chapter 10 scattering by a central potentialphillips/phys735/afnanscattering.pdf · chapter 10...

Chapter 10

Scattering by a Central Potential

Abstract : With the central role played by scattering in any measurements on aquantum system, in this chapter we introduce the concepts in scattering theorywith application to the simple square well problem as well as the Coulombproblem.

Most scattering experiments consist of a beam of particles incident on a stationarytarget, and a detector to measure the number of scattered particles in a given directionper unit time. In practice, the beam has many particles in it, but we will neglect anyinteraction between the particles. Similarly, the target has many particles in a confinedregion of space and again we will assume the interaction between the target particles canbe ignored. Although we should be considering a wave packet incident on the target, itcan be shown1 that we can get the same result for the scattering amplitude assuming theincident beam is described by a plane wave. In this way we reduce the complexity ofthe algebra in the formulation of the scattering problem. The plane incident wave withmomentum !ki is then given, up to a normalization, by

"inc(!r) ! ei!ki·!r . (10.1)

On the other hand, the scattered particles will be originating at the target and moving ina spherical wave. The amplitude of this scattered wave might depend on the scatteringangles. However, because of the cylindrical symmetry about the incident beam, the onlyangle dependence is the angle between the incident beam direction, ki, and the scatteredparticle’s direction of momentum kf . Here, the subscripts i and f refer to the initial andfinal states. We now can write the scattered wave in terms of a spherical wave, eikr/r, as

"scat ! f(ki, ki · kf )eikf r

r. (10.2)

1For a full treatment of scattering using wave packets consult Goldberger and Watson [28].

173

174 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

For elastic scattering, i.e. when the initial energy of the incident particle is equal to thefinal energy of the scattered particle, we have that |!ki| = |!kf | = k,2 and we can write thetime independent wave function outside the interaction region as

"(!r) ! ei!ki·!r + f(k, #)eikr

r, (10.3)

where cos # = ki · kf and f(k, #) is the amplitude of the scattered wave or the scatteringamplitude. This wave function is illustrated in Figure 10.1.

r

e i k . re i k r

Figure 10.1: Illustration of the asymptotic wave function for a scattering problem. It con-sists of an incident plane wave and an outgoing spherical wave to represent the scatteredparticles.

In the next two sections we will relate the probability for scattering in a given directionto the coe!cient of the outgoing spherical wave f(k, #). We will then relate the amplitudef(k, #) to the solution of the Schrodinger equation. In this way we establish the relationbetween the wave function and the quantities we measure in any scattering experiment.In the final section of this chapter we will consider the problem of scattering by a Coulombpotential, which will require special treatment because the interaction in this case has aninfinite range.

10.1 The Cross Section

In any scattering experiment, we measure the number of particles scattered at a givenangle #. If we divide the number of scattered particles per unit time by the flux of incidentparticles, we get the probability for a particle scattering at a given angle. This is referred

2This definition of elastic scattering assumes we are in the center of mass where !ki and !kf are therelative initial and final momentum (see Sec. 2.2).

10.1. THE CROSS SECTION 175

to as the di"erential cross section for that angle. In this section we derive the relationbetween the cross section and the scattering amplitude f(k, #).

The time dependent Schrodinger equation for the potential V (r) is given by

ih$#

$t=

!

" h2

2µ#2 + V (r)

"

#(!r, t) , (10.4)

where µ is the reduced mass (see Eq. (8.12)) of the incident and target particles. Thecomplex conjugate solution, #!(!r, t) satisfies the equation

" ih$#!

$t=

!

" h2

2µ#2 + V (r)

"

#!(!r, t) . (10.5)

Multiplying Eq. (10.4) by #! from the left, and Eq. (10.5) by # from the right andsubtracting, we get

ih$

$t(#!#) = " h2

2µ

##!#2#" (#2#!)#

$

= "# · h2

2µ( #!##" (##!)# ) . (10.6)

If we now define the density, % = #!#, then Eq. (10.6) can be written as

$%

$t+# ·!j = 0 , (10.7)

where the current !j is given by

!j =h

2µi(#!##" (##!)#)

=h

2µi("!#" " (#"!)") . (10.8)

Here, we have dropped the time dependence in the wave function since we are not dealingwith wave packets. In any scattering experiment the scattered particle is measured at adistance r which is large compared to the range of the interaction. This means that thedetector is in the asymptotic region where the wave function is given by

"(!r) = ei!ki·!r + f(k, #)eikr

r. (10.9)

Using Eq. (10.9) in Eq, (10.8), we get, after some algebra that involves writing # in


spherical polar coordinates,3 an expression of the form

!j =h!ki

µ+

hk

µ

er

r2|f(k, #)|2

+eikr(1"cos ") [· · ·] + e"ikr(1"cos ") [· · ·] . (10.10)

Since the detector is never put in the forward direction, i.e. # = 0, and it subtends afinite solid angle, we need to integrate over the angular range of the detector, i.e.,

%d# d& sin # g(#, &) e±ikr(1"cos ") ,

where g(#, &) is a smooth function that depends on the detector’s properties. For kr $ 1,we are integrating a highly oscillatory function, and the integral is zero according to theRiemann-Labegue lemma. We thus have for the current

!j =h!ki

µ+

h k

µ

er

r2|f(k, #)|2 . (10.11)

The first term represents the incident beam of particles, and is present even if there wasno scattering target. The radial flux of scattered particles is then given by

!j · er =hk

µ

|f(k, #)|r2

. (10.12)

D

θ

d A = r 2 d Ω

k i

Figure 10.2: Illustration of the scattering angle #, and the solid angle d$, subtended bythe detector D.

The number of particles crossing the area that subtends a solid angle d$ is given by(see Figure 10.2)

!j · er dA =hk

µ|f(k, #)|2 d$ . (10.13)

3In spherical polar coordinates we have

# = er"

"r+ e!

1r

"

"#+ e"

1r sin #

"

"$.

10.2. KINEMATICS 177

The di"erential cross section is the flux of scattered particles per unit area divided by theincident flux, hk/µ. Therefore we have that

d' = |f(k, #)|2 d$

and we haved'

d$= |f(k, #)|2 . (10.14)

This in fact is the result we would have expected from the construction of Eq. (10.3) forthe asymptotic wave function. We now have to relate the scattering amplitude f(k, #) tothe solution of the Schrodinger equation. One important observation we can make at thisstage is that the cross section measured experimentally depends on the form of the wavefunction for large r, i.e., outside the interaction region.

10.2 Kinematics

Consider the scattering of two particles where the potential between the particles is afunction of the relative distance between the particles, i.e. V (|!r1"!r2|). The Hamiltonianfor such a system is given by

H =p2

1

2m1+

p22

2m2+ V (|!r1 " !r2|) , (10.15)

where p21 = "h2#2

1 and p22 = "h2#2

2. We introduce relative and center of mass coordinatesand momenta (see Eqs. (8.6) and (8.7)), i.e.

!r = !r1 " !r2 and !R =m1!r1 + m2!r2

m1 + m2

(10.16)

!p =m2!p1 "m1!p2

m1 + m2and !P = !p1 + !p2 .

