Particle symbol rest 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 H2 1875.580

triton H3 2808.873

alpha (He4) 3727.315

or mass in MeV/c2

Imagine a narrow, well-collimated beam of mono-energetic particles

passing through a slab of matter

EoE

Eo

EoE

Energy loss

E1

E

0 beam direction,

Similarly, the initially well-defined beam direction

suffers at least small-angle scattering:

EoE

Energyloss

E1

E

If the target is thick, this implies that the overall mean energy loss thickness

For sufficiently high initial E0

(or thin enough targets)all particles get through.

EoE0

Electrons e

so light they can scatter madly, suffering large deflections & energy losses

Ionization Region Radiation Region

Z

MeVEc

600

Two regions are defined by an emperical “Critical Energy”

Z of target atoms

dEdx just like protons, etc

except range (penetration depth)is measured over actual path length

Bremsstrahlung(braking radiation)

Bremsstrahlung braking radiation

Photon energies radiated, E4

4

2

mc

E

Resulting in “all-or-nothing” depletion of the beam’s energy

0X/0

xeEE “radiation length”

after ~7X0 only 1/103 of the initial electron energy remains

If Bremsstrahlung photons energetic enough > 1 MeV

they can

pair produce

The mean free path of a (high energy) photon

through matter is

007

9XXX

ee

e-

e-

e-

e+

Bremsstrahlung

Pair productionPhoton or “gamma” ray

Counting interaction lengths

Photonsinteract within matter via 3 processes

1. The photo-electric effect2. Compton scattering3. Pair production

e-

1.

p1

p2

me

2.

)cos1( cm

h

e

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~keVPair production

impossibleand

Comptoncross section

dominates

atE=2mec2

pair production

turnson

Experimental ObjectsExperimental Objects

muonchambers

steel

HAD calorimeter

EM calorimetersolenoid

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

DØ 5500 tons120,000 digitized readout channels

/1 rpii

ieV

)r(

For “free” particles (unbounded in the “continuum”)

/1 rpif e

V)r(

f

the solutions to Schrödinger’s equation

with no potential

Sorry!…this V is a volume appearingfor normalization

rkiie

rkie

f

3* )(),( drrVkkF

V

iffi

iffi

M

3/)()( drrVe

rppifi

3)()( drrVe

rkkifi

3)()( drrVeqF rqi

q q

pi

pi q = ki kf =(pi-pf )/ħ

momentum transferthe momentumgiven up (lost)

by the scatteredparticle

00 ),( tdtttU

tttdt

di I )(H

We’ve found (your homework!) the time evolution ofa state from some initial (time, t0) unperturbed state

can in principal be described using:

complete commuting set of observables, e.g. En, 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 | t0.

tttttUt ),( 0000

give the probability amplitudes (which we’ll relate to the rates)

of the transitions | t0 |″ t during the interval ( t0, 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 !

00 ),( tdtttU HI(t) HI(t)Operator on both sides, by the Hamiltonian of the perturbing interaction:

Then integrate over (t0,t)

00 ),( tdtttU HI(t′) HI(t′)t0

t

dt′t′ t0

tdt′t′

tdtdttd

di

0

t0

tdt′

0

0

tdtdi t

t

dttti 00

0

td

td

tdtdi

dt′dt′

t'

t0

t

dtttitdttUtt

t 000I ) ()(H0

dttidtti 000

I 0 idtti

Which notice has lead us to an iterative equation for U

IU) (U)(H 0I0

iitdtttt

t

I

U U I

U(t to) = tdttti t

t ) (U)(H1 0I

0U

U

If at time t0=0 the system is in a definite energy eigenstate of H0

(intitial state is, for example, a well-defined beam)

Ho|En,t0> = En |En,to >then to first order

U(t to)|En,t0> = 0)(10

tEtdti

n

t

t HI

and the transition probability2

000 00

2

0 )( tdtEtHtEi

EUEt

Ifff

2

000 02)(

1tdtEtHtE

t

If ( for f 0 )

Note: probability to remain unchanged = 1 – P !!

recall: 00 )()( )(0

tHitHi etVeUtVUtH (homework!)

H0†=H0 (Hermitian!)

where each operator acts separately on:

00

0

0

0

tEe

etE

itH

tHif

2))(/(

000 02

2

00)(

1tdetEtVtEEUE tEEit

fff

†

So:

If we simplify the action (as we do impulse in momentum problems)to an average, effective potential V(t) during its action from (t0,t)

≈factor out

2

0

/

E

i

tt

t

tEie

2/

2

2

1 E

Etie

tEE

EE

tEVtEEUE f

f

efff

f 0

20

2

0002

0 cos-1 )(

2

0

))(/(2

0002

2

00

1tdetEVtEEUE

t tEEieffff

f

The probability of a transition to a particular final state |Ef t>

2

sin

||4 2

2

2tEE

EE

EVEP if

if

if

The total transition probability:

2

sin

||4 2

2

2tEE

EE

EVEP iN

iN

iN

Ntotal

If < EN|V|Ei > ~ constant over the narrowly allowed E

N iN

iN

iNtotalEE

tEE

EVEP 2

2

2 2sin

||4

tEE

EE

tEVtEEUE f

f

efff

f 0

20

2

0002

0 cos-1 )(

2

sin 4 02 tEE f

E=2h/t

for scattering, the final state particles are free, & actually

in the continuum

n=1

n=2

n=3

n=

N

2

iN

iN2

2

iNtotal EE2

tEEsin

E|V|E4P

2

i

i2

2

itotal E)N(E2

tE)N(Esin

dN E|V|)N(E4P

With the change of variables:

2

22

/2

sin2||4

tx

xdx

tdE

dNEVEP iNtotal

2/ )( tENEx i dNdN

dEtdx

2

dx

x

xt

dE

dNEVEP iNtotal 2

22 sin

2||4

2

i

i2

2

itotal E)N(E2

tE)N(Esin

dN E|V|)N(E4P

Notice the total transition probability t

dE

dNEVE

tP iNtotal

2||

2

and the transition rate

dE

dNEVEtPW iNtotal

2||2

/

n=1

n=2

n=3

n=

E

dN/dE

Does the densityof states varythrough thecontinuum?

vx

vy

vzClassically, for free particlesE = ½ mv2 = ½ m(vx

2 + vy2 + vz

2 )

Notice for any fixed E, m this definesa 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 = 4v2dv

Classically, for free particlesE = ½ mv2 = ½ m(vx

2 + vy2 + vz

2 )

dv mvdE mE

Ed

mv

Eddv

2

We just argued the number of accessible states (the “density of states”) is proportional to 4v2dv

mE

dE

m

ECdvvCdN

2

244 2

dEEm

CdN 2/1

2/32

4

dN

dE E1/2