Imagine many particles fired at and scattering from a small target

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14. Scattering
14A. Cross Section and Differential Cross Sec.
Cross Section

• Imagine many particles fired at and scattering
  from a small target
• We want to predict the rate ? at which particles
  are scattered
• It makes sense that it would be proportional to
  thenumber density n of the projectiles
• It makes sense that it would be proportional to
  the speedv of the projectiles
• Call the remaining proportionality constant ?
• The cross-section ? has units of area
• Classically, it is just the size of the target as
  viewed from the direction the projectiles are
  approaching
• This formula is the definition of ?, and hence is
  correct by definition
• ?v is the difference in speeds, even if
  everything is relativistic

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Flux and Differential Cross-Section

• The flux (in or out) is the number density times
  the relative velocity
• Imagine we are far from a small target
• It makes sense that scattered particles will be
  moving radially outwards
• Imagine a spherical detector radius r covering
  all possible angles
• The total rate for scattering will then be
• The cross-section is then
• If detector only covers some angles, restrict
  integral
• Useful to define the differential cross-section

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Lab Frame and Center of Mass Frame

• There are two types of collision experimentsthat
  commonly occur
• The lab frame, where thetarget is stationary and
  theprojectile collides with it
• The center of mass frame,where the incoming
  particleshave net zero momentum
• Usually easier to calculate in the cm frame
• But then we must relate these two frames
• Let the mass of the projectile be m and the
  target M
• Commonly write things in terms of the ratio
• The velocity of the center of mass compared to
  the lab frame can be found

v'
v'cm
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Choice of Axis and Description of Angles

• Denote the velocities beforehand by v and vcm
• Denote the velocities after by v' and v'cm
• We will assume that the initial direction isin
  the z-direction
• Then the scattering angles ? and ?L are thepolar
  angle of the direction of the outgoingparticles
  in the two frames
• The azimuthal angle ? and ?L will be the same
• However, the polar angles ? and ?L will not
  beequal, because of the change in velocities
  between the two frames
• The velocities in the two frames are related by
  the relative speed of the frames, ucm

v'
v
v'cm
vcm
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Comparing Angles in the Two Frames

• It follows that
• From conservation of momentum in the cm frame, it
  isclear that the final particles are back to
  back andtheir momenta must be equal and opposite
• In cm frame, conservation of energy implies
• Now, lets do some geometry
• The sum of two interior angles of atriangle
  equals the other exterior angle
• Now use the law of sines

v'
v
v'cm
vcm
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An Equivalent Formula

• Algebra can make this into a more useful formula

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Differential Cross-Section In the Two Frames

• The cross-section and the differentialcross-secti
  on can be computed in either the cm or the lab
  frame
• But it is usually easier in the cm frame
• Well do all calculations in the cm frame
• Because the density n, the rate ?, and the
  relative velocity ?v are all the same in both
  frames, the total cross-section ? will be the
  same
• But because the angles are different, the
  differential cross-section will not be the same

True even relativistically
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Quantum Mechanics and Cross-Section

• We need to figure out how to calculate these
  expression in QM
• Typically, we dont want to use multiple
  particles at once
• Fortunately, even for one particle, we have a
  prettygood idea of what the density and the flux
  are
• Number density corresponds to probability
  density
• Flux corresponds to probability current
• Our formula for differentialcross section
  becomes
• Note that since j appears in numeratorand
  denominator, no need to normalize ?

