Chapter 16 – Potential Scattering

part of Lectures on Quantum Mechanics

 

by Reinaldo Baretti Machín

 

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http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysics.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart2.htm 

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart3.htm

 

References:

1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.    

by    W. V. Houston

2.  Lectures On Quantum Mechanics (Lecture Notes & Supplements in Physics Ser.)) by Gordon Baym , chapter 9

3. Quantum Mechanics Non-Relativistic Theory, Third Edition: Volume 3 by L. D. Landau and L. M. Lifshitz ,Chapter  xvii

4. Fundamentals of Modern Physics by Robert Martin Eisberg (Hardcover - June 1961) , chapter 9.

5. Advanced Quantum Mechanics by J. J. Sakurai

6. Schaum's Outline of Quantum Mechanics, Second Edition (Schaum's Outline Series) by Yoav Peleg, Reuven Pnini, Elyahu Zaarur,   and Eugene Hecht  , chapter 15 .

 

In this note we treat elastic scattering. A flux of particles with well defined energies is incident upon a scattering center. The center of force interacts through a scalar potential U(r) without loss of energy of the incident particles.  No decay occurs and the particles maintain their identities.  The original incident plane wave is distorted. At large distances the scattered particles are detected, think of Rutherford experiment. The particle dispersion is dependent upon the the potentail U(r). The role of the theorist is to predict quantities like the the total cross section σ ~ m2 ,if a potential U(r) is assumed.

Fig 1. Plane waves are incident on the scattering center with potential U(r) and scattered.

 

 

a) the phase shift

The incident plane waves are solutions of

                                                      ∆2 Φ + k2 Φ  = 0                                ,                                (1)

where k = ( 2mE/h'2)  , or in our units k = (2E)1/2  length.

Assuming symmetry about the z  axis the solutions are independent of the azimuth angle φ and are expanded in terms of functions of the angular momentum quantum number l ,

                                              Φ(r,θ) = exp(ikz) = exp( ikrcos(θ) ) = ∑ l=0  Al Pl ( cos(θ) ) fl (r)  ,    (2)

whrere Pl ( cos(θ) ) are the Legendre polynomials.

The functions fl (r) satisfy the DE

                                                     d2 fl (r) /dr2 + (2/r) d fl (r)/dr + [ k2 - l( l+1) /r2 ] fl (r)    = 0        , (3)

with boundary conditions  lim fl (r) r →0 = r l  , lim fl (r) r → ∞ = 0 .

The coefficients Al can be evaluated from the asymptotic form of   Al Pl ( cos(θ) ) fl (r) ( see ref . 1)  ) but this will not be necessary for our present purpose.

Of utmost importance is the fact that fl (r) is related to the Bessel function of order ( l+1/2),

 

                                                        fl (r) = (π/(2kr))1/2  Jl +1/2 (kr)                                         .     (4)

The asymptotic from of fl (r) is

                                                                fl (r) ≈ [1/(kr)]  sin (kr - lπ/2)                            .     (5)

 

When a non - Coulombic potential energy U(r) is introduced in (3)  we have a new function Ll (r) satisfying  the new DE

                                  d2 L l (r) /dr2 + (2/r) d Ll (r)/dr + [ k2 -2U(r) - l( l+1) /r2 ] Ll (r)    = 0      ,       (6)

with boundary condition L l (∞) = 0 and     L l (r) near the origin will depend on the specific form of U(r). In the case of a Coulomb potential , L l (r) ~ r l near the origin.

Supposing that U(r) falls to zero for large r , one has the asymptotic form of Ll (r) ,

                                                    Ll (r) → [1/((kr)] sin ( kr - lπ/2 + ηl  )                ,                            (7)

differing from  the asymptotic fl (r) in (5), by a the phase constant ηl .

To find the phase shift ηl  , we integrate numerically eq (6) and compare successive  maxima with those of (5) .

Suppose that in the asymptotic region there is a maxima of Ll (r) at r2 and the corresponding maxima of fl (r) is at r1 .

Then    kr2 - lπ/2 + ηl  =   kr1 - lπ/2   and

                                                       ηl = - ( kr2 - kr1 ) = -k( r2 - r1 )                                       .         (8)

Example 1  :  Impenetrable spheres (ref. 1)  U(r) = ∞     r <a ; U(r) = 0 , r>a .

