Introductory Nuclear Physics

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PHY431 Lecture 4 1 2/27/2000 That pion exchange between nucleons is able to describe only the longer distance behavior of the force, and not the short-distance repulsion, nor the transition region to attraction, is highlighted in the article by Philip Anderson:
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PHY431 Lecture 4 2 2/27/2000 6.1. The Deuteron The deuteron is the only bound nucleon-nucleon state: the bound neutron-proton state. Its binding en- ergy is 2.23 MeV, and it has a total angular momentum j =1, and the deuteron parity is even, compactly denoted as J P =1 + . The deuteron consists of a neutron and proton circling their common center of mass. The spatial vari- ables describing the system are thus the distance of separation, and the polar and azimuthal angles that this distance vector makes with the coordinate axes. The neutron-proton system therefore has an orbital angular momentum (with respect to the center of mass), characterized by the orbital quantum number l . In addition, the nucleons each have spin s = [ s ( s +1)] ħ . From the rules of addition of spins, the total spin S of a system of two nucleons a and b has a quantum number S either S =0 (a singlet state), or S =1 (a triplet of states): 11 11 1 1 1 1 1 1,1 , , , 1, 0 , , , , , 1, 1 , , 22 22 2 2 2 2 2 11111 1111 0,0 , , , , (Singlet) 2 ab a ba b  =+ + = + + + =−   + (6.1) Because the Pauli principle does not allow two identical nucleons to coexist, the nucleons in a neutron- neutron or proton-proton L =0,1,2,. .. state should have anti-parallel spins and S =0. Experimentally, no nn or pp bound states are observed, and thus an anti-parallel spin state ( S =0) for neutron-proton state should also not bind. We are led to conclude that the deuteron consists of a parallel-spin S =1 neutron- proton system. The total angular momentum J of the deuteron is the vector sum of the orbital angular momentum L and the total spin S : J = L+S . Thus, with J =1 and S =1, the possible values for L are L =0,1,2: denoted by 3 S 1 , 3 P 1 , 3 D 1 respectively, in the usual spectroscopic notation: 2 S +1 L J . Because the deuteron parity is even, the orbital angular momentum quantum number also must be even: P =(-1) L , thus L =0, 2; with notation 3 S 1 , 3 D 1 respectively. The predicted magnetic moment of the deuteron in case L =0 comes purely from the nucleons’ moments: µ d predicted = p + n = (2.7925 µ N ) + (-1.9128 µ N ) = 0.8797 µ N , close to the experimental value: d expt =0.85742 µ N (the Nuclear magneton µ N e ħ /2 m p ). The existence of a non-zero quadrupole moment (0.286 fm 2 ) indicates that the L =2 state also contrib- utes a bit: the deuteron ground state is an admixture of 3 S 1 (96%) and 3 D 1 (4%). The classical quadru- pole moment is defined as: 22 2 2 2 (3 ) ( ) cos (3cos 1) ( ) Q dV z r e e r drd d r ρ θϕ θ =−= ∫∫ rr (6.2) which vanishes if ( r )= ( r ), i.e. spherically symmetric as in a L =0 state, because 2 (3 1) 0 xd x −= . The existence of an D-state admixture then implies that the strong nucleon potential cannot be completely spherically symmetric and is not a pure central force: a “tensor interaction” must be present! Such an
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This note was uploaded on 02/05/2008 for the course PHY 431 taught by Professor Rijssenbeek during the Spring '01 term at SUNY Stony Brook.

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Lecture 04 - PHY431 Lecture 4 1 2/27/2000 That pion...

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