Consider the Hamiltonian $H=H_{0}+V$, where $V$ is a small perturbation. If $H_{0}|n\rangle=E_{n}|n\rangle$, write down an expression for the eigenvalues of $H$, correct to second order in the perturbation, assuming the energy levels of $H_{0}$ are non-degenerate.

In a certain three-state system, $H_{0}$ and $V$ take the form

$$H_{0}=\left(\begin{array}{ccc} E_{1} & 0 & 0 \\ 0 & E_{2} & 0 \\ 0 & 0 & E_{3} \end{array}\right) \quad \text { and } \quad V=V_{0}\left(\begin{array}{ccc} 0 & \epsilon & \epsilon^{2} \\ \epsilon & 0 & 0 \\ \epsilon^{2} & 0 & 0 \end{array}\right)$$

with $V_{0}$ and $\epsilon$ real, positive constants and $\epsilon \ll 1$.

(a) Consider first the case $E_{1}=E_{2} \neq E_{3}$ and $\left|\epsilon V_{0} /\left(E_{3}-E_{2}\right)\right| \ll 1$. Use the results of degenerate perturbation theory to obtain the energy eigenvalues correct to order $\epsilon$.

(b) Now consider the different case $E_{3}=E_{2} \neq E_{1}$ and $\left|\epsilon V_{0} /\left(E_{2}-E_{1}\right)\right| \ll 1$. Use the results of non-degenerate perturbation theory to obtain the energy eigenvalues correct to order $\epsilon^{2}$. Why is it not necessary to use degenerate perturbation theory in this case?

(c) Obtain the exact energy eigenvalues in case (b), and compare these to your perturbative results by expanding to second order in $\epsilon$.