Hydrogen in a Strong B Field

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

$$\bgroup\color{black}H_{hf}={\cal A}\vec{S}\cdot\vec{I}\egroup$$

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

Recalling that we can write

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

the matrix elements can be easily computed. Note that the terms like $I_-S_+$ which change the state will give zero.

$$\bgroup\color{black}{\cal A}\left< +-\left\vert \vec{I}\cdot\... ...{1\over 2}I_+S_- +{1\over 2}I_-S_+ \right\vert +-\right>\egroup$$

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

$$\bgroup\color{black}\left< -+\left\vert H_{hf}\right\vert -+\right> = -{\cal A}{\hbar^2\over 4}.\egroup$$

$$\bgroup\color{black}\left< ++\left\vert H_{hf}\right\vert ++\right> = {{\cal A}\hbar^2\over 4}\egroup$$

$$\bgroup\color{black}\left< --\left\vert H_{hf}\right\vert --\right> = {{\cal A}\hbar^2\over 4}\egroup$$

We can write all of these in one simple formula that only depends on relative sign of $m_s$ and $m_i$.

$$\bgroup\color{black} E = E_{n00}+2\mu_BBm_s \pm {\cal A}{\hbar^2\over 4} = E_{n00}+2\mu_BBm_s +{\cal A}\hbar^2(m_sm_I)\egroup$$