To determine if an atom is a boson or a fermion I have to count the fermions that constitute the atom (protons, neutrons and electrons). My question is: How to theoretically (as opposed to experimentally) determine the total angular momentum of an atom? Can I just sum the spin of all the electrons and the nucleus spin, and the orbital angular momunta of the electron orbitals?

For example: A Hydrogen atom in the ground state of can be considered like a spin 1 particle?

  • $\begingroup$ In general, adding the spins does not work. For example, see hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nspin.html. I don't know how the total spin is computed, but one phenomenon to take into account is that two electrons on the same orbital necessarily have oppositive spins. $\endgroup$ – minar Aug 28 '13 at 19:33
  • $\begingroup$ Like addition of angular momentum in quantum mechanics textbook? $\endgroup$ – user26143 Aug 29 '13 at 1:29
  • $\begingroup$ @minar: two electrons on the same orbital necessarily have oppositive spins This only holds if they have zero orbital angualar momentum. $\endgroup$ – user4552 Aug 29 '13 at 1:56
  • $\begingroup$ @BenCrowell well not only. They only must have all quantum numbers equal - except spin. This can be any stationary quantum state, not only with zero orbital angular momentum. $\endgroup$ – Ruslan Aug 29 '13 at 13:35
  • $\begingroup$ @Ruslan: No, that's incorrect. Spin is treated the same as any other quantum number in the exclusion principle. $\endgroup$ – user4552 Aug 29 '13 at 15:22

Angular momentum is a vector, not a scalar. In the case of hydrogen, this makes it possible for the total spin to be either 1/2+1/2 or 1/2-1/2.

By the way, in most cases we don't care at all about the total spin of an atom, because the hyperfine coupling is very weak. We care about the total spin of the nucleus, and the total spin of the electrons.

It is in general beyond the state of the art to predict with perfect reliability, from first principles, the ground-state spins of all nuclei. I'm a nuclear physicist, not an atomic physicist, but I believe the prediction of the total electronic spins is much more tractable, and can basically be done using Hund's rules.

In nuclear physics, for an even-even nucleus, the total spin is always zero. For an odd nucleus, if you know the shell that the odd nucleon is in, then the rule is that all the angular momenta couple so as to cancel, except for that of the odd nucleon. I.e., angular momentum is minimized, which is the opposite of Hund's rules. In an odd-odd nucleus, you have to deal with the coupling between the two odd spins.

  • $\begingroup$ So what kind of field should I use in atomic physics? A scalar field, a 4-vectorial field or a spinor one? $\endgroup$ – m.mybo Aug 29 '13 at 9:21
  • $\begingroup$ @m.mybo: Electrons, neutrons, and protons are all spin-1/2 particles. Or are you asking about the wavefunction of the whole system? Its character would depend on the total spin. $\endgroup$ – user4552 Aug 29 '13 at 15:23
  • $\begingroup$ Yes I am asking about the whole system. My real question is: How could I treat a complex system of atoms with the formalism of quantum field theory? I have to know the spin of atoms to understand which kind of field I need $\endgroup$ – m.mybo Aug 29 '13 at 15:42
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    $\begingroup$ One approximation is to treat it as a Slater determinant of spin-1/2 fields. Another approximation is to treat the composite object as a single fundamental particle, in which case you would have to determine its total spin, and then that would determine what type of field you'd represent it as. For example, a helium atom has spin 0, so you'd represent it as a scalar field in the latter approximation. $\endgroup$ – user4552 Aug 29 '13 at 15:47
  • $\begingroup$ LoLwut? An ¹H atom at 1s¹ has a singlet spin-0 state and a triplet spin-1 state. What does it definitely have not is a total angular momentum of ½, and not only at 1s¹ since two fermions never combine into one fermion. $\endgroup$ – Incnis Mrsi Aug 20 '14 at 13:19

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