In terms of these new variables the Hamiltonian takes the form

H =P 2

2M+

p2

2µ+ V (r) , (10.17)

whereM = m1 + m2 and µ =

m1m2

m1 + m2. (10.18)

Here, µ is the reduced mass, and M the total mass. In the center of mass we have !P = 0,and the two-particle Hamiltonian reduces to the one-particle Hamiltonian with a reducedmass µ, i.e.

H =p2

2µ+ V (r) . (10.19)


We thus have reduced the two-particle problem to a one particle problem of mass µscattering from a potential V (r). The Schrodinger equation in the two-particle center ofmass is !

" h2

2µ#2 + V (r)

"

"(!r) = E"(!r) . (10.20)

10.3 The Square Well Potential

Before we proceed to a general discussion of the solution of Eq. (10.20), let us considerthe solution of Eq. (10.20) for the case of a simple square well of radius a, i.e.,

V (r) =

&'(

')

"V0 for r < a

0 for r > a. (10.21)

In the last chapter, we considered the bound state (i.e. E < 0) problem for this potential.We now have to consider the case of E > 0, i.e. the scattering problem. For r < a, thesolution of the radial equation is the same as for the bound state (see Eq. (9.19)), whichis

R#(r) = A j#((r) with (2 =2µ

h2 (E + V0) for r < a . (10.22)

For r > a, we have a linear combination of the two solutions of the radial equation, i.e.,

R#(r) = B j#(kr) + C n#(kr) , (10.23)

orR#(r) = B# h(+)

# (kr) + C # h(")# (kr) , (10.24)

where k2 = 2µE/h2. Since we are not considering bound states, the wave function R#(r)does not go to zero as r %&, otherwise, the current at the detector would be zero. Thismeans neither B nor C (B# or C #) is zero. In this case either combination is valid and wewill use both at di"erent times.

To get a feeling for the form of the scattering wave function, let us take ) = 0. In thiscase we have that

R0(r) =u0(r)

r, (10.25)

with

u0(r) =

&'(

')

A sin (r for r < a

B sin kr + C cos kr for r > a. (10.26)

An alternative way of writing the solution for r > a is

u0(r) = A# sin(kr + *) for r > a . (10.27)

10.3. THE SQUARE WELL POTENTIAL 179

ψ ( r )

r

V = 0

V ≠ 0

Figure 10.3: A comparison the radial scattering wave function in the presence and absenceof a potential V .

Note that both forms for the solution for r > a have two constants to be adjustedby the boundary condition which requires that the wave function and its derivative becontinuous at r = a. Before we determine these constants, let us examine the form of thewave function for the case of V0 = 0, and V0 '= 0 at large distances, i.e. for r > a. InFigure 10.3, we sketch both wave functions. We observe that for V0 '= 0 the wave lengthfor r < a is smaller than is the case for r > a. However, for V0 = 0, the wave length isthe same for all r. Thus the presence of the potential, shifts the wave function for r > arelative to the wave function for the case V0 = 0 by an amount *. In other words, theconstant * introduced in Eq. (10.27) depends on the parameters of the potential, in thiscase V0 and a. The inverse might also be possible, i.e., if we measure * we might be ableto determine the parameters of the potential. The * introduced in Eq. (10.27) is calledthe scattering Phase Shift.

To guarantee that the wave function and its derivative are continuous at r = a, wetake

A sin (a = A# sin(ka + *)

(A cos (a = kA# cos(ka + *) .

Therefore, to determine *, we take the ratio of the above two equations. This gives usthe result that

( cot (a = k cot(ka + *)

= kcot ka cot * " 1

cot * + cot ka. (10.28)

Solving this equation for cot * we get

k cot * =( cot (a + k tan ka

1" $k cot (a tan ka

. (10.29)


Having determined the phase shift * we can determine one of the other two constants(A or A#). This leaves one overall multiplicative constant to be determined. For boundstates, this constant was determined by the requirement that the wave function be nor-malized. However, the scattering wave function is not normalizable in the same manner asthe bound state, because the normalization integral is mathematically not well defined.4

This is a consequence of the fact that the wave function does not go to zero as r %&.

θk iz

Figure 10.4: The scattering plane in the two-body center of mass. The incident beam isalong the positive z-axis.

10.4 The Scattering Amplitude

Having considered the simple problem of S-wave (i.e. ) = 0) scattering by a square well,we now turn to the more general problem of two-particle scattering. In the two-bodycenter of mass, we have two particles with opposite momenta, initially along the z-axis(see Figure 10.4). Since the scattering takes place in a plane, we can eliminate the &dependence in this problem. Furthermore, the initial momentum defines a direction inspace, thus braking the isotropy of the space. This implies that the angular momentumof the system is not fixed as was the case for the bound state problem, but dependson the momentum in the initial state. In fact, the angular momentum, classically, isperpendicular to the scattering plane. In general, the wave function is a linear combinationof many such angular momenta. As we have chosen our z-axis to be along the directionof the incident momentum and therefore in the scattering plane, we expect the projectionof our angular momentum along the z-axis to be zero. This means that the wave functionfor a given angular momentum is of the form

R#(r) Y#0(#&) . (10.30)

But

Y#0(#, &) =

*2) + 1

4+P#(cos #) , (10.31)

4We will show in a later chapter that the scattering wave function has a %-function normalization.

10.4. THE SCATTERING AMPLITUDE 181

where cos # = r · ki. We now can write the general form of the wave function as

"(!r) =1

(2+)3/2

+

#

i#(2) + 1) "#(r) P#(cos #) . (10.32)

The choice of the factor of (2+)"3/2 i# (2)+1) is for later convenience in the normalization.At this stage we would like to point out that this choice for the normalization is not unique.

Since P#(cos #) is related to the spherical harmonics, it is an eigenstate of the angularmomentum operator square, L2. Making use of the orthogonality of the P#, we can showthat the radial Schrodinger equation for "#(r) is given by

d2"#

dr2+

2

r

d"#

dr" )() + 1)

r2"# + [k2 " U(r)]"#(r) = 0 , (10.33)

where

k2 =2µE

h2 and U(r) =2µ

h2 V (r) . (10.34)

We now assume that the potential satisfies the condition that; U(r)% 0 faster than r"1.In this case the radial equation for large r is given by

d2"#

dr2+

2

r

d"#

dr" )() + 1)

r2"#(r) + k2"#(r) = 0 (10.35)

which is the spherical Bessel’s equation.

Note: Our assumption that U(r)% 0 faster than r"1 excludes the Coulombpotential. From this point on we restrict ourselves to the class of potentialsthat satisfy the above condition. We will examine the Coulomb potential asa special case at the end of this chapter.

Before we write the general solution to Eq. (10.33), let us rewrite the solution to thesquare well potential in terms of the spherical Bessel function. We have from Eq. (10.25)and (10.27) that, for ) = 0,

"0(r) =A#

rsin(kr + *) =

A#

r[ sin kr cos * + cos kr sin * ]

= kA# [ cos * j0(kr) + sin * n0(kr) ] . (10.36)

This gives us the idea of writing the general solution to Eq. (10.35), for any angularmomentum, ), as

"#(r)% A# [ cos *# j#(kr) + sin *# n#(kr) ] for r %& . (10.37)


Since for r %& we have that (see Eq. (9.12))

j#(%) % 1

%sin

!

%" +)

2

"

(10.38)

n#(%) % 1

%cos

!

