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Nucleon Scattering
d d
d d
d d
| I,I3 | 1, 1
| 1,1
| 1 0
If the strong interactionis I3-invariant
These reactions must occur withequal strengths…equal probabilities…
equal CROSS SECTIONS
involve identical matrix elements
p + p d +
p + n d +
n + n d +
Now consider these (possible and observed) collisions
| 1, 1 >
| 1,-1 >
< 1, 1 |
< 1, 0 |
< 1,-1 |
<pp|L|d+> : <pn|L|d0> : <nn|L|d>
(|1,0 + |1,1)2
1
1 : : 1
ppd+ : pnd0 : nnd : = 2 : 1 : 2
1/2
Then the ratio of cross sections:
p p
n n
p p
Consider the scattering reactions:
The strong force does not discriminate between nucleon or pion charge.
What can we expect for the cross section of these three reactions?
If we enforce conservation of isospin we can only connect initial and final states of the same total I, I3
| I,I3 > | 3/2, 3/2 >
| 3/2, -3/2 >
But
<p|L|p> = M + M 3
1
3
2
2
1-
2
3
2
1-
2
1
this interaction involves two matrix elements
p p
n n
+ p + p
means combining:
1 ½ 1 ½
-1 -½ -1 -½
-1 ½ -1 ½
| 1 -1 > | 1/2, 1/2 > = |3/2, -1/2 > |1/2, -1/2 > )3
1
3
2
+ p + p
+ p + n
+ p + pa.
b.
c.
elastic scattering
but only one of the above can also participate in a
charge exchange process
1,1
1,-1
1,-1
½ ,½
½ ,½
½ ,½ 1, 0 ½ ,½
This IS observed!
So all strong interactions not SIMPLY charge independent.
I3 ISOSPIN independence is more general.
? ?
+ p + p
+ p + n
+ p + pa.b.c.
elastic scattering
charge exchange process
These three interactions involve the ISOSPIN spaces:
p p
n
23
23
21
2111
21
21
21
23
21
2111 -
21
21
21
23
21
2101
13
23
23
13
Recall: if
L 2= Mfi
221
21
21
21 L
21
23
21
23 L
23
23
23
23
L
M½
M3/2 same byI3-indep.
Let’s denote:
1,1
1,-1
1,-1
½ ,½
½ ,½
½ ,½ 1, 0 ½ ,½
+ p + p
+ p + n
+ p + pa.b.c.
elastic scattering
charge exchange process
p p
n
23
23
21
21
21
23
21
21
21
23
13
23
23
13
21
21
21
21 L
21
23
21
23 L
23
23
23
23
L
M½
M3/2
a.b.c.
a M3/2
2
b | M3/2 + M1/2|21
323
c | M3/2 M1/2|22
323
+ p + p
+ p + n
+ p + pa.b.c.
a M3/2
2
b | M3/2 + M1/2|21
323
c | M3/2 M1/2|22
323
a : b : c = : : M3/2 |M3/2 +2M1/2|22 19 |M3/2 M1/2|22
9
for the combined cross section of both processes22
2223
23 11
449
1 MMMM
22
2223
23 11
29
1 42 MMMM
22
223 1
69
1 3 MM
Now if M3/2=M1/2 then +p = -p total
but also -p0n= 0
2H target
S1 S2 S3 S4
beam
Total Cross
Section T
(10-27 cm2)
+ p
+ p
40 100 200 400Lab Energy of Pion Beam (MeV)
200180
160
140
120
100
80
60
40
20
0
Measured the depletion of pion beamrepeated with the tank full, emptyrepeated with + and - beam
for KE 195 MeV (the resonance of the 3/2-spin )
3 cb
a
a : b : c = : : M3/2 |M3/2 +2M1/2|22 19 |M3/2 M1/2|22
922
223 1
69
1 3 MM a : b + c = : M3/2
2
3 cb
a
223 1
MM
a : b : c = 9 : 1 : 2
Symmetry implies any transformation still satisfies the same Schrödinger equation, same Hamiltonian:
H dt
di (U) (U) H
dt
di U† U
U†H U= Hmeans we must demand:
[H ,U]= 0Which means that the operator U must be associated with a CONSERVED quantity!
Though U are UNITARY, not necessarily HERMITIAN, but remember:GieU where the G is Hermitian!
!2
)(1
2GiGiU
since you’ve already shown
[H ,U]= 0 [H ,G]= 0The GENERATOR of any SYMMETRY OPERATION is an
OPERATOR of a CONSERVED OBSERVABLE (quantum number!)
