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