1.2. CURRENTS AND OPTICAL THEOREM
February 11, 2015
Lecture IX
1.2
Currents and optical theorem
ψ = ψ 0 + ψs
(1.1)
ψ 0 represents the incoming free particle and is a solution to
H 0 ψ 0 = i~
∂
ψ0
∂t
2
p
∂
. Then Hψ = i~ ∂t
ψ where H = H 0 + V and
where H0 = 2m
The flux of scattered particles into area element da is
~
Im(ψs∗ ∇ψs ) · n̂da
m
k ~ |f |2 2
r dΩ
8π 3 m r2
js · n̂da =
=
The flux of incoming particles is
jinc =
The differential cross section is
~
k ~
Im(ψ0∗ ∇ψ0 ) =
m
8π 3 m
jscat · n̂da
dσ
=
= |f |2
dΩ
jinc
Then along with the divergence theorem,
Z
8π 3 m
js · n̂da =
σt =
~k
8π 3 m
~k
(1.3)
Z
∇ · js dv
By the continuity equation
Z
Z
∇ · js dv
=
∂
|ψs |2 dv
∂t
Altogether we find that
σt =
m
~k
Z
∂
|ψs |2 dv
∂t
Substituting Equation 1.1 we have
Z
8π 3 m ∂
σt =
dv |ψ0 |2 + |ψ|2 − 2<ψ0∗ ψ
~k ∂t
∗
Z
∂ψ0
∂ψ ∗
∂ψ0∗
8π 3 m
∗ ∂ψ0
∗ ∂ψ
∗ ∂ψ
dv
ψ0 + ψ0
+
ψ+ψ
− 2<(
ψ + ψ0
)
=
~k
∂t
∂t
∂t
∂t
∂t
∂t
1
(1.2)
1.2. CURRENTS AND OPTICAL THEOREM
=
=
=
=
=
Z
8π 3 m
i
<
dv(−H0 ψ0∗ ψ + ψ0∗ (H0 + V )ψ)
~k
~
Z
8π 3 m
i
−2
<
dv(ψ0∗ V ψ)
~k
~
8π 3 m 2
hk | V | ψi
Im
~k ~
8π 3 m 2 ~2
f (0)
Im
~k ~ 4π 2 m
4π
Im f (0)
k
−2
The total cross section is proportional to the imaginary part of the forward scattering amplitude.
The flux of scattered particles is balanced by the imaginary part of the forward amplitude, the
shadow.
There is another way: The total wave function
ψinc + ψscat = eikz + f (θ)
eikr
r
(1.4)
The flux density is
~
Im(ψ ∗ ∇ψ)
m
The total flux density in the radial direction is
−ikr
~
1 eikr
−ikz
∗e
ikr cos θ
jr =
Im (e
+f
)(ik cos θe
+ (ikf − 2 )
)r̂
m
r
r
r
j=
(1.5)
Since we are interested in r → ∞, only the first and two terms in the second brackets will remain.
Then
~
eikr
e−ikr
jr =
Im (e−ikz + f ∗
)(ik cos θeikr cos θ + ikf )
r̂
m
r
r
The interference term is
jrint
=
~
ik −ikr(cos θ−1)
Im
e
f + f ∗ e−ikr(1−cos θ cos θ r̂]
m
r
Next integrate jrint over a tiny cone in the forward direction to show
Z
~k 4π
Imf (0)
jrint r2 dΩ = −
m
k
forward cone
Z
forward cone
jrint r2 dΩ
=
~rk
2π
m
Z
β
Imi(e−ikr(cos θ−1) f + cos θf ∗ e−ikr(1−cos θ )d(cos θ)
0
2
(1.6)
1.2. CURRENTS AND OPTICAL THEOREM
Z
~rk β
Imi(eikr e−ikr cos θ f + f ∗ e−ikr eikr cos θ )d(cos θ)
m 0
~rk
eikr −ikr cos θ
e−ikr ikr cos θ β
= 2π
Imi(
e
f + f∗
e
) |0
m
−ikr
ikr
~rk
eikr −ikr cos β
e−ikr ikr cos β
= 2π
Imi(
(e
− 1)f + f ∗
(e
− 1))
m
−ikr
ikr
1
1
~rk
Im(
(e−ikr(cos β−1) − eikr )f + f ∗ (eikr(cos β−1) − e−ikr ))
= 2π
m
−kr
kr
∼ 2π
As long as θ 6= 0, the average of jrint over any small solid angle is zero because r → ∞. (Assume
f (θ) is a smooth function.)
