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:
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!