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

Download Scattering Theory II - Lecture Notes | PHY 4605 and more Study notes Physics in PDF only on Docsity! 9 Scattering Theory II 9.1 Partial wave analysis Expand ψ in spherical harmonics Y`m(θ, φ), derive 1D differential equa- tions 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 ∂ ∂y = −y ∂ ∂x + x ∂ ∂y , or L̂z = −ih̄ ∂ ∂φ (4) by constructing ∂∂θ , find also L̂x = ih̄(sin φ ∂ ∂θ + cot θ cos φ ∂ ∂φ ) (5) L̂y = −ih̄(cos φ ∂ ∂θ − cot θ sin φ ∂ ∂φ ) (6) and L̂+ = h̄e iφ( ∂ ∂θ + i cot θ ∂ ∂φ ), (7) L̂− = −h̄e−iφ( ∂ ∂θ − i cot θ ∂ ∂φ ), (8) L̂2 = L̂+L̂− + L2z + h̄L̂z (9) = −h̄2   1 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 sin2 θ ∂2 ∂φ2   (10) Reminder: spherical harmonics 1 Eigenstates of L̂2, L̂z: L̂ 2Y`m = h̄ 2Y`m, L̂zY`m = h̄mY`m Relation to Legendre functions: Y`m = P`me imφ (11) P`0(θ) = P`(cos θ) Legendre polynomial (12) Normalization: ∫ dΩ Y ∗`mY`′m′ = δ``′δ``′ (13) Compute Y`` from L̂+Y`` = 0: dP`` dθ = ` cot θP`` (14) or dP`` d sin θ = `P`` sin θ (15) which has soln. P`` ∝ sin` θ (16) For large ` 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 = 1 r ∂2 ∂r2 r − L̂ 2 h̄2r2 (17) = 1 r ∂2 ∂r2 r + 1 r2 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 r2 sin2 θ ∂2 ∂φ2 (18) 2 Now look at full ψ again, argue as follows: have shown (24) that f0 ∝ e±ikr/r, but can’t have e−ikr/r in scattered part of wave, since it cor- responds to incoming, not outgoing wave boundary condition. Yet as we see from (28), a term e−ikr/r is already part of incident plane wave. Coeffcient of e−ikr/r in ψ must therefore be the same as in (28), since just comes from plane wave. Since we’ve assumed higher-` components not scattered, these must also be same in ψ. Only thing which can be different in ψ from unperturbed eik·r is coefficient of e+ikr/r, which we will call η0. So we have deduced that, in s-wave approximation: ψ = 12i  −e −ikr kr + η0 eikr kr   + ∑ `>0,m g`mY`m (29) Now recall that for isotropic potentials we are considering, S.-eqn. decom- poses into separate equations (22) for the amplitudes f`m. This means each `m is like separate scattering problem, in particular probability flux must be conserved for each `,m separately. For ` = 0 case this means amount of probability flux into r = 0 (coefficient of e−ikr/r term, magni- tude squared) has to equal flux out of r = 0 (coefficient of e+ikr/r term, magnitude squared). This =⇒ |η0|2 = 1. (30) More generally, |η`m|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 η0 is complex phase: conventional to write η0 ≡ e2iδ0 (31) where the quantity δ0 called s-wave scattering phase shift. In our s-wave approximation, all other δ` are zero, & we have for r À r0 ψ ' eik·r + η0 − 1 2ik eikr kr , (32) obtained by subtracting (28) from (29). Now compare to our standard asymptotic form for ψ in a scattering problem, ψ ' eik·r +f (θ, φ)e+ikr/r. 5 Can immediately read off form of scattering amplitude, f (θ, φ) = ei2δ0 − 1 2ik , (33) or dσ dΩ = |f |2 = sin2 δ0 k2 isotropic, s-wave only (34) from which we immediately get total cross section σ = 4π k2 sin2 δ0 (35) Check to make sure this agrees with optical theorem! Imf = (1/2k)Re (e2iδ0 − 1) = (1/2k)(cos 2δ0 − 1) = (1/2k)2 sin2 δ0 = (1/k)k 2σ/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 δ0–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 δ0 from the potential for special case. 6 9.3 Hard sphere “Hard spere” potential strictly means V (r) = ∞ for r < r0, zero for r > r0. We can treat approximately more gen’l problems where sphere is not quite “hard”, i.e. V0 is finite and there is some small amplitude for the particle to be inside r0. From (22), “radial ` = 0 wave function” u0 ≡ rf0 satisfies − h̄ 2 2m ∂2u0 ∂r2 + V u0 = h̄2k2 2m u0 (37) For r > r0 we solved this prob. already: f0 = 1 2i  −e −ikr kr + eikr+2iδ0 kr   (38) Applying “boundary condition” u0(r0) ' 0 (exact if V0 = ∞), find e−ikr0 = eikr0+2iδ0 =⇒ δ0 = −kr0 (39) so from (34) have dσ dΩ = sin2 δ0 k2 = sin2 kr0 k2 ' r20 if kr0 ¿ 1 (40) so for low energies we recover nearly the classical hard sphere cross section, σ = ∫ dΩ dσ dΩ = 4πr20 > πr 2 0 ≡ σcl, (41) 7 f (θ, φ) = η0 − 1 2ik =⇒ dσ dΩ = |η0 − 1|2 4k2 (56) so s-wave approx. for total scattering cross section σs is σs = 4π dσ dΩ = π k2 |η0 − 1|2 (57) To summarize, in the s-wave approx. we have σa = π k2 (1− |η0|2) ; σs = πk2 |η0 − 1|2 (58) To get a little intuition for these expressions, ask when is absorption max- imum? When η0 =0, σa = π/k 2. In this case, σs = π/k 2 = σa. Even in the case of a totally absorbing target, there is same amount of scattering due to diffraction (shadow effect). 9.5 Higher-` partial waves To simplify discussion somewhat, let’s assume scattering potential is ax- ially symmetric along line of incident particle (e.g. sphere or football- shaped); therefore only m = 0 partial waves will be produced (no φ- dependence). Then let’s redo the discussion for ` = 0, but continue the expansion for higher `, but m = 0. I will just summarize basic results. Expand plane wave again, assuming k ‖ ẑ: 10 eikr cos θ = ∑ ` g`(kr)Y`0(θ) (59) Once again we can invert to get g`’s as in (26). Harder integral to do, but simplifies as r →∞ as usual, leaving result g`(kr) = −iπ1/2(2` + 1)1/2  (−1)`+1e −ikr kr + eikr kr   (60) so analog of (28) is eikr cos θ = −iπ1/2 ∑ ` Y`0(2` + 1) 1/2  (−1)`+1e −ikr kr + eikr kr   (61) Now game same as before: in each ` “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 = −iπ1/2 ∑ ` Y`0(2` + 1) 1/2  (−1)`+1e −ikr kr + η` eikr kr   (62) = eikr cos θ − iπ1/2 ∑ ` Y`0(2` + 1) 1/2(η` − 1)e ikr kr (63) Verify this checks with ` = 0 results (28-29). Each amplitude η` obeys |η`|2 = 1, (64) and reading off scattering amplitude from (63) we have f (θ, φ) = −iπ1/2 ∑ ` Y`0(2` + 1) 1/2(η` − 1)/k (65) and using orthonormality of Y`m’s and definitions dσ dΩ = |f |2, η` = e2iδ` σ = ∫ |f |2dΩ = π k2 ∑ ` (2` + 1)|η` − 1|2 (66) = 4π k2 ∑ ` (2` + 1) sin2 δ` (67) where δ` is the phase shift in the `th partial wave. 11