Looking at various particles regarding being fermions or bosons, it seems to me that fermions are something fundamentally different from bosons.

What I mean by "fundamentally different" is "as different as the electromagnetic force is to the strong nuclear force". The opposite, similar, is like "energy versus mass", or "photon particle versus wave".

Most notably, electrons, protons and neutrons are fermions, while photons are bosons.

Now, a difference that seems to me as fundamental as it gets is that the Pauli exclusion applies to fermions, but not to bosons.

On the other hand, there are very similar particles of which one is a fermion, and the other is a boson:

  • A hydrogen ion - a single proton - is, as noted above, a fermion.

in contrast

  • A deuterium ion - a nucleus consisting of a proton and a neutron - is a boson.

To me, and to chemists, deuterium is just a hydrogen isotope, nothing special. Or in other words, they are fundamentally similar.

Now, are fermions and bosons fundamentally different, or "not really different"?

Maybe the cases above are in some way unsuitable to be compared to each other?

  • 5
    $\begingroup$ You have to define what you mean by fundamentally different. $\endgroup$
    – JamalS
    Commented Jun 20, 2017 at 7:11
  • $\begingroup$ Good point! In a way, this is part of the question. Maybe my idea of "different" does not apply in this context somehow... I'll clarify. $\endgroup$ Commented Jun 20, 2017 at 7:17
  • 3
    $\begingroup$ Rhetorical question: Are odd and even numbers fundamentally different? $\endgroup$ Commented Jun 20, 2017 at 9:58
  • 2
    $\begingroup$ "as different as the electromagnetic force is to the weak force" -- bad choice; electromagnetism and the weak force have been shown to be a broken symmetry of a single force, electroweak $\endgroup$ Commented Jun 20, 2017 at 16:45
  • $\begingroup$ @RossPresser Thanks to catch that! Fixed (using strong).. $\endgroup$ Commented Jun 20, 2017 at 16:51

3 Answers 3


If two atoms differ only by the spin of their nuclei, then their individual properties will be almost identical, but their collective properties will be extremely different.

Chemists often consider individual atoms (or, more often, molecules). In the case of individual hydrogen and deuterium atoms, their electronic properties are identical (except for the hyperfine spitting of their electronic energy levels due to spin-spin coupling between the electrons and the nuclei). Practically speaking, the only important difference is the mass difference due to the extra neutron.

But once you put a bunch of them together and lower the temperature enough that quantum effects (specifically multiple-occupancy of energy levels) become significant, their extremely different many-body properties become manifest. For example, neutral hydrogen atoms are bosons and can in principle Bose-Einstein condense, while neutral deuterium atoms are fermions and will instead form a free Fermi gas (to a decent approximation), which has extremely different properties. That's why it's extremely important that cold-atom experimentalists trying to form Bose-Einstein condensates get the right isotopes of the atoms that they're trying to condense.


For chemist, ${}^3\mathrm{He}$ and ${}^4\mathrm{He}$ are just two isotope of the same element. But the fact that one is a fermion and the other is a bosons is super important in cryogenics:

If you take liquid ${}^4\mathrm{He}$ at 4.2K and you pump it, you'll get a superfluid of ${}^4\mathrm{He}$ with a superconducting/liquid temperature of 1.4K. This superfluid is remarquable (like the Foutain effect).

If you do the same with ${}^3\mathrm{He}$, you will not get this superfluid, but some other behavior (Fermi liquid under certain circumstance) that do not have the superfluid properties (like the foutain effect for instance).

Edit: Just to put number, here are the two phase diagram at low temperature of ${}^4\mathrm{He}$ and ${}^3\mathrm{He}$. As you can see, they are quite differente both in aspect and in range. It's to be noted, that the superfluid phases in ${}^3\mathrm{He}$ are pair-condensation and not Helium atom condensation.

Helium 4 phase diagram:

phase diagrame of 4He

Helium 3 phase diagram:

phase diagram of 3He

Source of the phase diagrams: LTL/Helsinki University of Technology.

  • $\begingroup$ It's worth noting that the behaviour of $^3\text{He}$ in low temperature indeed manifests the very Pauli principle (and the behaviour of $^4\text{He}$ the lack thereof) the OP mentions as the primary difference. $\endgroup$
    – The Vee
    Commented Jun 22, 2017 at 19:41

Let's take another pair of very similar particles - a helium 4 atom and a helium 3 atom. Under most circumstances there is hardly any difference between them, but once we start looking at a situation where the fermionic/bosonic nature matters, i.e. superfluidity, we discover there is a huge difference between them.

In principle hydrogen and deuterium would show similar differences though as far as I know neither has been conclusively observed to form a superfluid.

The fermion/boson distinction only matters in circumstances where there is a big difference between the Fermi-Dirac and Bose-Einstein distributions. In everyday life this is often not the case as at high energies both distributions approximate the Boltzmann distribution and both behave in a similar manner.


Not the answer you're looking for? Browse other questions tagged or ask your own question.