Nuclear Physics I (PHY 551)
Joanna Kiryluk
Spring Semester Lectures 2014
Department of Physics and Astronomy, Stony Brook University
Lecture 19:
4.  Nuclear Force
§  Low energy nucleon-nucleon scattering (phase shifts, l>0)
2nd Midterm April 17 (Friday 12pm P119) Lectures 10-18
Homework3 posted on the web page (due 04/22)
Lecture
Nucleon-Nucleon
18
Scattering Phase Shifts l ≠ 0
new
Detector
(lies far away from the scattering
centers, so r →∝ is justified)

k

k!
θ
ikr
e
ψ ( r, θ , ϕ ) #r→∝
##
→ eikz + f (θ )
r
Incident plane wave
Scattered spherical wave
Scattering amplitude
2
dσ
(θ, ϕ ) = f (θ, ϕ )
dΩ
Low energy elastic scattering: k = k !
z
Partial-wave analysis l ≠ 0
Schrodinger equation:

2 2 
−
∇ ψ (r ) + (V − E ) ψ (r ) = 0
2µ
and its solution:

1
ψ = ∑ almψlm (r ) = ∑ alm ul (r)Ylm (θ , ϕ )
r 
l,m
l,m

ψl (r )
d 2 u 2µ "
 2 l(l +1) %
+ 2 $ E −V (r) −
'u = 0
2
2
dr
 #
2µ r &
Boundary condition: ul (0) = 0
Axial symmetry of the problem eliminates dependence on φ (m=0)

u (r)
ψ = ∑ alψl (r ) = ∑ al l Pl (cosθ )
kr
l
l
Legendre polynomials
partial waves
3
Extra
The first few Legendre polynomials
Partial-wave analysis l ≠ 0

ul (r)
ψ = ∑ alψl (r ) = ∑ al
Pl (cosθ )
kr
l
l
partial waves

If V (r ) is spherically symmetric (central force), the angular momentum is conserved (since it
is a constant of motion). States with different l independently contribute to the scattering.
ψincident (z) = Aeikz
∝
∝
describes radial and
angular dependence
eikz = eikr cosθ = ∑ Cl (r)Pl (cosθ ) =∑ ( 2l +1) i l jl (kr)Pl (cosθ )
l=0
l=0
amplitude
phase factor
Pl (cosθ ) − Lagrange polynomials
jl (kr) − Bessel functions
nl (kr) − Neumann functions
Solutions of
Schrodinger equation
(radial)
7
Extra
large distances
Useful expression:
8
Partial-wave analysis l ≠ 0
ψincident (z) = Aeikz
ikz
e =e
∝
ikr cosθ
= ∑ ( 2l +1) i l jl (kr)Pl (cosθ )
l=0
i kr−lπ 2)
Asymptotic form of a plane wave:
sin(kr − lπ 2) e (
jl (kr) #r→∝
##
→
=
kr
−i( kr−lπ 2)
−e
2ikr
& ei(kr−lπ 2) e−i(kr−lπ 2) )
1 ∝
l
e #r→∝
##
→ ∑ ( 2l +1) i Pl (cosθ )(
−
+
2i l=0
kr *
' kr
ikz
incoming wave
outgoing wave
Each partial wave (l ) is a superposition of two spherical waves
Exercise:
A % eikr e−ikr (
l = 0, r →∝ ⇒ ψincident (r) =
−
'
*
2ik & r
r )
10
Partial-wave analysis l ≠ 0
ψincident (z) = Aeikz
& ei(kr−lπ 2) e−i(kr−lπ 2) )
1 ∝
l
e #r→∝
##
→ ∑ ( 2l +1) i Pl (cosθ )(
−
+
2i l=0
kr
kr
'
*
ikz
incoming wave
outgoing wave
Potential perturbation:
$ S (k)ei(kr−lπ 2) e−i(kr−lπ 2) '
1 ∝
l
ψ ~ ∑ ( 2l +1) i Pl (cosθ )& l
−
)
2i l=0
kr
kr
%
(
2iδ
Elastic scattering: Sl (k) = e l
(last lecture)

