Particle
symbol
rest energy in MeV
or mass in MeV/c2
electron
e
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
0.5109989
Imagine a narrow, well-collimated beam of mono-energetic particles
Eo
passing through a slab of matter
Eo
E
DE
Energy loss
E1
Eo
E
Similarly, the initially
well-defined beam direction
suffers at least
small-angle scattering:
0
D
beam direction, 
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.
Energy
loss
DE
E1
Eo
E
0
Eo
E
Electrons eso light they can scatter madly,
suffering large deflections & energy losses
Two regions are defined by an emperical “Critical Energy”
600MeV
Ec 
Z
Z of target atoms
Ionization Region
Radiation Region
dE
dx just like protons, etc
Bremsstrahlung
except range
(penetration depth)
is measured over
actual path length
(braking radiation)
Bremsstrahlung
braking radiation
4
 E 
4
Photon energies radiated, E  

2 
 mc 
Resulting in “all-or-nothing” depletion of the beam’s energy
E  E0e
- x / X0
“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
X  e  e -
9
 X0  X0
7
Photon or
“gamma” ray
Pair production
e+

e-
eeBremsstrahlung
Counting interaction lengths
Photons
interact within matter via 3 processes
1. The photo-electric effect
2. Compton scattering
3. Pair production
e-
1.

me
2.
p1
h
D 
(1 - cos )
mec


p2
E~keV
Pair production
impossible
and
Compton
cross section
dominates
at
E=2mec2
pair
production
turns
on
ħ /mc=1 corresponds to 511 keV
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.
Experimental Objects
muon
chambers
stee
l
HAD calorimeter
EM calorimeter
solenoid

tracking volume
Electrons and
photons deposit
most of their
energy in the
EM calorimeter
e
electrons & photons
muons
Quarks & gluons
do not exist (for
long) as free
particles
jet

, K, , L,  etc.
quarks & gluons
Due to
hadronization we
observe a
collimated spray of
particles (“jet”)
Neutrinos
escape without
detection
n
neutrinos
DØ 5500 tons
120,000 digitized readout channels
For “free” particles (unbounded in the “continuum”)


 
ipi r / 
1

e
i(r ) 
V
 
ip f r / 
1

e
f (r ) 
V
the solutions to
Schrödinger’s equation
with no potential
Sorry!…this V is
a volume appearing
for normalization
e
e
 
iki r
 
ik f r
M fi   f V  i
 
*
3
 F ( ki , k f )   f V ( r ) i dr
 e
  
i ( pi - p f )r / 
V ( r )dr
 

i ( ki - k f )r
 e
3
V ( r )dr
3


iqr
3
F ( q )   e V ( r )dr
momentum transfer
pi
pi
q = ki - kf =(pi-pf )/ħ
the momentum
given up (lost)
by the scattered
particle
We’ve found (your homework!) the time evolution of
a state from some initial (time, t0) unperturbed state
can in principal be described using:
U (t , t0 )     t d    t0
complete commuting set of observables, e.g. En, etc…
Where the | t are eigenstates
satisfying Schrödingers equation:
d
i  t  H I (t )  t
dt
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.
You’ve also shown the “matrix elements” of this operator
(the “overlap” of initial and potential “final” states)
  t0 U (t , t0 )   t0    t0   t
give the probability amplitudes (which we’ll relate to the rates)
of the transitions | t0 
|″ t  during the interval ( t0, t ).
to use this idea we need an expression representing U!
Operator on both sides, by the Hamiltonian of the perturbing interaction:
HI(t)U (t , t0 )   HI(t)   t d    t0
Then integrate over (t0,t)
t
t HI(t′) U (t′, t0 ) dt′  t  HI(t′)   t′ d    t0
0
t
dt′
0
d
   i
  t  d    t0 ddt′
t
t0
dt 
d   tt'
t
 i   ddt′
t
d   t0
dt′t 
d
t0
t
 i  
  t
  t0
d   t  d   t0
 i     t -   t0
  t
0
d 

t0
t
H I (t )U (t  t0 )dt   i     t -   t0   t0 d 
 i    t    t0 d  - i    t0   t0 d 
 i    t    t0 d  - iI
U
Which notice has lead us to an iterative equation for U
t



