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Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 441: Nuclear Physics 1
Chapter V: Scattering Chapter V: Scattering Theory Theory -- ApplicationApplication
This Chapter:1- Introduction - Importance of the scattering theory.2- Cross-Section.
3- Scattering by a Potential.
5- Scattering Stationary States.
6- Probability Currents.
7- Applications. Central PotentialReferences:1. Quantum Mechanics, Vol. 2 Chapter VIIIChapter VIII
C. Cohen-Tannoudji, B. Diu and F. Laloë2. Quantum Physics, Chapter 19
Stephen Gasiorowicz
4- The Optical Theorem.
Introduction
4
Scattering evokes a simple image:
“What is scattering?”
Separate objects which are far apart and moving towards each other collide after some time and then travel away from each other and eventually are far apart again. We don’t necessarily care about the details of the collision except insofar as we can predict from it where and how the objects will end up.This picture of scattering is the first one we physicists learn.
5
A beautiful example of the power of conservation laws
“In many cases the laws of conservation of momentum and energy alone can be used to obtain important results concerning the properties of various mechanical processes. It should be noted that these properties are independent of the particular type of interaction between the particles involved”L.D. Landau Mechanics (1976)
6
An incident beam of particles (1) strikes a target. Two devices (detectors) are placed in order to detect the reaction results.
General Scheme
Fig. 5-1
7
The final state can be composed of new entities. In nuclear physics the outgoing particle b and the Y are the result of a rearrangement of nucleons and the process is called “rearrangement collision”
Elastic and Inelastic Scattering
Another important case is when the final state is composed of the same particles as in the initial state. Here we call the process “elastic scattering”. Otherwise it is called “inelastic scattering”
8
Knowing the initial state of the reaction and measuring the final state gives an idea about the nature of the reaction.
Understanding the reaction
In quantum mechanics language the interaction potential of the process can be understood.
In general, a scattering experiment aims to verify a theoretical model about the reaction.
Importance of the scattering theory
10
Since Rutherford’s explanation of Geiger and Marsden results, physicists understood that more energetic particles than the ααααparticles were necessary to “probe” the nucleus.
Understanding Matter
In addition quantum mechanics is necessary in order to understand what happens when a nucleus is bombarded by such a projectile.
Collision or Scattering Theory is the frame used by Physicists for this purpose.
11
On the experimental level this led to build accelerators more and more capable of providing high energy particles (electrons, protons or alpha’s, heavy ions …………).
Accelerators
On the theoretical level this type of experiments led to major developments in quantum mechanics.
Accelerators
See Supplements 6
13
The first probes used for investigating the nuclear structure were high energy electrons.
Electron Scattering - Hofstadter
Hofstadter used the first SLAC (Stanford Linear accelerator) to investigate the charge distribution in atomic nuclei and afterwards the charge and magnetic moment distributions in the proton and neutron. The electron-scattering method was used to find the size and surface thickness parameters of nuclei.
Electron Scattering and the Structure of the Nucleus
See Supplements 7
15
In 1950 Hofstadter left Princeton to become Associate Professor of Physics at Stanford University where he initiated a program on the scattering of energetic electrons from the linear accelerator, invented by W. W. Hansen, which was then under construction.
Hofstadter
After 1953 electron-scattering measurements became Hofstadter's principal interest. With students and colleagues heinvestigated the charge distribution in atomic nuclei and afterwards the charge and magnetic moment distributions in the proton and neutron. The electron-scattering method was used to find the size and surface thickness parameters of nuclei.
Many of the principal results on the proton and neutron were obtained in the years 1954-1957. Since 1957 emphasis in the research program has been placed on making more precise studies of the nucleon form factors. This work is still in progress.
2- Cross Section
See Supplements 8
Physics DepartmentYarmouk University 21163 Irbid Jordan
Phys. 441: Nuclear Physics 1
Scattering by a Scattering by a PotentialPotential
Reference: Introductory Nuclear PhysicsChapter 3 (pages 54-48)K. S. Kane
19
Simple Mathematical Treatment
One can consider the wave function of the incident (free) particles to be
(((( )))) rkii
ierrrr ••••====ψψψψ (((( ))))ii kp
rh
r ====
Because of the interaction potential with the diffracting (nuclear) potential, the wave function representing the scattered particle can be written as:
(((( )))) rkif
ferrr
r ••••====ψψψψ (((( ))))ff kpr
hr ====
We shall start by a simple mathematical treatment of the problem in quantum mechanics.
