I want to know if deuterium is a fermion or boson. Please give me a descriptive answer.

I tried the formula that is the combination of protons and electrons which gives odd number but the answer is boson.


The deuterium nucleus is a boson, with spin $\hbar$ (and positive parity). Unlike other stable nuclei, deuterium doesn't have any bound excited states; however if it did they would also have integer spin.

The deuterium atom is a fermion, which may have spin $\frac12\hbar$ or $\frac32\hbar$, to be combined with the orbital angular momentum (which is zero in the ground state). However, the total atomic spin isn't a good quantum number in a nonzero magnetic field: since the electron and the nucleus have different magnetic moments, you can use the magnetic energy to tell which is polarized which way and can't generate a proper symmetric or antisymmetric mixture where the spin of each is superposed.

The deuterium molecule, like all diatomic molecules, is a boson. The electrons combine into an antisymmetric spin singlet. The two nuclei must combine symmetrically, because they are identical bosons. That requirement introduces a coupling between the nuclear spin and the molecular angular momentum quantum number $L$, because a molecule with orbital angular momentum $L$ has a sign change $(-1)^L$ under parity. This means the symmetric spin states, called "orthodeuterium" with spin 0 or spin $2\hbar$, may only have even $L$, while the antisymmetric state "paradeuterium" with spin $\hbar$ must have odd $L$. This means that when deuterium is cooled near absolute zero it's impossible to remove all the nuclear angular momentum; you generate pure orthodeuterium, but it has some spin-2 fraction that can't be removed.

This behavior is very different from ordinary dihydrogen molecules, where the nuclei are fermions. In that case the ground state is parahydrogen, with spin zero and even $L$, and cooling liquid or solid hydrogen eventually produces a material without any angular momentum. This makes a big difference in the heat capacities for cold hydrogen and cold deuterium, and also for their interaction with low-energy spin systems, most notably cold neutrons.

If you were to call up your gas supplier and order a cylinder of deuterium, you'd get $\mathrm{D_2}$ gas, which is bosons. And the reason that cold deuterium acts differently from cold hydrogen is that the deuterium nucleus obeys Bose-Einstein statistics, while the hydrogen nucleus obeys Fermi-Dirac statistics — the two molecules have the same electronic configuration. If you held a gun to my head and forced me to check "boson or fermion" without any other context, I'd say boson get shot in the middle of this little lecture.

  • $\begingroup$ > The electrons combine into an antisymmetric spin singlet. Well they can combine into triplet states too, but that would be an excited state. $\endgroup$ – Ruslan Dec 17 '14 at 10:32
  • $\begingroup$ Correct. But the electronic excitations take a lot more energy than the vibrational states, which in turn take a lot more energy that the rotational states I talked about. At low temperatures it's safe to think of hydrogen and deuterium molecules as simple rotors, segregated by nuclear spin. $\endgroup$ – rob Dec 17 '14 at 13:35

Protons, electrons, and neutrons too.

Hydrogen is a boson, deuterium is a fermion.

  • $\begingroup$ Sorry, Nanite. There are a lot know-littles on this deteriorating Stack pile. You're correct, of course. You've realized that hydrogen and deuterium denote the charge-neutral atom. Hydrogen is comprised of a single proton as the nucleus and a single electron in 'orbit' about it. Deuterium is comprised of the deuteron -- a nucleus comprised of a proton and a neutron -- and a single electron in orbit. Therefore, exchange of the particle labels in a two-hydrogen wave function results in the same wave function. While that of two deuterium results in a minus sign. Stop lowering this score, morons. $\endgroup$ – MarkWayne Nov 4 '16 at 22:42
  • $\begingroup$ In fact, if you have lowered the score you should undo do it and remove yourselves from the site. $\endgroup$ – MarkWayne Nov 4 '16 at 22:46
  • $\begingroup$ @MarkWayne thanks man. :) Well I guess in retrospect, it's possible people were also objecting to my short/curt answer. But I don't mind, I have lots of other answers that got more recognition than even I thought they deserve. $\endgroup$ – Nanite Nov 5 '16 at 9:41
  • $\begingroup$ Yes. Some more elaboration would have served you better. People don't really think about what they read much. They just react. And the quality of science discussion on SE-phys has declined severely in recent years. $\endgroup$ – MarkWayne Nov 15 '16 at 16:29

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