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Isospin

Isospin. 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. Charge independence. Accordingly, one can develop a formalism that encompasses this concept.

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Isospin

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  1. Isospin

  2. 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. Charge independence Accordingly, one can develop a formalism that encompasses this concept.

  3. 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) -- Isospin up Isospin down What insights can we gain by using this formalism?

  4. 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) In general, for nuclei - T3 = Z(1/2) + N(-1/2) T3 = (1/2) (Z  N)

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

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

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

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

  9. 6.09 1 8.06 1 5.17 1 T=1 T=1 5.69 1 T=1 0.0 0+ T=1 2.31 0+ 0.0 0+ 14O 14C (8,6) (6,8) 0.0 1+ T3 = +1 T3 = 1 14N (7,7) T3 = 0 Nuclear structure: A=14 T=0 Electromagnetic forces “turned off”

  10. Nuclear structure: A=14 Compare “identical” nucleon systems; different only in T3, T 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

  11. Ti = 0 Ti = 1 Ti = 0 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 of4He* ~20 MeV  4Hegs (T = 0) Therefore -- T = 0 states in 14N are only allowed.  (2.31 MeV) should be ~0 12C(a,d)14N12B(6Li,d)14N12B(7Li,3H)14N

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