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Consider a degenerate two-state system with states denoted by $|1\rangle$ and $|2\rangle$. If we apply a perturbation $H^\prime$, the first order correction to the energy is obtained by choosing two linear combinations of $|1\rangle$ and $|2\rangle$ that diagonalizes $H^\prime$. So can we say that the second order correction always vanish in this case because $H^\prime_{12}$ vanishes? I am disturbed by the denominator which blows up.

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  • $\begingroup$ I think the answer is simply yes, you construct the states so that the numerator is exactly zero, then in the derivation of the 2nd order perturbation you get 0=0 and can't divide by a denominator of 0 $\endgroup$ – KF Gauss Jun 11 at 19:43
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According to Sakurai, once you have the first order energy shifts, you no longer deal with a degenerate case and you can use the non-degenerate perturbation formulas to calculate higher order corrections. So, I suppose for the second order you need to use the new energy levels in the denominator and not the unperturbed ones.

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