Why is the spin magnetic moment of an electron equal to the orbital magnetic moment of a hydrogen atom?

The intrinsic magnetic moment of an electron is equal to the magnetic moment that we obtain by effects of orbital motion of a hydrogen electron in the lowest shell (1 Bohr magneton), using classical physics.

Is this merely a coincidence or is there any relation between the quantum mechanical spin moment and the classically obtained orbital moment? How do we reconcile the same value of the two different types of magnetic moments?

• You mean the spin factor $1/2$ cancels against the $g$ factor 2? That's purely numerology, and does not persist at higher order in perturbation theory. – Thomas Mar 5 '18 at 18:01
• @Thomas Yes that is what I mean. Even if it is purely coincidental, it is very intriguing to get the two values as same. Both the values are due to completely different reasons but are the same numerically. – Aniansh Mar 5 '18 at 18:53
• A hydrogen atom has only one electron? Compared to another electron why wouldn’t they be close? – Bill Alsept Mar 5 '18 at 19:27

What's more, the fact that to lowest order they agree is nothing but a coincidence. On first principles you can conclude that they are both an integer numbers of $\mu_B$. In this case it just so happens that this integer is the same, but this would not be true if you were considering a bound state with other particles (say, a positron/electron pair or a hydrogen-like atom but with a particle of higher spin).
• It's not an unexplained adjustable constant like the fine structure constant. It’s not even a mathematical coincidence like ${{\pi }^{2}}\approx 10$. – Bert Barrois Mar 7 '18 at 16:52
• @BertBarrois But note that I never said that it was an unexplained adjustable constant. Quite the contrary, I said that you can prove on first principles that it is an integer times $\mu_B$. I said that the coincidence is that it is the same integer. Again, this needn't be the case; it just so happen to be. – AccidentalFourierTransform Mar 7 '18 at 16:56