Consider first the case of waves with zero angular momentum , l = 0.

At  r= a the solution of (6) is required to satisfy  L0 (r=a ) = 0 since the potentail is impenetrable. We may set the initial slope equal to any value  , e.g . d L0 (r=a) /dr = 1.or 2.  Eq (6) is integrated using the finite difference method which is summed up in the declaration

( see FORTRAN code #1)

psi2= 2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr -  (k**2   -2.*v(r-dr)-al*(al+1.)/(r-dr)**2 )*psi1 )  .  (9)

 k,a,k*a= 1.41421354 1. 1.41421354

Let  E=1 , k =(2E)1/2  = 1.414  , a=1 , ka = 1.414  radians ( the unit of phase shift).

From Fig 1. we see that various consecutive maxima. Let r2 = 0.15465E+02   and r1 = 0.14432E+02  ( see Table 1) .From (8) we get

                      η0= -k( r2 - r1 )    = - 1.461  ≈ 21/2       ,  to four digits.                                  .              (10) 

 Further numerical  examination reveals  ,see the results of Table 2 with k=21/2 and  a=2 , that

   - k( r2 - r1 )=  k( 0.20860E+02 -0.18892E+02)=2.783  , approximately twice the result in (10). Also changing k will reveal that   η0 is proportional to k.

We conclude that , 

                                                           η0 = - ka                                                           . (11) 

Table 1. with k=21/2   a= 1

r,f(r),psi= 0.14136E+02 0.45499E-01 -0.13118E-01
r,f(r),psi= 0.14284E+02 0.48313E-01 -0.28063E-02
r,f(r),psi= 0.14432E+02 0.48994E-01  0.74148E-02
r,f(r),psi= 0.14579E+02 0.47556E-01 0.17110E-01
r,f(r),psi= 0.14727E+02 0.44102E-01 0.25876E-01
r,f(r),psi= 0.14874E+02 0.38822E-01 0.33355E-01
r,f(r),psi= 0.15022E+02 0.31976E-01 0.39253E-01
r,f(r),psi= 0.15170E+02 0.23889E-01 0.43350E-01
r,f(r),psi= 0.15317E+02 0.14931E-01 0.45503E-01
r,f(r),psi= 0.15465E+02 0.55018E-02 0.45658E-01
r,f(r),psi= 0.15612E+02 -0.39859E-02 0.43847E-01

       

Table 2.

r,f(r),psi= 0.18400E+02 0.29833E-01 -0.71841E-01
r,f(r),psi= 0.18564E+02 0.34297E-01 -0.75509E-01
r,f(r),psi= 0.18728E+02 0.36862E-01 -0.75104E-01
r,f(r),psi= 0.18892E+02 0.37425E-01 -0.70719E-01
r,f(r),psi= 0.19056E+02 0.35992E-01 -0.62653E-01
r,f(r),psi= 0.19220E+02 0.32673E-01 -0.51398E-01
r,f(r),psi= 0.19384E+02 0.27674E-01 -0.37603E-01
r,f(r),psi= 0.19548E+02 0.21290E-01 -0.22042E-01
r,f(r),psi= 0.19712E+02 0.13882E-01 -0.55694E-02
r,f(r),psi= 0.19876E+02 0.58583E-02 0.10927E-01
r,f(r),psi= 0.20040E+02 -0.23449E-02 0.26573E-01
r,f(r),psi= 0.20204E+02 -0.10290E-01 0.40553E-01
r,f(r),psi= 0.20368E+02 -0.17561E-01 0.52154E-01
r,f(r),psi= 0.20532E+02 -0.23783E-01 0.60799E-01
r,f(r),psi= 0.20696E+02 -0.28643E-01 0.66076E-01
r,f(r),psi= 0.20860E+02 -0.31905E-01 0.67758E-01
r,f(r),psi= 0.21024E+02 -0.33420E-01 0.65812E-01
r,f(r),psi= 0.21188E+02 -0.33136E-01 0.60399E-01
r,f(r),psi= 0.21352E+02 -0.31096E-01 0.51859E-01
r,f(r),psi= 0.21516E+02 -0.27434E-01 0.40693E-01

               

Fig 1. Comparison of the asymptotic f l (r) with  L l  (r) .

