Degenerate perturbation theory first-order correction signs and order terms

[sxc656]

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

 

[vela]

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

 

[sxc656]

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

Do you mean the plus and minus modulus $$(\lambda)$$ for the two eigenvalues? If so how does that give the minus sign? Thanks

 

[go quantum!]

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?

 

[sxc656]

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)