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According to this accepted answer, the most simple example of two atoms with degenerate ground states are boron and carbon. But hydrogen is the simplest of all which has the ground state configuration $^2{\rm S}_{1/2}$, characterized by $L=0, S=\frac{1}{2}, J=\frac{1}{2}$. Why is the hydrogen ground state not $2J+1=2$-fold degenerate?

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    $\begingroup$ The answer you've linked to is not accepted and does not mention either boron or carbon (and, also, it is incorrect). Presumably you intended to link to this answer on the same thread? $\endgroup$ – Emilio Pisanty May 4 '20 at 19:24
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The answer you (presumably) refer to is talking about other things, so the specifics are glossed over. (This was done in the expectation that readers who are aware of physics that require sharpening of those details would also be able to apply common sense. Perhaps that expectation was misguided.)

The claim there is that boron and carbon are the simplest examples of atoms with degenerate ground state, because their spatial degeneracy comes from having a partially-filled $p$ shell which can admit multiple different orientations; as such, their degeneracy can be explained using only electrostatics and QM, and without involving spin.

There is no claim there that they are the simplest atoms with degenerate ground states. This is because they're not: once spin is included, both hydrogen and lithium have spin degeneracy higher than 1.


Now, if you really want to split hairs beyond all reasonable limits, as the existing answer does, then the ground state of hydrogen ceases to be degenerate, as the electronic spin doublet couples with the nuclear spin doublet to make a hyperfine singlet ground state and a triplet first excited state.

But then again, in deuterium the nuclear spin is 1, so the hyperfine manifolds in the electronic ground state have total spin $F=\tfrac12$ and $F=\tfrac32$, with the former lower in energy and doubly degenerate.

So, back to you: does this make "hydrogen" the simplest atom to have a degenerate ground state? Or does it simply show the fact that it was a fairly ridiculous thing to worry about in the first place?

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Hyperfine splitting separates the electron spin parallel to the proton state from the electron spin antiparallel to that of the proton by the energy of the $21$cm line. I did not know that this did not happen in Boron and carbon. Ones lives and learns ....

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  • $\begingroup$ Boron and carbon really do not have hyperfine splitting of the ground state? $\endgroup$ – mithusengupta123 May 4 '20 at 17:01
  • $\begingroup$ Actually, the answer linked doesn't elaborate about the precision to which boron and carbon and other atoms have the same-energy $m=\mathrm{var}$ states. It might appear that they do have hyperfine splitting, but that answer simply was on a cruder approximation. $\endgroup$ – Ruslan May 4 '20 at 17:51
  • $\begingroup$ Boron and carbon obviously have hyperfine splitting. Why would you think (or claim) that they don't? $\endgroup$ – Emilio Pisanty May 4 '20 at 19:30
  • $\begingroup$ Mmmm. As Emilio says,you can't really have exactly degenerate states unless some symmetry forbids the coupling. $\endgroup$ – mike stone May 4 '20 at 23:34

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