 

Consider now the partial waves with angular momentum l ≥ 1 . If the impact parameter b ≥ a of the impenetrable sphere , the angular momentum is of the order

If  L ~ h'k b = kb  ≥ k a , then  ( l ( l+1) )1/2 ≥  ka and the particle will have little or no  interaction with  the impenetrable sphere.

 

The smallest angular momentum resulting in  a phase shift η satisfies  l2 +  l - 2 = 0 when k=21/2 and a=1  . Hence  l =1 or l = -2 .The negative value is not allowed.  In Fig 2. we plot the asymptotic behavior , with l=1 showing a small phase shift. 

Fig 2. Phase shift for angular momentum l = 1.

 

Fig 3. Phase shift for angular momentum l = 2.

 

Fortran code #1

c scattering by a rectangular potential Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 185
real k
data a, al,nstep /1.,1.,2000/
data e ,ntrial /1., 1/
v(u)= 0.
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
pi=2.*asin(1.)
rcutoff=12.*a
k=sqrt(2.*e)
dr=rcutoff/float(nstep)
kp=int(float(nstep)/100.)
kount=kp
do 10 ie=1,ntrial
psi0=0.
c deriv=sqrt(2./pi)*(1./a)
deriv=1.
psi1=psi0 + dr*deriv
ri=a
print*,'k,a,k*a=',k,a,k*a
print*,' '
print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2= 2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr)-al*(al+1.)/(r-dr)**2 )*psi1 )
if(i.eq.kount)then
print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
phase=asin(k*r*psi2)
c print 110, e,psi2
c eta0=k*r*(psi2-1.)+ al*pi/2.
c print*,'eta0=',eta0 ,-k*a
110 format('e,psi2,phase=',3(3x,e10.3))
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e11.4))
10 continue
stop
end


b) the cross section and the phase shift

We will now examine how the phase shifts  ηl are connected to  measurable quantities like the differential cross section , dσ/dΩ or the total cross section σT ~ area .   

Fig 4. A flux of particles is scattered from the potential U(r).

The flux has diemnsions  F ~ particles/(area-time) . The number of particles  scattered in the direction (θ,φ) described by the solid angle dΩ   per unit time is denoted by dn(θ,φ) .

The differential cross section is empircally defined by

                                          dσ( θ,φ ) /dΩ = dn(θ,φ) /(F dΩ)  ~ area /sr                             (  12  )

If we assume that the scattering is independent of φ , the asymptotic scattered wave is written in the form ( see Ref. 1 , chapter XII)

         f (θ) exp(ikr)/r    = ∑ l=0 (2l+1) (exp(2i ηl ) -1) Pl (cos(θ) ) exp(ikr)/(2ikr) ~ length      ,       (  13    )

where f (θ) ~ length, is called the scattering amplitude ( do not confuse f (θ) with the solutions of eq (3) ) . The right side of (  ) contains the phase shifts   ηl .  

Our first connection with  dσ/dΩ  comes from working with the square of the absolute value of    f (θ) exp(ikr)/r  . Multiplying it by

area differential  r2 dΩ  yields

                                              (1/r2 )   / f (θ) /2   r2 dΩ  =   /f (θ) /2   dΩ  ~ length2 ~ area        ,  (  14  )

and this the differential "area " dσ . Therefore the differential cross section is related to the scattering amplitude by 

                                                 dσ / dΩ   =  /f (θ) /2      ~ area                                                 , (15)   

and 

            f (θ) = ( 1/(2ik) ) ∑l=0 (2 l +1) Pl (cos( θ) ) {exp(2 i ηl ) -1 }                                         . (16)

The complex conjugate square on the right hand side of (13) is also multiplied by r2 dΩ = r2 sin(θ )dθ dφ . The φ integration yields a factor 2π. Use is made  the orthogonal property

π 0 Pl (cos(θ) ) Pl' (cos(θ) ) sin(θ) dθ = 2 δ l , l'  /(2l +1)  and  ( exp(2i ηl ) -1) (  exp(-2i ηl ) -1) = 4 sin2l ) . ( 17)  

This gives the relatively simple expression for the total cross section 

                                    σT =   (4π /k2) ∑l=0 (2 l +1)  sin2 ( ηl  ) ≡ ∑l=0  σl ~   area                     .      (18)            

                    

Example 2 : Find σT for the impenetrable sphere of example 1.We already found that with the given k and a , the zero angular momentum shift   η0 = -ka = - 21/2  , made the most important contribution.