%" +)

2

"

,

then

"#(r)%A#

krsin

!

kr " +)

2+ *#

"

for r %& . (10.39)

Here, A# and *# are the two constants to be determined by the normalization of the wavefunction and the continuity of the logarithmic derivative.5 We now would like to makeuse of Eqs. (10.39) and (10.32) to write the total asymptotic wave function in the formgiven by Eq. (10.3). In this way we hope to relate the scattering amplitude f(k, #) tothe phase shifts *#. To achieve this result we first write sin(kr " +)/2 + *#) in terms ofexponentials, to get

"#(r) % A#

kr

1

2i

,ei(kr"!"

2 +%") " e"i(kr"!"2 +%")

-

=A#

2ikre"i%"

,"e"i(kr"!"

2 ) + e2i%" ei(kr"!"2 )

-

=A#

2ikre"i%"

.

2i sin

!

kr " +)

2

"

+#e2i%" " 1

$ei(kr"!"

2 )/

. (10.40)

Using Eq. (10.39) we get the asymptotic radial wave function for a given ) to be of theform

"#(r)% A#e"i%"

!

j#(kr) + ("i)# 1

2ik

#e2i%" " 1

$ eikr

r

"

. (10.41)

Using this result in Eq. (10.32), and the fact that A# = ei%" , we get the scattering wavefunction for r %& to be

"(!r)% 1

(2+)3/2

!

ei!k·!r + f(k, #)eikr

r

"

, (10.42)

5The logarithmic derivative is given by

d

drlog &#(r) =

1&#(r)

d&#(r)dr

which is identical to the condition given in Eq. (10.28) for the square well with ' = 0.

10.4. THE SCATTERING AMPLITUDE 183

whereei!k·!r =

+

#

i# (2) + 1) j#(kr) P#(cos #) , (10.43)

and

f(k, #) =1

k

+

#

(2) + 1)

!e2i%" " 1

2i

"

P#(cos #) , (10.44)

where cos # = k · r. Now since the direction of the final momentum !kf is the radial

direction, i.e., r = kf , we have cos # = ki · kf , with !ki being the initial momentum of theparticles in the beam.

In Eq. (10.44), we have established the relation between the scattering amplitude andthe phase shifts. In this way we have completed the relation between the experimentallymeasured cross section and the wave function which is a solution to the Schrodingerequation. Finally, we can write Eq. (10.44) as

f(k, #) =1

k

+

#

(2) + 1) f#(k) P#(cos #) , (10.45)

where f#(k), the partial wave amplitude, is given by

f#(k) =1

2i

#e2i%" " 1

$= ei%" sin *# . (10.46)

In general, this partial wave amplitude is a complex number, i.e. it has a magnitude anda phase. With the help of Eq. (10.14) we can write the di"erential cross section in termsof the partial wave amplitude as

d'

d$=

1

k2

000+

#

(2) + 1) f#(k) P#(cos #)0002

. (10.47)

This gives the probability for an incident particle, with momentum k and energy E = k2

2µ ,to be scattered in the direction defined by the angle #. To get the total cross section, i.e.the probability of scattering in any direction, we have to integrate the di"erential crosssection over the 4+ solid angle, i.e.

'T =% !

d'

d$

"

d$

=1

k2

+

#

+

#!(2) + 1)(2)# + 1) f#(k)f !

#!(k)%

d$ P#(cos #)P#!(cos #) .

Using the orthogonality of the Legendre polynomials (i.e. Eq. (8.60)), we get

'T =4+

k2

+

#

(2) + 1) |f#(k)|2 =4+

k2

+

#

(2) + 1) sin2 *# . (10.48)


Comparing Eqs. (10.47) and (10.48), we observe that the di"erential cross section has moreinformation about the scattering amplitude than the total cross section. To illustrate this,consider the case when only two partial waves are important, the ) = 0 and 1. In thiscase we have for the di"erential cross section

d'

d$=

1

k2

#|f0|2 + 9|f1|2 cos2 # + 3(f0f

!1 + f !

0 f1) cos #$

, (10.49)

while for the total cross section we have

'T =4+

k2

#|f0|2 + 3|f1|2

$. (10.50)

Thus we see that the di"erential cross section will give us information on the relativephase of the ) = 0 and ) = 1 amplitudes. Here, we note that the in the event of S-wavescattering only, the di"erential cross section is angle independent, i.e. the di"erentialcross section is isotropic.

10.5 The Optical Theorem

This theorem, is a special case of a more general theorem that sets a nonlinear constrainton the scattering amplitude called unitarity. In particular it gives a relation between thescattering amplitude in the forward direction, f(k, # = 0), and the total cross section,'T . It results from the condition that probability should be conserved in the scatteringprocess.

From Eq. (10.45), we have that the forward scattering amplitude is given by

f(k, # = 0) =1

k

+

#

(2) + 1) f#(k) P#(1)

=1

k

+

#

(2) + 1) f#(k) (10.51)

since P#(1) = 1. Taking the imaginary part of this equation we get

Im [f(k, 0)] =1

k

+

#

(2) + 1) Im [f#(k)] . (10.52)

But we have from Eq. (10.46) that for real phase shifts

Im [f#(k)] = sin2 *# . (10.53)

Therefore, we can write

Im [f(k, 0)] =1

k

+

#

(2) + 1) sin2 *# . (10.54)

10.6. THE PHASE SHIFTS FOR TWO-BODY SCATTERING 185

Making use of Eq. (10.48), we can write

Im [f(k, 0)] =k

4+'T . (10.55)

This result is commonly known as the Optical Theorem, and relates the forward scatteringamplitude to the total cross section. Considering the fact that the total cross section isproportional to the scattering amplitude squared, then Eq. (10.55) imposes a non-linearconstraint on the scattering amplitude, f(k, #).

To illustrate the relation between unitarity and the optical theorem, we recall fromEq. (10.46) that the partial wave scattering amplitude is given by

f#(k) =1

2i

#e2i%" " 1

$

( 1

2i(S#(k)" 1) , (10.56)

where S#(k) is the partial wave S-matrix element. Solving Eq. (10.52) for S#(k), we get

S#(k) = 1 + 2if#(k) . (10.57)

Unitarity is the result of the fact that the S-matrix is unitary, i.e.,

S† S = I , (10.58)

which is obviously the case for S#(k) if the the phase shifts *# are real. Also, the unitarityof S#(k) gives us the result

S†# (k) S#(k) = 1 , (10.59)

orIm f#(k) = |f#(k)|2 . (10.60)

This non-linear relation for the partial wave scattering amplitude is identical to the opticaltheorem. We will see in a later chapter on formal scattering theory, Chapter 14, that thisresult can be derived from the Schrodinger equation directly.

