Nucleon Scattering

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