 Solution

In the original eigenbasis, $Q^{\prime }=\left(\begin{array}{cc}-1& iϵ\\ -iϵ& 1\end{array}\right)$. Insert the proposed first-order state for eigenvalue near minus one. The first component gives $c_1=0$; the second gives $-i{a}_1+a_2=-a_2$, hence $a_2=i{a}_1/2$. Similarly for the eigenvalue near one, $c_2=0$ and $b_2=i{b}_1/2$. Normalization gives $|a_1|=|b_1|=1$, with their phases freely selectable. Choosing the original-state phase conventions,The states are orthogonal through first order because the two cross terms cancel.

Exactly, $(Q^{\prime }{)}^2=(1+{ϵ}^2)I$, so putting $R=\sqrt{1+{ϵ}^2}$ givesFor an eigenvalue $q^{\prime }$, the first row requires the component ratio $v_2/v_1=-i(1+q^{\prime })/ϵ$, so the coefficient in the printed parametrization is $\boxed{B=1}$. With $D=\sqrt{(1+R{)}^2+{ϵ}^2}$, smooth normalized choices areThey have unit norm and zero inner product, and direct multiplication gives the two eigenvalue equations. For $ϵ\ne 0$, the printed scalar factors can be chosen $A_1=(1+R)/D$ and $A_2=iϵ/D$; the second is allowed a complex phase. The displayed ratio form is singular at zero, but the normalized vectors have the correct continuous limits there.

Expanding $R=1+{ϵ}^2/2+O({ϵ}^4)$ and $D=2+O({ϵ}^2)$ gives $|1^{\prime }⟩=|1⟩+iϵ|2⟩/2+O({ϵ}^2)$ and $|2^{\prime }⟩=|2⟩+iϵ|1⟩/2+O({ϵ}^2)$, while the eigenvalues have no first-order shift. This recovers all six perturbative coefficients with the same phase convention.