Download Lectures #18-19 Scattering: short

November 11, 13
Lectures #18-19
Scattering. Differential cross section.
Born approximation. Validity of the Born
approximation.
Solving scattering problems: examples.
Chapter 10, pages 378378-393, Jasprit Singh, Quantum Mechanics
Chapter 13, pages 587587-595, 608608-620, Bransden & Joachain,
Joachain,
Quantum Mechanics
Scattering: short-range potential
θ,φ
Detector
incident particles
mass: m
velocity: v
dΑ
z
origin
dΩ
range of the
potential V(r)
Problem: calculate the probability that the particle is scattered into a
small solid angle dΩ in the direction θ,φ.
1
Differential cross section
Problem: calculate the probability that the particle is scattered
into a small solid angle dΩ in the direction θ,φ.
How? By solving Schrödinger equation ... again.
This probability is expressed in terms of differential cross section
dσ
= number of particles scattered per unit time into d Ω in the
dΩ
direction θ ,φ / F ( flux of the incident particles) d Ω
The total cross section
σ tot = total number of particles scattered per second / F
is obtained as σ tot =
dσ
dΩ =
dΩ
2π
π
dφ dθ sin θ
0
0
dσ
.
dΩ
Note: the flux is a number of particles the unit time per unit area so
the dimensions of both
dσ
and σ tot are those of the area.
dΩ
Schrödinger equation ... again
−
Case 1. Free particles: well outside the
scattering potential range.
Hamiltonian: H0
Case 2. In the presence of the
scattering potential.
Hamiltonian: H=H0+V
(∇
2
2m
∇ 2ψ (r ) = Eψ (r )
2 2
k
1 2
mv =
2
2m
( ∇2 + k 2 )ψ (r) = 0
E=
2
+ k 2 )ψ (r ) =
2m
2
V (r ) ψ (r )
U (r)
(∇
2
+ k 2 )ψ (r ) = U (r )ψ (r )
2
Wave function at large r &
differential cross section
i k ⋅r
ψ (r ) 
r →∞ → e
+
f (θ , φ ) ikr
e
r
Outgoing
spherical wave
Incident beam of particles
(plane wave)
Note: potential must decrease faster than 1/r with r→∞ .
Differential cross section is given by
dσ
2
= f (θ , φ )
dΩ
kin = kout : elastic scattering
How to calculate differential
cross section?
1. Write the expression for the wave function
ψ (r ) .
2. Determine the asymptotic behavior of this wave function for r→∞ .
i k ⋅r
3. Compare it with ψ (r ) 
r →∞ → e
+
f (θ , φ ) ikr
e .
r
4. Determine f (θ , φ ) from this comparison.
5. Calculate differential cross section using
dσ
2
= f (θ , φ ) .
dΩ
Note: this general procedure is used to derive formula for the
differential cross section both using Born approximation and
method of partial waves.
3
Green’s function
(
)
2
2
The equation ∇ + k ψ (r ) = U (r )ψ (r ) can be solved using
Green’s function method:
ψ k (r ) = φk (r ) + G (r, r ') U (r ')ψ k (r ') dr ',
where φ (r ) is the solution of the free particle equation
(∇
2
+ k 2 ) φ (r ) = 0
and G (r, r ') is a Green’s function corresponding to
(∇
2
+ k 2 ) G (r, r ') = δ ( r − r ' ) .
Since we want to obtain the solution which has an asymptotic form
i k ⋅r
ψ (r ) 
r →∞ → e
+
f (θ , φ ) ikr
e
r
φ (r ) = ei k ⋅r .
Green’s function
ψ k (r ) = φk (r ) + G (r, r ') U (r ')ψ k (r ') dr '
ik r −r '
Note: see pages 609-613 of
Bransden & Joachain QM book
for derivation
1 e
G (r, r ') = −
4π r − r '
1
ψ k (r ) = φk (r ) −
4π
e
ik r −r '
r −r'
U (r ')ψ k (r ') dr '
4
Asymptotic behavior of ψ(r) r→∞
r
r − r ' = r 2 + 2r ⋅ r '+ r '2 ≈ r − rˆ ⋅ r '
r'
r − r ' ≈ r − r 'cos(r, r ') = r − r 'cos α
| r − r’|
r’
r
e
ik r −r '
≈ eik ( r −r 'cosα ) = eikr e−ikr 'cosα
= eikr e−i k 'r '
where k ' = k cos α = k r points in the
direction of the scattering particle
α = ∠(r, r ')
i k ⋅r
ψ k (r ) 
−
r →∞ → e
1
4π
e
ik r −r '
r −r'
eikr e−i k 'r '
U (r ')ψ k (r ') dr '
r
Differential cross-section
Matching
i k ⋅r
ψ k (r ) 
−
r →∞ → e
= ei k⋅r +
1
4π
eikr e−i k 'r '
U (r ')ψ k (r ') dr '
r
eikr
1
−
e−i k 'r 'U (r ')ψ k (r ') dr '
r
4π
with
i k ⋅r
ψ (r ) 
r →∞ → e
+
eikr
f (θ , φ )
r
we obtain
f (θ , φ ) = −
1
e−i k 'r 'U (r ')ψ k (r ') dr '
4π
5
Born approximation
f (θ , φ ) = −
1
1
e−i k 'r 'U (r ')ψ k (r ') dr ' = −
φk ' U ψ k
4π
4π
Main idea:
idea use perturbation theory to approximate ψ k (r ) .
