2. Isospin

3. Charge independence Accordingly, one can develop a formalism that encompasses this concept. Strong empirical empirical evidence to imply that: F n-n = F p-p = F n-p In the absence of the electromagnetic forces. Known as charge independence of the nuclear force.

4. 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, t 3 = + 1/2 and t 3 = - 1/2 (projections along a “3-axis”; axes 1,2,3) -- Isospin up Isospin down What insights can we gain by using this formalism?

5. Nuclear structure Consider first the simplist nuclear system, 2 H Proton: t = 1/2, t 3 = + 1/2 Neutron: t = 1/2, t 3 =  1/2 This gives states -- T d = 0  T d-3 = 0 (n,p) [singlet] T d = 1  T d-3 = +1 (p,p)  T d-3 = 0 (n,p) [triplet]  T d-3 =  1 (n,n) In general, for nuclei - T 3 = Z(1/2) + N(-1/2) T 3 = (1/2) (Z  N)

6. Nuclear structure: A=2 Data for 2 H : I d  = 1 + by parity mixture of s and d states measured

7. Nuclear structure: A=2 symmetric anti- symmetric is symmetric; OK because n, p are not identical particles. But, in the isospin formalism, they are identical particles!

8. Nuclear structure: A=2 must be anti-symmetric because n, p are identical particles. Therefore, must be anti-symmetric! symmetric anti- symmetric

9. Nuclear structure: A=2 Therefore, only the (n,p) 2-nucleon system can be a bound state! symmetric anti- symmetric (p, p) (n, n) (p, p) (n, p)

10. Nuclear structure: A=14 14 C 14 N 14 O 0.0 6.09 8.06 5.17 2.31 5.69 11 11 11 1+1+ 11 (6,8) T=1 0+0+ 0+0+ 0+0+ (7,7) (8,6) T 3 =  1 T 3 = +1 T 3 = 0 T=0 Electromagnetic forces “turned off”

11. Nuclear structure: A=14 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 Compare “identical” nucleon systems; different only in T 3, T

12. Nuclear reactions Nuclear (strong) interactions/reactions conserve T T 3 is conserved because numbers of nucleon types are conserved. d + 16 O  14 N + 4 He All 4 have T gs = 0. First excited state of 4 He * ~20 MeV  4 He gs (T = 0) Therefore -- T = 0 states in 14 N are only allowed.   (2.31 MeV) should be ~0 12 C( ,d) 14 N 12 B( 6 Li,d) 14 N 12 B( 7 Li, 3 H) 14 N T i = 0 T i = 1