# Does the second-order correction to degenerate perturbation theory vanish?

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.

• 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 – KF Gauss Jun 11 at 19:43