The scattering wave function is expanded in
powers of the interaction potential.
ψ k(0) (r ) = φk (r ) = ei k⋅r
Born series:
ψ k(1) (r ) = φk (r ) + G (r, r ') U (r ')ψ k(0) (r ') dr'
ψ k(1) (r ) = φk (r ) + G (r, r ') U (r ')ψ k( n −1) (r ') dr'
Born approximation
ψ k (r ) = φk (r ) + G (r, r ') U (r ')φk (r ') dr'
+ G (r, r ') U (r ')G (r ', r '')U (r '')φk (r '') dr'dr ''+ ...
We substitute this expansion into the formula for the scattering amplitude
f (θ , φ ) = −
f (θ , φ ) = −
1
φk ' U ψ k
4π
1
φk ' U + UGU + UGUGU + ... φk
4π
First term of this series is (first) Born
approximation to the scattering amplitude:
f B (θ , φ ) = −
1
φk ' U φk
4π
6
Born approximation:
just some transformations
1
1 2 m − i k 'r '
e
V (r ') ei k ⋅r 'dr '
φk ' U φk = −
4π
4π 2
m
ei (k −k ')⋅r ' V (r ') dr ' = −
ei q⋅r ' V (r ') dr '
2π 2
f B (θ , φ ) = −
=−
m
2π
2
q = k −k '
( k − k ')
2
θ = ∠(k , k ')
= k 2 + (k ') 2 − 2kk 'cos(k , k ')
= 2k 2 (1 − cos θ )
q = 2k sin
θ
2
Born approximation:
central potentials V(r)
z
q
f B (θ , φ ) = −
r
θ
f B (θ , φ ) = −
x
q = k −k '
q = 2k sin
θ
=−
2
θ = ∠(k , k ')
m
∞
2
∞
m
2π
dr ' ( r ' )
0
=−
2π
2
ei q⋅r ' V (r ') dr '
In the case of central potential V(r) we can
integrate over θ and φ .
We choose axis z to be in the direction of q.
y
φ
m
2
dr ' ( r ' )
0
1
2
2π
2m
dr r V (r ) sin(qr )
q 20
π
dφ ' sinθ ' ei q⋅r 'cosθ ' V (r ')
0
du ei q⋅r 'u = −
-1
∞
2
0
m
∞
2
dr ' ( r ' ) V (r ')
2
0
2sin(qr ')
qr '
u = cos θ '
7
Born approximation: central potential V(r)
How to solve problems: Example 1
Problem: elastic nucleon scattering from heavy nucleus can be
represented by a potential
−V0 r < R
V (r ) =
0 r>R
V0 > 0
Calculate the differential cross section in the lowest order in V.
Solution:
f B (θ , φ ) = −
f B (θ , φ ) = V0
∞
2m
dr r V (r ) sin(qr )
q 20
2
dσ
= f B (θ , φ )
dΩ
2µ
2µ 1
dr r sin(qr ) = V0 2 2 [sin(qR) − qR cos(qR )]
2
q 0
q
q
R
dσ 4 µ 2V02
2
= 6 4 [sin(qR) − qR cos(qR) ]
dΩ
q
q = 2k sin
θ
2
µ : reduced mass
Born approximation: central potential V(r)