Hence  σT  ≈  σ0 =  (4π /k2) sin2 ( ηl  ) = (4π /2 ) sin2( -21/2 ) = 4.448   area units .

Let  the range of U(r)  ,  a= 1.0E-15 m be our unit of length. Then a2 = 1.0E-30 m2 is our unit of area. Hence

                                                                    σT  = 4.448 area units = 4.448 E-30 m2                  .    (19)

 

 At low energies   / η0 / =  ka << 1   and  sin 2( ka  ) ≈ (ka)2 and  σT =  4π a2  . The potential scatters as a hard sphere of radius a and total area  4π a2.

Example 3:  FORTRAN code #2

The energy is E= 1 and  k= 21/2 , k' = 2.44948983.

al,a,e,D= 0. 2. 1. 2.
k,kprim eta0= 1.41421354 2.44948983 2.19999146

With a attractive potential a positive phase shift is produced given by

                    η0 =  2π  + atan( (k/kprime)*tan(kprime*a)) - k*a              .                                 (20)

The term 2π is introduced because the numerical code gives     0 < atan( (k/kprime)*tan(kprime*a)) < - π/2  , see ref 1 page 186.

Fig 5 . The attractive potential produces a positive shift , η > 0 .The function L(r) is pulled towards the origin. Compared with Fig 1.

 

The phase shift from (20) is 2.20 radians.  The last maxima of L(r) , in Fig 5 is at r2 ≈ 17.2  and the last maxima of f(r) is at

r1 ≈ 18.9 . Hence our numerical estimate of  η0   ≈ - k (r2 -  r1 ) = 2.40   radians . The difference with the analytical value , may be due to the fact that the wave functions are not plotted far enough from the origin.



 

FORTRAN code #2

c scattering by a rectangular potential Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 185
c attractive potentail well V=-D r< a
real k ,kprim
data a, D, al,nstep /2.,2. ,0.,5000/
data ri,e ,ntrial /0.,1., 1/
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
c eta0(k,kprim,a)=atan((k/kprim)*tan(kprim*a))-k*a
pi=2.*asin(1.)
k=sqrt(2.*e)
c rcutoff=(6.*pi+al*pi/2.)/k
rcutoff=20.
kprim=sqrt(2.*(e-v(0.,a,d)))
dr=rcutoff/float(nstep)
kp=int(float(nstep)/120.)
kount=kp
do 10 ie=1,ntrial
psi0=1.
c deriv=sqrt(2./pi)*(1./a)
c for al=0. , deriv=0.
deriv=0.
psi1=psi0 + dr*deriv
print*,'al,a,e,D=',al,a,e,D
print*,'k,kprim eta0=',k,kprim,eta0(k,kprim,a)
print*,' '
c print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2= 2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr,a,d)-al*(al+1.)/(r-dr)**2 )*psi1 )
if(i.eq.kount)then
if(r.ge.5.)print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e12.5))
10 continue
stop
end


function v(r,a,D)
if(r.le.a)v=-D
if(r.gt.a)v=0.
return
end

function eta0(k,kprim,a)
c eta0(k,kprim,a)=atan((k/kprim)*tan(kprim*a))-k*a
real k ,kprim
pi=2.*asin(1.)
term1=atan((k/kprim)*tan(kprim*a))
if(term1.lt.0.)term1=2.*pi + term1
eta0=term1-k*a
return
end
 

 

Example 3:  FORTRAN code #3

We explore the phase shift η0 with a  'low ' energy of  E =0.1.

We are reminded that the following angles are "equivalent". θ1 = abs ( η ) ,θ2 = π - abs ( η ) ,θ3 = π + abs ( η ) ,

θ4 = 2π - abs ( η ) .

            




 

 

Fig 6. The phase shift is  approximately  -k (r2 - r1 ) = 0.44721359( 11.5-17.2) = + 2.55 radians.

The equivalent angles are ,  0.593972325 2.5476203 3.73556519 5.68921328  ~ radians.