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Solution in the Asymptotic Region

• We now need to start solvingSchrödingers
  Equation
• First define the asymptotic wavenumber k and the
  scaled potential U by
• Schrödingers equation is now
• Assume potential vanishes (sufficiently quickly)
  at infinity
• Roughly speaking, it must vanish faster than 1/r
• For the incoming wave, we expect a plane wave in
  the z direction
• For the outgoing wave, makes sense to work in
  spherical coordinates
• Since we are taking r ? ?, drop any terms with
  1/r2

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Asymptotic Form and Probability Currents

• The gk corresponds to waves coming in from
  infinity
• Physically wrong, so ignore it
• The wave at infinity is therefore
• We will needprobability currents
• Probability current in
• Probability current out

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Cross Section from Asymptotic Form

• Now we can do the differential cross-section
• And then we can get the total cross-section
• Of course, we dont yet have a clue what fk is

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14B. The Born Approximation
The Basic Idea

• We are trying tosolve Schrodinger
• We know what ?in is, so write
• Substitute in
• Now, imagine that we knew the right-hand side
• We want to find a straightforward way to find ?
  if we know the right side
• We will use superposition
• Treat right hand side as a superposition of point
  sources
• First step replace right side by a point source
  at the origin and solve equation

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The Greens Function

• Replace right side by a point source
• To simplify, place it at the origin
• Call the result the Greens function G(r)
• Spherically symmetric problem shouldhave a
  spherically symmetric solution
• Write this out inspherical coordinates
• Away from theorigin, this is
• This has solutions
• But we are looking for outgoing waves, not
  incoming, so
• We still have to make sure it works at the origin
  
• This will tell us what ? is

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Getting G to Work at the Origin

• It is hard to check this equation at the origin,
  becauseeverything is infinite
• To avoid this problem, integrate over a sphere of
  tiny radius R
• Take the limit R ? 0
• We now know that
• We could easily have put the sourcedelta-function
  anywhere, so we generalize
• Laplacian is understood to affect r, not r'

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Solving the Actual Problem

• We solved our equation for asource that is an
  arbitrary point
• We now use this to solve the actual problem
• Multiply top equation by U(r')?(r')
• Integrateover r'
• We therefore see that
• Substitute back in to
• We therefore have

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The Born Approximation

• This looks useless
• You can find ? if you know ?
• Substitute this equation into itself repeatedly
• This is a perturbative expansion in U
• If you keep just the first integral, it is called
  the first Born approximation, or sometimes, the
  Born approximation
• We wont go past the first Born approximation

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Asymptotic Behavoir

• We need behavior at large rto calculate
  cross-section
• Define the change in wave number
• Compare with the general asymptotic form
• We therefore have

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Differential and Total Cross-Section

• We are now ready to get the cross section using
• You then get the total cross section from
• Often need to write K in Cartesian coordinates
• Also handy to work out K2

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Sample Problem
Calculate the differential and total
cross-section for scattering from the potential

• Work on the Fourier transform of the potential
• Since potential is spherically symmetric, we can
  pretend K is any direction
• For convenience, pick it in the z-direction for
  this integration
• Done thinking of K in z-direction
• Now get the differential cross-section
• Recall that

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Sample Problem (2)
Calculate the differential and total
cross-section for scattering from the potential

• Now get the total cross-section

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The Coulomb Potential

• Consider the Coulomb potential, given by
• We immediately get the differential cross
  section
• Rewrite in terms of the energy E ?2k2/2?
• Why this is a cheat
• We assumed potential falls off faster than 1/r,
  for ? 0 it does not
• Turns out this only causes a shift in the phase
  of ? at large r, so answer is right
• Total cross-section is infinite
• Comes from small angles
• Experimentally, there is a minimum angle where we
  can tell if it scattered

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14C. Method of Partial Waves
Spherically Symmetric Potentials

• Suppose our potential is not small, but is
  spherically symmetric
• We generally know how to find somesolutions to
  Schrödingers equation
• The radial function then satisfies
• Which we rewrite as
• Second order differential equation has two
  linearly independent solutions

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Small r Behavior

• For small r, the 1/r2 term tendsto dominate the
  U and k2 terms
• Two linearly independent solutions, that roughly
  go like
• The latter one is unacceptable, because we want ?
  finite
• This means for each l, there is only one
  acceptable solution, up to normalization
• We can always find these solutions, numerically
  if necessary
• Assume we have done so
• For large l, solution Rl tends to be very small
  near the origin
• Roughly, it vanishes if kr ltlt l
• If U(r) vanishes or is negligible for r gt r0, you
  can ignore U(r) if l gt kr0