1
2
The Form Factor
21
The Form FactorThe transition probability between the two states is defined by:
The form factor is written as a function of the momentum transfer, thus eq. 3 is written as:
23
Consider a differential element of the nucleus charge dQ (Fig. 5-2). Consider also an arbitrary referential system. Let be the vectors representing the position of the electron and the charge dQ respectively. We shall compute here the corresponding potential: Nucleus
rr
r ′′′′rrr ′′′′−−−−rr
dQ
The Interaction Potential
4
rr ′′′′rr&
(((( )))) (((( ))))rr
vdreZrrdV e
′′′′−−−−′′′′′′′′ρρρρ
εεεεππππ−−−−====′′′′ rr
rrrr
04,
2
if ρ ρ ρ ρe is the distribution of nuclear charge then (((( )))) v ′′′′′′′′ρρρρ==== rr
The transition probability is thus related to which is the Fourier Transform of the
density ρρρρ. (((( ))))r
25
The Form Factor – Central Potentialif for simplicity depends only on (not on θθθθ’ or φφφφ’) then:
r ′′′′r(((( ))))re ′′′′ρρρρr
and assuming an elastic scattering, i.e. fi pprr ====
Measuring q (the moment transfer) as a
function of αααα gives F(q) and by an inverse Fourier Transform ρρρρ is calculated
αααα====
2sin2
h
pq 10
(((( )))) (((( )))) rdrrrqq
qF e ′′′′′′′′′′′′ρρρρ′′′′ππππ==== ∫∫∫∫ sin4
9
26
Figure 5-3: Elastic scattering
Moment Transfer - Geometry
(((( ))))[[[[ ]]]]jαpiαpkkq fi
ˆsinˆcos1 ++++−−−−====−−−−====h
rrr
(((( ))))2
sin2cos12 αpαpqq
hh
r ====−−−−========
ippiˆ====
r
ii pkr
h
r 1====
jαpiαpp fˆsinˆcos ++++====
r
ff pkr
h
r 1====
qr
fkr
αααα
qr
ikr
fkr
−−−−
4- The Optical Theorem
28
Two ApproachesTwo approaches are used to study what happens in a scattering process
1) The wave packet approach
2) The scattering stationary states
29
The Wave Packet ApproachThe incident particle is described by a wave packet which should have two principal characteristics:
•It should be spatially large so it does not spread appreciably during the experiment,
•It must be large compared to the target dimensions but small compared to the dimensions of the laboratory where we observe the scattering.
30
The Wave Packet ApproachThe lateral dimensions are determined by the beam size.The outgoing particle is described by a wave packet which “flies” in some direction defined by the scattering angle θθθθ
This is what we have done in the previous paragraph.
31
Here we shall simply state facts you know from your previous study in quantum mechanics.
The solution of the Schrödinger equation in the
absence of a potential, (i.e. for a free classical
particle), is the plane wave form where:rkierr
••••
k is the wavenumber of the associated wave
packet, i.e.
Schrödinger Equation
c
Kcmpk
hh
222 ========λλλλππππ====
K is the kinetic energy of the particle of rest
mass mc2.
c
KcmKpk
hh
22 2++++========For a relativistic particle we have
11
12
32
For a free proton (mc2 = 938.28 MeV) of kinetic
energy K= 50 MeV we have:
The wavenumber kIn the examples below we compute the value
of k for a proton and an e- of kinetic energy
K = 50 MeV.
1247.01240
5028.9382 −−−−====××××××××
==== FFMeV
MeVMeVk
For a free electron (mc2 = 0.511 MeV) of kinetic
energy K= 50 MeV we have:
(((( )))) 12
041.01240
511.050250 −−−−≈≈≈≈××××××××++++==== F
FMeV
MeVMeVMeVk
Fk
44.252 ====ππππ====λλλλ⇒⇒⇒⇒
Fk
1.1572 ≈≈≈≈ππππ====λλλλ⇒⇒⇒⇒
33
The plane wave form describes a flux:Flux
mp
mk
mij
→→→→→→→→→→→→→→→→
========
ψψψψ∇∇∇∇ψψψψ−−−−ψψψψ∇∇∇∇ψψψψ==== hh **2
The classical equivalence of this flux is the number of incident particles per area unit per time unit.
Conservation of energy-matter implies that this flux is conserved.
