Hydrogen in a Strong B Field

We need to compute the matrix elements of the hyperfine perturbation using \bgroup\color{black}$\left\vert m_sm_i\right>$\egroup as a basis with energies \bgroup\color{black}$E=E_{n00}+2\mu_BBm_s$\egroup. The perturbation is

\begin{displaymath}\bgroup\color{black}H_{hf}={\cal A}{\vec{S}\cdot\vec{I}\over\hbar^2}\egroup\end{displaymath}

where \bgroup\color{black}${\cal A}={4\over 3}(Z\alpha)^4\left(m_e\over M_N\right)m_ec^2g_N{1\over n^3}$\egroup.

Recalling that we can write

\begin{displaymath}\bgroup\color{black}\vec{S}\cdot\vec{I} = I_zS_z+{1\over 2}I_+S_- +{1\over 2}I_-S_+,\egroup\end{displaymath}

the matrix elements can be easily computed. Note that the terms like \bgroup\color{black}$I_-S_+$\egroup which change the state will give zero.


\begin{displaymath}\bgroup\color{black}{{\cal A}\over\hbar^2}\left< +-\left\vert...
...{1\over 2}I_+S_- +{1\over 2}I_-S_+ \right\vert +-\right>\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} ={{\cal A}\over\hbar^2}\left< +-\left\vert I_zS_z\right\vert +-\right>
= -{{\cal A}\over 4}\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\left< -+\left\vert H_{hf}\right\vert -+\right> = -{{\cal A}\over 4}.\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\left< ++\left\vert H_{hf}\right\vert ++\right> = {{\cal A}\over 4}\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\left< -\left\vert H_{hf}\right\vert -\right> = {{\cal A}\over 4}\egroup\end{displaymath}

We can write all of these in one simple formula that only depends on relative sign of \bgroup\color{black}$m_s$\egroup and \bgroup\color{black}$m_i$\egroup.

\begin{displaymath}\bgroup\color{black} E = E_{n00}+2\mu_BBm_s \pm {{\cal A}\over 4} = E_{n00}+2\mu_BBm_s +{\cal A}(m_sm_I)\egroup\end{displaymath}

Jim Branson 2013-04-22