10.6 The Phase Shifts for Two-Body Scattering

So far we have determined the cross section in terms of the scattering amplitude or thephase shifts. However, the phase shifts are constants used in writing the asymptoticform of the wave function. In the case of a square well potential these phase shifts weredetermined by matching the logarithmic derivative of the radial wave function at the wellradius r = a. In general, we can follow the same procedure and integrate the di"erentialequation from the origin to a large enough radial distance r, which is larger than the


range of the potential, and then match the logarithmic derivative, as calculated fromthe asymptotic solution and the solution we get by integrating the radial Schrodingerequation. This is achieved by taking r0 to be such that r0 $ a, where a is the rangeof the potential, and solving the radial Schrodinger equation for "#(r) for r < r0 andcalculating the logarithmic derivative of "#(r), i.e.,

, =d

drlog "#(r)

000r=r0"&

, (10.61)

where - is infinitesimal. On the other hand, for r > r0, we have the asymptotic solution

"#(r) = cos *# ( j#(kr) + tan *# n#(kr) )

and its derivatived"#

dr= k cos *# ( j##(kr) + tan *# n#

#(kr) )

where j## and n## are the derivative of the spherical Bessel and Neumann functions. We

now can write the logarithmic derivative for r > r0 as

d

drlog "#(r)

000r=r0+&

= kj##(kr0) + tan *# n#

#(kr0)

j#(kr0) + tan *# n#(kr0)

= , . (10.62)

This equation can be solved for the phase shift, or tan *#, to give

tan *# =k j##(kr0)" , j#(kr0)

, n#(kr0)" k n##(kr0)

. (10.63)

Here, we observe that given ,, we can determine the phase shift *#.

Note: Here , is a function of r0, and r0 should be chosen large enough sothat *# or tan *# is independent of r0.

From the above results we can study the behavior of *# at low energies, i.e. k % 0. Forkr0 ) ), we can write the spherical Bessel and Neumann functions and their derivativeas

j#(kr0)%(kr0)#

(2) + 1)!!and n#(kr0)%

(2)" 1)!!

(kr0)#+1

(10.64)

j##(kr0)% ")(kr0)#"1

(2) + 1)!!and n#

#(kr0)%() + 1)(2)" 1)!!

(kr0)#+2.

We then can write the phase shift for kr0 ) ) as

tan *# %(kr0)2#+1

(2) + 1)!! (2)" 1)!!

)" ,r0

) + 1 + ,r0, (10.65)

10.7. COULOMB SCATTERING 187

or

tan *# ! k2#+1 for kr0 ) ) . (10.66)

From this result we may deduce that for small wave number k, i.e. low energy, sin *# !k2#+1. If we now write the cross section as

'T =4+

k2

+

#

(2) + 1) sin2 *# (+

#

'# , (10.67)

then the partial wave cross section, '#, is given by

'# =4+

k2(2) + 1) sin2 *#

% (const.) k4# for kr0 ) ) . (10.68)

From this result we may conclude that for k % 0 we have

'# =

1const. for ) = 00 for ) '= 0

. (10.69)

Thus at low energies, we expect the ) = 0 partial wave to dominate the cross section.

10.7 Coulomb Scattering

The analysis in this chapter has so far been restricted to finite range potentials. Thisexcludes the Coulomb potential which is considered to be infinite in range. Because ofthe central role played by the Coulomb potential in both atomic and molecular physics,and the fact that all accelerators produce beams of charged particles which are scatteredby targets often made of charged particles (e.g. nuclei), a discussion of scattering theorythat excludes the Coulomb problem is incomplete. The aim of this section is to derivethe amplitude for the scattering of two charged particles, and from that, extract theRutherford cross section.

Consider for the present, the scattering of two charged particles with charges Ze andZ #e#, with the Coulomb potential between the particles given by

V (!r) =ZZ #ee#

r. (10.70)

The Schrodinger equation for this potential is then given by

!

" h2

2µ#2 +

ZZ #ee#

r

"

"(!r ) = E "(!r ) , (10.71)


where µ is the reduced mass of the two particles. We can rewrite this equation, aftermultiplication by "2µ

h2 , as!

#2 + k2 " 2,k

r

"

"(!r ) = 0 , (10.72)

where

k2 =2µE

h2 and , =µZZ #ee#

h2k. (10.73)

We are going to consider the solution of this equation for E > 0, i.e., we would like toderive the scattering wave function "(!r ), and from its asymptotic behavior extract thescattering amplitude.

For any scattering experiment, we can define the direction of the incident beam to bethe z-axis. The presence of this preferred direction in space breaks the symmetry (thatspace is isotropic) and it gives us a problem which has cylindrical symmetry. As a resultof this new symmetry we expect the wave function and the scattering amplitude to beindependent of the angle &, as illustrated in Eq. (10.32). Furthermore, the wave functionasymptotically should have an incident beam given up to a normalization by

eikz ,

and a scattered beam that is proportional to a spherical out going wave, i.e.,

eikr

r.

This suggests that the solution to Eq. (10.72) can be written as

"(!r ) = eikz g(r " z) ,

and excludes the possibility of having a function of the form

"(!r) = eikz g(r + z) ,

since the latter leads to an incoming spherical wave given by

e"ikr

r.

The fact that the boundary condition on this scattering problem requires the total wavefunction to be a function of z and r"z, suggests that the optimum choice for a coordinatesystem for solving the Schrodinger equation is the parabolic coordinate given by

. = r " z = r(1" cos #)

/ = r + z = r(1 + cos #) (10.74)

& = & .

10.7. COULOMB SCATTERING 189

In this new coordinate system, our total scattering wave function is a product of anincident plane wave and a function of ., i.e.,

"(!r) = eik('"()/2 g(.) . (10.75)

The Laplacian, #2, in this coordinate system is given by

#2 =4

. + /

1$

$.

!

.$

$.

"

+$

$/

!

/$

$/

" 2

+1

./

$2

$&2. (10.76)

We now can write the Schrodinger equation in this coordinate system as a partial di"er-ential equation in two variables, since we have no & dependence. Furthermore, for thewave function with the structure given in Eq. (10.75), we have

#2"(!r) =4 eik('"xi)/2

/ + .

.

.d2g

d.2+ (1" ik.)

dg

d." k2

4(/ + .) g

/

. (10.77)

This allows us to rewrite Eq. (10.72) for g(.) as

.d2g

d.2+ (1" ik.)

dg

d." ,kg(.) = 0 , (10.78)

which is the Confluent Hypergeometric equation, with the solution given by

g(.) = F ("i,|1|ik.) , (10.79)

where the Confluent Hypergeometric function F ((|0|z) is given in Eq. (8.35) as an infiniteseries. We thus can write the total wave function for the scattering of two charged particlesas

"(!r) = eikz g(r " z)

= eikz F ("i,|1|ik(r " z)) . (10.80)

To get the asymptotic form of this wave function, and thus determine the scatteringamplitude, we need to know the behavior of F ((|0|%) for large %. We have that6

F ((|0|%)% %(0)

1("%)"$

%(0 " ()+

%$") e*

%(()

2

as |%|%& . (10.81)

With this result for the asymptotic behavior of the Confluent Hypergeometric equation,we can write the total scattering wave function for r %& as

"(!r) = eikz F ("i,|1|ik.)

