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8 Scattering Theory I 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; Á far away from scattering center. Solve S.-eqn. with boundary condition that at t = ¡1 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 0 1 ikr B ik¢r e C ¡i!t à (r; t) ' @e + f (θ; Á) A e ; (1) k k r whereh! ¯ =h ¯2k2=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 ikr u = fke =r (2) varies as 1=r, so intensity of scattered wave falls off as juj2 » 1=r2 as it must. To verify this, construct probability current ¡ih¯ j = (ärà ¡ Ãrä): (3) 2m For scattered flux density (replace à by u above), find ¡ih¯ h¯k r j = (u¤ru ¡ uru¤) = jf(θ; Á)j2 (4) scatt 2m m r3 dσ Differential scattering cross section dΩ 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; Á from incident direc- tion. A therefore subtends ±Ω = A=r2 steradians. Recall classically dN dσ = (nv) ¢ ¢ ±Ω (5) dt dΩ differential rate of detecting incident subtended = ¢ scattering ¢ (6) particles in A flux density solid angle cross sec. From eq. (4) above, have dN h¯k r hk¯ dσ A = j A = jf(θ; Á)j2A = ¢ ¢ (7) dt scatt m r3 m dΩ r2 so dσ (θ; Á) = jf(θ; Á)j2 (8) dΩ Total scattering cross-section σ Total cross-section defined by integrating over all angles: Z dσ Z σ = dΩ = dΩjf(θ; Á)j2 (9) dΩ 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 ¢ ¼r2, total σ = ¼a2. 2 8.2 Optical theorem A conservation law–relates diminished amplitude of unscattered beam to flux scattered out of beam. Incident flux density in Fourier component eikz of incident beam is hk¯ v = (10) classical m 2 Role of classical density of particles n played by jÃbeamj = 1. From (9) net scattering rate is hk¯ Z nvσ = dΩjf j2 (11) m k so outgoing unscattered beam must be diminished by this amount. “Out- going” means eikz plus forward scattering part of scattered wave, f(0) à = eikz + eikr (12) r For angles close to forward scattering, p x2 r = z2 + x2 ' z + (13) 2z 1 so at large z, 0 1 kx2 ikz B f(0) i C à » e @1 + e 2z A (14) z Now want to compute probability flux in forward direction. Rapidly vary- ing part of (14) is eikz, so rà ' ikÃzˆ (15) 0 10 1 ¤ kx2 kx2 ¤ ¤ B f (0) ¡i CB f(0) i C à rà ' ikà à = ik @1 + e 2z A@1 + e 2z A (16) z z 0 1 kx2 B 2 i C ' ik @1 + Re f(0)e 2z A (17) z 1 Peebles calls this Ãf for forward, but I’m too lazy, so you have to remember I’m always talking about small angles 3 So to lowest order flux density in forward direction is ih¯ j = ¡ (ärà ¡ c:c:) (18) 2m0 1 kx2 hk¯ B 2 i C ' @1 + Re f(0)e 2z A (19) m z The first term is the same as the incident flux density, the second is the reduction of the outgoing flux by forward scattering. If we go out far enough along z, even for small angles the x can get large, so we need to integrate all the scattered flux (2nd term) in the forward direction:2 0 1 0 1 Z 2 B reduction of flux C 1 2 ikx @ A = ¡vcl 2¼xdx Re @f(0)e 2z A (21) by forward scattering 0 z 2 2z see footnote ! = v ¼ Imf(0) ´ v σ (22) cl z k cl since vσ is the definition of the total flux removed from the beam by scattering! So 4¼ σ = k Imf(0) “Optical theorem” (23) Interpretation: Incident plane wave brings in probability current density in z direction. Some of it gets scattered into various directions. Must give rise to decrease in current density behind target (θ = 0) due to destructive interference of incident plane wave and scattered wave in forward direction. So total flux scattered (vσ) related to forward scattering amplitude f(0). 