I am supremely confused when something has spin or when it does not. For example, atomic Hydrogen has 4 fermions, three quarks to make a proton, and 1 electron. There is an even number of fermions, and each fermion has a 1/2 spin. Since there are an even number of fermions, the total spin value is an integer. This spin number is the "intrinsic" spin number that cannot be changed, but its orientation "up" or "down" can be changed.
For atomic Hydrogen, it is a Boson because it has integer spin, however it also has a single electron. I read on physics forums, http://www.physicsforums.com/showthread.php?t=69992, that the spin of atom comes from the electrons and not its nucleus. I also read on here, How to find that a molecule has zero spin?, that the spin of atomic Hydrogen is 1/2! The answer says atomic Hydrogen has spin 1/2 because it ignores the nuclear spin.
This is one thing that is confusing me. Shouldn't atomic Hydrogen have an integer spin because of the nuclear component? So does atomic Hydrogen have spin and is affected by a magnetic field? Nuclear spins are affected by magnetic fields, but they aren't as affected as electrons according to the discussion on physics forums.
Why do we ignore nuclear spin sometimes? Also, can someone help me out here with all the possibilities?
Is there a Boson with an half integer spin value? (Surely, there must not be) However atomic Hydrogen is one of those cases! (It seems…) (Why don't we cancel out the nuclear spin with the electron spin?)
Say we have another atom that is a Boson, It has unpaired electrons in different orbitals, so what determines whether or not electrons fill in orbitals as spin up or down? Does spin down nuclear spin cancel out a electron up spin?
Best Answer
In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms.
A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a fermion. Bosons always have an integer spin; fermions always have a half-integer spin. By the spin-statistics theorem, the wave function of two bosons is invariant under their exchange but it is antisymmetric under the exchange of two identical fermions. These two rules are theorems for elementary particles and assuming this theorem, it's also trivial to prove these statements for composite particles.
In particular, for a neutral atom, the numbers of protons and electrons are equal so their total parity is even. That's why only neutrons matter. An isotope with an even number of neutrons is a boson (the whole atom: e.g. helium-4); an isotope with an odd number of neutrons is a fermion (the whole atom: e.g. helium-3).
That doesn't mean that composite bosons exhibit all the same physical phenomena like superfluidity that we may observe with some bosons.
The magnetic moment of a charged "elementary enough" particle scales like $1/m$ where $m$ is the mass of the particle. That's why the magnetic moment of the protons, neutrons, and nuclei are about 1,000-2,000 times smaller than the magnetic moments of the electrons. That's why the nuclear spins are largely negligible for the behavior of the atom in a magnetic field.
This is no contradiction because the whole atoms have a much larger magnetic moments than the nuclei separately – because of the neutrons: atoms are not "elementary enough" in this definition. Both the electrons' spins and their orbital angular momentum contribute to an atom's magnetic moment. Also, there exist a higher number of states because an atom is a typical example of the "addition of several angular momenta". The tensor product Hilbert space may be decomposed as a direct sum of Hilbert spaces with fixed values of the total angular momentum. The degeneracy and the magnetic moment of these components depend on the total angular momentum i.e. on the relative orientation of the nucleus-based and electron-related angular momenta.
In effect, the spectral lines of the whole atom exhibit the so-called hyperfine structure. Up to some approximation, the nuclear spin may be totally ignored. But when the considerations from the previous paragraph are properly account for, each spectral line is actually split to several nearby (3 or so orders of magnitude closer to each other) finer spectral lines, each of which corresponds to a different value of the total angular momentum of the whole atom (or, equivalently, a different value of $\vec J_{\rm electrons}\cdot \vec J_{\rm nucleus}$).