% e+,/2

%(1 + i,)

1

ei(kz+, log k(r"z)) +%(1 + i,)

i%("i,)

e"i, log sin2 "/2

2k sin2 #/2

ei(kr", log 2kr

r

2

( e+,/2

%(1 + i,)

.

ei(kz+, log k(r"z)) + f(k, #)ei(kr", log 2kr)

r

/

, (10.82)

6See Abramowitz and Stegun [24] Eq. 13.5.1


where the scattering amplitude f(k, #) is given by 7

f(k, #) = ",%(1 + i,)

%(1" i,)

e"i, log sin2 "/2

2k sin2 #/2

= " ,

2k sin2 #/2ei('0", log sin2 "/2) , (10.83)

where/0 = arg %(1 + i,) . (10.84)

The corresponding di"erential cross section is then given by

d'

d$=

,2

4k2 sin4 #/2. (10.85)

This cross section is commonly known as the Rutherford cross section. Here we observethat:

1. The di"erential cross section for the scattering of two charged particles, d-d! % &

as # % 0. This in fact is the case both for electron scattering on atoms and protonscattering o" a nucleus.

2. The total cross section is also infinity. This is a result of the fact that the Coulombpotential has an infinite range.

3. The asymptotic wave function as presented in Eq. (10.82) consists of an incidentplane wave and an outgoing spherical scattering wave. However, if we comparethis result with the equation we got for finite range potentials, i.e., Eq. (10.42), weobserve that both the plane wave and the scattering wave are modified by a factorproportional to ,, which is basically the product of the charges on the two particles.This distortion of the incident plane wave and the scattered spherical wave are theresult of the infinite range of the Coulomb potential.

10.8 Problems

1. Calculate the S-wave phase shift (i.e. ) = 0) for neutron proton scattering at centerof mass energies of 5 and 10 MeV given that the potential between a neutron and

7In writing this result we have made use of the fact that

(1" cos #) = 2 sin2 #/2 ,

and"z !("z) = !(1" z) .

10.8. PROBLEMS 191

proton is a square well of radius r0 = 2.51 fm. and the depth is V0 = 17.8 MeV.Calculate the corresponding cross section.

Note: 1 fm. = 10"13 cm, and h2/2µ = 41.47 MeV fm2.

2. Calculate the cross section for scattering o" a hard sphere of radius R at very lowenergies, i.e. in the limit where kR) 1.

Hint: A hard sphere can be represented by the potential

V (r) =

1+& r * R0 r > R

.

3. Show that for P -wave () = 1) scattering by an attractive square well potential ofradius a and depth V0, the phase shifts satisfy the equation

(2 [ka cot(ka + *1)" 1] = k2 ((a cot (a" 1) ,

where k2 = 2µh2 E and (2 = 2µ

h2 (E + V0). For neutron proton scattering we can take

V0 = 36.2 MeV. and a = 2.02 fm. Taking h2

2µ = 41.47 MeV fm2, calculate the phaseshift at E = 10 MeV.

4. Using the orthogonality of the Legendre polynomials, show that if the scatteringwave function is written as

"(!r) =1

(2+)3/2

+

#

i#(2) + 1)"#(r) P#(cos #) ,

then "#(r) satisfies the radial Schrodinger equation.

5. Show that for complex phase shifts the total cross section as calculated from the Op-tical Theorem is larger than the total elastic cross section obtained from integratingthe di"erential cross section.

Hint: To prove the above you can restrict your argument to one partial wave, e.g.) = 0.

6. The amplitude of S-wave scattering at at low energies is given by

f0 = " ak

1 + iak " 12arek2

where a and re are real constants, and k is the momentum.

(a) Show that this amplitude satisfies unitarity.

(b) Show that the corresponding cross section goes to a constant as the energygoes to zero, i.e. k % 0.

Chapter 14

Scattering Theory; Revisited

In Chapter 2 we showed how we can calculate the cross section for two particle scatteringin terms of the scattering amplitude f(k, !), and how this scattering amplitude is relatedto the phase shifts "!. In particular, we found that we needed to solve the Schrodingerequation for the wave function #!(r) for all r and then extract the phase shifts, knowingthat for r !"

#!(r) !1

krsin(kr # 1

2$% + "!) .

In other words, the asymptotic wave function determines the scattering amplitude and,therefore cross section, yet we need to calculate the wave function for all r. Since thewave function is not the observable we measure, we would like to set up an equation forthe scattering amplitude f(k, !) which is the observable.

In this chapter we will first derive an equation, the Lippmann-Schwinger equation,which can be solved for the scattering amplitude. We will find that this equation allowsus not only to derive the Optical Theorem, but also to consider perturbation expansionfor the scattering amplitude in the form of the Born series. Finally, we will consider aspecial class of potentials, called separable potentials, where we can study the propertiesof the scattering amplitude analytically.

14.1 Formal Theory of Scattering

To derive an equation for the scattering amplitude f(k, !), we need to convert the Schrodingerequation plus boundary condition into an integral equation which incorporates theseboundary conditions. Consider the Hamiltonian

H = H0 + V , (14.1)

where H0 is taken to be the kinetic energy in the center of mass, i.e., H0 = p2

2µ with µ thereduced mass of the system given by

µ =m1m2

m1 + m2.

293

294 CHAPTER 14. SCATTERING THEORY; REVISITED

Here m1 and m2 are the masses of the two particles in the collision. In Eq. (14.1), V isthe interaction between the two particles. We also introduce the eigenstates of H0 to be

H0|&"k$ = E|&"k$ , (14.2)

where E = h2k2

2µ . These states are labeled by the initial momentum 'p = h'k. In coordinaterepresentation, the state |&"k$ is a plane wave, i.e.,

%'r |&"k$ =1

(2$)3/2ei"k·"r . (14.3)

We note here that this state is identical to the eigenstate of the momentum operatoras given in Eq. (11.93), i.e. %'r |'k$ = %'r |&"k$. The Schrodinger equation for the fullHamiltonian H can now be written as

(E #H0) |#$ = V |#$ . (14.4)

This can be thought of as an inhomogeneous di!erential equation if we take it in coordinaterepresentation, and in this case, the solution can be written as the sum of a particularsolution plus a solution to the homogeneous equation. The particular solution is given by

|#$ =1

E #H0V |#$ . (14.5)

In writing the operator (E #H0)!1, we have assumed that the operator (E #H0) has nozero eigenvalues. In fact, as we will find, the operator (E#H0) does have zero eigenvaluesand therefore its inverse which appears in Eq. (14.5) is singular. We will show next how theboundary conditions in the Schrodinger equation are used to overcome these singularities.The general solution of Eq. (14.4) is then given by

|#"k$ = |&"k$+1

E #H0V |#"k$ . (14.6)

We have labeled our solution |#"k$ by the vector 'k to indicate that the momentum of the

incident beam is h'k. This in turn means that the energy of the incident beam in thecenter of mass is given by E = h2k2

2µ .

To examine the singularities of (E #H0)!1, we write Eq. (14.6) in coordinate repre-sentation as

%'r |#"k$ = %'r |&"k$+!

d3r %'r |(E #H0)!1|'r "$ %'r "|V |#"k$ . (14.7)

We are now in a position to examine the singularities of the Green’s function.1 Makinguse of the fact that the complete set of eigenstates of the momentum operator are also

1This Green’s function is identical to the Green’s function encountered in the theory of di!erentialequations. After all, we are solving the Schrodinger equation which is a second order di!erential equation.

