The quantum-mechanical observable $Q$ has just two orthonormal eigenstates $|1\rangle$ and $|2\rangle$ with eigenvalues $-1$ and 1 , respectively. The operator $Q^{\prime}$ is defined by $Q^{\prime}=Q+\epsilon T$, where

$$T=\left(\begin{array}{cc} 0 & i \\ -i & 0 \end{array}\right)$$

Defining orthonormal eigenstates of $Q^{\prime}$ to be $\left|1^{\prime}\right\rangle$ and $\left|2^{\prime}\right\rangle$ with eigenvalues $q_{1}^{\prime}$, $q_{2}^{\prime}$, respectively, consider a perturbation to first order in $\epsilon \in \mathbb{R}$ for the states

$$\left|1^{\prime}\right\rangle=a_{1}|1\rangle+a_{2} \epsilon|2\rangle, \quad\left|2^{\prime}\right\rangle=b_{1}|2\rangle+b_{2} \epsilon|1\rangle,$$

where $a_{1}, a_{2}, b_{1}, b_{2}$ are complex coefficients. The real eigenvalues are also expanded to first order in $\epsilon$ :

$$q_{1}^{\prime}=-1+c_{1} \epsilon, \quad q_{2}^{\prime}=1+c_{2} \epsilon$$

From first principles, find $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$.

Working exactly to all orders, find the real eigenvalues $q_{1}^{\prime}, q_{2}^{\prime}$ directly. Show that the exact eigenvectors of $Q^{\prime}$ may be taken to be of the form

$$A_{j}(\epsilon)\left(\begin{array}{c} 1 \\ -i\left(1+B q_{j}^{\prime}\right) / \epsilon \end{array}\right)$$

finding $A_{j}(\epsilon)$ and the real numerical coefficient $B$ in the process.

By expanding the exact expressions, again find $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$, verifying the perturbation theory results above.