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According to Griffiths, the degenerate perturbation theory says that the first-order corrections to the energies are the eigenvalues of the perturbation matrix. Griffiths solves for the eigenvalues in the unperturbed energy eigenbasis. But, because the eigenvalues of a matrix are independent of the choice of basis, the eigenvalues are just the eigenvalues of the perturbation matrix $\delta H$ on any basis.

However, I encounter a problem where $$ H_0=\frac{1}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \quad \text{and} \quad \delta H= \begin{pmatrix} 0 & 0 & \frac{1}{2} & \frac{1}{4} \\ 0 & 0 & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & 0 & 0 \end{pmatrix}. $$

The eigenvalues of the perturbation matrix $\delta H$ equals to $-\frac{3}{4}, \frac{3}{4}, -\frac{1}{4}, \frac{1}{4}$. But the solution says the first order corrections to the energies are $-\frac{1}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$. Why they are different?

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The relevant matrix that you need to diagonalize is not $\delta H$ itself, but its projection onto each one of the degenerate subspaces of $H_0$. Since the $E=1$ subspace is generated by $v_1=(1,0,0,0),v_2=(0,0,0,1)$ the matrix you need to diagonalize is then $$\delta H_1 =\pmatrix{0 &\frac{1}{4}\\\frac{1}{4}&0}$$ (and similarly for $-1$) which will give the right corrections.

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