14.1. FORMAL THEORY OF SCATTERING 295

eigenstates of H0, we can write the Green’s function as

%'r |(E #H0)!1|'r "$ =

!d3k %'r |(E #H0)

!1|'k $ %'k |'r "$

=!

d3k%'r |'k $ %'k |'r "$

E # h2k2

2µ

, (14.8)

since the eigenvalue of H0 is h2k2

2µ . Taking the energy E, which is the energy of the initialincident particle in the two-body center mass, to be related to the incident momentum,hk0, then the integral in Eq. (14.8) can be written as

%'r |(E #H0)!1|'r "$ =

1

(2$)3

2µ

h2

!d3k

ei"k·("r!"r !)

k20 # k2

=1

(2$)3

2µ

h2

#!

0

dkk2

k20 # k2

#!

0

d! sin ! eik|"r!"r !| cos $

2#!

0

d&

=1

(2$)2

2µ

h2

#!

0

dkk2

k20 # k2

eik|"r!"r !| # e!ik|"r!"r !|

ik|'r # 'r "|

=1

i(2$)2

2µ

h2

1

|'r # 'r "|

+#!

!#

dk keik|"r!"r !|

k20 # k2

. (14.9)

However this integral is not defined for k20 > 0, (i.e. E > 0) because the integrand has a

pole at k = ±k0, which is on the integration path. To overcome this problem, we need togo around these poles. This can be achieved by taking2

k20 ! k2

0 ± i( , (14.10)

where ( is an infinitesimal quantity. We then perform the integral and take the limit of( ! 0. As we will see, the choice of sign in k2

0 ± i( will determine the boundary condition.Let us take the positive sign, i.e., k2

0 + i(, for the present. Then our Green’s function hasthe integral representation given by

%'r |(E #H0)!1|'r "$ =

1

i(2$)2

2µ

h2

1

|'r # 'r "|

+#!

!#

dk keik|"r!"r !|

k20 + i(# k2

. (14.11)

The integrand now has two poles at

k = ±(k0 + i() .

2This is one choice of integration path. Other choices will entail taking the integration path below orabove both poles. These correspond to other boundary conditions.


We now can perform the integral in Eq. (14.11) by making use of Cauchy’s Theorem. Fork in the upper half of the complex k-plane, i.e., Im(k) > 0, the integrand has the propertythat it goes to zero as |k|!" . This condition allows us to convert the integral over theinterval #" to +" to an integral along a contour C in Figure 14.1 where the semi-circlecorresponds to |k|!" . Because the integrand is zero on the infinite semi-circle it doesnot contribute to the value of the Green’s function. This allows us to write the Green’sfunction as

k

C

k0 + iε

-k0 - iε

Figure 14.1: The contour of integration used to calculate the Green’s function with anoutgoing spherical wave.

%'r |(E #H0)!1|'r "$ =

1

i(2$)2

2µ

h2

1

|'r # 'r "|

"

C

dk k eik|"r!"r !|

(k0 + i(# k)(k0 + i( + k)

= # 1

4$

2µ

h2

eik0|"r!"r !|

|'r # 'r "| , (14.12)

where we have made use of Cauchy’s Residue Theorem3 and have taken the limit ( ! 0.Had we chosen k2

0 ! k20 # i(, the poles in the integrand would have been at

k = ±(k0 # i() .

In this case the contour C would have enclosed the pole at k = #k0 + i(, and we wouldget the the factor of

e!ik0|"r!"r !|

3Cauchy’s theorem allows us to write the integral over a closed contour in terms of the sum over theresidues of all the poles in the integrand that are inside the contour, i.e.,

"

C

dk f(k) = 2!i#

n

Res[f(k)] ,

where n runs over all the poles inside the contour C.


in Eq. (14.12) instead ofeik0|"r!"r !| .

This we will see corresponds to spherical waves which are moving towards the scatteringcenter, while the Green’s function in Eq. (14.12) corresponds to spherical outgoing waves.These two Green’s functions can now be written as

%'r | (E± #H0)!1|'r "$ = lim

%$0%'r | (E ± i(#H0)

!1|'r "$

= & 1

4$

2µ

h2

e±ik0|"r!"r !|

|'r # 'r "| . (14.13)

With this result in hand we can write the scattering wave function, in coordinate rep-resentation, by substituting the Green’s function %'r | (E+ # H0)!1|'r "$ in Eq. (14.7) toget

%'r |#(+)"k0$ = %'r |&"k0

$ # 1

4$

2µ

h2

!d3r"

eik0|"r!"r !|

|'r # 'r "| %'r"|V |#(+)

"k0$ , (14.14)

where (+) superscript on the wave function indicates that the boundary condition cor-responds to an outgoing spherical wave. To extract the scattering amplitude out of thiswave function we need to take the limit as r ! " to determine the coe"cient of theoutgoing spherical wave and compare that equation with Eq. (10.3) to determine thescattering amplitude. To take the limit as r !" we make use of the fact that

|'r # 'r "| ='

r2 + r"2 # 2'r · 'r "

= r

$%

&1 +

'r"

r

(2

# 2'r · 'r "

r2

)*

+

12

! r # r · 'r " for r !" ,

where r is a unit vector in the 'r direction. We now can write the asymptotic wave function(i.e., r !") as

%'r |#(+)"k0$ ! 1

(2$)3/2ei"k0·"r # 1

4$

2µ

h2

eik0r

r

!d3r" e!ik0r·"r ! %'r |V |#(+)

"k0$ . (14.15)

Since the unit vector r is in the same direction as the final momentum of the scatteredparticle, and the magnitude of the initial and final momentum are the same due to energyconservation, we can take the final momentum to be 'kf = k0r. Making use of this resultand the fact that

%&"kf|'r "$ =

1

(2$)3/2e!i"kf ·"r !

,

we can write the asymptotic wave function as

%'r |#(+)"ki$ ! 1

(2$)3/2

,

ei"ki·"r # 4$2µ

h2 %&"kf|V |#(+)

"ki$ eikr

r

-

. (14.16)


In writing the above scattering wave function, we have labeled the initial momentum, i.e.,the momentum of the incident beam, by 'ki, while the final momentum is labeled as 'kf .Conservation of energy then requires that the magnitude of the initial and final momentumbe equal if the scattering is elastic, i.e., |'ki| = |'kf | ( k. Comparing Eqs. (14.16) and (10.3)allows us to write the scattering amplitude in terms of the matrix elements of the potentialas

f(k, !) = #4$2µ

h2 %&"kf|V |#(+)

"ki$ , (14.17)

where cos ! = kf · ki. Since the scattering amplitude is basically an observable, we willintroduce a corresponding operator which we will call the T -matrix, and which is definedby the relation

V |#(+)"ki$ ( T (E+)|&"ki

$ , (14.18)

where the boundary condition is now specified by labeling the T -matrix with E+ = E+i(.We now can write the scattering amplitude in terms of the T -matrix as

f(k, !) = #4$2µ

h2 %&"kf|T (E+)|&"ki

$ ( 14.19a)

= #4$2µ

h2 %'kf |T (E+)|'ki$ . ( 14.19b)

Therefore, the problem of finding an equation for the scattering amplitude has beenreduced to the problem of finding an equation for the T -matrix. Making use of thedefinition of the T -matrix as given in Eq. (14.18), we can write the scattering wavefunction given in Eq. (14.6) as

|#(+)"ki$ = |&"ki

$+ G0(E+) T (E+)|&"ki

$ , (14.20)

where G0(E+) = (E+ # H0)!1 is the Green’s function in operator form. This Green’sfunction is often referred to as the free-particle Green’s function since it is the Green’sfunction for the Schrodinger equation in the absence of any interaction. We now multiplythis equation from the left by the potential V , to get

