Solution distribution of degenerate perturbation secular EQ

[zhanhai]

When two states |k> and |k'> degenerate, a perturbation H' would lead to an energy split of <k|H'|k'>. As the number of degenrate states increases, the order of the secular equation rises correspondingly (and the equation could hardly be solved ?)

My question is: is there any knowledge of the distribution of the energy levels of the "good" states (linear combination of the original degenerate states) when the number of degenerate states is big? Are they mostly in the range of E₀±<k|H'|k'>? Or, is there any knowledge of their range of spread ?

Thank you.

 

[eljose79]

Thank you for your question. When two states degenerate, it means that they have the same energy level. This can occur when the Hamiltonian operator has a symmetry that leads to the degeneracy. When a perturbation H' is applied, it breaks the symmetry and causes the degenerate states to split in energy.

As the number of degenerate states increases, the order of the secular equation also increases. This can make it more difficult to solve the equation, but it is still possible to find the energy levels of the system. However, the complexity of the secular equation may make it more challenging to determine the exact energy levels.

In terms of the distribution of energy levels of the "good" states (linear combination of the original degenerate states), it will depend on the specific system and the nature of the perturbation H'. Generally, the energy levels of the "good" states will be in the range of E0±<k|H'|k'>, but the exact range of spread will depend on the strength and type of the perturbation.

I hope this helps answer your question. If you have any further inquiries, please do not hesitate to ask.