In the limit β → 0, and as r → ∞, we use the average value for e±irk , namely zero, we get
Z
~k
1
1
jrint r2 dΩ = 2π rIm(
(1)f + f ∗ (1))
m
−kr
kr
forward cone
~k 1
Im(f ∗ − f )
= 2π
mk
~k 1
Imf (0)
= −4π
mk
~k 1
= −4π
Imf (0)]
mk
In evaluating the upper limit in the θ integration, assume that the limit of a function that
oscillates as its argument approaches infinity is equal to its average value.
The total probability current in the region behind the target produces a depletion of particles.
It must be that the product of the incident flux and the total cross section is equivalent to hwat is
depleted in the forward direction. Therefore
Z
~k 1
~k
jrint r2 dΩ = −4π
Imf (0) =
σt
mk
m
f orward
4π
Imf (0)
→ σt =
k
1.2.1
Optical Theorem
We begin with the basic Lippmann/Schwinger equation
| ψ ± i = | φi +
1
V | ψ± i
E − H0 ± i
(1.7)
The scattering amplitude is
f (k, k0 ) = −
4π 2 m
hφ(k) | T | φ(k0 )i
~2
3
(1.8)
1.2. CURRENTS AND OPTICAL THEOREM
where T | φi = V | ψ ± i by definition. Then
0
f (k, k ) =
hψ ± | − hψ ± |V
=
1
V | ψ± i
E − H0 ± i
1
hψ ± |V | ψ ± i − hψ ± |V
V | ψ± i
E − H0 ± i
The imaginary part of the forward scattering amplitude is
Imf (k, k)
=
Imf (k, k)
=
Imf (k, k)
=
=
=
8π 3 2m
hk | T | ki
4π ~2
Z Z
3
8π 2m
1
−Im
hφ(k) |T | φk0 ihφk0 |
| φk00 ihφk00 |T | φ(k)id3 k 0 d3 k 00
2
4π ~
E − H0 ± i
Z
8π 3 2m 2m
1
d3 k 0 hφ(k) |T | φk0 i 00 2
−Im
hφk0 |T | φ(k)i
2
2
4π ~ ~
k − k 0 2 ± i
Z
1
8π 3 2m 2m
2
dk 0 k 0 dΩhφ(k) |T | φk0 i 00 2
hφk0 |T | φ(k)i
−Im
4π ~2 ~2
k − k 0 2 ± i
Z
8π 3 2m 2m
1
2
q
q
−Im
dk 0 k 0 dΩhφ(k) |T | φk0 i
hφk0 |T | φ(k)i
4π ~2 ~2
i
(k 00 1 ±
+ k 0 )(k 00 1 ± i − k 0 )
−
k00 2
3
k002
Z
1
8π 2m 2m
2
dk 0 k 0 dΩhφ(k) |T | φk0 i 00
hφk0 |T | φ(k)i
4π ~2 ~2
(k ± i + k 0 )(k 00 ± i − k 0 ))
Z
1 8π 3 8π 3 2m 2m
1
2
∼ −Imiπ 2
k 00 dΩhφ(k) |T | φk00 i 00 hφk00 |T | φ(k)i
2π 4π 4π ~2 ~2
2k
Z
ik 00
∼ −Im
dΩf ∗ (k, k00 )f (k00 , k)
4π
k 00
∼ − σt
4π
∼
−Im
Shadowing
We write the solution to Schrodinger’s equation as ψ + as the sum of an incoming plane wave that
extends over all space, and an outgoing spherical wave with angular distribution represented as
f (θ). Consider scattering from a hard sphere. The forward direction along the axis of the incoming
wave is shadowed. In that region the probability density |ψ + |2 = 0. There must be destructive
interference between the incoming plane wave and the scattering amplitude in the forward direction.
So the scattering amplitude in the forward direction cannot be zero.
More generally we write
4π
σtot =
Imfelastic (0)
(1.9)
k
Only states scattered elastically in the forward direction will have the same energy as the incident
state, which is required if there is to be destructive interference. Also the depletion of the forward
flux must account for all scattered states elastic or inelastic.
