Hyperfine Splitting in a Weak B Field

Since the field is weak we work in the states \bgroup\color{black}$\vert fm_f\rangle$\egroup in which the hyperfine perturbation is diagonal and compute the matrix elements for \bgroup\color{black}$W_z=\mu_BB\sigma_z$\egroup. But to do the computation, we will have to write those states in terms of \bgroup\color{black}$\vert m_s m_i\rangle$\egroup which we will abbreviate like \bgroup\color{black}$\vert+-\rangle$\egroup, which means the electron's spin is up and the proton's spin is down.

& \sigma_z\left\vert 11\right>=\sigma_z\left\vert++\right>=\le...
...vert+-\right> + \left\vert-+\right>\right) =\left\vert 10\right>

Now since the three \bgroup\color{black}$(f=1)$\egroup states are degenerate, we have to make sure all the matrix elements between those states are zero, otherwise we should bite the bullet and do the full problem as in the intermediate field case. The \bgroup\color{black}$f=1$\egroup matrix is diagonal, as we could have guessed.

\begin{displaymath}\bgroup\color{black}\mu_BB\left(\matrix{1&0&0\cr 0&0&0\cr 0&0&-1}\right)\egroup\end{displaymath}

The only nonzero connection between states is between \bgroup\color{black}$f=1$\egroup and \bgroup\color{black}$f=0$\egroup and we are assuming the hyperfine splitting between these states is large compared to the matrix element.

So the full answer is


which is correct for both \bgroup\color{black}$f$\egroup states.

Jim Branson 2013-04-22