How to find energy levels with approx up to $\lambda^2$? The problem is from stationary perturbation theory. Given a Hamiltonian $H = H_0 + \lambda V$ where $\lambda \ll 1$, $H_0 = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$, $a = b$, $V = \begin{bmatrix} \alpha & \beta \\ \beta & \gamma \end{bmatrix}$. Find the energy levels of the system to within $\lambda^2$. Here we solve the secular equation, we find from it the first-order corrections to the energy (they look like two solutions of the quadratic equation). These corrections have the accuracy of $\lambda^1$. How to find second-order corrections? I understand that we go to the basis of the correct functions of the zeroth approximation, the degeneracy is removed, and further we can apply the perturbation theory formulas for the nondegenerate case. But this approach seems wrong to me, because 1) there is no $\lambda^2$ 2) the degenerate energy level split into two close ones - how to handle it?
Please explain in detail how to solve this problem.
 A: The difficulty is that, if $a=b$, then it is not possible to make an expansion in terms of the original eigenstates or energies.  If $H=H_0+\lambda V$, the effective expansion parameter is not $\lambda$ but rather $\lambda/(a-b)$, i.e. the assumption for an expansion to be valid is that 
$$
\frac{\lambda}{a-b}\ll 1\, ,
$$
which obviously fails when $a=b$.  
When this happens, one cannot proceed by expansion.  Finding eigenvalues and eigenvectors is a process of making a similarity transformation, i.e finding a transformation $T$ so that 
$T(H_0+\lambda V)T^{-1}$ becomes diagonal.   Since the unperturbed Hamiltonian is $H_0=a\hat 1$, we have $TH_0T^{-1}=H_0$ so the transformation $T$ is completely determined by the perturbation $V$ and is independent of the factor $\lambda$.   
The eigenvalues and eigenvectors are determined from $V$ alone, independent of $\lambda$. 
Thus, there is no such thing as a "first order" or a "second order" since nothing in $T$ depends on $\lambda$, i.e. the exact eigenstates do not depend on $\lambda$.    
For instance, take $H_0=a\left(\begin{array}{cc} 1&0 \\ 0& 1\end{array}\right)$ and $V= \left(\begin{array}{cc} 1&1 \\ 1 &-1\end{array}\right)$. The eigenvectors for $H_0+\lambda V$ are 
$$
\left(\begin{array}{c}
\frac{1-\sqrt{2}}{\sqrt{4-2 \sqrt{2}}}\\
\frac{1}{\sqrt{4-2 \sqrt{2}}}
\end{array}\right)\, ,\qquad 
\left(\begin{array}{c}\frac{1+\sqrt{2}}{\sqrt{2 \left(2+\sqrt{2}\right)}}
\\
\frac{1}{\sqrt{2\left(2+\sqrt{2}\right)}}\end{array}\right)
$$
independent of $\lambda$.  
If $\eta_\pm$ are the exact eigenvalues of $V$ (for which obviously you cannot find an expansion in terms of $\lambda$ as $\lambda$ does not enter in $V$, then the exact eigenvalues of $H_0+\lambda V$ will be $a+\lambda \eta_\pm$.  In the example above, they are $a\pm \sqrt{2}\lambda$.  You can see they are linear in $\lambda$ and unrelated in size to the diagonal entries of $V$.  Obviously since the exact answer is linear in $\lambda$, there is no meaning to calculating a term in $\lambda^2$.
