Lec 6_Scattering Theory
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Transcript of Lec 6_Scattering Theory
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Scattering Theory
P. Thangadurai
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Scattering and Cross Section
Importance of Scattering
Much of our understanding about the structure of matter is extracted from the scattering of
particles.
Without scattering, the structure of the microphysical world would haveremained inaccessible to humans.
It is through scattering experiments that important building blocks of matter, such as the
atomic nucleus, the nucleons, and the various quarks, have been discovered.
Importance of Scattering
In a scattering experiment, one observes the collisions between a beam of incident particles
and a target material.
The total number of collisions over the duration of the experiment is proportional to the
total number of incident particles and to the number of target particles per unit area in the
path of the beam.
In these experiments, one counts the collision products that come out of the target.After scattering, some will go undisturbed those particles that do not interact with the target
continue
their motion (undisturbed) in the forward direction, but those that interact with the target
get scattered (deflected) at some angle as depicted in Figure 11.1.
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The relationship between d/d and the total cross section is obvious:
--------(B)
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The scattering amplitudef (, ) plays a central role in the theory of scattering, since it
determines the differential cross section.
To see this, let us first introduce the incident and scattered flux densities:
Scattering Amplitude and Differential Cross Section
In this case behaves as afree particle before collision and hence can be described by aplane
wave (k0 and k are the wave vectors of incident and scattered waves respectively)
--------(1)
--------(2)
--------(3)
--------(4)
Inserting (3) into (1),0
0
0
0
.
.*
.
.*
( )
( )
( ) ( )
( ) ( )
ik r
incik r
incik r
incik r
inc
r Ae
r Ae
r ik Ae
r ik Ae
* *2
inc inc inc inc inc
iJ
* *2
sc sc sc sc sc
iJ
0 .( ) ik rinc r Ae .
( ) ( , )i k r
sc
er Af
r
After the scattering has taken place,
the total wave consists of a
superposition of the incident plane
wave (3) and the scattered wave (4):
0
..( ) ( ) ( ) ( , )
ik rik r
inc sc
er r r A e f
r
--------(4a)
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Eqn. (1) becomes
0 0 0 0
* *
. . . .
2 2
0 0
2
0
2 0
2 0
2
( ) ( )2
( ) ( )2
( 2 )
2
inc inc inc inc inc
ik r ik r ik r ik r inc
inc inc
iJ
iJ Ae ik Ae Ae ik Ae
iA ik A ik
iik A
kA
kJ J A
* *.
.*
. .
2
. .*
2
.
2
( ) ( , )
( ) ( , )
( ) ( ) ( , ) ( , )
( ) ( ) ( , ) ( , )
( , ) ( ) ( , )2
sc sc sc sc sc
ik r
sc
ik r
sc
ik r ik r
sc
ik r ik r
sc
ik r ik
sc
iJ
er Af
re
r Afr
e er ik Af Af
r re e
r ik Af Af r r
i e eJ Af ik Af
r
. . . . .
2 2
2 2 2 2 2 2 2 2
2 3 2 3
2 2 2 2
2
( , ) ( , ) ( ) ( , ) ( , )
1 1 1 1( ) ( , ) ( , ) ( ) ( , ) ( , )2
1 1( ) ( , ) ( ) ( , )
2
r ik r ik r ik r ik r
sc
e e e eAf Af ik Af Af
r r r r r
iJ ik A f A f ik A f A fr r r r
iik A f ik A f
r r
2
2 2
2
2 2
2
2 2
2
12 ( , )
21
( , )
( , )sc sc
iikA f
rk
A fr
kJ J A fr
Eqn. (2) becomes
Inserting (4) into (2),
--------(5a)
--------(5b)
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Now, we may recall that the number dN (, ) of particles scattered into an element of
solidangle d in the direction (, )and passing through a surface element dA = r2d
per unittime is given as follows (see Eqn.1):
whereJsc
is the scattered flux (or incident current density); it is equal to the number of incident
particles per area per unit time.
2( , ) scdN J r d --------(6)
Using equantion (5b) in (6), we get
2
2 2 2
2
2 2
( , )
( , )
( , )( , )
sc
dN
J rdk
A f rr
dN kA f
d
--------(7)
Substitute Eq. (7) and Jincfrom fromEq.(5a) in (1)
2 2
2 0
2
0
( , ) 1 ( , )
1( , )
( , )
( , )
inc
d dNd J d
kA f
kA
d k
fd k
--------(8)
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Since the normalization factorA does not contribute to the differential cross section, we
will be taking it equal to one.