 

Fortran code of Example #3

c scattering by a rectangular potential Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 185
c attractive potentail well V = -D r< a
real k ,kprim
data a, D, al,nstep /2.,2. ,0.,5000/
data ri,e ,ntrial /0.,.1, 1/
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
pi=2.*asin(1.)
k=sqrt(2.*e)
c rcutoff=(6.*pi+al*pi/2.)/k
rcutoff=20.
kprim=sqrt(2.*(e-v(0.,a,d)))
term1=atan((k/kprim)*tan(kprim*a))
if(term1.ge.0.)theta=term1
if(term1.lt.0.)theta=pi-abs(term1)
eta0=theta-k*a
theta1=abs(eta0)
theta2=pi-theta1
theta3=pi+theta1
theta4=2.*pi-theta1
dr=rcutoff/float(nstep)
kp=int(float(nstep)/120.)
kount=kp
do 10 ie=1,ntrial
psi0=1.
c deriv=sqrt(2./pi)*(1./a)
c for al=0. , deriv=0.
deriv=0.
psi1=psi0 + dr*deriv
print*,'al,a,e,D=',al,a,e,D
print*,'k,kprim =',k,kprim
print*,'"equivalent angles"=',theta1,theta2,theta3,theta4
print*,' '
c print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2= 2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr,a,d)-al*(al+1.)/(r-dr)**2 )*psi1 )
if(i.eq.kount)then
if(r.ge.5.)print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e12.5))
10 continue
stop
end


function v(r,a,D)
if(r.le.a)v=-D
if(r.gt.a)v=0.
return
end

 


Scattering by a Coulomb Field

 

by Reinaldo Baretti Machín , Alfonso Baretti Huertas, Reinaldo J. Baretti Huertas

 

 Free counter and web stats

Home page

http://www1.uprh.edu/rbaretti

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysics.htm

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart2.htm 

http://www1.uprh.edu/rbaretti/methodsoftheoreticalphysicsPart3.htm

 

References:

1. Principles of quantum mechanics; nonrelativistic wave mechanics with illustrative applications.    

by    W. V. Houston , Chapter XII .

2.  Lectures On Quantum Mechanics (Lecture Notes & Supplements in Physics Ser.)) by Gordon Baym , chapter 9

3. Quantum Mechanics Non-Relativistic Theory, Third Edition: Volume 3 by L. D. Landau and L. M. Lifshitz ,Chapter  xvii

4. Fundamentals of Modern Physics by Robert Martin Eisberg (Hardcover - June 1961) , chapter 9.

5. Advanced Quantum Mechanics by J. J. Sakurai

6. Schaum's Outline of Quantum Mechanics, Second Edition (Schaum's Outline Series) by Yoav Peleg, Reuven Pnini, Elyahu Zaarur,   and Eugene Hecht , chapter 15 .

7.  http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Born_approximation

 

Incoming charge = Z' e   nuclear charge   = + Ze  the potential energy is     V(r) = kcoulomb  (Z Z' e2 )/r  

The function Ll (r)   satisfies ,

 

                      (1/r2 ) d  { r2 (d L/dr )  }  /dr   + { k2 - (2 μ/h'2 ) V - l (l+1)/r2  } L (r)  = 0                        .  (1)

A length scale for this equation can be defined as    

Lscale  = h'2 /( μ kcoulomb  e2  )                                                                                                    .  (2)

If we had an electron in the hydrogen atom the parameters are,   μ = 9.1E-31 kg , Z=1 and  Z' =1, resulting in the well known atomic scale

                                Lscale =  5.24 E-11 m  (Bohr radius)                                                       . (3)

In the case of an alpha particle dispersed by a gold nuclei  μ = 4(1.67E-27) = 6.68E-27 kg , and the length scale is many orders of magnitude smaller

                                 Lscale =  7.14 E-15  m                                                                        .   (4)

This magnitude is of the order of the nuclear radius which is approximately R = 1.2E-15 A1/3   ~ meters   , A is the mass number.

A = 200,  for gold   the nuclear radius is  R = 7.02E-15 m   .

The energy scale is ,

Escale = h'2 /(μ L2 scale ) =  3.23 E-14 J =  0.202  MeV                                                    .     (5)

With this choice eq(1)  is simply

                                         (1/r2 ) d  { r2 (d L/dr )  }  /dr   + { k2 -  2V - l (l+1)/r2  } L (r)  = 0       ,          (6)

with the potential energy     V(r) =  Z Z'/r.

Eq (1) is solved by finite differences. The initial value of r is given by  ri = Z Z'/E  where E is the energy of the particle.

