1. Particlesymbolrest energy in MeV electron e 0.5109989 muon  105.658357 neutral pion  0 134.9766 charged pion   139.57018 proton p 938.27200 neutron n 939.56533 deuteron H 2 1875.580 triton H 3 2808.873 alpha  (He 4 ) 3727.315 or mass in MeV/c 2

2. Imagine a narrow, well-collimated beam of mono-energetic particles passing through a slab of matter EoEo E EoEo

3. EoEo E Energy loss E1E1 EE

4. 0 beam direction,   Similarly, the initially well-defined beam direction suffers at least small-angle scattering:

5. EoEo E Energy loss E1E1 EE If the target is thick, this implies that the overall mean energy loss  thickness For sufficiently high initial E 0 (or thin enough targets) all particles get through.

6. EoEo E 0

7. Electrons e  so light they can scatter madly, suffering large deflections & energy losses Ionization Region Radiation Region Two regions are defined by an emperical “Critical Energy” Z of target atoms dE dx just like protons, etc except range ( penetration depth ) is measured over actual path length Bremsstrahlung (braking radiation)

8. Bremsstrahlung braking radiation Photon energies radiated, E  Resulting in “all-or-nothing” depletion of the beam’s energy “radiation length” after ~7X 0 only 1/10 3 of the initial electron energy remains

9. If Bremsstrahlung photons energetic enough > 1 MeV they can pair produce The mean free path of a (high energy) photon through matter is

10. e-e- e-e- e-e- e+e+  Bremsstrahlung Pair production Photon or “gamma” ray Counting interaction lengths

11. Photons interact within matter via 3 processes 1.The photo-electric effect 2.Compton scattering 3.Pair production  e-e- 1.   p1p1 p2p2 meme 2.

12. Total absorption coefficients of  rays by lead and aluminum as a function of energy. W. Heitler, The Quantum Theory of Radiation, The Clarendon Press, Oxford, 1936. ħ  /mc=1 corresponds to 511 keV E  ~keV Pair production impossible and Compton cross section dominates at E=2m e c 2 pair production turns on

13. Experimental Objects muon chambers stee l HAD calorimeter EM calorimeter solenoid jet e  electrons & photons quarks & gluons neutrinos  K,  etc. tracking volume Quarks & gluons do not exist (for long) as free particles Due to hadronization we observe a collimated spray of particles (“jet”) Neutrinos escape without detection Electrons and photons deposit most of their energy in the EM calorimeter  muons

14. DØ 5500 tons 120,000 digitized readout channels

17. For “free” particles (unbounded in the “continuum”) the solutions to Schrödinger’s equation with no potential Sorry!…this V is a volume appearing for normalization

18. q q pipi pipi q = k i  k f =(p i -p f )/ħ momentum transfer the momentum given up (lost) by the scattered particle

19. We’ve found (your homework!) the time evolution of a state from some initial (time, t 0 ) unperturbed state can in principal be described using: complete commuting set of observables, e.g. E n, etc… Where the |  t  are eigenstates satisfying Schrödingers equation: Since the set is “complete” we can even express the final state of a system in terms of the complete representation of the initial, unperturbed eigenstates |  t 0 .

20. give the probability amplitudes (which we’ll relate to the rates) of the transitions |  t 0  |  ″ t  during the interval ( t 0, t ). You’ve also shown the “matrix elements” of this operator (the “overlap” of initial and potential “final” states) to use this idea we need an expression representing U !

21. HI(t)HI(t)HI(t)HI(t) Operator on both sides, by the Hamiltonian of the perturbing interaction: Then integrate over (t 0,t) HI(t′)HI(t′) HI(t′)HI(t′)  t0t0 t dt′ t′t′  t0t0 t t′t′  t0t0 t t't'  t 0 t

22. Which notice has lead us to an iterative equation for U I U U I U(t t o ) =U U

23. If at time t 0 =0 the system is in a definite energy eigenstate of H 0 ( intitial state is, for example, a well-defined beam ) H o |E n,t 0 > = E n |E n,t o > then to first order U(t t o )|E n,t 0 > = and the transition probability ( for f  0 ) Note: probability to remain unchanged = 1 – P !!

24. recall: ( homework !) H 0 † = H 0 (Hermitian!) where each operator acts separately on: † So:

25. If we simplify the action (as we do impulse in momentum problems) to an average, effective potential V(t) during its action from (t 0,t) ≈ factor out

26. The probability of a transition to a particular final state |E f t> The total transition probability: If ~ constant over the narrowly allowed  E

27.  E =2 h /  t

28. for scattering, the final state particles are free, & actually in the continuum n=1 n=2 n=3 n=n= 

29. With the change of variables:  

30. Notice the total transition probability  t and the transition rate

31. n=1 n=2 n=3 n=n= E dN/dE Does the density of states vary through the continuum?

32. vxvx vyvy vzvz Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for any fixed E, m this defines a sphere of velocity points all which give the same kinetic energy. The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere). dV = 4  v 2 dv

33. Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) We just argued the number of accessible states (the “density of states”) is proportional to 4  v 2 dv dN dE  E 1/2