For elastic scattering k0 is equal to k; hence (8) reduces to
2( , )( , )
df
d
The problem of determining the differential cross section d/dtherefore reduces to that
of obtaining the scattering amplitudef (, )
--------(9)
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Scattering Amplitude,f( , )Consider elastic scattering of two particles of mass m1 and m2 (spinless, non-relativistic). We
are going to show here that we can obtain the differential cross section for this case in the CM
frame from an asymptotic form of the solution of the Schrdinger equation (Eq. 10).
Let us first focus on the determination off( , ); it can be obtained from the solutions of
Schrodinger Eqn.
22
( ) ( ) ( ) ( )2
r V r r E r m
--------(10)
This eqn is a reduced form of eqn explaining the elastic scattering of two particles of massm1 and m2 (spinless, non-relativistic). Scattering between two particles is thus reduced to
solving this equation (See page 622 of Zettili Book).
Eq (10) can be rewritten as
2 2 2
2 2
2 2( ) ( ) ( ) ( ), wherek r V r r k E
--------(11)
The general solution to this equation consists of a sum of two components: a general
solution to the homogeneous equation:
2 2
homo( ) ( ) 0k r --------(12)
and a particular solution to Eqn. (11). Note that homo(r) is nothing but the incident
plane as given in Eq. (4)
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As for the particular solution to Eq.(11), we can express it in terms ofGreens function. Thus,
the general solution of Eq. (11) is given by
3
2
2( ) ( ) ( ') ( ') ( ') 'incr r G r r V r r d r
--------(13)
is Greens function corresponding to theoperator on the left-hand side of (12).
0 .( ) i k rinc r Ae where and ( ')G r r
Greens function is obtained by solving the point source equation and the Greens function for
the scattered wave (outgoing wave) is given as'
1
( ') 4 '
ik r r e
G r r r r
--------(14)
Inserting this in eqn (13) we get,'
3
2( ) ( ) ( ') ( ') '
2 '
ik r r
inc
er r V r r d r
r r
--------(15)
0
.. ( , ) as r
i k ri k r ee f
r --------(16)
. ' 32, ( ') ( ') '2ik rf e V r r d r
Where, --------(17)
The differential cross sectionis then given by
2 22 . ' 3
2 4, ( ') ( ') '4
i k rdf e V r r d rd
-----(18)
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The First Born Approximation
If the potential V(r) is weak enough, it will distort only slightly the incident plane wave. The
first Born approximation consists then of approximating the scattered wave function (r)
by a plane wave.
This approximation corresponds to the first iteration of (15); that is, (r) is given by:'
3
2( ) ( ) ( ') ( ') '2 '
ik r r
inc inc
e
r r V r r d r r r
--------(19)
Thus, using (17) and (18), we can write the scattering amplitude and the differential cross
section in the first Born approximation as follows:
. ' 3
2
. ' 3
2
, ( ') ( ') '2
, ( ') '2
i k r
inc
i q r
f e V r r d r
f e V r d r
0
0 0
0
. '
0
. ' . ' . ' . '
( - ). ' . '
( ')
Here -
Therefore, ( ') ( ')( ') ( ')
i k r
inc
i k r i k r i k r i k r
i k k r i q r
r e
q k k
e V r e e V r ee V r e V r
------(20)
2 22 . ' 3
2 4, ( ') '
4
i q rdf e V r d r
d
The differential cross section
------(21)
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whereq = k0 kand q is the momentum transfer; k0 andk are the linear
momenta of the incident and scattered particles, respectively.
In elastic scattering, the magnitudes ofk0 = k(Figure 11.6); hence
0
2 2
0 0
2 2 2
2
2 2
-
2 cos
2 cos
2 (1 cos )
2 2sin ( cos 2 1 2sin )2
2 sin2
q q k k
k k kk
k k k
k
k
q k
Modulus ofq ------(22)
------(23)
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If the potential is spherically symmetric, and choosing the z-axis
along q (Figure 11.6), then
( ')V r ( ') ( )V r V r
and therefore the integral part of Eq. (21) becomes
. ' 'cos 'q r qr
. ' 3 'cos ' 2
2
2 'cos '
0 0 0
( ') ' ( ') ' 'sin ' ' '
' ( ') ' sin ' ' '
i q r iqr
iqr
e V r d r e V r r dr d d
r V r dr e d d
2
0
' 2d
'cos '
0
11 ''cos ' '
0 1 1
' '
'
Take cos 'sin ' '
sin ' 'when =0, x = 1 and = , x = -1
'
'1
'1
2 sin( ')'
12sin( ')
'
iqr
iqr xiqr iqr x
iqr iqr
e d
xdx ddx d
ee d e dx
iqr
e eiqr
i qriqr
qrqr
20 0
1 2' ( ') 2sin( ') ' ' ( ')sin( ') ''
r V r qr dr r V r qr dr qr q
= 2 1
2sin( ')'
qrqr
. ' 3
0
4( ') ' ' ( ')sin( ') 'iq re V r d r r V r qr dr
q
Therefore
------(24)
3 2Volume element in 3-D spherical coordinates
sind r r drd d
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Inserting (24) into (20) we obtain
20
4' ( ')sin( ') '
2f r V r qr dr
q
------(25)
Inserting (25) into (21) we obtain
22
2
2 4
0
4 ' ( ')sin( ') 'd f r V r qr drd q
In summary,
The Schrdinger equation (11.30) is solved with first-order Born approximation (where thepotential V(r) is weak enough that the scattered wave function is only slightly different
from the incident plane wave)
The scattering amplitude (Eqn.25)and the differential cross section (Eqn.25) were obtained
for for a spherically symmetric potential.