We assume L (ri) = 0 and dL(ri ) /dr =1 since the Columbic wall at the origin is like an infinite wall.

 

Example 1.  Let Z = 79 , Z'= 2   , E = 35 = 7.07 MeV

ri,lambda,rcutoff,nstep= 4.51428556 0.750984311 9.02857113 5000
al,e,k= 0. 35. 8.36660004

 

Fig 1.  The L0 (r) partial wave is pushed outwards, in comparison with f0(r) ,by the repulsive Coulomb potential.

The region plotted in Fig.1 is not  asymptotic. Nevertheless some order of magnitude for the phase of L(r2) can be ascertained. (see ref 1 page 187)

                                          φL = kr2 -(ZZ') ln (2kr2) - lπ/2 + ηl                            .                                 (7)

The term  (ZZ') ln (2kr2) is independent of the angular momentum and would appear in all expressions of  ηl  as

exp (-(ZZ') ln (2kr2) i) . Thus when the scattering amplitude is squared /fsc(θ )/2 these terms are equal to one.

The asymptotic phase of fl (r1) is

                                          φf = kr1  - lπ/2                                                        .                              (8)

We equate these two phases for two successive maxima. They actually differ by an unknown  multiple of 2π. This is not important since at the end we are dealing with

the function sin (η) = sin(η + m 2π) where m is an integer.

Let r2 = 7.95 be the position of a maxima in  L(r) , and r1 = 7.67 the position of a maxima of f(r) .

                                       η0 = k(r1 - r2 ) =  - 2.34 radians                                          .       (9)

 

 

We calculate next the phase shift  ηl  for l =1. 

 

    

 

 

 

FORTRAN code

c Coulomb scattering Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 186 -187
c attractive potentail well V = -D r< a
real k ,kprim ,lambda
data al, z, zprime /0.,79.,2./
data e ,nstep,ntrial /35.,5000, 1 /
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
V(r)= z*zprime/r
pi=2.*asin(1.)
ri=z*zprime/e
k=sqrt(2.*e)
lambda=2.*pi/k
rcutoff=2.*ri
ctheoretical eta, r2 maxima postion for L(r) , r1 maxima position f(r)
c eta= -k*(r2-r1) + C*log(2.*k*r2) ; C=Z*zprime
dr=(rcutoff-ri)/float(nstep)
c nstep=int((rcutoff-ri)/dr)
kp=int(float(nstep)/200.)
kount=kp
do 10 ie=1,ntrial
psi0=0.
c for al=0. , deriv=0.
deriv=.5
psi1=psi0 + dr*deriv
print*,'ri,lambda,rcutoff,nstep=',ri,lambda,rcutoff,nstep
print*,'al,e,k=',al,e,k
print*,' '
c print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2=2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr)-al*(al+1.)/(r-dr)**2 )*psi1 )
c if(i.le.40)print 150, r,fl(r) ,psi2
if(i.eq.kount)then
print 150, r,fl(r) ,psi2
c if(r.ge.rcutoff/2.)print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e10.3))
10 continue
stop
end



FORTRAN code for al=1.

c Coulomb scattering Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 186 -187
c attractive potentail well V = -D r< a 12 oct 2010
real k ,kprim ,lambda
data al, z, zprime /1.,79.,2./
data e ,nstep,ntrial /35.,6000, 1 /
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
V(r)= z*zprime/r
pi=2.*asin(1.)
c ri=z*zprime/e when al=0. , ri=9.03 when al=1.
c ri=z*zprime/e
ri=9.03
k=sqrt(2.*e)
lambda=2.*pi/k
rcutoff= 11.7
ctheoretical eta, r2 maxima postion for L(r) , r1 maxima position f(r)
c eta= -k*(r2-r1) + C*log(2.*k*r2) ; C=Z*zprime
dr=(rcutoff-ri)/float(nstep)
c nstep=int((rcutoff-ri)/dr)
kp=int(float(nstep)/50.)
kount=kp
c we consider the the repulsive Coulomb potentail as an impentrarbel
c wall starting at ri=z*zprime/e psi0=0. deriv =arbitrary postive e.g 1.
do 10 ie=1,ntrial
psi0=0.
c for al=0. , deriv=0.
deriv=.1
psi1=psi0 + dr*deriv
print*,'ri,lambda,rcutoff,nstep=',ri,lambda,rcutoff,nstep
print*,'al,e,E(Mev),k=',al,e,e*.202, k
print*,' '
c print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2=2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr)-al*(al+1.)/(r-dr)**2 )*psi1 )
if(i.eq.kount)then
print 150, r,fl(r) ,psi2
c if(r.lt..5*rcutoff)print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e10.3))
10 continue
stop
end



