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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
∂y
, or
Lˆz = −ih ¯ ∂ ∂φ
by constructing (^) ∂θ∂ , find also
Lˆx = i¯h(sin φ ∂ ∂θ
∂φ
Lˆy = −ih¯(cos φ ∂ ∂θ
− cot θ sin φ
∂φ
and
Lˆ+ = ¯heiφ( ∂ ∂θ
∂φ
Lˆ− = −he¯ −iφ( ∂ ∂θ
− i cot θ
∂φ
Lˆ^2 = Lˆ+ Lˆ− + L^2 z + ¯h Lˆz (9)
= −h¯^2
^1 sin θ
∂θ sin θ
∂θ
sin^2 θ
∂^2
∂φ^2
(^) (10)
Reminder: spherical harmonics
Eigenstates of Lˆ^2 , Lˆz: Lˆ^2 Ym = ¯h^2 Ym, LˆzYm = ¯hmYm
Relation to Legendre functions:
Ym = Pmeimφ^ (11) P0 (θ) = P(cos θ) Legendre polynomial (12)
Normalization: (^) ∫ dΩ Y (^) m∗ Y′m′^ = δ′δ′^ (13)
Compute Yfrom Lˆ+Y = 0:
dP`` dθ
= ` cot θP`` (14)
or dP`` d sin θ
`P``
sin θ
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 =
r
∂^2
∂r^2
r −
Lˆ^2
¯h^2 r^2
r
∂^2
∂r^2
r +
r^2 sin θ
∂θ
sin θ
∂θ
r^2 sin^2 θ
∂^2
∂φ^2
thing inside ( ) is fctn. of r only! Since Ym are independent, get one separate equation for each, m (“partial wave”): − h¯
2 2 m
r
∂^2
∂r^2
r +
h¯^2 ( + 1) 2 mr^2
(^) f`m = h¯
(^2) k 2 2 m
f`m (22)
9.2 s-wave scattering
Low energies scattering more isotropic, may approximate cross section by considering only ` = 0 partial wave. Suppose potential has hard finite range, V = 0 for r > r 0. Outside r 0 get just
d^2 dr^2
rf 0 (r) = −k^2 rf 0 (r) (23)
solution
f 0 ∝ e±ikr r
outgoing/incoming sph. waves (24)
Plane wave expansion in Y`m’s. Let’s go back and reexamine sph. harm. expansion of original plane wave!
eik·r^ = ∑ `m
gm(r)Ym (25)
and use orthonormality condition (13) for Ym’s to project out gm’s. Find by multiplying by Y 00 , integrating over dΩ, ∫ dΩeik·r^ = 4π
sin kr kr
= g 0 (r) · 4 π ·
√^1
4 π
or
g 0 =
4 π 2 i
^ e ikr kr
e−ikr kr
(^) (27)
so (unperturbed!) plane wave may be written
eik·r^ = 21 i
−e−ikr kr
eikr kr
(^) + ∑ `> 0 ,m
gmYm (28)
Now look at full ψ again, argue as follows: have shown (24) that f 0 ∝ 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 η 0. So we have deduced that, in s-wave approximation:
ψ = 21 i
−e−ikr kr
eikr kr
(^) + ∑ `> 0 ,m
gmYm (29)
Now recall that for isotropic potentials we are considering, S.-eqn. decom- poses into separate equations (22) for the amplitudes fm. This means eachm 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 =⇒
|η 0 |^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 η 0 is complex phase: conventional to write
η 0 ≡ e^2 iδ^0 (31)
where the quantity δ 0 called s-wave scattering phase shift. In our s-wave approximation, all other δ` are zero, & we have for r ¿ r 0
ψ ' eik·r^ + η 0 − 1 2 ik
eikr kr
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.
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 ` = 0 wave function” u 0 ≡ rf 0 satisfies
−
h¯^2 2 m
∂^2 u 0 ∂r^2
h¯^2 k^2 2 m
u 0 (37)
For r > r 0 we solved this prob. already:
f 0 =
2 i
−e−ikr kr
eikr+2iδ^0 kr
(^) (38)
Applying “boundary condition” u 0 (r 0 ) ' 0 (exact if V 0 = ∞), find
e−ikr^0 = eikr^0 +2iδ^0 =⇒ δ 0 = −kr 0 (39)
so from (34) have
dσ dΩ
sin^2 δ 0 k^2
sin^2 kr 0 k^2
' r^20 if kr 0 ø 1 (40)
so for low energies we recover nearly the classical hard sphere cross section,
σ =
∫ dΩ dσ dΩ
= 4πr 02 > πr 02 ≡ σcl, (41)
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︷︷ )︸ →^ (Z, N^ + 1)^ +^ γ^ (42) neutron nucleus Z protons N neutrons
Outside nucleus, neutron wave fctn satisfies free S.-eqn., so must be able to represent wave fctn. outside as
ψn =
2 i
−e −ikr kr
eikr kr
(^) + ∑ `> 0
Consider net flux of probability out of sphere F , radius r:
F =
∫ r^2 dΩ j · rˆ (44)
j = −
ih¯ 2 m
(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 contribu- tions from all partial waves . In particular for s-wave part we’ll need ∇ψ|=0 to calculate j|`=0:
∇
2 i
−e−ikr kr
eikr kr
(^) = k 2 rˆ
^ e−ikr kr
eikr kr
(^) + O
1 r^2
(^) (46)
since last terms are negligible at r → ∞, find
f (θ, φ) = η 0 − 1 2 ik
dσ dΩ
|η 0 − 1 |^2 4 k^2
so s-wave approx. for total scattering cross section σs is
σs = 4π dσ dΩ
π k^2
|η 0 − 1 |^2 (57)
To summarize, in the s-wave approx. we have
σa = (^) kπ 2 (1 − |η 0 |^2 ) ; σs = (^) kπ 2 |η 0 − 1 |^2 (58)
To get a little intuition for these expressions, ask when is absorption max- imum? When η 0 =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 ‖ zˆ:
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) = −iπ^1 /^2 (2 + 1)^1 /^2
(−1)`+1 e−ikr kr
eikr kr
(60)
so analog of (28) is
eikr^ cos^ θ^ = −iπ^1 /^2
∑ `
Y0 (2 + 1)^1 /^2
(−1)`+1 e−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 = −iπ^1 /^2
∑ `
Y0 (2 + 1)^1 /^2
(−1)`+1 e−ikr kr
eikr kr
(^) (62)
= eikr^ cos^ θ^ − iπ^1 /^2
∑ `
Y0 (2 + 1)^1 /^2 (η` − 1)
eikr kr
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) = −iπ^1 /^2 ∑ `
Y0 (2 + 1)^1 /^2 (η` − 1)/k (65)
and using orthonormality of Ym’s and definitions (^) ddσΩ = |f |^2 , η = e^2 iδ`
σ =
∫ |f |^2 dΩ =
π k^2
∑ `
(2+ 1)|η − 1 |^2 (66)
= 4 π k^2
∑ `
(2+ 1) sin^2 δ (67)
where δis the phase shift in theth partial wave.
