Scattering Theory I - Lecture Notes | PHY 4605, Study notes of Physics

Download Scattering Theory I - Lecture Notes | PHY 4605 and more Study notes Physics in PDF only on Docsity! 8 Scattering Theory I 8.1 Kinematics Problem: wave packet incident on fixed scattering center V (r) with finite range. Goal: find probability particle is scattered into angle θ, φ far away from scattering center. Solve S.-eqn. with boundary condition that at t = −∞ the wave function is a wave packet incident on scattering ctr. Decompose packet into component waves eik·r, then either this wave is scattered or it’s not. If it’s scattered, at large distances we expect a spherical wave. So at large r our soln. should have form of linear combs. of ψk(r, t) '  eik·r + fk(θ, φ) eikr r   e−iωt, (1) where h̄ω = h̄2k2/2m is the energy of the asymptotic plane wave before scattering or after it has scattered (elastically). Intuitively 1st term in (1) is unscattered part, 2nd term is scattered wave with angular dependence fk. fk called scattering amplitude. 2nd term u = fke ikr/r (2) varies as 1/r, so intensity of scattered wave falls off as |u|2 ∼ 1/r2 as it must. To verify this, construct probability current j = −ih̄ 2m (ψ∗∇ψ − ψ∇ψ∗). (3) For scattered flux density (replace ψ by u above), find jscatt = −ih̄ 2m (u∗∇u− u∇u∗) = h̄k m r r3 |f (θ, φ)|2 (4) Differential scattering cross section dσdΩ Def: given incident flux nv particles per unit area and unit time. (n is density and v speed of particles) 1 Particles collected in detector of area A at angles θ, φ from incident direc- tion. A therefore subtends δΩ = A/r2 steradians. Recall classically dN dt = (nv) · dσ dΩ · δΩ (5) rate of detecting particles in A = incident flux density · differential scattering cross sec. · subtended solid angle (6) From eq. (4) above, have dN dt = jscattA = h̄k m r r3 |f (θ, φ)|2A = h̄k m · dσ dΩ · A r2 (7) so dσ dΩ (θ, φ) = |f (θ, φ)|2 (8) Total scattering cross-section σ Total cross-section defined by integrating over all angles: σ = ∫ dΩ dσ dΩ = ∫ dΩ|f (θ, φ)|2 (9) Reminder–classical analog & origin of name “cross-section”: Consider n pt. particles/vol. incident on sphere, radius a, flux nv. Particle scatters if its impact parameter is less than a, so net scattered flux is nv · πr2, total σ = πa2. 2 8.3 Born approximation. Valid for weak scattering or fast particles! Want to solve S.’s eqn for scattering potential V (r) with boundary condi- tion ψ → eik·r + f e ikr r (24) for suff. large r. We’re looking first for solutions with E = h̄2k2 2m “scattering states” (25) i.e., not bound states where the particle is trapped by the target. So S.-eqn. looks like ∇2ψ + k2ψ = 2mV h̄2 ψ ≡ ²Uψ (26) Now seek power series expansion of ψ (in powers of scattering potential): ψ(r) = eik·r + ²ψ(1) + ²2ψ2 + . . . (27) plugging into (26) gives ²(∇2ψ(1) + k2ψ(1)) + ²2(∇2ψ(2) + k2ψ(2)) = = ²Ueik·r + ²2Uψ(1) + . . . (28) as usual equate powers of ²: ∇2ψ(1) + k2ψ(1) = Ueik·r (29) ∇2ψ(2) + k2ψ(2) = Uψ(1), etc. (30) ... (31) See that we get sequence of coupled eqns., each coupled to previous one on rhs. This is called Born’s series. If we want to solve to linear order in ², take only (29). Solve by Fourier transform; and define 5 ψk = ∫ ψ(1)(r)e−ik·rd3r (32) and the inverse ψ(1)(r) = ∫ ψke ik·r d 3k (2π)3 (33) So multiply (29) by e−ik ′·r and integrate lhs by parts4 to get (k2 − k′2)ψk′ = ∫ U(r)ei(k−k ′)·rd3r (35) which yields immediately ψ(1)(r) = 1 (2π)3 ∫ d3k′ eik ′·r k2 − k′2 ∫ d3r′U(r′)ei(k−k ′)·r′ (36) = 1 (2π)3 ∫ d3r′U(r′)eik·r ′ ∫ d3k′ eik ′·(r−r′) k2 − k′2︸ ︷︷ ︸ (37) I(r− r′) Now we spend some tedious but hopefully useful time showing how to calculate integrals of this type. First evaluate I(r− r′). Use polar coords., polar axis along y ≡ r− r′, θ is angle between y and k′: k′ · y = k′y cos θ (38) d3k′ = k′2dk′ sin θdθdφ. (39) So angular part of I is ∫ 2π 0 dφ ∫ π 0 sin θdθ eik ′y cos θ = 2π ∫ 1 −1 d cos θeik ′y cos θ (40) = 4π sin k′y k′y (41) so left with I = 4π ∫ ∞ 0 k′2dk′ k2 − k′2 sin k′y k′y (42) Trick: note integrand is even in k′, write I as I = 2π i ∫ ∞ ∞ k′dk′ k2 − k′2 eik ′y y because cos part odd in k′, therefore doesn’t contribute (43) 4Explicitly, use ∫ (∇2ψ(r))e−ik′·rd3r = −k′2ψk′ (34) 6 This is an improper integral due to the singularity at k = ±k′. However we use physics to regularize: so far we haven’t used condition that we want w. fctn. ψ at ∞ to look like scattered wave. This corresponds in integral scattering formulation to specifying contour for I in complex plane. One possibility is to handle singularities by following contour shown below. Use Cauchy integral formula,5 complete the contour in upper half-plane, shrink the contour to the pole at +k, to get: I = 2π i eiky y · 1 2 ∮ dk′ k − k′︸ ︷︷ ︸ = −2π2 e iky y (45) − 2πi (46) This procedure may seem rather arbitrary at 1st sight, but a look at other choices suggests it’s right thing to do. If we take contour above both poles, get zero! If we take contour opposite to figure (above k and below −k) or below both poles, get a contribution I ∝ e−iky. Recall y = |r− r′; if r is in asymptotic region, r À r′, I ∝ e−ikr, which is ingoing, not outgoing scattered wave. So choice of contour effectively implements boundary condition. Continuing, find ψ(1)(r) = − 1 4π ∫ d3r′U(r′)eik·r ′ eik|r−r ′| |r− r′| (47) Finally, look at ψ in asymptotic region, r À r′: use |r− r′| ' r + r̂ · r′ (48) eik|r−r ′| |r− r′| | ' 1 r eikre−ikr̂·r ′ (49) define ks ≡ kr̂ (50) to write ψ(1)(r) ' − 1 4π (∫ d3r′U(r′)ei(k−ks)·r ′ ) eikr r (51) 5If you don’t like the integral formula, note this choice is equivalent to replacing (29) by ∇2ψ(1) + (k2 + i²k)ψ(1) = Ueik·r (44) i.e. shifting the zeros of the integrand of I to k′ = ±(k + i²/2). 7