4
1.3. PARTIAL WAVE ANALYSIS
1.3
Partial Wave Analysis
We have described scattering in terms of an incoming plane wave, a momentum eigenket, and and
outgoing spherical wave, also with definite momentum. We now consider the basis of free particle
states with definite energy and angular momentum (rather than linear momentum) that look like
| E, l, mi. These are eigenkets of of H0 , L2 , and Lz . We would like to expand our plane wave in
terms of these spherical waves like so
X
| ki =
| E, l, mihE, l, m | ki
(1.10)
l,m
Then we can write the scattering amplitude
f (k0 , k)
=
=
1 2m 3 0
8π hk | T | ki
4π ~2
Z
X
XZ
1 2m 3
0
−
8π
dE
dE
hk0 | E, l0 , m0 i hE, l0 , m0 | T | E, l, mi hE, l, m | ki
2
4π ~
0
0
−
l ,m
l,m
If the scattering potential is spherically symmetric, T is a scalar operator, and by WE, l = l0 , m =
m0 , and hE, l, m | T | E, l, mi is independent of m. Then
Z Z
X
1 2m 3
8π
dEdE 0
hk0 | E 0 , l, mi hE 0 , l, m | T | E, l, mi hE, l, m | ki
f (k0 , k) = −
2
4π ~
l,m
Z Z
X
1 2m 3
= −
8π
dEdE 0
hk0 | E 0 , l, miTl hE, l, m | ki
(1.11)
4π ~2
l,m
(1.12)
The ”spherical” scattering amplitude conserves angular momentum.
Now let’s figure out hk | E, l, mi. Consider the state | kẑi.
hkẑ | Lz | E, l, mi =
0 (m 6= 0)
→ hkẑ | E, l, mi =
0 (m 6= 0)
q
Also hkẑ | E, l, m = 0i is independent of θ, φ,so hk, ẑ | E, l, m = 0i = 2l+1
4π gl (k). We can transform
the z-direction momentum ket into an arbitrary direction by a rotation.
| ki = D(α = φ, β = θ, 0)| kẑi
Then
hk | E, l, mi =
=
hkẑ | D | E, l, mi
X
hkẑ | E, l0 , m0 = 0i hE, l0 , m0 = 0 | D | E, l, mi
l0
r
=
X
l0
2l + 1
l
gl (k)D0,m
4π
5
(1.13)
1.3. PARTIAL WAVE ANALYSIS
r
=
X
=
X
l0
r
2l + 1
4π
gl (k)
Y m (θ, φ)
4π
2l + 1 l
gl (k)Ylm (θ, φ)
l
One more thing.
hk | H0 − E | E, l, mi = hk | E, l, mi(
→ hk | E, l, mi
→ gl (k)
~2 k 2
− E) = 0
2m
~2 k 2
− E)
2m
~2 k 2
= Nl δ(
− E)
2m
∝ δ(
To determine Nl let’s try to normalize.
Z
d3 khE 0 , l0 , m0 | kihk | E, l, mi
hE 0 , l0 , m0 | E, l, mi =
k 00 dk 00 dΩNl∗0 Nl Ylm
Ylm δ(
0
Z
k 00 dk 00 |Nl |2 δ(
=
=
0∗
2
=
→ gl
=
2
~2 k 00
~2 k 00
− E 0 )δ(
− E)
2m
2m
2
2
2 00
k 00 m 00
~2 k 00
2 ~ k
0
dE
|N
|
δ(
−
E
)δ(
− E)
l
~2
2m
2m
k 00 m
|Nl |2 δ(E − E 0 )δll0 δmm0
~2
2
~
~2 k 00
√
δ(E −
)
2m
k 00 m
Z
=
~2 k 00
~2 k 00
− E 0 )δ(
− E)
2m
2m
2
2
2
2
Z
From which we get
hk | E, lm,i = √
1.3.1
~
~2 k 2
Ylm (θ, φ)δ(E −
)
2m
km
(1.14)
ρ0 → ππ
The ρ meson is spin 1 and it decays to two spin 0 pions. Suppose that the ρ is in the l = 1, m = 1
state, where there is some z-axis defined by something. The final state has the same angular
momentum quantum numbers and the amplitude to find a π with momentum in the k̂ direction is
hk | E, l, mi ∝ Y11 (k̂) ∝ sin θ
The angular distribution of the π is
|Y11 |2 ∼ sin2 θ
If we imagine producing ρ in e+ e− collisions where electrons and positrons are polarized so that
jz = +1 along the z-axis defined by the direction of the positron beam, then
dσ
(θ) ∝ sin2 θ
dΩ
6
1.3. PARTIAL WAVE ANALYSIS
1.3.2
Back to partial wave expansion