 

The Born Approximation (see ref. 7)

Write Schrodinger equation as

\left(\frac{\hbar^2}{2m} \nabla^2 + E  \right) \psi(\mathbf{r}) = V(\mathbf{r})\psi(\mathbf{r})\,.                                                                                          (12) 

where the right side acts as a perturbation.     

 

where E is given and V(r) is the scattering potential. The plane waves along The Z axis ,  Φk =exp(ikz)  , are solutions of

           - (h'2 /2m) 2 Φk =  E Φk                                                                                        (13)

 

An approximation to the solution is

   Ψk (r) =   exp(ikz)  - (2m/h'2 ) ∫ d3 r' Gk (r , r ' ) V(r ' ) Ψk (r' )                                  (14-a)

which is equated to

                           Ψk  (r)  = exp( ikz ) + (1/r) exp(ikr) f (θ)                            .              (14-b)

                 

where Gk (r , r ' ) is the Green function of the operator L=   2 + k2 , k2 = 2E , letting  m=1 ,h'=1 . G  satisfies

                                   ( 2 + k2 ) G = - δ (r - r')                                                ,           (15)

with the same boundary conditions as the scattered waves.

The expressions for G are 

                     Gk (r , r ' ) = -(1/4π) {1/abs( r - r' )} exp[ (+/- ) ik abs( r-r' ) ]       .       (16)

But we pick only the solution with the  positive phase  exp( ik /r-r'/ ) .

Approximations:

a) let u(r) =exp(i k z' )

b) assume  r>> r' and  let 1/abs( r -r' ) =  1/r      bringing it out of the integral  

c) the phase  k / r - r '/ ≈  k (r - r' cos(α) )  , see the figure below.                                            

                                                

Fig 5 . Visualization of the incident wave vector k0 , the scattered vector k , the difference K= k -k0  , the unit vector  ez , and the vectors r  and r'.

 From the figure we can see that    k0 z' = k0 r'   and   k cos(α)  k  =  k r'  . Substituion of all approximations yields

 

                       (1/r) exp(ikr) f (θ) = (1/4π)(2m/h'2 )  exp(ikr)/r) ∫ exp( i K ∙r' ) V(r') dτ'           ,        ( 17  )

where K = k - k0 = scattered wave vector - incident wave vector. The magnitude of capital  K = 2 k sin (θ/2) , see Fig 5.

The scattering amplitude is then

                        f (θ) =   (2m/h'2 ) ∫ exp{ i K r' cos ( θ) } V(r') sin (θ) dθ dφ r'2 dr'        .             ( 18 )  

Integration over φ produces a factor of 2π , leaving the expression

                    f (θ) =   ( m/h'2 ) ∫ exp{ i K r' cos ( θ) } V(r') sin (θ) dθ  r'2 dr'                    (19)

The integral

 ∫ exp{ i K r' cos ( θ) } sin (θ) dθ = (1/(Kr') )(i) (-2i) sin (Kr') = (2/(Kr')) sin (Kr')        .     (20)

 

We finally arive at , 

                                     f (θ) =  (2 m/h'2 ) (1/K) ∫ V(r') sin (Kr')   r' dr'             .   (21)

Example : Find f( θ) for the screened potential  V(r) = V0 exp(-α r)  .

The integral is 

                               ∫ V0 exp(-αr') sin (Kr')  r' dr'  =  2 α K V0 /( K2 + α2 )2         ,    (22)

hence

                                       f (θ) =  (4 α V0 m/h'2 ) /  ( K2 + α2 )2   ~  length            .     (23)   

MAXIMA code

 

 

             

Fig 6. Plot of   α /  ( K2 + α2 )2   vs K , α = 1.

 

Notice that the range of r' in (22)  was from 0 to infinity while it was assumed in the derivation above that  r' << r .

