9 Scattering Theory II 9.1 Partial wave analysis Expand ψ in spherical harmonics Y lm (θ, φ), derive 1D differential equations for expansion coefficients. Spherical coordinates: x = r sin θ cos φ (1) y = r sin θ sin φ (2) z = r cos θ (3) from which follow by application of chain rule relations φ = x φ x + y φ ˆL z = i h φ by constructing θ, find also y = y x + x y, or (4) and ˆL x ˆL y = i h(sin φ θ = i h(cos φ θ + cot θ cos φ φ ) (5) cot θ sin φ φ ) (6) ˆL + = he iφ ( θ + i cot θ ), φ (7) ˆL = he iφ ( θ i cot θ ), φ (8) ˆL 2 = ˆL + ˆL + L 2 z + h ˆL z (9) = h 2 1 sin θ Reminder: spherical harmonics θ sin θ θ + 1 sin 2 θ 2 φ 2 (10) 1
Eigenstates of ˆL 2, ˆL z : ˆL2 Y lm = h 2 Y lm, ˆLz Y lm = hmy lm Relation to Legendre functions: Y lm = P lm e imφ (11) P l0 (θ) = P l (cos θ) Legendre polynomial (12) Normalization: dω Y lmy l m = δ ll δ ll (13) Compute Y ll from ˆL + Y ll = 0: which has soln. or dp ll = l cot θp ll (14) dθ dp ll d sin θ = lp ll (15) sin θ P ll sin l θ (16) For large l this looks like and L / φ acting on this yields something like Recall we also showed that (discussion of radial eqn. for H-atom way back when): 2 2 = 1 2 ˆL r r2r h 2 r 2 (17) = 1 2 r r 2r + 1 r 2 sin θ θ sin θ θ + 1 2 r 2 sin 2 θ φ 2 (18) 2
where last step follows from (10). So want to solve h2 2m 2 + V (r) ψ = h2 k 2 2m ψ (19) Since Y lm form complete set, expand for any ψ and insert in S.-eqn (19) to get lm h2 2m ψ = l,m f lm (r)y lm (θ, φ) (20) 1 2 r r 2rf lm + h2 l(l + 1) f 2mr 2 lm +V (r)f lm h2 k 2 2m f lm = 0 Y lm (21) Case of isotropic potential V (r) = V (r). Under this assumption, every- 3
thing inside ( ) is fctn. of r only! Since Y lm are independent, get one separate equation for each l, m ( partial wave ): h2 1 2 2m r r 2r + h2 l(l + 1) + V (r) 2mr 2 f lm = h2 k 2 2m f lm (22) 9.2 s-wave scattering Low energies scattering more isotropic, may approximate cross section by considering only l = 0 partial wave. Suppose potential has hard finite range, V = 0 for r > r 0. Outside r 0 get just d 2 dr 2rf 0(r) = k 2 rf 0 (r) (23) solution f 0 e±i outgoing/incoming sph. waves (24) r Plane wave expansion in Y lm s. Let s go back and reexamine sph. harm. expansion of original plane wave! e ik r = lm g lm (r)y lm (25) and use orthonormality condition (13) for Y lm s to project out g lm s. Find by multiplying by Y 00, integrating over dω, dωe ik r sin 1 = 4π = g 0 (r) 4π (26) 4π or 4π e i g 0 = 2i e i (27) so (unperturbed!) plane wave may be written e ik r = 2i 1 e i + ei + l>0,m g lm Y lm (28) 4
Now look at full ψ again, argue as follows: have shown (24) that f 0 e ±i /r, but can t have e i /r in scattered part of wave, since it corresponds to incoming, not outgoing wave boundary condition. Yet as we see from (28), a term e i /r is already part of incident plane wave. Coeffcient of e i /r in ψ must therefore be the same as in (28), since just comes from plane wave. Since we ve assumed higher-l components not scattered, these must also be same in ψ. Only thing which can be different in ψ from unperturbed e ik r is coefficient of e +i /r, which we will call η 0. So we have deduced that, in s-wave approximation: ψ = 1 2i e i + η 0 e i + l>0,m g lm Y lm (29) Now recall that for isotropic potentials we are considering, S.