Symmetries and Conservation Laws II

Isospin, Strangeness, G-parity

Introduction to Elementary Particle Physics

Diego Bettoni
Anno Accademico 2010-2011
 Outline

• Isospin
– Definition, conservation
– Isospin in the N system
• Strangeness
• G-parity
 Isospin

m p  938.27 MeV mn  939.57 MeV

m p  mn
Heisenberg (1932):
Proton and neutron considered as different charge substates of one
particle, the Nucleon.
A nucleon is ascribed a quantum number, isospin, conserved in the
strong interaction, not conserved in electromagnetic interactions.
Nucleon is assigned isospin I  12

I 3   12 p Q 1
  I3
I 3   12 n e 2
The nucleon has an internal degree of freedom with two allowed states
( the proton and the neutron) which are not distinguished by the nuclear
force.
Let us write the nucleon states as I , I 3

p , 1 1
2 2 n  , 1
2
1
2
For a two-nucleon system we have therefore:

  (1,1)  12 , 12 12 , 12

  (1,0)  2  2 , 2 2 , 2  
Triplet
(symmetric)
1 1 1 1 1 1
2 , 12 1
2 , 12
  (1,1)  1 , 1 1 , 1
 2 2 2 2

Singlet
(antisymmetric)
 (0,0)  
1
2
1
2 , 12 1
2 , 12  12 , 12 1
2 , 12 
Example: deuteron (S-wave pn bound state)

   ( spazio )   ( spin )   (isospin )

(1) l  1 (1) S 1  1 (1) I 1
(l  0) ( S  1)

(1) I 1  1  I  0

 is the wave function for two identical fermions (two nucleons),

hence it must be globally antisymmetric. This implies that the
deuteron must have zero isospin:

Id  0
As an example let us consider the two reactions
p  p    d
( I  1)
p  n   d 0

Since Id=0 in each case the final state has isospin 1.

Let us now consider the initial states:
pp  1,1
np  1
2
 1,0  0,0 
The cross section
  ampiezza   I , I 3 A I , I 3
2 2

I
Isospin conservation implies
I  I   1 I 3  I 3
np0d
2
 1 
The reaction proceeds with probability   with respect to
 2

  pp    d 
pp+d hence:
2
 np   d 
0
 Isospin in the N System

The  meson exists in three charge states of roughly the same mass:

m   139.57 MeV
m 0  134.98 MeV
Consequently it is assigned I=1, with the charge given by Q/e=I3.

   1,1  0  1,0    1,1

Q B
For the  B=0:  I3 
e 2
 For the N system the total isospin can be either I=1/2 or I=3/2

p p I 3
2 I 1
2
pure
 n   n I=3/2
1
1
I3 3 1
 12  23 1
 12
2 2 2
p p 2
p 1
  p   0n combination of
 n 1 2

 n   n
I=1/2 and I=3/2 3 3

0p 2
 1
 n   0 p 3 3

 0n 2
3
1
3
The coefficients in the linear combinations, i.e.
the relative weights of the 1/2 and 3/2 amplitudes,
p 1
3  2
3
are given by Clebsch-Gordan coefficients  n 1

  n  1,1  12 , 12 3
2 , 12  1
3  n  2
3 0p
 1 3
3 2 , 12  2 1
3 2 , 12  1
3 1,1  12 , 12  2
3 1,0  12 , 12
(1) p p
Elastic scattering
( 2)   p    p
(3)   p   0n Charge exchange
2 2
 f Hi
H1 if it acts between states of
 M if
I=1/2
H=
H3 if it acts between states of I=3/2
let M 1  I  12 H 1 I  1
2

M3  I  3
2 H3 I  3
2

1  K M 3
2
(1)
i  f  1 3
, 12  2 1
, 12
3 2 3 2
1 :  2 :  3 
2
(2)  2  K f ( H1  H 3 ) i
M 3 : 19 M 3  2 M 1 : 92 M 3  M 1
2 2 2

 2  K 13 M 3  23 M 1
2

(3) i  1 3
3 2 , 12  2 1
3 2 , 12 M 3  M 1  1 :  2 :  3  9 : 1 : 2
f  2 3
3 2 , 12  1 1
3 2 , 12 M 1  M 3  1 :  2 :  3  0 : 2 : 1
2
3  K 2
9 M3  2
9 M1
 p Total Cross Section

 (1236)   120 MeV


3 3
J 
P
I (3,3)
2 2

2J 1   ab  cd
 (E ) 
( 2 s1  1)( 2 s 2  1) k 2 2 a+b  R  c+d
(E  M R ) 
2
4
 Strangeness S

Strange particles are copiously produced in strong interactions

They have a long lifetime, typical of a weak decay.
S quantum number: strangeness conserved in strong and
electromagnetic interactions, not conserved in weak interactions.
Example:    p    K 0
p     2.6  1010 s

  p  
I = 0, because the  has no charged counterparts
1
I 0 2 1
I3 0 1
1    p    K0
2
1 1
I 1 2 0 2

I3 1 1
2 0  12
 0  Q
 I3 
1 Q BS (Gell-Mann Nishijima)
K ,K
e 2
 I3 
e 2
Q 1
K 0, K   I3  Y = B+S hypercharge
e 2

Using the Gell-Mann Nishijima formula strangeness

is assigned together with isospin.
Example.: n, p S 0 I  1 2

 S  1 I  0
K 0, K  S 1 I  1
2

K , K 0 S  1 I  1
2

Example of strangeness    p    K   0     e.m.

conservation: S 0 0 1 1 S 1 1 0
K  p   0 I 1 1
2 1 1
2   n    weak
S 1 0 1 0 I3 1 1
2 1 1
2 S 1 0 0
I3  12  12 0 0
      weak
S  2 1 0
 G-parity G

i I 2
G  Ce
Rotation of  around the 2 axis in isospin space followed by
charge conjugation. iI
I 3 
  I 3 
e 2 C
I3
Consider an isospin state (I,I3=0): under isospin rotations this state
behaves like Y0l(,) (under rotations in ordinary space)
The rotation around the 2 axis implies:
       
Yl 0  ( 1)l Yl 0
therefore
 ( I ,0)  ( 1) I  ( I ,0)
Example: for a nucleon-antinucleon state the effect of C is to give a
factor (-1)l+s (just is in the case of positronium). Therefore:
G  ( NN )  ( 1)l  s  I  ( NN )
This formula has general validity, not limited to the I3=0 case.

For the  G  
   

G  
G 0  0
For the 0 C=+1 (0), the rotation gives (-1)I=-1(I=1) so that G= -1.
G 0  1
It is the practice to assign the phases so that all members of an isospin
triplet have the same G-parity as the neutral member.

G   with C    

 
Since the C operation reverses the sign of the baryon number B, the
eigenstates of G-parity must have baryon number zero B=0.
G is a multiplicative quantum number, so for a system of n 
G=(-1)n
   G  1
   G  1 B.R.  89%
   G f  1 B.R.  2.2%

  C=+1 which, with I=0, yields G=+1.

  viola P
   viola G  e.m.