Mesons
isospin mass chargeParticle I3 MeV/c2 states Q
nucleon 938.280 p +1 939.573 n 0
pion 139.569 + +1 134.964 0 0 139.569 1
delta 1232. ++ +2 + +1 0 0
1
rho 770. + +1 0 0
1
eta 548.8 0 0
+1/21/2
+1 01
0+1 01
+3/2+1/21/23/2
Spin-0
BaryonsSpin-1/2
Spin-3/2
omega 783.0 0 00
Q = I3 + ½Y “hypercharge” or BARYON NUMBER
because =1 for baryons 0 for mesons
1947 Rochester and Butler
cloud chamber cosmic ray event
of a neutral object decaying into two pions
K0 +
1947 Rochester and Butler cloud chamber cosmic ray event
of a neutral object decaying into two pions
K0 +
1949 C. F. Powell photographic emulsion event
K+
m = 497.72 MeV
m = 493.67 MeV
p
p + m=1115.6 MeVmp=938.27 MeV
1950 Carl Anderson (Cal Tech)
1952 Brookhaven Cosmotron 1st modern accelerator
artificially creating these particles for study
1954 6.2-GeV p synchrotron Lawrence,Berkeley
1960 28-GeV p synchrotron CERN, Geneva 33-GeV p synchrotron Brookhaven Lab
1962 6-GeV e synchrotron Cambridge
1963 12.5-GeV p synchrotron Argonne Lab
1964 6.5-GeV p synchrotron DESY,Germany
1966 21-GeV e Linac SLAC (Standford)
Spin-0 Pseudoscalar Mesons
nucleon 938.280 p +1 939.573 n 0
pion 139.569 + +1 134.964 0 0 139.569 1
rho 770. + +1 0 0
1
eta 548.8 0 0
+1/21/2
+1 01
0+1 01
Spin-1/2 Baryons
omega 783.0 0 00
isospin mass charge Particle I3 MeV/c2 states Q
lambda 1115.6 0
Sigma 1385. + +1 0 0
1
+1 01
Cascade 1533. + +1 1
+1/21/2
0
kaon 493.67 K+ +1 497.72 K0 0
+1/21/2
kaon 497.72 K0 0 493.67 K 1
+1/21/2
Delta 1232. ++ +2 + +1 0 0
1
+3/2+1/21/23/2
Spin-3/2 Baryons
isospin mass chargeParticle I3 MeV/c2 states Q
Sigma-star 1385. + +1
0 0
1
+1 01
Cascade-star 1533. *+ +1
* 1
+1/21/2
pdg.lbl.gov/pdgmail
FRANK & EARNEST
These new heavier particle states were produced as copiously as s
in nuclear collisions (and in fact decay into s)
all evidence of STRONG INTERACTIONS
these new states decayed slowly like the weak decaysp n + e + +
which decay via neutrinos(accepted as the “signature” of a weak decay)
but unlike
STRONG production/decay phenomena like nuclear resonances
(all with final decay products, like the ) which decay “instantly”,
i.e., as readily as they are produced
ELECTROMAGNETIC production/decay phenomena
atomic (electron) resonances (all with decay)
or
What else was about them?
Observed
+ p+ K+ + K + K +
NEVER Observed
+ p+ + + K + n +
all still conserve mass, charge, isospin
“Associated production”
Also NEVER observe: + p+ +
but DO see: + p+ + +
K+
K
K
K
+1+1111
Q = I3 + ½Y YB+S
1952-53 (Pais, Gell-man) “Strangeness”
Spin-0 Pseudoscalar Mesons
nucleon 938.280 p +1 939.573 n 0
pion 139.569 + +1 134.964 0 0 139.569 1
rho 770. + +1 0 0
1
eta 548.8 0 0
+1/21/2
+1 01
0+1 01
Spin-1/2 Baryons
omega 783.0 0 00
isospin mass charge StrangenessParticle I3 MeV/c2 states Q S
lambda 1115.6 0
Sigma 1385. + +1 0 0
1
+1 01
Cascade 1533. + +1 1
+1/21/2
0
kaon 493.67 K+ +1 497.72 K0 0
+1/21/2
kaon 497.72 K0 0 493.67 K 1
+1/21/2
000
000
0
0
00
+11
11
1 1122
1
SU(2) Combining SPIN or ISOSPIN ½ objects gives new statesdescribed by the DIRECT PRODUCT REPRESENTATION
built from two 2-dim irreducible representations: one 2(½)+1 and another 2(½)+1 yielding a 4-dim space.
isospinspace
+½
½
=
=
which we noted reduces to 2 2 = 1 3
the isospin 0singlet state
( 12
ispin=1triplet
SU(2)- Spin added a new variable to the parameter space defining all state functions
- it introduced a degeneracy to the states already identified; each eigenstate became associated with a 2+1 multiplet of additional states
- the new eigenvalues were integers, restricted to a range (- to + ) and separated in integral steps
- only one of its 3 operators, J3, was diagonal, giving distinct eigenvalues. The remaining operators, J1 and J2, actually mixed states.
- however, a pair of ladder operators could constructed: J+= J1 + iJ2 and J= J1 - iJ2
which stepped between eigenstates of a given multiplet.
n
-1/2 +1/2 -1 0 1
010
100
000
6
000
001
010
1
000
00
00
2 i
i
001
000
100
4
00
000
00
5
i
i
00
00
000
7
i
i
000
010
001
3
3200
0310
0031
8
The SU(3) Generators are Gi = ½i
just like the Gi = ½i are for SU(2)
The ½ distinguishesUNITARY from ORTHOGONAL
operators.
i appear in the
SU(2) subspacesin block diagonal
form.3’s diagonal entries
are just the eigenvaluesof the isospin projection.
8 is ALSO diagonal! It’s eigenvalues must represent a NEW QUANTUM number!
Notice, like hypercharge (a linear combination of conserved quantities),8 is a linear combinations of 2 diagonal matrices: 2 SU(2) subspaces.
In exactly the same way you found the complete multipletsrepresenting angular momentum/spin, we can define
T± G1± iG2
U± G6± iG7
V± G4± iG5
The remaining matrices MIX states.
000
001
010
1
000
00
00
2 i
i
000
010
001
3
T±, T3 are isospin operators
By slightly redefining our variables we can associate the eigenvalues of
8 with HYPERCHARGE.
831
832 )()( GY
3200
0310
0031
3
18