I've applied second order time-independent degenerate perturbation theory corrections to the energy with the method presented in Modern Quantum Mechanics by J.J. Sakurai.

I shortly summarize this method:

I diagonalised my perturbation $V$, write $H_0$ (the unperturbed Hamiltonian) in the same basis and I find the eigenkets of $H_0$ to be $\left|l^{\left(0\right)}\right\rangle$.

Now I look at the degenerate subspace $D$, where $\left|l^{\left(0\right)}\right\rangle$ has only 2 vectors and calculate the first order correction:

$$\Delta_{l}^{\left(1\right)}=\left\langle l^{\left(0\right)}\right|V \left|l^{\left(0\right)}\right\rangle$$

This becomes a 2 by 2 matrix (double degenerate) for which I find the eigenvalues. These are the corrections to the energies.

and the second order correction.

$$\Delta_{l}^{\left(2\right)}=\sum_{k\notin D}\frac{\left|V_{kl}^{2}\right|}{E_{D}^{\left(0\right)}-E_{k}^{\left(0\right)}}$$

  • Where $k$ are state that are not in the degenerate subspace
  • $E_{k}^{\left(0\right)}$ is the energy of state $k$.
  • $E_{D}^{\left(0\right)}$ is the energy of the degenerate unperturbed state.

These are my results: Results of perturbation theory

The "exact" results are obtained by numerically solving the Schrödinger equation.

Now my issue is the following, the correction to the groundstate is very good (as can be seen in the image).

The corrections to the second and thirds energies are not as good, this is due to the fact that they lie close to each other and thus the denominator in the second order correction gets to big, which is not good (so I've been told).

How can I correct for this issue?

If anything in my question is unclear, please let me know!

  • $\begingroup$ you did not present the hamiltonian and perturbation in your question. $\endgroup$
    – Ali Moh
    Commented Mar 12, 2015 at 11:06
  • $\begingroup$ I think (fairly sure) that the solution to this problem is completely independent of the form of the Hamiltonian in question. I consciously left out these (rather complicated) equations. $\endgroup$
    – johnbaltis
    Commented Mar 12, 2015 at 12:15
  • $\begingroup$ I don't see how "the energies are too close to each other" can have anything to do with the issue other than stabilitiy issues in the numerics - unless this just implies (for mathematical reasons) that you will have large corrections by high order terms, which would only leave one conclusion: Include higher order terms in the calculation... But maybe my thinking is wrong? $\endgroup$
    – Martin
    Commented Mar 12, 2015 at 18:56

1 Answer 1


The second and the third energy level act like an "almost" degeneracy and therefore it doesn't work. One of the conditions for perturbation theory to work is that the matrix elements of the perturbation can't be bigger than the spacing between the energy levels, which is the case here.

A solution would be to introduce a real degeneracy for this "almost" degeneracy and treat the difference as a perturbation. This can be done as follows, write the Hamiltonian in a diagonal basis $ H=diag(E_1, E_2, E_3, E_4, \cdots)$ to $ H=diag(E_1, E_{23}, E_{23}, E_4, \cdots)$ with an extra perturbation $V=diag(0,-\Delta, \Delta,0,\cdots$, where $E_{23}$ is the average between $E_2$ and $E_3$ and $\Delta$ the differences between the average and the actual energy level.

I implemented this and it can be seen that this solves the problem.

new solution


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