Substituting into Equation 1.11 we have
Z
Z
XX
1 2m 3
f (k0 , k) = −
8π
dE
dE 0
hk0 | E 0 , l0 , m0 ihE 0 , l0 , m0 |Tl | E, l, mihE, l, m | ki
2
4π ~
0
0
l,m l ,m
Z
Z
X
1 2m 3
dE dE 0
8π
hk0 | E 0 , l, miTl hE, l, m | ki
= −
4π ~2
l,m
Z
Z
2
X
~2 k 0
~2 k 2
1 2m 3
~
~
m 0
0
m∗
0
√
√
Y
(k
)δ(E
−
Y
(k)δ(E
−
= −
8π
dE
dE
)T
)
l
l
l
4π ~2
2m
2m
k0 m
km
l,m
= −
1 2m 3 X ~2 m 0 m ∗
8π
Y (k )Yl (k) Tl
4π ~2
km l
l,m
4π 2 X m 0 m ∗
= −
Yl (k )Yl (k) Tl
k
l,m
q
Let k = |k|ẑ so that θ = 0, φ =? and then Ylm (k) = 2l+1
4π δm0 . So only m = 0 contributes. Then
q
2l+1
0
Yl0 (k0 ) =
4π Pl (cos θ) where θ is the angle between k and k . The scattering amplitude becomes
f (k0 , k) = −
4π 2 X 2l + 1
πX
Pl (θ)Tl = −
(2l + 1)Pl (cos θ)Tl
k
4π
k
l
(1.15)
l
Define fl (k) = − πTlk(E) and
f (k0 , k) =
X
(2l + 1)Pl (cos θ)fl (k)
(1.16)
l
fl (k) is amplitude to scatter an incident particle with angular momentum ~l or impact parameter
b such that kb = l. Remember that the outgoing solution to the SE far outside the range of the
potential is
1
eikr
+
ikz
ψ
=
) e + f (θ)
r
(2π)3/2
7
1.3. PARTIAL WAVE ANALYSIS
1.3.3
Expansion of plane wave as spherical waves
The radial part of the free particle Schrodinger equation is
~2 d2 u
l(l + 1)~2
−
+ V (r) +
u = Eu
2m dr2
2mr2
The solution to the free particle Schrodinger equation in spherical coordinates is
hx | E, l, mi = cl jl (kr)Ylm (r̂).
Next expand the plane wave as a linear combination of incoming and outgoing spherical waves.
XZ
eik·x
=
dEhx | E, l, mihE, l, m | ki
hx | ki =
(2π)3/2
l,m
XZ
~
~2 k 2 m
dEcl jl (kr)Ylm (r̂) √
=
δ(E −
)Y (k̂)
2m l
mk
l,m
=
X (2l + 1)
4π
l
where we use the addition theorem
X
Pk (k̂ · r̂) √
Ylm (r̂)Ylm ∗ k̂) =
m
Turns out that cl =
q
l
i
~
2mk
π
~
cl jl (kr)
mk
2l + 1
Pl (k̂ · r̂)
4π
so that
eik·x =
X
(2l + 1)il jl (kr)Pl (k̂ · r̂)
l
As r → ∞,
eik·x →
X
ei(kr)) − e−i(kr−(lπ))
(2l + 1)
Pl (cos θ)
2ikr
(1.17)
l
1.3.4
Partial wave expansion
Now
+
=
ψ+
=
ψ
=
1
eikr
ikz
) e + f (θ)
r
(2π)3/2
X
1
√
il (2l + 1)Al (r)Pl (cos θ)
8π 3 l
1 X l
√
i (2l + 1)(c1l h1l (r) + c2l h2l (r))Pl (cos θ)
8π 3 l
(1.18)
(1.19)
(1.20)
Remember that for large r,
h1l →
ei(kr−(lπ/2)
,
ikr
h2l →
8
e−i(kr−(lπ/2)
ikr
(1.21)
1.3. PARTIAL WAVE ANALYSIS
becomes, using
ei(kr−(lπ/2)) − e−i(kr−(lπ/2))
2ikr
Anyway, we can write, the general solution to the Schrodinger equation in the partial wave basis,
far from the scattering potential as
jl (kr) → (large r) →
= ψ+ = √
ψelastic
1
X
8π 3
il (2l + 1)
l
ηl ei(kr−(lπ/2)) − cl e−i(kr−(lπ/2))
2ikr
X
ηl ei(kr)) − cl e−i(kr−(lπ))
= ψ+ = √
(2l + 1)
2ikr
8π 3 l
1
Then since
ψelastic −
1
1
eikr
ix·k
e
=
f
(θ)
r
(2π)3/2
(2π)3/2
we know that cl = 1 so that the ingoing wave is the same. Therefore
ψelastic = ψ + = √
1
1 X l
i (2l + 1)(ηl ei(kr−lπ/2) − e−i(kr−lπ/2) )
8π 3 2ikr
l
Probablity conservation requres |ηl | < 1. If |ηl | = 1, the scattering is pure elastic and each partial
wave gets some phase shift. If ηl = 0 the scattering for that partial wave is purely inelastic.
