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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asked Jun 11, 2019 at 4:10
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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 GaussCommented Jun 11, 2019 at 19:43
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2 Answers
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For degenerate levels the first order correction is obtained by the exact diagionalization of the Hamiltonian for the degenerate states. If all the states are degenerate, belonging to the same energy eigenvalue, then this is equivalent to the exact diagonalization of the Hamiltonian - no perturbation theory is needed (It would be necessary, if we have other states or multiple degenerate energies.)
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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.
