The Kanamori hamiltonian, if the coefficients satisfy a certain relationship, can be seen to be rotationally invariant. Its symmetry is $U(1)_C\times SU(2)_S\times SO(3)_O$ (I add the subscripts $C,S,O$ to indicate that these are the symmetries associated respectively with charge, spin and orbitals). In this many-body hamiltonian basically we have diagonal terms, spin-flip terms and pair-hopping terms. If we ignore the pair-hopping terms, the hamiltonian can be seen to take the form $c_1\hat{N}(\hat{N}-1)+c_2 \vert \hat{S} \vert ^2,$ where $\hat{N}$ is the number operator and $\hat{S}$ is the spin operator. With this form, it is clear that the Hamiltonian is rotationally invariant and that the states $\vert n, s,m,\alpha \rangle$ are eigenstates of the hamiltonian, where here $n$ the number of electrons in the atom, $s$ the total spin, $m$ the z-component of spin and $\alpha$ is an index labeling the orbital degeneracy. Details can be found here, on page 6. However, if we include the pair-hopping terms, when writing the hamiltonian in this style a new term appears which is $\vert \hat{L} \vert^2,$ where, if there are $M$ levels in the atom, $\hat{L}$ is a vector of $M$ components whose $j$-th component is $$\sum_{k,l,\sigma} \epsilon_{jkl}\hat{c}^\dagger_{k \sigma}\hat{c}_{l \sigma}$$ (the $\epsilon_{jkl}$ is the Levi-Civita symbol).

In the paper I linked, they call this orbital isospin. My problem is that I don't understand what is this operator in physical terms. I don't understand its eigenstates and eigenvalues and I don't understand either why is it rotationally invariant. For example, states satisfying the firs Hund rule, which are of the form $\vert 2s ,s ,m, \alpha ,\rangle$ (in the case the number of electrons is less or equal to the number of levels in the atom; that is, we have at much a half-filled atom) are eigenstates of the hamiltonian, but I don't see why are they eigenstates of $\vert \hat{L} \vert ^2.$ This could be proved by an explicit (and very lengthy) algebraic computation, but should be much easier if I understood what is $\hat{L}.$ I thought at first that it was related to orbital angular momentum, but when I write the orbital angular momentum operator in second quantization with respect to a basis of localized orbitals (as is the case) what I get is something more complicated which involves integrals of the wavefunctions of the basis, of course.

  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Sep 7 '18 at 9:06

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