1. 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 the number density n of the projectiles It makes sense that it would be proportional to the speed |v| 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 14. Scattering Cross Section 14A. Cross Section and Differential Cross Sec.

2. 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:

3. Lab Frame and Center of Mass Frame There are two types of collision experiments that commonly occur: The lab frame, where the target is stationary and the projectile collides with it The center of mass frame, where the incoming particles have 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 LL v cm  v'v' v' cm

4. Choice of Axis and Description of Angles Denote the velocities beforehand by v and v cm Denote the velocities after by v' and v' cm We will assume that the initial direction is in the z-direction Then the scattering angles  and  L are the polar angle of the direction of the outgoing particles in the two frames The azimuthal angle  and  L will be the same However, the polar angles  and  L will not be equal, 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, u cm : v LL v cm  v' v' cm

5. Comparing Angles in the Two Frames It follows that From conservation of momentum in the cm frame, it is clear that the final particles are back to back and their momenta must be equal and opposite In cm frame, conservation of energy implies Now, let’s do some geometry: The sum of two interior angles of a triangle equals the other exterior angle: Now use the law of sines: v LL v cm  v' v' cm v' v' cm u cm LL   –  L

6. An Equivalent Formula Algebra can make this into a more useful formula

7. Differential Cross-Section In the Two Frames The cross-section and the differential cross-section can be computed in either the cm or the lab frame –But it is usually easier in the cm frame –We’ll 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

8. Quantum Mechanics and Cross-Section We need to figure out how to calculate these expression in QM Typically, we don’t want to use multiple particles at once Fortunately, even for one particle, we have a pretty good idea of what the density and the flux are: –Number density corresponds to probability density: –Flux corresponds to probability current: Our formula for differential cross section becomes: Note that since j appears in numerator and denominator, no need to normalize 

9. We now need to start solving Schrödinger’s Equation: First define the asymptotic wave number k and the scaled potential U by Schrödinger’s 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/r 2 : Solution in the Asymptotic Region

10. The g k corresponds to waves coming in from infinity –Physically wrong, so ignore it The wave at infinity is therefore We will need probability currents: Probability current in: Probability current out: Asymptotic Form and Probability Currents

11. Now we can do the differential cross-section: And then we can get the total cross-section Of course, we don’t yet have a clue what f k is Cross Section from Asymptotic Form

12. We are trying to solve 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 The Basic Idea 14B. The Born Approximation

13. Replace right side by a point source –To simplify, place it at the origin –Call the result the Green’s function G(r): Spherically symmetric problem should have a spherically symmetric solution: Write this out in spherical coordinates Away from the origin, 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 The Green’s Function

14. It is hard to check this equation at the origin, because everything 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 source delta-function anywhere, so we generalize Laplacian is understood to affect r, not r' Getting G to Work at the Origin

15. We solved our equation for a “source” that is an arbitrary point We now use this to solve the actual problem Multiply top equation by U(r')  (r') Integrate over r': We therefore see that Substitute back in to We therefore have: Solving the Actual Problem

16. 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 won’t go past the first Born approximation The Born Approximation

17. We need behavior at large r to calculate cross-section Define the change in wave number Compare with the general asymptotic form: We therefore have: Asymptotic Behavoir

18. 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 K 2 : Differential and Total Cross-Section

19. 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

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

21. Consider the Coulomb potential, given by We immediately get the differential cross section: Rewrite in terms of the energy E =  2 k 2 /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 The Coulomb Potential

22. Suppose our potential is not small, but is spherically symmetric We generally know how to find some solutions to Schrödinger’s equation: The radial function then satisfies Which we rewrite as: Second order differential equation has two linearly independent solutions Spherically Symmetric Potentials 14C. Method of Partial Waves

23. For small r, the 1/r 2 term tends to dominate the U and k 2 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 R l tends to be very small near the origin –Roughly, it vanishes if kr << l If U(r) vanishes or is negligible for r > r 0, you can ignore U(r) if l > kr 0 Small r Behavior

24. 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 Large r Behavior

25. l = 0 l = 1 l = 2 l = 3 jl(r)nl(r)jl(r)nl(r) The spherical Bessel functions are closely related to regular Bessel functions The j l ’s are small at x = 0, and the n l ’s diverge We most want their asymptotic behavior at large x: Spherical Bessel Functions

26. We assume we have the radial wave functions R l, up to normalization At large r, we know it takes the form Write these constants in the form The phase shift  l is determined by the behavior of the R l ’s The phase shift is what you need to finish this analysis –It vanishes for large l, because U(r) is irrelevant and n l 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 Asymptotic Solution

27. 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 e ikz in terms of spherical harmonics at large r Our Remaining Goal

28. Since spherical harmonics are complete, everything can be written in terms of them, including e ikz : The functions c l m can be found using orthogonality of the Y l m ’s: The Y l m ’s go like e im , so the  integral is easy: Integrate by parts on cos  repeatedly We know these functions, so e ikz in Spherical Harmonics at Large r

29. Announcements 3/2 ASSIGNMENTS DayReadHomework Todaynone14.3 Wednesdaynonenone Friday no classno class Test Thursday 1:45, Olin 102 Covers through 13 Reviews on Monday and Wed. Homework Solutions Chapter 12

30. We have We want Where We have to get the e -ikr terms to match, so try Substitute this in: Compare to e ikz : We therefore have: Putting it Together...

31. Some algebra: The differential cross section: The total cross-section: The Differential and Total Cross-Section

32. Find U(r): Now, for each value of l: –Find appropriate inner boundary conditions (typically R l (0) = 0) –Solve numerically or analytically: –At large r, match your solution to: –Deduce phase shift: –Stop when  l gets small, or l >> kr 0, where r 0 is where the potential gets small Differential cross-section: Total cross-section: The Procedure

33. 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 R l (a) = 0 We need to solve outside: Solution of this is known: Boundary condition: Now we attempt to find phase shift:

34. 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 <<1 Just keep l = 0 Look up spherical Bessel functions: Get the phase shift: Now find differential cross section: We assumed ka small, so

35. Interestingly, the cross-section is equal to the surface area (NOT the silhouette 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 Comments on Cross Sections

36. Limits on Total Cross Sections Note that if a particular value of l dominates the cross-section, then there is a limit on the cross section: Before the discovery of the Higgs boson, it was pointed out that without the Higgs boson, the cross-section for WW scattering was predicted to be dominated by l = 0, and it grows as k 2 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/c 2 –No lose theorem Higgs discovered in 2012 at 126 GeV/c 2