see hint)
∫forward conejrintr2dΩ=−(μℏk)k4πImf(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 ℏk/μ and the cross section σ, we regain the optical theorem. [Hint: Since θ is small, set sinθ≈θ, cosθ=1 or 1−θ2/2 using the judgment. In evaluating the upper limit in the θ 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 and dσ/dΩL is
dΩLdσ∣∣θ0=dΩdσ∣∣2θ04cosθ0
(2) Show that θL⩽π/2 by using just energy and momentum conservation.
(3) For unequal mass scattering, show that
tanθL=cosθ+(m1/m2)sinθ
where m2 is the target mass.
Exercise 19.6.2 Derive Eq. (19.6.21) using Eq. (19.3.16) for fc(θ).
Exercise 19.6.3 Assuming f=f(θ) show that (dσ/dΩ)π/2=0 for fermions in the triplet state.