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1968-70 Georges Charpak develops the multiwire proportional chamber
1992 Charpak receives the Nobel Prize in Physics for his invention
100 MHz
50 MHz 0. 0.4 0.8 1.2 1.6 2.0 2.4 2.8
0. 0.4 0.8 1.2 1.6 2.0 2.4 2.8
800
600
400
200
0
1000
800
600
400
200
0
20 m dia 2 mm spacing
argon-isobutane
spatial resolutions < 1mm possible
DO
DO
DØ 5500 tons120,000 digitized readout channels
– 2T super conducting solenoid– disk/barrel silicon detector– 8 layers of scintillating fiber tracker– preshower detectors
Shielding
New Solenoid & Tracking:Silicon, SciFi, Preshowers + New Electronics, Trigger, DAQ
Forward ScintillatorForward Mini-drift chambers
The Detector in various stages of assembly
5500 tons
120,000 digitized readout channels
Fermilab, Batavia, Illinois
CERN, Geneva, Switzerland
Protons
Anti-protons
/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 irppi
fi
3)( )( drrVe irkki
fi
3)()( drrVeqF irqi
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
tEE
EE
tEVtEEUE f
f
efff
f 0
20
2
0002
0 cos-1 )(
2
sin 4 02 tEE f
E=2h/t
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
/
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