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Isospin
Charge independence
Strong empirical empirical evidence to imply that:
Fn-n = Fp-p = Fn-p
In the absence of the electromagnetic forces.
Known as charge independence of the nuclear force.
Accordingly, one can develop a formalism
that encompasses this concept.
Charge independence
For example, we can consider the neutron and proton to be
identical particles (nucleons) that can be found in two states.
In analogy with ordinary spin, we can consider the
nucleon to be a particle with “isospin” t = 1/2 and two
possible states, t3 = + 1/2 and t3 = - 1/2 (projections
along a “3-axis”; axes 1,2,3) --
 1 
p
2 


n
1
 2 

Isospin up
Isospin down
t  t(t  1)
What insights can we gain by using this formalism?

Nuclear structure
Consider first the simplist nuclear system, 2H
Proton: t = 1/2, t3 = + 1/2
Neutron: t = 1/2, t3 =  1/2
This gives states -Td = 0  Td-3 = 0 (n,p) [singlet]
Td = 1  Td-3 = +1 (p,p)
 Td-3 = 0 (n,p) [triplet]
 Td-3 = 1 (n,n)
t p  t n  Td  t p  t n
1 1
1 1
  Td  
2 2
2 2
0  Td  1
In general, for nuclei -
T3 = Z(1/2) + N(-1/2)
T3 = (1/2) (Z  N)
Nuclear structure: A=2
Data for 2H : Id = 1+
s p  sn  S  s p  sn
S IS
1 1
1 1
 S 
2 2
2 2
0  S 1
0  1 0   1
d
d
d

by parity
1  1  1   0,1, 2

1
 N gsn  gs p  0.879804 N if  0

2
 0.8574376 N
measured
 A 0  0  A 2  2 A 0  0.96 A 2  0.04
mixture of s and d states

Nuclear structure: A=2
 d  rY0 ,   (I  1)
 I  1,mI  1  1 2  1 2
 I  1,mI  1  1 2  1 2
symmetric




1
 I  1,mI  0 
1 2  1 2  1 2  1 2
2
1
 I  0,mI  0 
1 2  1 2  1 2  1 2
2
antisymmetric
d is symmetric; OK because n, p are not identical particles.
But, in the isospin formalism, they are identical particles!
Nuclear structure: A=2
 d  rY0 ,   (I  1) T,T3 
d must be anti-symmetric because n, p are identical particles.
Therefore,  T,T3  must be anti-symmetric!
 T  1,T3  1   1 2   1 2
 T  1,T3  1   1 2   1 2
symmetric





1
 T  1,T3  0 
 1 2   1 2   1 2   1 2
2
1
 T  0,T3  0 
 1 2   1 2   1 2   1 2
2
antisymmetric
Nuclear structure: A=2
Therefore, only the (n,p) 2-nucleon system
can be a bound state!
(p, p)  T  1,T3  1   1 2   1 2
(n, n)  T  1,T3  1   1 2   1 2
symmetric




1
 1 2   1 2   1 2   1 2
(p, p)  T  1,T3  0 
2
1

T

0,T

0

 1 2   1 2   1 2   1 2


(n, p)
3
2
antisymmetric
Nuclear structure: A=14
6.09
1
8.06
1
T=1
0+
0.0
T=1
T=1
5.69
1
2.31
0+
T3 = 1
1
0.0
0+
T=0
T=1
14O
14C
(6,8)
5.17
0.0
14N
(7,7)
T3 = 0
1+
(8,6)
T3 = +1
Electromagnetic
forces “turned off”
Nuclear structure: A=14
Compare “identical” nucleon systems; different only in T3, T
T  T3  T  T  T3
Empirically, T will tend to be its smallest allowed value
--- look at triad structure -And, consider isospin physics as applied to --- electromagnetic de-excitation
--  decay
-- nuclear reactions
Nuclear reactions
Nuclear (strong) interactions/reactions conserve T
T3 is conserved because numbers of nucleon types are conserved.
d + 16O 14N + 4He
All 4 have Tgs = 0.
First excited state of 4He* ~20 MeV  4Hegs (T = 0)
Therefore -- T = 0 states in 14N are only allowed.
 (2.31 MeV) should be ~0
12C(,d)14N
Ti = 0
12B(6Li,d)14N
Ti = 0
12B(7Li,3H)14N
Ti = 1