And we know that
ηl − 1
= fl
2ik
#
"
ikr
X
eikr
1
e − e−i(kr−lπ)
+
+ f (θ)
ψ
=
(2l + 1)Pl (cos θ)
2ikr
r
(2π)3/2 ) l
"
#
X
X
1
eikr − e−i(kr−lπ)
eikr
=
(2l + 1)Pl (cos θ)
+
(2l + 1)Pl (cos θ)fl (k)
2ikr
r
(2π)3/2 ) l
l
"
#
ikr X
X
e
e−i(kr−lπ)
1
(2l + 1)Pl (cos θ)(1 + 2ikfl (θ))
+
(2l + 1)Pl (cos θ)
=
2ikr
2ikr
(2π)3/2 )
l
l
Unitarity requires that flux is conserved for each angular momentum state. Outgoing flux is no
more than incoming. Therefore
|1 + 2ikfl (θ)| = |ηl | ≤ 1
(1.22)
and equal to one for elastic scattering.
For elastic scattering we define a phase shift
1 + 2ikfl (θ) = e2iδl
The elastic partial wave amplitude
fl (θ)
=
e2iδl − 1
eiδl sin δl
=
2ik
k
9
(1.23)
1.3. PARTIAL WAVE ANALYSIS
and
f (θ)
=
1X
(2l + 1)eiδl sin δl Pl (cos θ)
k
(1.24)
l
The total cross section is
Z
X
σt =
dΩ
(2l + 1)2 |fl |2 Pl2 (cos θ)
(1.25)
l
=
4π
X
(2l + 1)|fl |2
(1.26)
l
=
π X
π X
(2l + 1)|ηl − 1|2 = 2
(2l + 1)(|ηl |2 + 1 − 2<(ηl ))
2
k
k
l
(1.27)
l
(1.28)
If the scattering is elastic then the total cross section is
Z
X
1
dΩ
(2l + 1)2 sin2 δl Pl2 (cos θ)
σela =
2
k
l
4π X
=
(2l + 1) sin2 δl
k2
l
where ηl = e2iδl . Suppose that there is an inelastic component, so that the magnitude of the
outgoing wave at momentum k in ψscat is less than the magnitude in the incoming plane wave.
Then |ηl | < 1. The inelastic cross section is the piece lost from the outgoing, namely 1 − |ηl |2 .
Therefore
X
(1 − |ηl |2 )
π
σinelastic = 4π
(2l + 1)
= 2 (2l + 1)(1 − |ηl |2 )
(1.29)
2
|2ik|
k
l
And the optical theorem ?
X
ηl − 1
(2l + 1)
Pl (cos θ)
2ik
l
X
ηl − 1
Imf (0) = −
(2l + 1)<
2k
l
X
2π
σtot = − 2
(2l + 1)<(ηl − 1)
k
f (θ)
=
l
The inelastic cross section is the difference of the total and the elastic
σine
= σtot − σelas
2π X
π X
= − 2
(2l + 1)<(ηl − 1) − 2
(2l + 1)(|ηl |2 + 1 − 2<(ηl ))
k
k
l
l
π X
2
=
(2l + 1)(1 − |ηl | )
k2
l
10