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Large r Behavior

• Assume the potential falls off quickly at large r
  then ignore U(r)
• Define x kr, then
• This equation has two known exact linearly
  independent solutions, called spherical Bessel
  functions
• Most general solution, at large r, is therefore a
  linear combination of these

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Spherical Bessel Functions

• The spherical Bessel functions are closely
  related to regular Bessel functions
• The jls are small at x 0, andthe nls diverge
• We most want their asymptoticbehavior at large x

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Asymptotic Solution

• We assume we have the radial wave functions Rl,
  up to normalization
• At large r, we knowit takes the form
• Write these constants in the form
• The phase shift ?l is determined by the behavior
  of the Rls
• The phase shift is what you need to finish this
  analysis
• It vanishes for large l, because U(r) is
  irrelevant and nl badly behaved
• The amplitude A is arbitrary
• For large r, we now know what our solution looks
  like
• Use asymptotic form of spherical Bessel functions

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Our Remaining Goal

• Write this in terms of exponentials
• This expression has waves coming in and going out
  in all directions
• We want a wave that looks like
• Most general solution will be a linear
  combination
• Want to find a combination to make it look like
  what we want
• First step write eikz in terms of spherical
  harmonics at large r

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eikz in Spherical Harmonics at Large r

• Since spherical harmonics are complete,
  everythingcan be written in terms of them,
  including eikz
• The functions clm can be found using
  orthogonality of the Ylms
• The Ylms go like eim?, sothe ? integral is
  easy
• Integrate byparts oncos?repeatedly
• We know thesefunctions, so

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Putting it Together . . .

• We have
• We want
• Where
• We have to get the e-ikrterms to match, so try
• Substitute this in
• Compare to eikz
• We therefore have

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The Differential and Total Cross-Section

• Some algebra
• The differentialcross section
• The total cross-section

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The Procedure

• Find U(r)
• Now, for each value of l
• Find appropriate inner boundary conditions
  (typically Rl(0) 0)
• Solve numerically or analytically
• At large r, match your solution to
• Deduce phase shift
• Stop when ?l gets small, or l gtgt kr0, where r0 is
  where the potential gets small
• Differential cross-section
• Total cross-section

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Sample Problem
Calculate the differential and total
cross-section from a hard sphere of radius a,
with potential as given at right, where ka is
small

• Perturbation theory cannot work, because the
  potential is infinite
• U(r) V(r), because it is zero or infinity
• Boundary condition must be that Rl(a) 0
• We need to solve outside
• Solution of this is known
• Boundary condition
• Now we attempt to find phase shift

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Sample Problem (2)
Calculate the differential and total
cross-section from a hard sphere of radius a,
with potential as given at right, where ka is
small

• Now use approximation ka small
• This means we need to keep up to l ka ltlt1
• Just keep l 0
• Look up spherical Bessel functions
• Get the phase shift
• Now find differentialcross section
• We assumed ka small, so

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Comments on Cross Sections

• Interestingly, the cross-section is equal to the
  surface area (NOT thesilhouette area)
• Generally described as diffraction allows
  particles to scatter off of all sides
• Note that whenever the scale of the potential is
  small, scattering is dominated by l 1 (s-wave
  scattering)
• All angles are scattered equally

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Limits on Total Cross Sections

• Note that if a particular value of l dominates
  thecross-section, then there is a limit on the
  cross section
• Before the discovery of the Higgs boson, it was
  pointed out that withoutthe Higgs boson, the
  cross-section for WW scattering was predictedto
  be dominated by l 0, and it grows as k2
• Higgs boson cancels part of amplitude, and
  suppresses the cross section, once you get at or
  near the Higgs mass
• Predicted Higgs, or something had to be lighter
  than 1000 GeV/c2
• No lose theorem
• Higgs discovered in 2012 at 126 GeV/c2