20
2' ( ')sin( ') 'f r V r qr dr
q
------(26)
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The first Born approximation is valid whenever the wave function (r) is only slightly different
from the incident plane wave
That is, whenever the second term in Eq. (19) is very small compared to the first:
'
3
2( ) ( ) ( ') ( ') '
2 '
ik r r
inc inc
er r V r r d r
r r
--------(19)'
23
2( ') ( ') ' ( )
2 '
ik r r
inc inc
eV r r d r r
r r
0
0 0
.
2. .
Since ( )
( ) 1
i k r
inc
i k r i k r
inc
r e
r e e
'
3
2( ') ( ') ' 1
2 '
ik r r
inc
eV r r d r
r r
In elastic scattering k0 = k and assuming that the scattering
potential is largest near r = 0, we have (following the same
integration procedure followed for Eqn. (24) and putting r =
0),
--------(20)
--------(21)
' 'cos '
2
0 0
' ( ') ' sin ' ' 1ikr ikr
r e V r dr e d
'cos '
0
11 ''cos ' '
0 1 1
' '
'
Take cos 'sin ' '
sin ' 'when =0, x = 1 and = , x = -1
''
1
'
ikr
ikr xikr ikr x
ikr ikr
e d
xdx ddx d
ee d e dx
ikr
e eikr
--------(22)
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Since the energy of the incident particle is proportional to k (it is purely kinetic, )
we infer from (22) that
the Born approximation is valid for large incident energies and weak scattering potentials.
That is, when the average interaction energy between the incident particle and the
scattering potential is much smaller than the particles incident kinetic energy, the scatteredwave can be considered to be a plane wave.
2 2E k
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So far we have considered only an approximate calculation of the differential
cross section where the interaction between the projectile particle and the
scattering potential V(r) is consideredsmall compared with the energy of the
incident particle.
In this section we are going to calculate the cross section without placing any
limitation on the strength ofV(r)
Partial Wave Analysis for Elastic Scattering
We assume here the potential to be spherically symmetric.
(In the special case of a central potential V(r), the orbital angular momentum L of the
particles is a constant of motion. Therefore, there exist stationary states with well
defined angular momentum: that is, eigen states commone to H, L2, Lz. We shall call the
wave functions associated with these states PARTIAL WAVES. )
The angular momentum of the incident particle will therefore be conserved; a particle
scattering from a central potential willhave the same angular momentum before and after
collision.
Assuming that the incident plane wave is in the z-direction and hence
cos
( )ikr
inc r e
--------(22)
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we may express it in terms of a superposition of angular momentum eigenstates, each with a
definite angular momentum number l
. cos
0
(2 1) ( ) (cos )i k r ikr ll l
l
e e i l j kr P
--------(23)
We can then examine how each of the partial waves is distorted by V(r) after the particlescatters from the potential. [jl(kr) is the Bessel Function]
The most general solution of the Schrdinger equation (10) is
Consider the Schrdinger equation in CM frame: (Eq. 10)
22
( ) ( ) ( ) ( )2
r V r r E r m
--------(10)
( ) ( ) ( , )lm kl lmlm
r C R r Y --------(24)
Since V(r ) is central, the system issymmetrical (rotationally invariant)about thez-axis. The scattered wave function must not then depend on the azimuthal
angle ; m =0. Thus, as Yl0 Pl(cos), thescattered wave functionEq.(24)becomes
0
( , ) ( ) (cos )l kl ll
r a R r P
--------(25)
Each term in (25), which is known as apartial wave, is a joint eigenfunction of L2and LZ.
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where Rkl(r ) obeys the followingradial equation
2 2
2 2 2 2
( 1) 2 2( ( )) ( )( ( )) , where
d l l m E k rRkl r V r rRkl r k
dr r
--------(26)
0
..
