I Have a hamilton matrix in perturbation form ($H=H_{o}+H'$):
$ H= \begin{bmatrix}1&0&0\\0&3&0\\0&0&-2\end{bmatrix}+\begin{bmatrix}0&c&0\\c&0&0\\0&0&c\end{bmatrix} $
I found the exact eigenvalues the traditional way: $det(A-\lambda I)=0$ yielding
$ \lambda=c-2$
$ \lambda=2-(c^2+1)^{1/2}$
$ \lambda=2+(c^2+1)^{1/2}$
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How would I find the eigenvalues using 2nd order perturbation theory?
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For the first order. Let $\lambda(c)=f(c)$ and $H=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$. Then $f(c)=f(0)+(d\lambda/dc)(0)c+o(c)$. Note that $\det(A-\lambda(c)I_3+cH)=0$. By the implicit function theorem, $(d\lambda/dc)(0)=\dfrac{tr(H.adj(A-\lambda(0)I_3))}{tr(adj(A-\lambda(0)I_3))}$.
When $\lambda(0)=-2$, $\lambda(c)=-2+\dfrac{15}{15}c+o(c)=-2+c+o(c)$.
When $\lambda(0)=1$, $\lambda(c)=1+o(c)$.
When $\lambda(0)=3$, $\lambda(c)=3+o(c)$.
To obtain the second order, it "remains" to calculate the derivative with respect to $c$ in $c=0$ of the function $\dfrac{tr(H.adj(A-\lambda(c)I_3+cH))}{tr(adj(A-\lambda(c)I_3+cH))}$. Good luck!