I am a bit confused about the terminology concerning identical fermions. In quantum mechanics, identical fermions need to obey certain anticommutation relations i.e. have an antisymmetric total wavefunction (product of spatial and spin parts) under the exchange of two particles. But when exactly are two fermions considered to be 'identical'?
As far as I understand, two electrons are identical in the sense that they for example have the same electric charge, mass and spin quantum number $s=\frac{1}{2}$. (These are shared by all electrons.) How about the secondary spin (or spin projection) quantum number $m_s=\pm \frac{1}{2}$? If in an atom we have two electrons with the same principal quantum number, orbital quantum number and magnetic quantum number, but different $m_s$, are these two electrons 'identical' or not? The electrons are definitely identical if they additionally have the same $m_s$, but the Pauli exclusion principle forbids that. How does the consideration of the quantum numbers fit with the electrons having the same electric charge, mass, and spin $s$, which are independent of atomic orbitals?
Furthermore, are 'identical' and 'indistinguishable' fermions different things or not? If 'identical' particles cannot be told apart by an experiment, how about the Stern-Gerlach experiment? This links back to the above question whether $m_s=\pm \frac{1}{2}$ electrons are identical. My partial answer is that if we have electrons labelled 1 and 2, we cannot say which of the electrons has $m_s=-\frac{1}{2}$ and which has $m_s=\frac{1}{2}$. So in an experiment, you cannot say that 'The electron with $m_s=-\frac{1}{2}$ is electron 1' even if you were able to separate the $m_s=\pm \frac{1}{2}$ states. Or something along those lines. In other words, electrons are identical regardless of the quantum numbers.
Further confusion arises when we go beyond electrons. I have learned that atoms with an odd number of neutrons in an 'optical lattice' can be used to mimic electrons in a metal. Then the role of $m_s=\pm \frac{1}{2}$ is played by different hyperfine states of the atom as far as I can tell. If I now have atoms of the same mass (same species) in two different hyperfine states corresponding to spin-up and spin-down electrons, how are these atoms classified? Are they 'identical', 'indistinguishable', 'distinguishable', or what? Do these atoms obey the fermionic anticommutation relations?
I thought I understood what's going on i.e. two atoms of the same species are identical but can be in different internal states representing the spin projections and thus the atoms obey the usual fermionic anticommutation relations of spin-$\frac{1}{2}$ electrons. But then I found this paper (https://arxiv.org/abs/1111.2727) which says that atoms in the $| F=\frac{1}{2}, m_F=\pm \frac{1}{2} \rangle$ states of $^6$Li are distinguishable. How can distinguishable atoms play the role of electrons? How can they obey the anticommutation relations of spin-$\frac{1}{2}$ electrons? If I have understood correctly, there are no symmetry requirements for distinguishable particles in quantum mechanics. Or are the authors just possibly using sloppy language by calling two identical atoms in different internal states distinguishable? I would call two atoms of different species, e.g. $^6$Li and $^{40}$K, distinguishable.
Sorry for the long text, but hopefully I made my confusion sufficiently clear. Any clarifications are appreciated.