13
j has the dimension of velocity. It represents the “total” flux, i.e. the number of incident particles per area unit per time unit × × × × volume
34
We usually choose to define the z axis (quantization axis). Thus the large r behavior of this solution can be written as a linear combination of an incoming spherical wave
and an outgoing one, i.e.
→→→→k
The “Outgoing” Wave Packet
(((( ))))(((( )))) (((( ))))
(((( ))))θθθθ
−−−−++++⇒⇒⇒⇒ ∑∑∑∑
∞∞∞∞
====
ππππ−−−−••••ππππ−−−−•••••••• cos12
2 0
22
ll
lrkilrki-rki P
re
rel
kie
rrrrrr
14
35
The following figure (5-4) shows a schematic layout of a scattering experiment.
θθθθ: the scattering angle is the laboratory angle
Experimental Layout - Fluxes
Fig. 5-4
Incident flux ji
Scattered flux jf
36
In the presence of a radial potential the previous solution of Schrödinger equation is simply altered to a function whose asymptotic form is:
Effects of a Radial Potential
(((( )))) (((( ))))(((( ))))
(((( ))))(((( ))))
(((( ))))θθθθ
−−−−++++⇒⇒⇒⇒ψψψψ ∑∑∑∑
∞∞∞∞
====
ππππ−−−−••••ππππ−−−−••••cos12
2 0
22
ll
lrki
l
lrki-
Pr
ekSr
elkir
rrrr
r14’
The asymptotic form may be rewritten in the form:
(((( )))) (((( )))) (((( )))) (((( ))))r
ePki
kSler
rki
ll
lrki ∑∑∑∑∞∞∞∞
====
••••
θθθθ
−−−−++++++++⇒⇒⇒⇒ψψψψ0
cos2
112
rrr15
Exercise: Check Eq. 15
37
It is required for Eq. 14 to be consistent with the mathematical frame of wave mechanics
(See supplement 9) that |Sl(k)| = 1.
The Phase Shift δδδδ
Sl(k) in a standard notation can be written as(((( ))))ki2 le δδδδ
where the real function δδδδ(k) is called the phase shift.It translates the fact that the asymptotic form of the scattered wave differs from that of a free particle by a shift in phase of the argument (Think waves and interference).
38
Eq. 15 represents nothing else but the sum of a plane wave + an outgoing spherical wave!
In the following we shall use the function
with
ψψψψ = An incoming Plane Wave + An outgoing Spherical Wave
(((( )))) (((( )))) (((( )))) (((( ))))∑∑∑∑∞∞∞∞
====θθθθ++++====θθθθ
0
cos12l
ll Pkflf 16
(((( )))) (((( )))) (((( )))) (((( ))))r
ePki
kSler
rki
ll
lrki ∑∑∑∑∞∞∞∞
====
••••
θθθθ
−−−−++++++++⇒⇒⇒⇒ψψψψ0
cos2
112
rrr15
(((( )))) (((( ))))ki
kSkf l
l 21−−−−==== 17
Flux for the Asymptotic Solution
40
The flux for the asymptotic form is:
16
17
Flux for the Asymptotic Solution
ψψψψ∇∇∇∇ψψψψ−−−−ψψψψ∇∇∇∇ψψψψ====→→→→→→→→
**2 mi
j fh
(((( )))) (((( ))))
(((( )))) (((( ))))
θθθθ++++∇∇∇∇
θθθθ++++−−−−
θθθθ++++∇∇∇∇
θθθθ++++
====••••→→→→••••
••••→→→→••••
*
*
2
refe
refe
refe
refe
mij
rkirki
rkirki
rkirki
rkirki
frrrr
rrrr
h
(((( )))) (((( ))))
−−−−
θθθθ++++∇∇∇∇
θθθθ++++==== ••••→→→→•••• cc
refe
refe
mi
rkirki
rkirki
rrrrh*
2
41
We shall skip some steps in order to calculate jf and leave this as a homework.
2)Then calculating the gradient.
1) First using .θθθθ====•••• cosrkrkrr
3) Neglecting the terms in r3 since they are dominated by the 1/r2 for large r.This is largely justified by the fact that we are interested in the flux far away from the interaction region (where the potential is localized), i.e. at large r (see fig. 5-4)
Calculating jf
42
Calculating jf - Steps
4) Finally considering a scattering angle θθθθ#0.