V |#(+)"ki$ = V |&"ki

$+ V G0(E+) T (E+)|&"ki

$ . (14.21)

Applying the definition of the T -matrix, Eq. (14.18), to the left hand side of Eq. (14.21)gives us the equation

T (E+)|&"ki$ = V |&"ki

$+ V G0(E+) T (E+)|&"ki

$ . (14.22)

Since this is valid for any state |&"ki$, we can write an operator equation for the T -matrix

which is of the formT (E+) = V + V G0(E

+) T (E+) . (14.23)


We have taken the operator T to be a function of the energy E+ since the Green’s functionG0(E+) is a function of the energy, and we need to know the Green’s function to determinethe T -matrix. Also, by specifying the energy in the form of E+, we have specified theboundary condition that we have spherically outgoing waves. Equation (14.23) is knownas the Lippmann-Schwinger equation, and is equivalent to the Schrodinger equation in-cluding the boundary conditions. The solution of this equation will give us the scatteringamplitude directly and thus the cross section.

To solve Eq. (14.23), we need to write the equation in a given representation. Inthis case the natural representation is the momentum representation4 which gives us%'k |T (E+)|'k "$ and then the scattering amplitude, as measured in elastic scattering, isgiven by

%'k |T (E+)|'k "$ with |'k| = |'k "| and E =h2k2

2µ. (14.24)

The Lippmann-Schwinger equation, Eq. (14.23), can be written in momentum space as

%'k |T (E+)|'k "$ = %'k |V |'k "$+ %'k |V G0(E+) T (E+)|'k "$

= %'k |V |'k "$+!

d3k""%'k |V |'k ""$ %'k ""|T (E+)|'k "$

E + i(# h2k!!2

2µ

. (14.25)

In writing the second line of the above equation we have made use of the fact thatthe eigenstates of the momentum operator are complete, and that these states are alsoeigenstates of H0 with eigenvalue h2k!!2

2µ . In Eq. (14.25), we have an integral equation inthree-dimensions which is very di"cult to solve as it stands. To reduce the dimensionalityof the equation, we need to partial wave expand the matrix elements of the potential Vand the T -matrix T (E+). This expansion involves the introduction of eigenstates of thetotal angular momentum and its projection along the z-axis. In the case of the scatteringof spin-less particles, the total angular momentum is the orbital angular momentum. Forthe potential V , the partial wave expansion involves writing the matrix elements of V as5

%'k|V |'k "$ = %k; k|V |k"; k"$=

#

!m

%k|%m$ %%m, k|V |k", %m$ %%m|k"$

4Note that the eigenstates of the momentum operator and the free Hamiltonian H0 are identical withour choice of normalization, i.e.,

|"!k$ = |#k $ .

5In writing Eq. (14.26), we are making use of the notation

|#k $ = |k; k$ = |k$ |k$

to separate the radial from the angular variables.


(#

!m

%k|%m$ %k|V!|k"$ %%m|k"$ . (14.26)

In writing the first line of the above equation we have separated the radial from theangular part of the momentum eigenstates. We have also taken into consideration thatthe potential is spherically symmetric by taking the matrix elements of V to be diagonalin % and independent of m, i.e., since [H,L2] = [H,Lz] = 0, we have

%%m, k|V |k", %"m"$ = "!!! "mm" %k|V!|k"$ . (14.27)

The independence of the matrix elements from m is a result of the fact that the radialwave functions for a spherically symmetric potential are independent of m. In Eq. (14.26)the bracket %k|%m$ is nothing other than the spherical harmonic, i.e.,

%k|%m$ = Y!m(k) .

In other words the expansion in Eq. (14.26) is similar to that in Eq. (10.45) if we takeinto consideration that the Legendre polynomial can be written in terms of sphericalharmonics using the addition theorem, i.e.,

#

m

Y!m(k) Y %!m(k") =

2% + 1

4$P!(cos !) . (14.28)

In this case we can write the potential in momentum representation as

%'k |V |'k "$ =1

4$

#

!

(2% + 1) V!(k, k") P!(cos !) , (14.29)

with

V!(k, k") = %k|V!|k"$ , (14.30)

and cos ! = k · k". In a similar manner we can write a partial wave expansion for theT -matrix which is of the form

%'k |T (E+)|'k "$ =#

!m

%k|%m$ %k|T!(E+)|k"$ %%m|k"$

=1

4$

#

!

(2% + 1) T!(k, k"; E+) P!(cos !) . (14.31)

With these partial wave expansions for the potential V and the T -matrix T (E+), andusing the orthogonality of the spherical harmonics, i.e.,

!dk %%m|k$ %k|%"m"$ = "!!! "mm! , (14.32)


we can write the Lippmann-Schwinger equation as a one dimensional integral equation ofthe form

%k|T!(E+)|k"$ = %k|V!|k"$+

#!

0

dk""k""2%k|V!|k""$ %k""|T!(E+)|k"$

E + i(# h2k!!2

2µ

, ( 14.33a)

or

T!(k, k"; E+) = V!(k, k") +

#!

0

dk""k""2 V!(k, k"")

E + i(# h2k!!2

2µ

T!(k"", k"; E+) . ( 14.33b)

This equation can be solved numerically on any present day computer. This is achievedby replacing the integral by a sum and in this way we turn the integral equation into aset of linear algebraic equations. In converting the above integral equation into a set ofalgebraic equations we should note that the energy E and the initial momentum k" areparameters and play no role in the solution of the equation. In fact, to get the physicalamplitude we should take

E =h2k2

0

2µand k" = k0

where k0 is referred to as the on-shell momentum.By comparing the partial wave expansion of the scattering amplitude, f(k, !), as given

in Eq. (10.45) with the expansion of the T -matrix as given in Eq. (14.31), and makinguse of Eq. (14.19), we can write the partial wave scattering amplitude f!(k) in terms ofthe partial wave T -matrix T!(k0, k0; E+) as

f!(k0) =1

2i

.e2i&! # 1

/= #$µk0

h2 T!(k0, k0; E+)

( #$µk0

h2 T!(k0) . (14.34)

To solve Eq. (14.33) for the scattering amplitude, we need to determine the inputpartial wave potential V!(k, k"), and the Green’s function G0(E+) in momentum repre-sentation. To determine the partial wave potential we need to first write the potential inmomentum representation in terms of the potential in coordinate space as we have usedit in previous chapters. All the potentials we have encountered to date are diagonal incoordinate representation, i.e.,

%'r |V |'r "$ = "('r # 'r ") V ('r) . (14.35)

These potentials are known as local potentials in contrast to non-local potentials that arenot diagonal in coordinate representation. For local potentials, we have in momentumspace

%'k |V |'k "$ =!

d3r d3r" %'k |'r $ %'r |V |'r "$ %'r "|'k "$

=!

d3r %'k |'r $V ('r) %'r |'k "$ . (14.36)


Making use of the partial wave expansion of the eigenstates of the momentum operator,i.e.,

%'r |'k $ =1

(2$)3/2ei"k·"r

=

02

$

#

!m

i! j!(kr) Y!m(r) Y %!m(k) , (14.37)

and the orthogonality of the spherical harmonics in Eq. (14.36) allows us to write thepotential in momentum space as

%'k |V |'k "$ =#

!m

Y!m(k) V!(k, k") Y %!m(k") , (14.38)

where

V!(k, k") =2

$

#!

