Splitting of the Hydrogen Ground State

The ground state of Hydrogen has a spin $1\over2$ electron coupled to a spin $1\over2$ proton, giving total angular momentum state of $f=0,1$. We have computed in first order perturbation theory that

$$\bgroup\color{black} \Delta E = {2\over 3} (Z\alpha)^4 \left(... ...mc^2) g_N {1\over{n^3}} \left(f(f+1)-{3\over 2}\right). \egroup$$

The energy difference between the two hyperfine levels determines the wave length of the radiation emitted in hyperfine transitions.

$$\bgroup\color{black} \Delta E_{f=1} - \Delta E_{f=0} = {4\ove... ...^4 \left({m\over{M_N}}\right) (mc^2) g_N {1\over{n^3}} \egroup$$

For $n=1$ Hydrogen, this gives

$$\bgroup\color{black} \Delta E_{f=1} - \Delta E_{f=0} = {4\ov... ...t) (.51\times 10^6)(5.56)=5.84\times 10^{-6}\mbox{ eV} \egroup$$

Recall that at room temperature, $k_Bt$ is about ${1\over 40}$ eV, so the states have about equal population at room temperature. Even at a few degrees Kelvin, the upper state is populated so that transitions are possible. The wavelength is well known.

$$\bgroup\color{black}\lambda = 2\pi{\hbar c\over E} = 2\pi{197... ...imes 10^{-6}}}\AA = 2\times 10^9 \AA = 21.2\mbox{ cm} \egroup$$

This transition is seen in interstellar gas. The $f=1$ state is excited by collisions. Electromagnetic transitions are slow because of the selection rule $\Delta\ell=\pm 1$ we will learn later, and because of the small energy difference. The $f=1$ state does emit a photon to de-excite and those photons have a long mean free path in the gas.