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A Russel-Saunders Coupling in Extended Oxides

  1. Mar 1, 2016 #1
    Hey all,

    I am a graduate student (in chemistry) working on oxide crystals. Our group has a SQUID magnetometer which we use for magnetic property measurements. The other day a fellow student and myself got into a discussion about LS coupling and crystal electric fields. I know that CEF will quench the LS coupling, making the magnetic properties a function of only the electron spin contribution. So, in the presence of weak CEF effects, LS coupling becomes more predominant and is seen when we investigate the magnetic moment.

    My question is, what (if any) other factors influence the strength of LS coupling in oxide crystals? I was taught that Russel Saunders coupling (J = L + S) was always present, and I thought that the CEF was the only factor that quenches the coupling, but my colleague asserts that there must be other factors (material specific) that quenches LS coupling. I am open to this idea, but I'd like to know what those factors are.

    Thanks in advance!
     
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  3. Mar 3, 2016 #2

    TeethWhitener

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    I'm not a magnetism guy (so take what I say with a big pinch of salt), but it would be interesting to know what your colleague thinks these other factors would be. As far as I know the final expression for spin-orbit interaction is only dependent on the electric field that the electron sees. Maybe materials with strongly correlated electrons? I could believe that. We generally assume that ##\mathbf{L}## and ##\mathbf{S}## are good quantum numbers, but I imagine in a strongly correlated material this would no longer necessarily be true. Maybe this would have a bigger effect on the local ##\mathbf{E}## field than the electrostatic crystal fields.
     
  4. Mar 3, 2016 #3
    Thanks for the reply. We are synthetic chemists trying to understand magnetism, so any little bit helps. He has a bit less background in quantum than I do, so unfortunately he doesn't have a guess as to what these mysterious other factors are. Could strong magnetic coupling (i.e. superexchange) of the electrons be enough to influence the electric field in your opinion? I don't recall the exact plot, but if I remember correctly this material has magnetic ordering at fairly high temperature (maybe 150 K?) implying that there is fairly strong magnetic interaction between electrons.
     
  5. Mar 3, 2016 #4

    TeethWhitener

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    I don't know if superexchange would influence the (crystal) electric field so much, but I could certainly see a scenario where the spins of different electrons coupled more strongly than the spin and orbital angular momenta of the same electron. But I'm only basing this on experience with atomic angular momentum coupling, where you have LS vs. jj coupling (where LS here refers to total spin + total orbital angular momentum on all the electrons and jj refers to an individual electron's spin + orbital angular momentum). The point is that there may be a scenario where ##l## and ##s## are not good quantum numbers (in other words, the angular momentum operators ##\mathbf{L}## and ##\mathbf{S}## do not commute with the Hamiltonian ##\mathbf{H}##). In the atomic case, this happens with heavy atoms with strong spin-orbit effects, so that the spin and orbital quantum numbers are not good and you're forced to use the total angular momentum operator ##\mathbf{J}##. Again, I'm not sure how relevant all this is to a solid state problem.
     
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