0

dr r2

#!

0

dr" r"2 j!(kr) V!(r, r") j!(k

"r") ( 14.39a)

=2

$

#!

0

dr r2 j!(kr) V!(r) j!(k"r) . ( 14.39b)

Here, Eq. (14.39b) is to be used for the special case when the potential is local andcentral, in which case V!(r) = V (r). Thus for any local central potential, Eq. (14.39b)defines the momentum space partial wave potential for use in the integral equation givenin Eq. (14.33). The solution of this Lippmann-Schwinger equation then gives us thescattering amplitude for a given angular momentum %. To get the di!erential cross section,we need to solve Eq. (14.33) for all partial waves that are important. For finite rangepotentials, the maximum value of % can be estimated by taking the classical argumentthat the maximum angular momentum is the cross product of the incident momentum,times the maximum impact parameter, which in this case is the range of the potential,i.e., %max = r0k, where r0 is the range of the potential.6

14.2 The Born Approximation

The solution of the Lippmann-Schwinger equation, Eq. (14.33), is often not simple toget, and one may need to resort to approximation methods. One approximation oftenused at high energies or weak potentials in both atomic and nuclear physics is the Bornapproximation. This is the first term in a series similar to the perturbation series we

6Here, we should note that for the case of the Coulomb potential, where the range of the interactionis infinite, the partial wave sum should be examined carefully.

14.2. THE BORN APPROXIMATION 303

developed in the last chapter. To get the series, we iterate Eq. (14.23) to get, in operatorform, the Born series given by

T (E±) = V + V G0(E±) V + V G0(E

±) V G0(E±) V + · · · . (14.40)

This is a power series in the potential, and therefore for a potential that is weak, theseries converges. Here, as in the case of perturbation theory for bound states, we candefine H0 to include some of the interaction and in this way, the remaining interaction(H # H0) is weak enough for the Born series to converge. However, in this case, to getthe amplitude or T -matrix, we need to take the matrix element of (H #H0) with respectto the solution of the Schrodinger equation with the Hamiltonian H0. Also the Green’sfunction G0(E±) has to be replaced by the Green’s function for the Hamiltonian H0 whichincludes the interaction. This procedure has been implemented for Coulomb plus shortrange potential, where we know the solution for the Coulomb Hamiltonian analytically.

Another condition under which the above Born series converges is when the energy Eis high.7In that case the series is again a power series in E!1, and the first few terms ofthe series give a good approximation.

The Born approximation is used when the first term in the Born series is taken to bethe amplitude for scattering. For this case, we have

%'k |T (E+)|'k "$ ) %'k |V |'k "$ ( %'k |TB|'k "$ . (14.41)

For the case of a local central potential, the Born approximation reduces to

%'kf |TB|'ki$ = %&"kf|TB|&"ki

$

=!

d3r &%"kfV (r) &"ki

=1

(2$)3

!d3r ei"q·"r V (r) , (14.42)

where 'q = 'ki # 'kf is the momentum transfer in the scattering.As an example of the application of the Born approximation, let us consider the

problem of scattering by a Coulomb potential, and in particular the potential due to thenucleus with charge Ze, i.e.

VC(r) = #Ze2

r. (14.43)

Here the Born approximation is given by

%&"kf|TB|&"ki

$ = # Ze2

(2$)3

!d3r ei"q·"r 1

r

7By high energy we mean high, relative to the depth of the potential V , or the energy of bound statesin the potential V . For example, for electron scattering in atomic physics, high is more than 103 eV,while for proton scattering in nuclear physics, high is more than 103 MeV.


= # Ze2

(2$)2

#!

0

dr r

+1!

!1

dx eiqrx

= # Ze2

(2$)2

2

q

#!

0

dr sin qr . (14.44)

This integral, as it stands is not well defined. However, if we replace the Coulomb potentialby the screened Coulomb potential, i.e.,

V (r) = #Ze2 e!ar

r,

the integral becomes well defined, and then we can take the limit as a ! 0 to get theBorn amplitude for the Coulomb potential, i.e.8

%&"kf|TB|&"ki

$ = #Ze2

2$2

1

qlima$0

#!

0

dr e!ar sin qr

= #Ze2

2$2

1

qlima$0

q

q2 + a2

= #Ze2

2$2

1

q2, (14.45)

where the momentum transfer square is given by

q2 = 2k2(1# cos !) , (14.46)

and cos ! = kf · ki, and |'kf | = |'ki| = k. The scattering amplitude can now be written as

f(k, !) =Ze2µ

h2

1

k2(1# cos !). (14.47)

This scattering amplitude is real, and to that extent does not satisfy unitarity. This istrue for the Born approximation for any real potential. With this result, we can calculatethe di!erential cross section, which is given by

d)

d#= |f(k, !)|2

=Z2e4µ2

h4

1

k4(1# cos !)2

=Z2e4

16E2 sin4(!/2), (14.48)

8We have made use of the integral!!

0

dx e"ax sin qx =q

a2 + q2

to get the Born amplitude for Coulomb scattering.

14.3. ELECTRON ATOM SCATTERING 305

which is the Rutherford cross section we got in Eq. (10.85) This cross section goes toinfinity when cos ! = 1, i.e., for ! = 0 and ! = $. In fact, the total cross section is infinity.This is the result of the fact that the Coulomb potential is infinite in range. In nature, wealways have screened potentials in the sense that proton-proton scattering is a problem oftwo charge particles when the two hydrogen atoms overlap. For large distances we haveneutral hydrogen atoms. Although the Born amplitude for the Coulomb potential is realand does not satisfy unitarity, the cross section we get is the exact classical Rutherfordcross section.

14.3 Electron Atom scattering

As a second application of the Born approximation, let us consider electron atom scat-tering, and in particular, electron scattering from hydrogen. This approximation is verygood at high energies where the second term in the Born series, which goes as 1

E becomesnegligible. The potential that describes the interaction between the incident electron andthe target atom is the sum of two terms. The first term describes the interaction of theelectron with the point nucleus, is attractive, and of the form

V1(r) = #Z

r, (14.49)

where Z is the charge on the nucleus. Here we are using atomic units. The secondterm corresponds to the interaction of the incident electron with the charge distributionresulting from the bound electrons in the atom. This is of the from

V2('r ) =!

d3r"*('r ")

|'r # 'r "| , (14.50)

where *('r ") is the charge distribution of the bound electrons and is given by

*('r ) =Z#

n=1

|&n('r )|2 . (14.51)

Here, we note that the sum runs over the Z electrons each in a state &n('r ), n = 1, · · · , Z.For the case of electron hydrogen scattering, Z = 1 and the bound state electron wave

function can be taken as the ground state of the hydrogen atom, i.e., &100('r ). In this caseV2 is given by

V2('r ) =!

d3r"|&100('r ")|2

|'r # 'r "| , (14.52)

with

&100('r ) = R10(r) Y00(r) =1'4$

R10(r) , (14.53)

chapter 10 scattering by a central potentialphillips/phys735/afnanscattering.pdf · chapter 10...

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