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LS (Russell-Sauders) coupling and j-j coupling.

  1. Mar 23, 2014 #1
    I am not clear on why different models of coupling are useful in different situations.

    In general, J = L + S

    For small atoms, LS coupling dominates the angular momentum related component of the Hamiltonian. That means we calculate:

    Ltotal = l1 + l2 + ....
    Stotal = s1 + s2 + ....

    and then,

    Jtotal = Ltotal + Stotal.

    For larger atoms we calculate j = l + s for each electron before summing:

    Jtotal = j1 + j2 + ....

    So for smaller it atoms it seems like we can think of all the electrons as one single particle with the combined properties of all the electrons. This suggests that the electrons do not interact with each other at all. I mean that the spin and the angular momentum do not effect one another.

    For the larger atoms, we think in terms of total angular momentum. This suggests that the spin and angular momentum of electrons may effect other electrons but that if we know the total angular momentum of each single electron then we know everything we need to know.


    Am I drawing reasonable conclusions here?
     
  2. jcsd
  3. Mar 24, 2014 #2

    DrClaude

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    Staff: Mentor

    The different coupling schemes are best understood from the point of view of the central field (CF) approximation and perturbation theory.

    Take the full electronic Hamiltonian and split it into two parts:
    $$
    \begin{align}
    \hat{H}_\mathrm{e} &= \sum_{i=1}^{N} \left( - \frac{\hbar^2}{2m_\mathrm{e}} \nabla_i^2 - \frac{Z e^2}{4\pi \epsilon_0 r_i} + \sum_{j>i} \frac{e^2}{4\pi \epsilon_0 r_{ij}} \right) \\
    &= \sum_{i=1}^{N} \left[ - \frac{\hbar^2}{2m_\mathrm{e}} \nabla_i^2 - \frac{Z e^2}{4\pi \epsilon_0 r_i} + S(r_i) \right] + \sum_{i=1}^{N} \left[ -S(r_i) + \sum_{j>i} \frac{e^2}{4\pi \epsilon_0 r_{ij}} \right] \\
    &= \hat{H}_\mathrm{CF} + \hat{H}_\mathrm{res}
    \end{align}
    $$
    In the CFA, you take the best possible ##S(r)## such the residual electrostatic part, ##\hat{H}_\mathrm{res}## is as small possible, and can be seen as a perturbation. Solving ##\hat{H}_\mathrm{CF}## is easy since the potential is a central field: you are basically taking one electron at a time and approximating the other electrons as a spherical shell of negative charge around the nucleus.

    When solving the full Hamiltonian, you need to consider also spin-orbit coupling as an additional perturbation, so
    $$
    \hat{H} = \hat{H}_\mathrm{CF} + \hat{H}_\mathrm{res} + \hat{H}_\mathrm{SO}
    $$
    where both ##\hat{H}_\mathrm{res}## and ##\hat{H}_\mathrm{SO} ## are to be seen as perturbations. We then need to look at the relative strength of these perturbations.

    In the limit where ##E_\mathrm{res} \gg E_\mathrm{SO}##, the residual electrostatic interaction dominates and will couple the electrons to one another (into ##L## and ##S##) before the spin-orbit interaction couples ##L## and ##S##. This is Russell-Saunders coupling.

    In the limit where ##E_\mathrm{SO} \gg E_\mathrm{res}##, spin-orbit coupling acts first and couples ##l## and ##s## of each electron into ##j##, and ##\hat{H}_\mathrm{res}## couples them into ##J##. This is ##jj##-coupling.

    For light atoms, ##E_\mathrm{res} \gg E_\mathrm{SO}## is a (very) good approximation. For heavier atoms, you have ##E_\mathrm{res} \sim E_\mathrm{SO}##, so neither approximation is very good, and only for the heaviest atoms, like mercury, do you get a level structure that is close to the one predicted by ##jj##-coupling.
     
  4. Mar 24, 2014 #3
    Hi DrClaude, I don't think I follow your reply fully.

    What is S(r) in this case. Is it just a random function used to make the Hamiltonian easier to solve or is it the energy associated with the spin of the electrons?

    You say we choose S(r) such that Hres is as small as possible. Why can't we choose it to make H_{res} equal zero.
     
  5. Mar 24, 2014 #4

    DrClaude

    User Avatar

    Staff: Mentor

    It is not random. You can see it as the spherically symmetric part of the charge distribution. It has nothing to do with electron spin: it is purely electrostatic.

    Because its not. There is no way you can write ##\sum_{j>i} e^2/(4 \pi \epsilon_0 r_{ij})## exactly as a central potential. The idea is that you want to extract a central potential ##S(r)## from the electron-electron interaction, while limiting as much as what is leftover, such that the residual part is as small a perturbation as possible.
     
  6. Mar 24, 2014 #5
    Ok, so Hres contains a sum of all the potentials of all electrons in relation to all the other electrons, and this will not be spherically symmetric. We then subtract the nearest central potential we can make, S(r)?

    Hcf then becomes the total equivalent central potential (with kinetic energy of electrons aswell).

    Hres represents the perturbation from this central potential that all the electrons cause.

    How can I tell from the equation that Hres is dominant for low Z?
     
  7. Mar 25, 2014 #6

    DrClaude

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    Staff: Mentor

    You can't, at least with out calculating actual solutions. Keep in mind that the use of LS-coupling vs jj-coupling has an empirical basis. It allows to classify the electronic states that appear in actual atomic spectra.
     
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