2This is a mathematically very ill-defined procedure, and you shouldn’t believe it until you see we get the same result by phase shift analysis later. But the integration goes as follows: Z 1 Z i1 2 2z 2iz 2xdx eikx =(2z) = du eu = ¡ euji1 (20) ik k 0 0 0 We now want to argue that we can neglect the exponential at its upper limit because it is rapidly oscillating, i.e. we want to be able to set ei1 ! 0. Peebles argues physically: that a real beam will consist of a finite spread of incident directions and cut off the increase in the rate of oscillations as z ! 1. That the oscillations at 1 then average out to zero is guaranteed by the Riemann-Lesbegue lemma3. 4 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 eikr à ! eik¢r + f (24) r for suff. large r. We’re looking first for solutions with h¯2k2 E = “scattering states” (25) 2m i.e., not bound states where the particle is trapped by the target. So S.-eqn. looks like 2mV r2à + k2à = à ´ ²Uà (26) h¯2 Now seek power series expansion of à (in powers of scattering potential): Ã(r) = eik¢r + ²Ã(1) + ²2Ã2 + ::: (27) plugging into (26) gives ²(r2Ã(1) + k2Ã(1)) + ²2(r2Ã(2) + k2Ã(2)) = = ²Ueik¢r + ²2UÃ(1) + ::: (28) as usual equate powers of ²: r2Ã(1) + k2Ã(1) = Ueik¢r (29) r2Ã(2) + k2Ã(2) = UÃ(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 Z (1) ¡ik¢r 3 Ãk = à (r)e d r (32) and the inverse Z d3k Ã(1)(r) = à eik¢r (33) k (2¼)3 So multiply (29) by e¡ik0¢r and integrate lhs by parts4 to get Z 2 02 i(k¡k0)¢r 3 (k ¡ k )Ãk0 = U(r)e d r (35) which yields immediately ik0¢r 1 Z e Z 0 0 Ã(1)(r) = d3k0 d3r0U(r0)ei(k¡k )¢r (36) (2¼)3 k2 ¡ k02 ik0¢(r¡r0) 1 Z 0 Z e = d3r0U(r0)eik¢r d3k0 (37) 3 2 02 (2¼) | k{z ¡ k } I(r ¡ r0) Now we spend some tedious but hopefully useful time showing how to calculate integrals of this type. First evaluate I(r ¡ r0). Use polar coords., polar axis along y ´ r ¡ r0, θ is angle between y and k0: k0 ¢ y = k0y cos θ (38) d3k0 = k02dk0 sin θdθdÁ: (39) So angular part of I is Z 2¼ Z ¼ Z 1 dÁ sin θdθ eik0y cos θ = 2¼ d cos θeik0y cos θ (40) 0 0 ¡1 sin k0y = 4¼ (41) k0y so left with Z 1 k02dk0 sin k0y I = 4¼ 2 (42) 0 k2 ¡ k0 k0y Trick: note integrand is even in k0, write I as Z 0 2¼ 1 k0dk0 eik y because cos part odd in k0, I = 2 (43) i 1 k2 ¡ k0 y therefore doesn’t contribute 4Explicitly, use Z 2 ¡ik0¢r 3 02 (r Ã(r))e d r = ¡k Ãk0 (34) 6 This is an improper integral due to the singularity at k = §k0. However we use physics to regularize: so far we haven’t used condition that we want w. fctn. à at 1 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: iky I 0 iky 2¼ e 1 dk 2 e I = ¢ 0 = ¡2¼ (45) i y 2 | k{z¡ k} y ¡ 2¼i (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 = jr ¡ r0; if r is in asymptotic region, r À r0, I / e¡ikr, which is ingoing, not outgoing scattered wave. So choice of contour effectively implements boundary condition. Continuing, find Z ikjr¡r0j 1 0 e Ã(1)(r) = ¡ d3r0U(r0)eik¢r (47) 4¼ jr ¡ r0j Finally, look at à in asymptotic region, r À r0: use jr ¡ r0j ' r +r ˆ ¢ r0 (48) ikjr¡r0j e 1 0 j ' eikre¡ikrˆ¢r (49) jr ¡ r0j r define ks ´ krˆ (50) to write µZ ¶ ikr 1 0 e Ã(1)(r) '¡ d3r0U(r0)ei(k¡ks)¢r (51) 4¼ r 5If you don’t like the integral formula, note this choice is equivalent to replacing (29) by r2Ã(1) + (k2 + i²k)Ã(1) = Ueik¢r (44) i.e.
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