 

Example:  Use the Born approximation the scattering amplitude produced by  a screened Coulomb potential 

V(r) = A exp(-α r) /r  ., where A = kcoulomb Z Z' e2 . The scattering amplitude is

 

   f (θ) =  (2 m/h'2 ) (1/K) ∫ { A exp(-α r) /r } sin (Kr')   r' dr'

           = (2 m/h'2 ) (1/K) {K A/ ( K2 + α2 ) }

           = (2 m/h'2 ) A / ( K2 + α2 )   .

Letting   α  go to zero, the Rutherford scattering is retrieved. But this is not entirely legitimate since it was assumed the V(r') was of short range in the approximations used to derive  Born formula.                                                    

 

 

MAXIMA code

 

 

     

 

 

 

 

FORTRAN code

c Coulomb scattering Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 186 -187
c attractive potentail well V = -D r< a
real k ,kprim ,lambda
data al, z, zprime /0.,79.,2./
data e ,nstep,ntrial /35.,5000, 1 /
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
V(r)= z*zprime/r
pi=2.*asin(1.)
ri=z*zprime/e
k=sqrt(2.*e)
lambda=2.*pi/k
rcutoff=2.*ri
ctheoretical eta, r2 maxima postion for L(r) , r1 maxima position f(r)
c eta= -k*(r2-r1) + C*log(2.*k*r2) ; C=Z*zprime
dr=(rcutoff-ri)/float(nstep)
c nstep=int((rcutoff-ri)/dr)
kp=int(float(nstep)/200.)
kount=kp
do 10 ie=1,ntrial
psi0=0.
c for al=0. , deriv=0.
deriv=.5
psi1=psi0 + dr*deriv
print*,'ri,lambda,rcutoff,nstep=',ri,lambda,rcutoff,nstep
print*,'al,e,k=',al,e,k
print*,' '
c print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2=2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr)-al*(al+1.)/(r-dr)**2 )*psi1 )
c if(i.le.40)print 150, r,fl(r) ,psi2
if(i.eq.kount)then
print 150, r,fl(r) ,psi2
c if(r.ge.rcutoff/2.)print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e10.3))
10 continue
stop
end



FORTRAN code for al=1.

c Coulomb scattering Ref.W. V. Houston Principles of
c Quantum Mechanics, Dover Publ. , page 186 -187
c attractive potentail well V = -D r< a 12 oct 2010
real k ,kprim ,lambda
data al, z, zprime /1.,79.,2./
data e ,nstep,ntrial /35.,6000, 1 /
fl(r)=(1./(k*r))*sin(k*r-al*pi/2.)
V(r)= z*zprime/r
pi=2.*asin(1.)
c ri=z*zprime/e when al=0. , ri=9.03 when al=1.
c ri=z*zprime/e
ri=9.03
k=sqrt(2.*e)
lambda=2.*pi/k
rcutoff= 11.7
ctheoretical eta, r2 maxima postion for L(r) , r1 maxima position f(r)
c eta= -k*(r2-r1) + C*log(2.*k*r2) ; C=Z*zprime
dr=(rcutoff-ri)/float(nstep)
c nstep=int((rcutoff-ri)/dr)
kp=int(float(nstep)/50.)
kount=kp
c we consider the the repulsive Coulomb potentail as an impentrarbel
c wall starting at ri=z*zprime/e psi0=0. deriv =arbitrary postive e.g 1.
do 10 ie=1,ntrial
psi0=0.
c for al=0. , deriv=0.
deriv=.1
psi1=psi0 + dr*deriv
print*,'ri,lambda,rcutoff,nstep=',ri,lambda,rcutoff,nstep
print*,'al,e,E(Mev),k=',al,e,e*.202, k
print*,' '
c print 150,ri,fl(ri),psi1
do 20 i=2,nstep
r=ri+float(i)*dr
psi2=2.*psi1-psi0+dr**2*( -(2./(r-dr))*(psi1-psi0)/dr - (k**2
$ -2.*v(r-dr)-al*(al+1.)/(r-dr)**2 )*psi1 )
if(i.eq.kount)then
print 150, r,fl(r) ,psi2
c if(r.lt..5*rcutoff)print 150 ,r, fl(r),psi2
kount=kount+kp
endif
psi0=psi1
psi1=psi2
20 continue
100 format('r,psi=',2(3x,e10.3))
150 format('r,f(r),psi=',3(3x,e10.3))
10 continue
stop
end