This is also justified by the fact that the detector is always placed at an angle and in a measurement (Rutherford) we want the integrated flux over a small but finite solid angle (Rutherford scattering).5) Finally using Riemann-Lebesgue Lemma when integrating over the small but finite solid angle.
See Supplement 9 and Problem 7 of chapter 19 - in Gasiorowicz
43
jf
The result is the following expression for jf
is the unit vector in the direction of ri→→→→r
18(((( ))))
2
2ˆ
r
fi
mk
mk
j rfθθθθ
++++====→→→→
hh
44
jf - Discussion
18(((( ))))
2
2ˆ
r
fi
mk
mk
j rfθθθθ
++++====→→→→
hh
First remark: it is obvious that in the absence of a “scattering” potential, the 2nd
term on the right in eq. 18 vanishes and we retrieve the expression of the incident flux (eq. 13)
45
Discussion – The First Term1
18(((( ))))
2
2ˆ
r
fi
mk
mk
j rfθθθθ
++++====→→→→
hh
For the radial flux, i.e the first term
gives a contribution . mk
ir
→→→→
•••• hˆ
θθθθcosmkh
In a wave-packet treatment, would be
multiplied by a function which defines the
lateral dimensions of the beam.
mk→→→→
h
46
Discussion – The First Term2
This contribution is only significant in the forward region within a finite region of the z-axis. And since the detector is placed far outside this region we can consider that the first term does not contribute to the radial flux in the asymptotic region.
Asymptotic region
Cross Section
48
Flux → → → → Number of Particles
Translating the asymptotic flux jf into number of outgoing scattered particles (per time unit) we write:
19(((( ))))
ΩΩΩΩθθθθ
≈≈≈≈••••==== drr
f
mk
dAijdN rf2
2
2
ˆ h
Important remark: the factor r2 vanishes and this is, a posteriori, a justification of having neglected the terms in r3.
49
Differential Cross SectionThe differential cross section is defined by:
20
(((( ))))(((( )))) 2
22
2
θθθθ====ΩΩΩΩ
ΩΩΩΩθθθθ
≈≈≈≈ΩΩΩΩ
====ΩΩΩΩσσσσ
fd
mk
drr
f
mk
djdN
dd
i h
h
Compare with the classical definition of the differential cross section in supplement 8.
50
If the potential has a spin dependence, as it is the case for the nucleon-nucleon interaction then there may be an azimuthal dependence. Eq. 20 is generalized in this case and we write:
Spin Dependence
21(((( )))) 2,φφφφθθθθ====
ΩΩΩΩσσσσ
fdd
51
The total cross section of the scattering process is obtained by integrating over all possible values of θθθθ, i.e. of dΩΩΩΩ
Comparing with the expression of the total cross section;
we have:
Taking the imaginary part of eq. 27, we have:
The Optical Theorem2
(((( ))))(((( )))) (((( )))) (((( ))))∑∑∑∑∞∞∞∞
====δδδδ++++====
0
2sin1210Iml
klk
f 28
23(((( )))) (((( ))))∑∑∑∑
∞∞∞∞
====δδδδππππ====σσσσ
0
22 sin
4
lltot k
kk
(((( )))) (((( ))))kkf totσσσσππππ
====4
0Im 29
55
It is true even when inelastic processes can occur, as this may be the case in nuclear reactions.
Equation 29 is known as the optical theorem.
The Optical Theorem3
29(((( )))) (((( ))))kkf totσσσσππππ
====4
0Im
56
In wave language, this removal is the result of destructive interference which occurs in the forward region between the incident and the elastically scattered wave.
The optical theorem appears clearly when using the wave (De Broglie) language since it follows the fact that the total cross section represents the removal of flux from the incident beam..
The Optical Theorem & Wave Mechanics
Finally and for completeness one have to mention that the wave treatment does explain the imaginary part. This is generally true and explained in a more comprehensive quantum scattering theory.
5555----5555 Scattering Stationary Scattering Stationary Scattering Stationary Scattering Stationary States States States States –––– Other ApproachOther ApproachOther ApproachOther Approach5555----5555 Scattering Stationary Scattering Stationary Scattering Stationary Scattering Stationary States States States States –––– Other ApproachOther ApproachOther ApproachOther Approach
End of Lecture 18
Next Lecture
Reference: Paragraphs B and CQuantum Mechanics, Vol. 2 Chapter VIIIC. Cohen-Tannoudji, B. Diu and F. Laloë