( ) ( ) ( ) ( , )i k r
i k r
inc sc
er r r A e f
r
A substitution of (23) [for into
--------(4a)
with =0 (and k=k0for elastic scattering) gives
( )inc r
.
0
( , ) ( ) ( ) (2 1) ( ) (cos ) ( )
i k r
linc sc l l
l
er r r i l j kr P f r
--------(27)
The scattered wave function is given, on the one hand, by (25) and, on the other hand, by (27).
Consider the limit r
1) Since in almost all scattering experiments detectors are located at distances from thetarget that are much larger than the size of the target itself.
The limit of the Bessel functionjl(kr) for large values of r is given by
sin( / 2)( ) ( )l
kr lj kr r
kr
--------(28)
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the asymptotic form of (27) is given by
.
0
sin( / 2)( , ) ( ) ( ) (2 1) (cos ) ( )
ik rl
inc sc l
l
kr l er r r i l P f
kr r
--------(29)
one can write (29) as
/2 /2
since, sin( / 2) [( 1) ] / 2
because, ( ) ( )
l ikr l ikr
il i l l
kr l e i e i
e e i
0
2
0 0
[( 1) ] / 2( , ) (2 1) (cos ) ( )
1( , ) (2 1) (cos ) ( ) ( )(2 1) (cos )2 2
l ikr l ikr ikr l
l
likr ikr
l l ll l
l l
e i e i er i l P f
kr r
e er i l P f i i l Pikr r ik
------(30)
2) To find the asymptotic form of (25), we need first to determine the asymptotic form of the
radial function Rkl(r ). At large values of r, the scattering potential is effectivelyzero radial
equation (26) becomes
2
2( ( )) 0
dk rRkl r
dr
------(31)
The general solution of this equation is given by a linear combination of the spherical
Bessel and Neumann functions
--------(29a)
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we can write the asymptotic limit of the scattered wave function (25) as
0
sin( / 2 )( , ) (cos ) ( )ll l
l
kr lr a P r
kr
------(32)
l is called the phase shift of the lth partial wave.
The phase shift lmeasures the distortion ofRkl(r ) from the free solutionjl(kr)
due to the presence of the potential V(r )
This wave function (32) is known as a distorted plane wave, which differs from a
plane wave by the phase shifts l
With the same argument, in (29a), Eq (32) can be rewritten as,
0 0
( , ) (cos ) ( ) (cos )2 2
ikr ikr l i l l i l
l l l l
l l
e er a i e P a i e P
ikr ikr
------(33)Comparing Eqn. (33) with (30)
2
0 0
1( , ) (2 1) (cos ) ( ) ( )(2 1) (cos )
2 2
ikr ikr l l l
l l
l l
e er i l P f i i l P
ikr r ik
------(30)
2 (2 1)
(2 1)
l i l l
ll i l
l
a i e i l
a i l e
------(34)
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Substituting al in (33)
2
0 0
( , ) (2 1) (cos ) (2 1) ( ) (cos )2 2
ikr ikr l i l l i l l i l l
l l
l l
e er i l e i e P i l e i P
ikr ikr
------(35)
Equating the coefficient of in (30) with (35), we have
ikr
er
2
0 0
1 1( ) ( )(2 1) (cos ) (2 1) ( ) (cos )
2 2
l l l l i l
l l
l l
f i i l P l i i e Pik ik
------(36)
From (30) From (33)
2, ( 1) / 2 ( ) / 2(cos sin cos sin ) / 2(2 sin ) / 2sin
i l i l i l i l
i l
i l
i l
Since e i e e e ie l i l l i l i
e i l i
e l
Rearranging (36)
2 2
0 0 0
2
0
1 1( ) ( ) (2 1) ( ) (cos ) ( )(2 1) (cos ) [since ( ) ( ) 1]
2 21
( ) (2 1) (cos )( 1)2
l l i l l l l l l
l l l
l l l
i l
ll
f f l i i e P i i l P i i iik ik
f l P eik
0
1( ) (2 1) sin (cos )i l
l
l
f l e lPk
------(37)
where fl( ) is denoted as thepartial wave amplitude.
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From (37) we obtain the differential cross sections
'2 ( )
' '2
0 ' 0
1(2 1)(2 ' 1) sin sin (cos ) (cos )l l
i
l l l l
l l
df l l e P P
d k
The differential cross section (38) consists of a superposition of terms withdifferent angular momenta; this gives rise to interferencepatterns betweendifferent partial wavescorresponding to different values of l.
The interference terms go away in the total cross section when the integralover is carried out.
Note that when V=0everywhere, all the phase shifts lvanish, and hencethe partial(38) is zero.
------(38)
**********