The electrons in the hydrogen molecule experience the same potential and are thus in the same state, so the Pauli exclusion forces them to have opposite spins.
Since the protons are identical by symmetry, are these in the same state too? If they were, then the spins would be forced opposite and orthohydrogen would not exist.
So in what way am I thinking about this wrong? Is it that the ground state for the protons is degenerate?
The wavefunctions of the electrons in $H_2$ molecule is overlapping a lot, i.e. occupying the same space, and so it is very strongly needed to obey fermion statistics. The protons (core ions in other systems) are instead rather well-separated, and so the ``true ground state" of some fermion statistics obeying version of the protons will be separated from the low-lying excited states of the protons doing something else by a very tiny excitation energy.
This, in turn, makes the rate of relaxation, and the rate of oscillation, between these states very very slow, and then we will be able to observe in experiments that ortho- and para-hydrogen exists. It kinda looks to us as if they obey Maxwell-Boltzmann statistics rather than Fermi-Dirac statistics, if we are not going to wait figuratively forever for the MB statistics parts to decay away. It does not help that FD statistics would fill in from the bottom up, and so the difference will be really difficult to discern.
There is quite a lot of physics where the ground state alone is not sufficient for explaining actual real life experiences.
Orthohydrogen is observed, so this isn't really a physics question. Parahydrogen is the ground state of $H_2$. The ortho states are slightly higher (~14 meV) in energy. So, is only the ground state "dihydrogen", or do you consider excited states? Nature doesn't care what labels you use.
