# Can multiple Helium-4 nuclei occupy the same place because the nuclei are bosonic?

**Short answer**: no.

There are still electrons and quarks inside atoms, which are fermions, and they obey the Pauli exclusion principle. Quantum degeneracy pressure is indeed what stops the gravitational collapse of most stars (most = apart form those which become black holes).

**Longer story**:

First of all, the mainstream concept of adding each atomic sub-component's angular momentum quantum number to get a total spin number, and then using that to classify the atom as 'integer' or 'half-integer' and thus obeying Bose-Einstein or Fermi-Dirac (respectively) statistics is a simplification. Which usually works, sure, but still a simplification.

Take two $^4$He atoms, call them $a$ and $b$. Write the total wavefunction of the combined ($2$ atom) system $\Psi(\mathbf{r}_{a_1}, \mathbf{r}_{a_2}, \dots \mathbf{r}_{a_N}, \mathbf{r}_{b_1}, \mathbf{r}_{b_2}, \dots \mathbf{r}_{b_N})$, where $N$ is the number of constituents of each atom (quarks, electrons, etc.).

The spin-statistics theorem actually does not talk about Bose-Einstein or Fermi-Dirac (which are just "simple" results for non-interacting systems), but it just says that these two atoms are *bosons* if, upon exchanging the label of *each* constituent $a_i \leftrightarrow b_j \,\,\forall i, j$, you get the *same* wavefunction $\Psi$ as you started with, down to the same sign (would be an overall $-1$ for fermions).

Now, to carry out that permutation operation, you would need to take into account the situation where *one* quark of $a$ is swapped with *one* quark of $b$, *one* electron of $a$ swapped with *one* electron of $b$ and so forth. However, quarks' tunnelling from one nucleus to the other costs a lot of energy and is therefore not energetically allowed in usual day-to-day energy régimes. Same thing with stripping individual electrons (unless they are already shared in a molecular bond, in which case you'd be taking about *gerade* and *ungerade* symmetrical oribtals). So in the end you only consider swapping *all* the electrons and *all* the quarks of $a$ with *all* the electrons and *all* the quarks of $b$. Which is nothing other than exchanging the position of the two atoms. So, because you have energetically ruled out any sub-atomic permutation operation, you can just treat each atom as independent, and then proceed with "the sum of all the internal spin quantum numbers".

The moral of the story from above is that while you almost always can sum the internal spin quantum numbers to get the atom's total spin, in *some* cases that may not be useful (e.g. when an electron is shared in a molecular bond, or when you have access to energy scales comparable to nuclear binding energies). In these cases, you would have to revert to the original definition of the spin-statistics theorem, which is just an (anti-)symmetrisation of the wavefunction upon permutation of indices.

On the other hand, a Helium atom is a composite object. Made of fermions, which will indeed experience Pauli repulsion. As long as your energy is lower than what would be needed to probe the internal structure of the atom, then you can consider the actual atom as the foundational building block. This is the same as saying that the spatial extent of the wavefunction of the atom (de Broglie wavelength $\lambda_{\mathrm{dB}} \propto 1/E \propto 1/\sqrt{T}$) is larger than the physical extent of the atom $\sim 1Å$ (or of the nucleus if you want to know about nuclei $\sim 1$ fm). Where $E$ is the energy and $T$ the temperature.