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Basis for Spin Orbit Coupling

  1. Oct 23, 2015 #1
    So we obtain the perturbation Hamiltonian H as something proportional to S.L/r3 and the first order energy shift is then the expectation value of this perturbation Hamiltonian in the state that is being perturbed.

    So let a general gross structure state that we are perturbing be |n l ml s ms >. Finding the expectation of 1/r3 is fine using this state. However we must recast the other part as S.L=(J2-L2-S2)/2. This is where I'm losing what's going on with the maths and books don't tend to bother explaining it. So the state I have chosen is not an eigenket of J2. What I believe is happening is that we express the above state as a linear combination of eigenkets of the form |n j mj l, s> using the techniques of addition of angular momentum (because then we have it in terms of eigenkets of the squared operators which makes life easier). However this doesn't give the correct answer because the linear combination gives multiple terms whereas there should only be one.

    Can anybody see where I have made a mathematical error? Thank you :)
     
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  3. Oct 24, 2015 #2

    blue_leaf77

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    Consider the fact that ##\mathbf{J} = \mathbf{L} + \mathbf{S}##, if you calculate the scalar product ##|\mathbf{J}|^2##, what do you get in the right hand side?
    Yes, that's the right direction. You don't have to go through all the trouble of using the addition of angular momenta theorem if you use table of Clebsch-Gordan coefficients.
    May be it will help if you post your calculation.
     
    Last edited: Oct 24, 2015
  4. Oct 24, 2015 #3
    So what books seem to use is that (for dimensionless angular momentum operators)
    <n l ml s ms|S.L|n l ml s ms>∝<n l ml s ms|J2-S2-L2|n l ml s ms>=j(j+1)-l(l+1)-s(s+1).

    However my idea is that we should be writing
    |n l ml s ms> as an expansion in the states |n j mj l s> which gives something of the form |n l ml s ms>=a1|n j=l+s mj=ml+ms l s>+a2|n j=l+s-1 mj=ml+ms l s>+... as we can expand in states between j=l+s and j=l-s. Now doing the expectation value gives
    <n l ml s ms|S.L|n l ml s ms>∝(a1*<n j=l+s mj=ml+ms l s|+a2*<n j=l+s-1 mj=ml+ms l s|+...)(J2-S2-L2)(a1|n j=l+s mj=ml+ms l s>+a2|n j=l+s-1 mj=ml+ms l s>+...).

    Now this gives a massive mess - we get j(j+1)-l(l+1)-s(s+1) but this has to be summed over all the possible kets in the expectation value, i.e
    |a1|2[(l+s)(l+s+1)-l(l+1)-s(s+1)]+|a2|2[(l+s-1)(l+s)-l(l+1)-s(s+1)]+...
     
  5. Oct 24, 2015 #4

    blue_leaf77

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    If the right most expression is what is written in your book, then it must be implying that the state for which the expectation value was calculated is already the |n j mj l s> state.
     
  6. Oct 24, 2015 #5
    Yes it seems to, but that doesn't make sense because we are perturbing the gross structure states which are the set of |n l ml s ms>, so these should be the states used to calculate the expectation value. I've seen degenerate perturbation theory mentioned in a few places but I can't quite see how it would fit in here.
     
  7. Oct 24, 2015 #6

    blue_leaf77

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    What is required in the perturbation theory to be the zeroth-order wavefunction is that it must be eigenstate of the unperturbed Hamiltonian, in this sense the ##|n,l,m_l,m_s\rangle## obviously complies with that requirement. But this set of states does not diagonalize the perturbation Hamiltonian, which is proportional to ##\mathbf{L}\cdot \mathbf{S}##, moreover it is degenerate (one energy eigenvalue is shared by multiple different states). Degeneracy poses problem in perturbation theory because of the diverging value of some series in the derivation, the way to get around this (read the appropriate chapter in your textbook) is to use another set of states in which the perturbation is diagonal, yet at the same time is still eigenstate of the unperturbed Hamiltonian. In the current problem such a state is the ##|n,j,m_j,l,s\rangle##, for a given values of the quantum number contained in the ket there, it is a superposition of ##|n,l,m_l,m_s\rangle## states of different ##m_l## and ##m_s## but constant ##n##, ##l##, and ##s##, which means it is indeed eigenfunction of the unperturbed Hamiltonian. So, the expectation value of the perturbation should be computed in this new set of states.
     
  8. Oct 24, 2015 #7

    blue_leaf77

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    The state ##| n, l, m_l, s, m_s \rangle## is a superposition of ##| n, j, m_j, l,s\rangle## with different ##j##'s and ##m_j##'s, so I don't think we can say that ##| n, l, m_l, s, m_s \rangle## is an eigenstate of ##\mathbf{J}^2## and its ##z## component.
     
  9. Oct 24, 2015 #8

    DrClaude

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    Don't know what I was thinking. It is obviously not an eigenstate of J2. But it is an eigenstate of Jz, with ##m_j = m_l + m_s## (the triangle rule applies).
     
  10. Oct 24, 2015 #9

    blue_leaf77

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    Ah yes, it's eigenstate of ##J_z## but not of ##J^2##.
     
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