The ground of hydrogen atom in the Coulumb potential is
$$
E_0 = - 13.6057 eV.
$$
Its wavefunction
$$
\psi(\vec{r}) = \mathbb{N} \exp\left(-\frac{r}{a_B}\right).\,\,\,\text{ where } a_B = 0.529177 \dot A
$$
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, $S\cdot I$. 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, $A \approx 6\times 10^{-6} eV$. The splitting can only by observed by ESR (electron spin resonance, a microwave absorption technique).
