McIntyre, quantum mechanics,pg360

Suppose states $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$ are degenerate eigenstates of unperturbed Hamiltonian $H$.

The first-order perturbation equation we want to solve is $$ \left.\left(H_{0}-E_{n}^{(0)}\right)\left|n^{(1)}\right\rangle=\left(E_{n}^{(1)}-H^{\prime}\right) \mid n^{(0)} \right >) $$

But the energy degeneracy of these two states creates an ambiguity. Both $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$ satisfy the zeroth-order energy eigenvalue equation for the energy $E_{2}^{(0)}$, but so does any linear combination of the two states. If we are trying to find the energy correction to the state with zeroth-order energy $E_{2}^{(0)}$, how do we know whether to use the state $\left|2^{(0)}\right\rangle$ or the state $\left|3^{(0)}\right\rangle$ in the perturbation equation?

Why can't I use $\left|2^{(0)}\right\rangle$ and $\left|3^{(0)}\right\rangle$individually in the perturbation equation and carry on?


1 Answer 1


There are an infinite number of linear combinations of $|2^{(0)}\rangle$ and $|3^{(0)}\rangle$ which are eigenstates of $H_0$. However, only two specific linear combinations are eigenstates of $H'$. The main difference between degenerate perturbation theory and nondegenerate perturbation theory is that you need to find those special linear combinations that are eigenstates of $H'$. Once you have them, then the rest of the calculation proceeds in the same way as nondegenerate perturbation theory.

  • $\begingroup$ Did you mean to say that we need to find those eigenstates of $H_0$ that upon perturbation change to eigenstates of $H'$ ? $\endgroup$
    – Kashmiri
    Commented Apr 9, 2022 at 17:20
  • $\begingroup$ @Kashmiri No, you are looking for states that are simultaneously eigenstates of $H_0$ and $H'$. $\endgroup$
    – Andrew
    Commented Apr 9, 2022 at 17:35
  • $\begingroup$ By $H'$ do you mean the perturbation Hamiltonian or the total new Hamiltonian? $\endgroup$
    – Kashmiri
    Commented Apr 9, 2022 at 23:26
  • $\begingroup$ The new Hamiltonian. But it also doesn't matter, since if $|\alpha\rangle$ is an eigenstate of $H_0$ and $H'$, it is also an eigenstate of the peturbation Hamiltonain $\delta H = H' - H_0$. $\endgroup$
    – Andrew
    Commented Apr 9, 2022 at 23:28
  • $\begingroup$ The text book chooses an arbitrary state in the degenerate subspace which is eigenstate of $H_0$ ,then adds correction terms to it in increasing order. This new state is said to be the eigenstate of the total new Hamiltonian and we want to find this state. Why do you say that this state has to be eigenstate of both the perturbed and unperturbed Hamiltonian? $\endgroup$
    – Kashmiri
    Commented Apr 10, 2022 at 0:31

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