1000 Solved Problems in Modern Physics

3.3 Solutions 235

oreikrf(θ)=

∑

Blpl(cosθ)

[

ei(kr−

π 2 l+δl)

−e−i(kr−

π 2 l+δl)]

2 ik

−

∑

il(2l+1)pl(cosθ)

[

ei(kr−

π 2 l)

−e−i(kr−

π 2 l)]

2 ik

Equating coefficients ofe−ikr

0 =−

∑( 1

2 ik

)

Blpl(cosθ)

[

e−i(−πl/^2 +δl)+

∑

il(2l+1)Pl(cosθ)

2 ik

eiπl/^2

]

Therefore

Bl=il(2l+1)eiδl (12)

Equating coefficients ofeikr, and using the value ofBl

f(θ)=

(

1

2 ik

)[∑

il(2l+1)eiδlpl(cosθ)ei(−

πl

2 +δl)

−

∑

il(2l+1)pl(cosθ)e−

πl

2

]

=

(

1

2 ik

)∑

il(2l+1)Pl(cosθ)e−iπl/^2

[

e^2 iδl− 1

]

(13)

Using (10), formula (12) becomes

f(θ)=

1

2 ik

∑

(2l+1)(e^2 iδl−1)Pl(cosθ)

The above method is called the method of partial wave analysis. The sum-

mation over various integral values oflmeans physically summing over var-

ious values of angular momenta associated with various partial waves. The

quantityδlis understood to be the phase shift when the potential is present. At

low energies only a fewlvalues would be adequate to describe the scattering.

3.105 The differential cross-section for the scattering of identical particles of spin
sisgivenby

σ(θ)∗=|f(θ∗)|^2 +|f(π−θ∗)|^2 +

[

(−1)^2 s

2 s+ 1

]

2 Re[f(θ∗)f∗(π−θ∗)] (1)

where f is assumed to be independent of the azimuth angleφ. The angles refer

to the CM-system. The first two terms on RHS are given by the Rutherford

scattering, one for the scattered particle and the other for the target particle.

In the CMS the identical particles are oppositely directed and the detector

cannot tell one from the other. The third term on the RHS is due to quantum

mechanical interference and does not occur in the classical formula. Now for

alpha-alpha scatterings=0 and (1) reduces to

σ(θ∗)=|f(θ∗)|^2 +|f(π−θ∗)|^2 + 2 Re[f(θ∗)f∗(π−θ∗)].

Furthermore if the scattering atθ∗= 90 ◦is considered then obviouslyf(π−

θ∗)= f(θ∗) and the lab angleθ= 45 ◦. In that case classicallyσL(45◦)=

2 |f(90◦)|^2 CMwhile quantum mechanically