9
Scattering Theory II
9.1
Partial wave analysis
Expand ψ in spherical harmonics Y`m(θ, φ), derive 1D differential equations for expansion coefficients. Spherical coordinates:
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
(1)
(2)
(3)
from which follow by application of chain rule relations
∂
∂x ∂
∂y ∂
∂
∂
=
+
= −y
+x ,
∂φ
∂φ ∂x ∂φ ∂y
∂x
∂y
∂
L̂z = −ih̄
∂φ
by constructing
∂
∂θ ,
or
(4)
find also
∂
∂
+ cot θ cos φ )
∂θ
∂φ
∂
∂
L̂y = −ih̄(cos φ − cot θ sin φ )
∂θ
∂φ
L̂x = ih̄(sin φ
(5)
(6)
and
∂
∂
+ i cot θ ),
∂θ
∂φ
∂
∂
L̂− = −h̄e−iφ( − i cot θ ),
∂θ
∂φ
L̂2 = L̂+L̂− + L2z + h̄L̂z


2
1
∂
∂
∂
1



= −h̄2 
sin θ + 2
2
sin θ ∂θ
∂θ sin θ ∂φ
L̂+ = h̄eiφ(
Reminder: spherical harmonics
1
(7)
(8)
(9)
(10)
Eigenstates of L̂2, L̂z :
L̂2Y`m = h̄2Y`m, L̂z Y`m = h̄mY`m
Relation to Legendre functions:
Y`m = P`meimφ
P`0(θ) = P`(cos θ) Legendre polynomial
Normalization:
Z
∗
dΩ Y`m
Y`0m0 = δ``0 δ``0
(11)
(12)
(13)
Compute Y`` from L̂+Y`` = 0:
dP``
= ` cot θP``
dθ
dP``
`P``
or
=
d sin θ
sin θ
(14)
(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):
L̂2
1 ∂2
r− 2 2
∇ =
r ∂r2
h̄ r
2
1
∂
∂2
1∂
∂
1
r+ 2
=
sin θ + 2 2
r ∂r2
r sin θ ∂θ
∂θ r sin θ ∂φ2
2
2
(17)
(18)
where last step follows from (10). So want to solve




2
h̄ 2
h̄2k 2

ψ
− ∇ + V (r) ψ =
2m
2m
(19)
Since Y`m form complete set, expand for any ψ
ψ=
X
`,m
f`m(r)Y`m(θ, φ)
(20)
and insert in S.-eqn (19) to get

X

`m
h̄2 1 ∂ 2
h̄2`(` + 1)
−
rf`m +
f`m
2m r ∂r2
2mr2

h̄2k 2
 Y`m
+V (r)f`m −
f`m = 0
2m
(21)
Case of isotropic potential V (r) = V (r). Under this assumption, every3
thing inside ( ) is fctn. of r only! Since Y`m are independent, get one
separate equation for each `, m (“partial wave”):


h̄2 1 ∂ 2
h̄2`(` + 1)
h̄2k 2


−
r+
+ V (r) f`m =
f`m
2m r ∂r2
2mr2
2m
9.2
(22)
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 > r0. Outside r0 get just
d2
rf0(r) = −k 2rf0(r)
(23)
2
dr
solution
e±ikr
f0 ∝
outgoing/incoming sph. waves
(24)
r
Plane wave expansion in Y`m’s. Let’s go back and reexamine sph. harm.
expansion of original plane wave!
eik·r =
X
`m
g`m(r)Y`m
(25)
and use orthonormality condition (13) for Y`m’s to project out g`m’s. Find
by multiplying by Y00, integrating over dΩ,
Z
1
sin kr
dΩeik·r = 4π
= g0(r) · 4π · √
(26)
kr
4π
or
√  ikr

4π  e
e−ikr 


(27)
g0 =
−
2i
kr
kr
so (unperturbed!) plane wave may be written


eik·r
−ikr
eikr 
X
 e
1
+
g`mY`m
= 2i −
+
kr
kr
`>0,m
4
(28)
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 corresponds 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:

−ikr
ikr

e
e 
X
1
−
+
ψ = 2i
+ η0
g`mY`m
kr
kr
`>0,m
(29)
Now recall that for isotropic potentials we are considering, S.-eqn. decomposes 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, magnitude 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
η0 − 1 eikr
ik·r
ψ'e +
,
(32)
2ik 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.
5
Can immediately read off form of scattering amplitude,
ei2δ0 − 1
f (θ, φ) =
,
2ik
or
dσ = |f |2 = sin2 δ0 isotropic, s-wave only
dΩ
k2
(33)
(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
h̄2 ∂ 2u0
h̄2k 2
−
+ V u0 =
u0
(37)
2m ∂r2
2m
For r > r0 we solved this prob. already:


1  e−ikr eikr+2iδ0 

f0 = −
+
(38)
2i
kr
kr
Applying “boundary condition” u0(r0) ' 0 (exact if V0 = ∞), find
e−ikr0 = eikr0+2iδ0 =⇒ δ0 = −kr0
(39)
so from (34) have
dσ sin2 δ0 sin2 kr0
=
=
' r02 if kr0 ¿ 1
(40)
2
2
dΩ
k
k
so for low energies we recover nearly the classical hard sphere cross section,
Z
dσ
σ = dΩ
= 4πr02 > πr02 ≡ σcl ,
(41)
dΩ
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}
neutron
+ (Z,
N)
| {z }
→
(Z, N + 1) + γ
(42)
nucleus
Z protons
N neutrons
Outside nucleus, neutron wave fctn satisfies free S.-eqn., so must be able
to represent wave fctn. outside as


eikr  X
1  e−ikr
+
+ η0
ψn = −
...
(43)
2i
kr
kr
`>0
Consider net flux of probability out of sphere F , radius r:
Z
r2dΩ j · r̂
ih̄
j = − (ψ ∗∇ψ − c.c.)
2m
F =
(44)
(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 `. In particular for s-wave part we’ll need
∇ψ|`=0 to calculate j|`=0:






−ikr
−ikr
ikr
ikr
1  e
1
e  k e
e 

 =
+O
∇ −
(46)
+ η0
r̂ 
+ η0
2
2i
kr
kr
2
kr
kr
r
since last terms are negligible at r → ∞, find
8


r̂ · j|`=0

−ikr
−ikr
−ih̄ ik  −eikr
eikr 
 e
∗e


=
· · 
+ η0
+ η0
2m 4
kr
kr
kr
kr
−c.c.]
¶
h̄k 1 µ
2
2ikr
∗ −2ikr
=
−1 + |η0| − η0e
+ η0 e
− c.c.
2m (kr)2
−h̄
=
(1 − |η0|2)
2
4mkr
(47)
(48)
(49)
Wait—didn’t we say |η0|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, |η0|2 ≤ 1.
giving total negative F ≡ net outgoing rate. Say particles being absorbed
at rate1
Z
dNa
h̄
= −F = r2dΩr̂ · j = 4πr2 ·
(1 − |η0|2)
2
dt
4kmr
πv
= 2 (1 − |η0|2), v ≡ h̄k/m = classical velocity
k



absorption
incident





= 
cross section
flux density
≡ σanv
= σav here, inc. beam obeys1 n = |eikz |2 = 1
(50)
(51)
(52)
(53)
(54)
Comparing (51), the result, with (54), the definition of absorption cross
section σa which parallels our earlier definition of scattering cross-section,
find
π
(55)
σa = 2 (1 − |η0|2)
k
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 eikr 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
η0 − 1
dσ |η0 − 1|2
f (θ, φ) =
=⇒
=
2ik
dΩ
4k 2
so s-wave approx. for total scattering cross section σs is
dσ
π
σs = 4π
= 2 |η0 − 1|2
dΩ k
To summarize, in the s-wave approx. we have
σa =
π
(1
k2
− |η0|2)
;
σs =
π
|η
k2 0
− 1|2
(56)
(57)
(58)
To get a little intuition for these expressions, ask when is absorption maximum? 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 axially symmetric along line of incident particle (e.g. sphere or footballshaped); therefore only m = 0 partial waves will be produced (no φdependence). 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 k ẑ:
10
eikr cos θ =
X
`
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


−ikr
ikr
e
e

`+1
(−1)

g`(kr) = −iπ 1/2(2` + 1)1/2 
+
(60)
kr
kr
so analog of (28) is


eikr cos θ
−ikr
eikr 
1/2 X
1/2 
`+1 e

+
= −iπ
Y`0(2` + 1) (−1)
kr
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


−ikr
eikr 
1/2 
`+1 e

Y`0(2` + 1) (−1)
+ η`
ψ = −iπ
kr
kr
`
eikr
1/2
ikr cos θ
1/2 X
Y`0(2` + 1) (η` − 1)
= e
− iπ
kr
`
1/2 X
(62)
(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) = −iπ 1/2
X
`
Y`0(2` + 1)1/2(η` − 1)/k
and using orthonormality of Y`m’s and definitions
σ =
=
Z
|f |2dΩ =
dσ
dΩ
= |f |2, η` = e2iδ`
π X
(2` + 1)|η` − 1|2
2
k `
4π X
2
(2`
+
1)
sin
δ`
k2 `
where δ` is the phase shift in the `th partial wave.
11
(65)
(66)
(67)