u (r)
ψ = ∑ alψl (r ) = ∑ al l Pl (cosθ )
kr
l
l
⇒ al = i l (2l +1)eiδl
ul (r) ~ sin(kr − lπ 2 + δl )
eikr
1 ∝
ψ ~ e + f (θ )
⇒ f (θ ) = ∑ ( 2l +1) eiδl sin δl Pl (cosθ )
r
k l=0
lim (δl ) = 0
ikz
l→∝
Homework3
13
Cross Section
l≠0
1 ∝
f (θ ) = ∑ ( 2l +1) eiδl sin δl Pl (cosθ )
k l=0
2
dσ
(θ, ϕ ) = f (θ, ϕ ) = f (θ, ϕ ) f * (θ, ϕ )
dΩ
In our case there is no dependence on ϕ:
2
dσ
θ
,
ϕ
=
f
θ
( ) ( ) = f (θ ) f * (θ )
dΩ
Exercise: Calculate the total elastic scattering cross section:
4π ∝
l ≠ 0 ⇒ σ = 2 ∑ ( 2l +1) sin 2 δl
k l=0
eiδ0 sin δ0
4π sin 2 δ0
l = 0 ⇒ f (θ ) =
and σ =
k
k2
16
The Optical Theorem
1 ∝
f (θ ) = ∑ ( 2l +1) eiδl sin δl Pl (cosθ )
k l=0
4π ∝
σ = 2 ∑ ( 2l +1) sin 2 δl
k l=0
1 ∝
Im f ( 0 ) = ∑ ( 2l +1) sin 2 δl
k l=0
σ=
4π
Im f ( 0 )
k
!!!
Optical theorem connects the total cross section with the scattering
amplitude at scattering θ = 0 .
Note: the optical theorem is valid for elastic and inelastic processes.
18
Low Energy (Elastic) Scattering
1 ∝
Only first few terms f (θ ) = ∑ ( 2l +1) eiδl sin δl Pl (cosθ ) are important.
k l=0
§  Scattering at low energies: only a few of the lower order partial waves
can contribute.
§  For realistic nuclear potentials l is not conserved.
§  A partial-wave expansion remains to be useful as only a limited number
of l-values can be admixed by the non-central potential.
19
Low Energy (Elastic) Scattering
1 ∝
Only first few terms f (θ ) = ∑ ( 2l +1) eiδl sin δl Pl (cosθ ) are important.
k l=0
If E < 20MeV then we can assume:
eiδ0 sin δ0
l = 0 ⇒ f (θ ) =
k
2
4π sin δ0
σ=
k2
If E → 0 the scattering amplitude is finite if δ0 → 0 .
The scattering length (“a”) is a better parameter than a phase shift:
δ0
= −a
E→0 k
lim f (θ ) = lim
E→0
2
" sin δ0 %
2
' = 4π a
limσ = lim 4π $
E→0
E→0
# k &
k = 2µ E /  2
if E → 0 ⇒ k → 0
u0 (r) ~ sin(kr + δ0 ) "E→0
""
→ kr + δ0 ≈ k(r − a) Wavefunction
Straight line
[ul (r) ~ sin(kr − lπ 2 + δl )]
22
Reminder
Scattering Cross Section (p+n)
E ≤ 10keV
V0 = 35MeV (deuteron potential)
k1 = 2µ (E +V0 ) /  2 ≈ 0.92 fm −1
k2 cot(k2 b + δ ) = k1 cot(k1b) ≡ -α
2
cosk2 b + (α k2 ) sin k2 b
k2 = 2µ E /  2 ≤ 0.016 fm −1
sin δ =
b ~ 2 fm
σ=
4π
cosk2 b + (α k2 ) sin k2 b)
2 (
α
σ=
4π
1+ α b) ~ 4.6 [ barn ]
2 (
α
1+ (α k2 )
2
α ≈ 0.2 fm −1
k22 << α 2 and k2 b << 1
σ
~ 6 ×10 −13 cm
limσ = 4π a ⇒ a =
E→0
4π
2
23
energy E
r0
u0 (r) ~ sin(kr + δ0 ) "E→0
""
→ kr + δ0 ≈ k(r − a)
distance r
-V0
[ul (r) ~ sin(kr − lπ
2 + δl )]
Radial wave function u(r) vs distance r
u(r) maximum inside
potential range (bound state)
Negative derivative for r<b
no bound state
E very small
b
b
Attractive potential: The sign of the scattering length “a” shows if
a resonance occurs with a negative bound state or a positive
(virtual) energy
25
Scattering Length – Interpretation
Potential:
§  hard sphere of radius R,
§  infinitely repulsive for r<R
Quantum Mechanics
A comparison with:
gives for the phase shift obtained for the scattering from a hard sphere:
Negative phase shift arises from a repulsive potential.
When the potential is repulsive, the scattering length is positive
(negative phase shift)
Unpolarized Scattering Cross Section (p+n)
Discrepancy between the measured
cross section and calculated one
explained by Wigner, who proposed that
the nuclear force depends on the spin.
3
1
σ = σt + σs
4
4
Very low energies:
Probabilities of p+n being in triplet (spins parallel) or singlet
spin state (spins antiparallel)
3
1
σ = 4π a 2 = σ t + σ s = π (3at2 + as2 )
4
4
where at (as) is the scattering length for the triplet (singlet) potentials.
σ = π (3at2 + as2 ) expresses incoherent combination of singlet and triplet
scatterings (no interference term)
28
Effective Range
If E < 20MeV then we can assume l = 0
4π
σ = 2 sin 2 δ0
- the total cross section
k
(
§
E→0:
)
δ0
= −a
E→0 k
lim f (θ ) = lim
E→0
σ = 4π a 2
The influence of the potential is represented by 1 parameter: the scattering length a.
§
(Optional: derivation see textbook)
E ~ small :
1 1
Effective range (dimension of length, ~ potential range)
kcotδ = − + k 2 reff
a 2
&
4π #
1
4π a 2
σ= 2%
(⇒σ =
2
2
k $ 1+ cot δ0 '
#
&
1
a 2 k 2 + %1− ak 2 reff (
$ 2
'
The influence of the potential is represented by 2 parameters:
(1) the effective range reff and (2) the scattering length a.
29
Effective Range
E ~ small : σ =
Textbook: section 3.8
4π a 2
" 1
%
a k + $1− ak 2 reff '
# 2
&
2
2 2
Note: two potentials (singlet and triplet)
3
σ=
4
4π at2
1
4π as2
+
2
2
4
"
%
"
%
1
1
at2 k 2 + $1− at2 k 2 rt '
as2 k 2 + $1− as2 k 2 rs '
# 2
&
# 2
&
4 parameters: at , rt , as , rs
n+p results:
at = 5.423± 0.005 [ fm ]
as = −23.71± 0.001[ fm ]
n + H 2 scattering
H2
singlet (parahydrogen) or triplet (orthohydrogen)
rt ~ 1.76 [ fm ]
- determined from the deuteron
binding energy
rs ~ 2.56 [ fm ]
- determined from the best fit to the
measured cross sections:
30