Hyperfine Splitting

The ground of hydrogen atom in the Coulumb potential is $$ E_0 = - 13.6057 eV. $$ Its wavefunction $$ psi(vec{r}) = mathbb{N} expleft(-frac{r}{a_B}right).,,,text{ where } a_B = 0.529177 dot A $$ The degeneracy of the ground state is two electron spin states: $vert uparrow rangle$ , and $vert downarrow rangle$

The above description is our common knowledge about the ground state of $H(1S)$. But the above saying omitted the states of proton in the nuleus of hydrogen. There, the proton has spin $I = frac{1}{2}$. Taking the proton's spin into account, the degeneracy of $H(1s)$ is $4$:

$$ vert S_3 I_3 rangle = vert uparrow uparrow rangle;,vert uparrow downarrow rangle;,vert downarrow uparrow rangle;,vert downarrow downarrow rangle $$

Therefore, is we consider the interaction between the electron spin and the proton spin (the hyperfine interaction) into consideration, $Scdot I$. Using additive of angular momentum $$ vert S= frac{1}{2}rangle oplus vert I =frac{1}{2}rangle Longrightarrow j= 0,; ,,, 1. $$

The singlet state $j=0$ gives one single state, and the triplet state $j=1$ gives three degenerstate: begin{align} vert j=0, m rangle = & ,, vert 0, 0 rangle; ,,Longrightarrow Delta E = -3A vert j=1, m rangle = & ,,vert 1, 1 rangle, ,vert 1, 0 rangle,,vert 1, -1 rangle,,Longrightarrow Delta E = +A end{align}

The pertubation of hyperfine interaction conserves the total sum of energy, therefore as the triplet states rise an amount of energy $A$ to $13.6057+A$, the singlet lower the energy by $3A$ to $13.657 - 3A$. The value of $A$ is very small, $A approx 6times 10^{-6} eV$. The splitting can only by observed by ESR (electron spin resonance, a microwave absorption technique).