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For nuclei, I know that it is the $J^{\pi}$ that is usually measured/calculated, which is the spin-parity. I don't see "orbital angular momentum" of a nucleus very often. Now my notion of spin vs. orbital angular momentum is that spin is entirely instrisic, while orbital angular momentum is more of a classical characteristic (by movement/orbit in a bit more of a literal sense).

Do nuclei even have orbital angular momentum? Is their magnetic moment-like properties only derived from its spin-parity, or am I mistaken? What's really throwing me off is that spin is always in half-integer units, whereas orbital angular momentum is 'always' in whole-integer units.

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  • $\begingroup$ Couldn't it be that you confuse orbital angular momentum of the whole nucleus, with the orbital angular momentum of nucleons within the shell-model of nucleus? $\endgroup$ – Sofia Mar 9 '15 at 18:38
  • $\begingroup$ That is very possible. $\endgroup$ – Arturo don Juan Mar 9 '15 at 23:11
  • $\begingroup$ Could you explain what you mean by that? $\endgroup$ – Arturo don Juan Mar 9 '15 at 23:12
  • $\begingroup$ See Bill N's answer. $\endgroup$ – Sofia Mar 9 '15 at 23:14
  • $\begingroup$ Oh no I mean like how does one determine the orbital angular momentum of an entire nucleus going around, let's say, another entire nucleus? $\endgroup$ – Arturo don Juan Mar 9 '15 at 23:16
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For a famous example of a nucleus with internal orbital angular momentum, consider the deuteron. Considerations of exchange symmetry, spin, and isospin demand that the deuteron have unit spin, rather than zero spin. However the pion-nucleon interaction, gleaned from neutron-proton scattering and deuteron formation, suggests that about 4% of the deuteron wavefunction is $D$-wave, with orbital angular momentum $L=2$. (The $P$-wave component, with $L=1$, would have negative parity, $(-1)^L$; the deuteron's parity is positive.)

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The angular momentum of the nucleus is the combined contribution of the spin-orbit angular momenta of the constituent particles. In order for an entity to have orbital angular momentum of its own it must some conceptual orbit: electrons in the atom, protons and neutrons in the nucleus, atoms in a molecule. That's why the angular momentum of a nucleus is generally referred to as spin: there's no definable orbit when talking about a single nucleus.

The success of the nuclear shell model strongly supports that each nucleon has a spin-orbit ($\vec{j}$) angular momentum which it contributes, and like particles combine first. Each $j$ quantum number will be a half-integer, with like $|lj>$ pairs combining (in the energy ground state) to give zero spin contribution. All nuclei with even proton count (Z) and even neutron count (N=A-Z) have $0^+$ spin and parity in the ground state, the protons combining to zero and the neutrons combining to zero. All even-odds and odd-evens (referring to Z and N) have odd-half integer spin with a parity determined by the $l$ of the extra (odd) nucleon. Odd-odd nuclei have integer, usually non-zero, spin. The two half integer quantum numbers (from the odd proton and the odd neutron) combine for an integer.

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  • $\begingroup$ But the user didn't tell you whether he refers to a global orbiting of the nucleus, or to internal dynamics of the nucleus, i.e. of the constituents. $\endgroup$ – Sofia Mar 9 '15 at 19:09
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In a Hydrogen atom, the "orbital angular momentum of electron" is in fact the relative orbital angular momentum. The nucleus (proton) turns around the common atomic center too, but in a smaller orbit. The atomic electron and the nucleus do not have certain individual orbital angular momenta. They are in mixed states. See my explanations here and here.

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It's made up of many particles. I would assume some of them "orbit" the center. I know many electron orbitals have angular momentum. The orbitals of nuclei are less well-understood, but I would expect the same is true.

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