Intermediate Field

Now we will work the full problem with no assumptions about which perturbation is stronger. This is really not that hard so if we were just doing this problem on the homework, this assumption free method would be the one to use. The reason we work the problem all three ways is as an example of how to apply degenerate state perturbation theory to other problems.

We continue on as in the last section but work in the states of \bgroup\color{black}$\vert fm_f\rangle$\egroup. The matrix for \bgroup\color{black}$\langle fm_f\vert H_{hf}+H_B\vert f'm_f'\rangle$\egroup is

\begin{displaymath}\bgroup\color{black}\matrix{1&1\cr 1&-1\cr 1&0\cr 0&0}
... \mu_BB \cr
0&0& \mu_BB & {-3{\cal A}\over 4}}\right). \egroup\end{displaymath}

The top part is already diagonal so we only need to work in bottom right 2 by 2 matrix, solving the eigenvalue problem.

\begin{displaymath}\bgroup\color{black}\left(\matrix{ A&B\cr B&-3A}\right) \left...
\matrix{A\equiv {{\cal A}\over 4} \cr B\equiv \mu_BB} \egroup\end{displaymath}

Setting the determinant equal to zero, we get


\begin{displaymath}\bgroup\color{black} E^2+2AE-3A^2-B^2=0\egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black} E = {-2A\pm\sqrt{4A^2+4(3A^2+B^2)}\over 2}=-A\pm\sqrt{A^2+(3A^2+B^2)}\egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black} =-A\pm\sqrt{4A^2+B^2}\egroup\end{displaymath}

The eigenvalues for the \bgroup\color{black}$m_f=0$\egroup states, which mix differently as a function of the field strength, are

\begin{displaymath}\bgroup\color{black} E = -{{\cal A}\over 4}\pm\sqrt{ \left({{\cal A}\over 2}\right)^2
+ \left(\mu_BB\right)^2 }. \egroup\end{displaymath}

The eigenvalues for the other two states which remain eigenstates independent of the field strength are

\begin{displaymath}\bgroup\color{black} {{\cal A}\over 4}+\mu_BB \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} {{\cal A}\over 4}-\mu_BB .\egroup\end{displaymath}

Jim Branson 2013-04-22