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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

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  A_l P_l ( cos(θ) ) f_l (r)  ,    (2)

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

The functions f_l (r) satisfy the DE

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

with boundary conditions  lim f_l (r) _{r →0} = r ^l  , lim f_l (r) _{r → ∞} = 0 .

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

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

                                                        f_l (r) = (π/(2kr))^1/2  J_{l +1/2} (kr)                                         .     (4)

The asymptotic from of f_l (r) is

                                                                f_l (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 L_l (r) satisfying  the new DE

                                  d² L _l (r) /dr² + (2/r) d L_l (r)/dr + [ k² -2U(r) - l( l+1) /r² ] L_l (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 L_l (r) ,

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

differing from  the asymptotic f_l (r) in (5), by a the phase constant η_l .

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

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

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

                                                       η_l = - ( 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  ≈ 2^1/2       ,  to four digits.                                  .              (10) 

 Further numerical  examination reveals  ,see the results of Table 2 with k=2^1/2 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=2^1/2   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=2^1/2 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  η_l are connected to  measurable quantities like the differential cross section , dσ/dΩ or the total cross section σ_T ~ 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 η_l ) -1) P_l (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   η_l .  

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) P_l (cos( θ) ) {exp(2 i η_l ) -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 P_l (cos(θ) ) P_l' (cos(θ) ) sin(θ) dθ = 2 δ _{l , l'}  /(2l +1)  and  ( exp(2i η_l ) -1) (  exp(-2i η_l ) -1) = 4 sin² (η_l ) . ( 17)  

This gives the relatively simple expression for the total cross section 

                                    σ_T =   (4π /k²) ∑_l=0 (2 l +1)  sin² ( η_l  ) ≡ ∑_l=0  σ_l ~   area                     .      (18)            

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

Hence  σ_T  ≈  σ₀ =  (4π /k²) sin² ( η_l  ) = (4π /2 ) sin²( -2^1/2 ) = 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

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

 At low energies   / η₀ / =  ka << 1   and  sin ²( ka  ) ≈ (ka)² and  σ_T =  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= 2^1/2 , 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.

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