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Transcript
L22.P1
Lecture 22
Scattering
Classical scattering theory
Problem: given the scattering parameter b, calculate the scattering angle
Example: elastic hard-sphere scattering
Our target is a billiard ball of radius R,
the incident particle is a ball that
bounces elastically. The impact
parameter is
The scattering angle is
Therefore, the scattering angle is:
Lecture 22 Page 1
L22.P2
Scattering
More general problem: particles incident within an infinitesimal patch of crosssectional area
are scattering into an infinitesimal solid angle
.
The quantity
is called the differential (scattering) cross-section.
Total cross section is defined as the integral of D( ) over all solid angles:
Class exercise: Find the differential and total cross-sections for hard-sphere
scattering.
Lecture 22 Page 2
L22.P3
If we have a beam of incident particles, with uniform intensity (luminosity)
number of incident particles per
unit area per unit time
The member of particles entering
per unit time is
Lecture 22 Page 3
L22.P4
Quantum scattering theory
Our problem: incident plane wave
traveling in Z direction encounters a scattering potential that produces outgoing spherical
wave:
Therefore, the solutions of the Schrödinger equation have the general form:
energy of the incident
particles
The quantity
is called scattering amplitude.
It is the probability of scattering in a given direction
How is it connected to the differential cross-section?
Lecture 22 Page 4
.
L22.P5
Volume dV of incident beam (see above) passes through area d
in time dt.
The probability that the particle with speed v passes through this area d
is
This must be equal to the probability that the particle scatters into the solid angle
Therefore
Therefore, to solve the scattering problem, we need to calculate the scattering
amplitude
Lecture 22 Page 5
L22.P6
Partial wave analysis
Our potential is spherically symmetric
The solutions of the Schrödinger equation are
spherical harmonic
(radial Schrödinger equation)
Very large r
and "centrifugal contribution" is negligible
Radial equation becomes
The general solution is
outgoing spherical wave
incoming
spherical wave
Lecture 22 Page 6