Degenerate perturbation theory

In summary, you found the two eigenvalues of the perturbation matrix to be either plus or minus modulus of lamdba. The groundstate is minus lamda.
  • #1
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Homework Statement


Hi, i have put the question, my attempt and actual answer in the attached picture. My answer is not quite right; firstly why is the second term a minus lambda, and where does the O(lamdba^2) come from?


Homework Equations





The Attempt at a Solution


 

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  • #2
What are the two eigenvalues you found when you diagonalized the perturbation? You should have found two. Once you know both, I think you'll see why the minus sign is there in the answer.

The [tex]O(\lambda^2)[/tex] just means that the any further correction to the energy is of order [tex]\lambda^2[/tex] or higher. There are no other corrections proportional to [tex]|\lambda|[/tex].
 
  • #3
vela said:
What are the two eigenvalues you found when you diagonalized the perturbation? You should have found two. Once you know both, I think you'll see why the minus sign is there in the answer.

The [tex]O(\lambda^2)[/tex] just means that the any further correction to the energy is of order [tex]\lambda^2[/tex] or higher. There are no other corrections proportional to [tex]|\lambda|[/tex].

Do you mean the plus and minus modulus [tex](\lambda)[/tex] for the two eigenvalues? If so how does that give the minus sign? Thanks
 
  • #4
You found the eigenvalues of the perturbation matrix to be plus or minus modulus of lamdba. So, the initial two-fold degenerate state acted by the perturbation will no longer be degenerate it will give rise to two different states (eigenvectors in first order of H) with different energies. Which is now the groud state?
 
  • #5
go quantum! said:
You found the eigenvalues of the perturbation matrix to be plus or minus modulus of lamdba. So, the initial two-fold degenerate state acted by the perturbation will no longer be degenerate it will give rise to two different states (eigenvectors in first order of H) with different energies. Which is now the groud state?

Ah, the groundstate is minus lamda. Thanks (to all)
 

FAQ: Degenerate perturbation theory

1. What is degenerate perturbation theory?

Degenerate perturbation theory is a method used in quantum mechanics to calculate the energy levels and wavefunctions of a system that has multiple degenerate states. Degenerate states are those with the same energy level but different wavefunctions.

2. When is degenerate perturbation theory used?

Degenerate perturbation theory is used when the perturbation is large and the degeneracy of the system cannot be lifted by first-order perturbation theory. It is also used when the degenerate states cannot be distinguished by symmetry arguments.

3. What are the key assumptions of degenerate perturbation theory?

The key assumptions of degenerate perturbation theory are that the perturbation is small compared to the energy difference between the degenerate states, and that the degenerate states are non-degenerate in the absence of perturbation.

4. How does degenerate perturbation theory differ from non-degenerate perturbation theory?

Degenerate perturbation theory takes into account the mixing of degenerate states, while non-degenerate perturbation theory assumes that there is no degeneracy in the system. In degenerate perturbation theory, the perturbation matrix is diagonalized to find the energy corrections, while in non-degenerate perturbation theory, the perturbation matrix is used to calculate the energy corrections directly.

5. What are the limitations of degenerate perturbation theory?

One limitation of degenerate perturbation theory is that it can only be applied to systems with a small perturbation. It also assumes that the perturbation does not change the degeneracy of the system. Additionally, degenerate perturbation theory may not give accurate results for systems with highly degenerate states or when the perturbation is large compared to the energy difference between the degenerate states.

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