-eqn. decomposes into separate equations (22) for the amplitudes f lm. This means each lm is like separate scattering problem, in particular probability flux must be conserved for each l, m separately. For l = 0 case this means amount of probability flux into r = 0 (coefficient of e i /r term, magnitude squared) has to equal flux out of r = 0 (coefficient of e +i /r term, magnitude squared). This = η 0 2 = 1. (30) More generally, η lm 2 = 1! In plane wave (28), this condition is fulfilled by having coefficients be ±1; in perturbed ψ entire effect of scattering subsumed in fact that η 0 is complex phase: conventional to write η 0 e 2iδ 0 (31) where the quantity δ 0 called s-wave scattering phase shift. In our s-wave approximation, all other δ l are zero, & we have for r r 0 ψ e ik r + η 0 1 e i 2ik, (32) obtained by subtracting (28) from (29). Now compare to our standard asymptotic form for ψ in a scattering problem, ψ e ik r +f(θ, φ)e +i /r. 5
Can immediately read off form of scattering amplitude, or f(θ, φ) = ei2δ 0 1, (33) 2ik dσ dω = f 2 = sin2 δ 0 k 2 isotropic, s-wave only (34) from which we immediately get total cross section σ = 4π k 2 sin 2 δ 0 (35) Check to make sure this agrees with optical theorem! Imf = (1/2k)Re (e 2iδ 0 1) = (1/2k)(cos 2δ 0 1) = (1/2k)2 sin 2 δ 0 = (1/k)k 2 σ/4π = kσ/4π (36) which is optical theorem. I used Eq. (35) in 2nd line. Note the only effect of the scattering in the asymptotic region is to retard the wave by a fixed phase shift δ 0 the spherical fronts come a little bit behind or in front of the plane wave due to the potential around the scatterer. We can now try to calculate δ 0 from the potential for special case. 6
9.3 Hard sphere Hard spere potential strictly means V (r) = for r < r 0, zero for r > r 0. We can treat approximately more gen l problems where sphere is not quite hard, i.e. V 0 is finite and there is some small amplitude for the particle to be inside r 0. From (22), radial l = 0 wave function u 0 rf 0 satisfies h2 2 u 0 2m r + V u 2 0 = h2 k 2 2m u 0 (37) For r > r 0 we solved this prob. already: f 0 = 1 2i e i + ei+2iδ0 (38) Applying boundary condition u 0 (r 0 ) 0 (exact if V 0 = ), find so from (34) have e i 0 = e i 0+2iδ 0 = δ 0 = 0 (39) dσ dω = sin2 δ 0 k 2 = sin2 0 k 2 r 2 0 if 0 1 (40) so for low energies we recover nearly the classical hard sphere cross section, σ = dω dσ dω = 4πr2 0 > πr 2 0 σ cl, (41) 7
larger by factor of 4 than classical geometrical cross section. In general diffraction effects extend beyond edge of geometrical shadow to produce extra cross section. 9.4 Absorption Particle can be absorbed in scattering process have not allowed for this so far. Consider nuclear reaction }{{} n + (Z, N) }{{} neutron nucleus Z protons Nneutrons (Z, N + 1) + γ (42) Outside nucleus, neutron wave fctn satisfies free S.-eqn., so must be able to represent wave fctn. outside as ψ n = 1 e i 2i + η e i 0 +... (43) l>0 Consider net flux of probability out of sphere F, radius r: F = r 2 dω j ˆr (44) j = i h 2m (ψ ψ c.c.) (45) where c.c. means complex conjugate. Imagine substituting the spherical harmonic expansion into (44); no cross terms would occur because of the normalization thrm., so F may be written as a sum of distinct contributions from all partial waves l. In particular for s-wave part we ll need ψ l=0 to calculate j l=0 : 1 e i 2i + η e i 0 = k 2 ˆr e i + η e i 0 + O 1 r 2 (46) since last terms are negligible at r, find 8
