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How can you work out the average perturbation, from a normal hamiltonian, of all states that rely on the quantum numbers of s = __ and l = __, with the perturbation being proportional to the product of S.L, which if my memory serves is the J quantum number? i.e how can you work out all the average perturbation states of all the s=1/2 states? Thank you in advance, been puzzling over this for days. I've read that the S^2, L^2 and J^2 operators all commute with H, the normal hamiltonian.

H = p^2 /2m + V(r)

So if they commute then the eigenstates are zero, but how do you get past this?

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  • $\begingroup$ Hey, I may be able to help with this. Can you clarify what you mean by "average perturbation"? Do you mean the corrections to the eigen functions and/or eigen energies when applying a perturbing Hamiltonian proportional to $S\cdot L$ (spin-orbit coupling)? $\endgroup$
    – dhudsmith
    Nov 24, 2015 at 17:18
  • $\begingroup$ So you'd write down <H' subscript(nlj)> is the first order energy correction. Each will have a corresponding j value. And you can use all this information to determine the average perturbation of all states which relates to S and L. Where the average perturbation is the expectation value, I assume. $\endgroup$ Nov 24, 2015 at 18:12
  • $\begingroup$ I guess OP means the first-order energy correction due to spin-orbit coupling. You can either use standard degenerate perturbation theory in each degenerate subspace labelled by the quantum number $n$ or you could directly go to the basis $\left| l,s,j,m_j \right>$ in which the perturbation is diagonal and use $\mathbf L \cdot \mathbf S = \frac{1}{2} \left( \mathbf J^2 - \mathbf L^2 - \mathbf S^2 \right)$. Btw, if two operators commute it just means that there exists a common basis where both are diagonal. $\endgroup$
    – Praan
    Nov 25, 2015 at 15:07
  • $\begingroup$ So if two operators commute with the hamiltonian how do you find the the common eigenstate? $\endgroup$ Nov 25, 2015 at 16:26

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