particlesymbolrest energy in mev electron e 0.5109989 muon 105.658357 neutral pion 0 134.9766...
Post on 19-Dec-2015
219 Views
Preview:
TRANSCRIPT
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
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.
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
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
/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
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
/
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
top related