How to solve problems: Example 2
Problem: a particle of mass m is scattered by a potential
V (r ) = V0 e− r / a
a>0
1. Calculate the differential cross section in the lowest order in V.
2. Calculate the total cross section.
3. Define the criteria for the validity of Born approximation
B
Solution: 1. f (θ , φ ) = −
∞
2m
dr r {V0e − r / a } sin(qr )
2
q 0
> assume(a>0);
> Q:=simplify(int(r*exp(-r/a)*sin(q*r),r=0...infinity));
> DQ:=((2*m*V/(q*h^2))*Q)^2;
2 2 6
16 m V a~
DQ :=
16 m V a
dσ
= [ DQ] =
4
2
dΩ
(1 + q 2 a 2 )
2
2
0
6
4
2 2
h ( 1 + q a~ )
Q :=
2 q a~
3
2 2
( 1 + q a~ )
2
4
8
Born approximation: central potentials V(r)
How to solve problems: Example 2
Solution: 2. The total cross-section is given by the integral
σ tot =
dσ
dΩ =
dΩ
q = 2k sin
2π
0
π
dφ dθ sin θ
0
θ
π
−4
dσ 16 m 2V02 a 6
=
2π dθ sin θ (1 + 4k 2 a 2 sin(θ / 2) )
2
dΩ
0
> q:=2*k*sin(theta/2);
> DQ;
> R:=2*Pi*int(DQ*sin(theta),theta=0...Pi);
> factor(R);
2
σ tot =
2 2 6
2 2
4 4
64 π m V a~ ( 3 + 12 k a~ + 16 k a~ )
4
2 2
3 h ( 1 + 4 k a~ )
3
Validity of the Born approximation
This method is based on treating the scattering potential as a perturbation.
Therefore, for this approach to be valid, the correction to the wave
(1)
function which is introduced by a potential (our first order correction ∆ψ k (r ) )
must be small in comparison to the wave function in the
(0)
absence of the potential (in our case ψ k (r ) ).
Using this statement as a guide we use the following criteria for the
validity of the Born approximation:
∆ψ k(1) (0)
ψ k(0) (0)
1
9
Validity of the Born approximation
We need to evaluate this expression:
∆ψ k(1) (0)
ψ k(0) (0)
1
ψ k(0) (r ) = ei k ⋅r → ψ k(0) (0) = 1
1
∆ψ (r ) = −
4π
(1)
k
∆ψ k(1) (0) = −
1
4π
e
ik r −r '
r −r'
U (r ')ψ k (r ') dr '
eikr '
U (r ') eik⋅r ' dr '
r'
∆ψ k(1) (0)
m
=
(0)
ψ k (0)
2π 2
eikr '
V (r ') eik⋅r ' dr '
r'
1
Validity of the Born approximation:
central potential
∆ψ k(1) (0)
m
=
(0)
ψ k (0)
2π 2
eikr '
V (r ') eik ⋅r ' dr '
r'
1
Now we consider this condition for the case of the central potential
2π
=
2
m
2π
∞
π
eikr '
m
eikr '
2
V (r ') eik⋅r ' dr ' =
dr
r
d
φ
d
θ
θ
V (r ') eikr 'cosθ '
'
(
')
'
'
sin
'
r'
2π 2 0
r
'
0
0
m
∞
∞
dr ' (r ') 2
2
0
eikr '
2sin(kr ') 2m
V (r ')
= 2 dr ' eikr 'V (r ')sin(kr ')
r'
kr '
k 0
Validity of the Born approximation
condition for the central potential
∞
2m
dr eikrV (r )sin(kr )
k 20
1
10
Validity of the Born approximation
How to solve problems: back to Example 2
Problem: a particle of mass m is scattered by a potential
V (r ) = V0 e− r / a
a>0
3. Define the criteria for the validity of the Born approximation
Solution: 3.
∞
2mV0
dr eikr e− r / a sin(kr )
2
k
0
1
2 m V a~
2
> assume(k>0);
2
2 2
h
1 + 4 k~ a~
> assume(a>0);
> (2*m*V/(k*h^2))*abs(int(exp(I*k*r)*exp(-r/a)*sin(k*r),r=0...infinity));
2m V0 a 2
2
1
1 + 4k 2 a 2
Validity of the Born approximation
How to solve problems: back to Example 2
2m V0 a 2
2
1 + 4k 2 a 2
1
1→
1.Low k limit (slow particles) ka
2
V0
2ma 2
2
k
2.High k limit (fast particles) ka 1 →
V0
ma
Note that the validity of the Born approximation is considerably
2
extended in this case as V0
(ka )
and ka
ma 2
1
(compare with the other condition)
11
Validity of the Born approximation
(1)
(0)
General condition: | ∆ψ | | ψ |
The results derived above may also be obtained for an arbitrary potential if
we take the |V0| to be the average value of the potential and a to be the
range over which the potential is significant.
Case 1. Potential is sufficiently weak or sufficiently localized (or the particle
speed is slow enough). ik r −r '
≈1
e
| ∆ψ (1) | ≤
m
2π
2
m
a2 m
| V (r ') | (0)
(0)
| ψ (r ') | dr ' ≈
|
V
|
|
ψ
|
4
π
= 2 | V0 | | ψ (0) | a 2
0
2π 2
2
r −r'
∆ψ (1)
ψ (0)
≈
m
2
1→
| V0 | a 2
2
| V0 |
ma 2
2
Case 2. Fast particles ka
1
V0
k
v
.
=
ma m
12