Splitting of the Hydrogen Ground State

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

\begin{displaymath}\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\end{displaymath}

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

\begin{displaymath}\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\end{displaymath}

For \bgroup\color{black}$n=1$\egroup Hydrogen, this gives

\begin{displaymath}\bgroup\color{black} \Delta E_{f=1} - \Delta E_{f=0} =
(.51\times 10^6)(5.56)=5.84\times 10^{-6}\mbox{ eV} \egroup\end{displaymath}

Recall that at room temperature, \bgroup\color{black}$k_Bt$\egroup is about \bgroup\color{black}${1\over 40}$\egroup 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.

\begin{displaymath}\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\end{displaymath}

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

Jim Branson 2013-04-22