Chapter 6: Problem 16
Suppose that the potential energy of a particle on a ring depends on the angle \(\varphi\) as \(H^{(1)}=\varepsilon \sin ^{2} \varphi .\) Calculate the first- order corrections to the energy of the degenerate \(m_{l}=\pm 1\) states, and find the correct linear combinations for the perturbation calculation. Find the second-order correction to the energy. Hint. This is an example of degenerate-state perturbation theory, and so find the correct linear combinations by solving eqn 6.42 after deducing the energies from the roots of the secular determinant. For the matrix elements, express \(\sin \varphi\) as \((1 / 2 \mathrm{i})\left(\mathrm{e}^{\mathrm{i} \varphi}-\mathrm{e}^{-\mathrm{i} \varphi}\right)\) When evaluating eqn \(6.42,\) do not forget the \(m_{1}=0\) state lying beneath the degenerate pair. The energies are equal to \(m_{l}^{2} \hbar^{2} / 2 m r^{2} ;\) use \(\psi_{m_{l}}=(1 / 2 \pi)^{1 / 2} \mathrm{e}^{i m_{\mu} \varphi}\) for the unperturbed states.