H
(
t
)
U
(
t
t
)
d
t
 iU - iI
I
0
t0
i t
U(tto) = 1 - t H I (t ) U(t  t0 )dt 
 0
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
i t
U(tto)|En,t0> = 1 - t H I (t )dt  Ent0
 0
and the transition probability
E f U E0
( for f  0 )
2
i t
  f 0 -  E f t0 H I (t ) E0t0 dt 
 0
1
 2


t
0
E f t0 H I (t ) E0t0 dt 
Note: probability to remain unchanged = 1 – P !!
2
2
recall:
†
H 0 (t )  U V (t )U  e
 i tH 0 
V (t )e
- i tH 0 
H0†=H0 (Hermitian!)
where each operator acts separately on:
E f t0 e itH0
e
So:
E f U E0
2
1
 2


t
0
E f t0 V (t ) E0t0 e
(homework!)
-itH0 

E0t0
( i /  )( E f - E0 ) t 
dt 
2
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)
≈
E f U E0
2
factor out
1
 2 E f t0 Veff E0t0

2
t
e
0
( i /  )( E f - E0 ) t 
dt 
2
E f U E0
2

E f t0 Veff E0t0
( E f - E0 )
2
2
i iDEt  /  t t
e
t 0
DE
2
2
iDEt / 
e
-1
2
DE 
E f - E0 

1 - cos
t 



2
The probability of a transition to a particular final state |Ef t>
P4
 E f | V | Ei 
E
2
- Ei 
sin
2
f
2
E
f
2
The total transition probability:
Ptotal   4
N
 E N | V | Ei 
2
EN - Ei 
2
- Ei  t
sin
2
EN - Ei  t
2
If < EN|V|Ei > ~ constant over the narrowly allowed DE
Ptotal  4  EN | V | Ei 
2

N
sin
2
EN - Ei t
2
2
EN - Ei 
E f U E0
2

E f t0 Veff E0t0
( E f - E0 )
2
2
E f - E0 

1 - cos
t 




E f - E0  t 

2
 4 sin

2


DE=2h/Dt
for scattering, the final state
particles are free, & actually
in the continuum
- Ei t
sin
2
2
Ptotal  4  EN | V | Ei  
2
N
EN - Ei 
2
E
n=
N
n=3

Ptotal  4  E( N) | V | Ei   dN
2
sin
2
E( N) - E t
2
2
E( N) - Ei 
i
n=2
n=1

Ptotal  4  E( N) | V | Ei   dN
2
sin
2
E( N) - E t
i
2
2
E( N) - Ei 
With the change of variables:
t dE
dx 
dN
2 dN
x  E ( N ) - Ei  t / 2

2
dN
2

sin
x


Ptotal  4  EN | V | Ei  
dx

2
dE
t


2x / t


2

2
 dN  t sin x
Ptotal  4  E N | V | Ei  
dx

2

 dE  2 - x
2

Notice the total transition probability 
t
2 t
2  dN 
Ptotal 
 E N | V | Ei  


 dE 
and the transition rate
2
2  dN 
W  Ptotal / t 
 EN | V | Ei  


 dE 
Does the density
of states vary
through the
continuum?
n=
dN/dE
n=3
E
n=2
n=1
Classically, for free particles
E = ½ mv2 = ½ m(vx2 + vy2 + vz2 )
vz
vy
Notice for any fixed E, m this defines
a sphere of velocity points all which
give the same kinetic energy.
vx
The number of “states” accessible by that energy are within the
infinitesimal volume (a shell a thickness dv on that sphere).
dV = 4v2dv
Classically, for free particles
E = ½ mv2 = ½ m(vx2 + vy2 + vz2 )
dE  mv dv
dE
dE
dv 

mv
2mE
We just argued the number of accessible states (the “density of states”)
is proportional to
4v2dv
 2 E  dE 
dN  C 4v dv  C 4 


 m  2mE 
 4C
 1/ 2
dN
dN   3/ 2 2  E dE
 E1/2
dE
m

2