ˆr j l=0 = i h 2m ik 4 e + η e i e 0 + η e i 0 (47) c.c.] = hk 1 ( 1 + η0 2 η 2m () 2 0 e 2i + η0e 2i c.c. ) (48) = h 4m 2(1 η 0 2 ) (49) Wait didn t we say η 0 2 = 1 shouldn t F vanish? No, that was true only if we assumed flux in = flux out = total flux zero. Here have absorption, outgoing flux must be less than incoming flux, η 0 2 1. giving total negative F net outgoing rate. Say particles being absorbed at rate 1 dn a dt = F = r 2 dωˆr j = 4πr 2 h 4kmr 2(1 η 0 2 ) (50) = πv k (1 η 0 2 ), v hk/m = classical velocity (51) 2 = absorption incident (52) cross section flux density σ a nv (53) = σ a v here, inc. beam obeys 1 n = e ikz 2 = 1 (54) Comparing (51), the result, with (54), the definition of absorption cross section σ a which parallels our earlier definition of scattering cross-section, find σ a = π k 2(1 η 0 2 ) (55) We ve discussed absorption, how about scattering? From Eq. (32) we have differential cross section 1 Note on normalization: if you check the dimensions here you will be confused. j should have dimensions of current density, number per time per area. But instead it has dimensions of a velocity. This is the fudge mentioned earlier, that we are being careless about wave function normalization because plane waves aren t normalizable in the usual way. We can be more careful, making everything much more involved, or we can remember that the plane wave e i and all the terms which go with it should have a 1/ V with them, where V is the system volume. Then j has dimensions of [v]/[v] = t 1 L 2 ) as it should, and F is really a rate, [F ] = t 1 as it should be. 9
f(θ, φ) = η 0 1 2ik dσ = dω = η 0 1 2 (56) 4k 2 so s-wave approx. for total scattering cross section σ s is σ s = 4π dσ dω = π k 2 η 0 1 2 (57) To summarize, in the s-wave approx. we have σ a = π k 2 (1 η 0 2 ) ; σ s = π k 2 η 0 1 2 (58) To get a little intuition for these expressions, ask when is absorption maximum? When η 0 =0, σ a = π/k 2. In this case, σ s = π/k 2 = σ a. Even in the case of a totally absorbing target, there is same amount of scattering due to diffraction (shadow effect). 9.5 Higher-l partial waves To simplify discussion somewhat, let s assume scattering potential is axially symmetric along line of incident particle (e.g. sphere or footballshaped); therefore only m = 0 partial waves will be produced (no φ- dependence). Then let s redo the discussion for l = 0, but continue the expansion for higher l, but m = 0. I will just summarize basic results. Expand plane wave again, assuming k ẑ: 10
e i cos θ = l g l ()Y l0 (θ) (59) Once again we can invert to get g l s as in (26). Harder integral to do, but simplifies as r as usual, leaving result g l () = iπ 1/2 (2l + 1) 1/2 ( 1) l+1e i + ei (60) so analog of (28) is e i cos θ = iπ 1/2 l Y l0 (2l + 1) 1/2 ( 1) l+1e i + ei (61) Now game same as before: in each l channel, ingoing wave must be unaffected by scattering, outgoing wave can be modified such that flux conserved. The full wave fctn. (analog of (29) must then look like ψ = iπ 1/2 l Y l0 (2l + 1) 1/2 = e i cos θ iπ 1/2 l ( 1) l+1e i + η l Y l0 (2l + 1) 1/2 (η l 1) ei e i (62) Verify this checks with l = 0 results (28-29). Each amplitude η l obeys (63) η l 2 = 1, (64) and reading off scattering amplitude from (63) we have f(θ, φ) = iπ 1/2 l Y l0 (2l + 1) 1/2 (η l 1)/k (65) and using orthonormality of Y lm s and definitions dσ dω = f 2, η l = e 2iδ l σ = f 2 dω = π (2l + 1) η k 2 l 1 2 (66) = 4π k 2 l where δ l is the phase shift in the lth partial wave. l (2l + 1) sin 2 δ l (67) 11