Solomon A. Owerre, M. B. Paranjape
We study the simple Hamiltonian, $H=-K(S_{1z}^2 +S_{2z}^2)+ \lambda\vec S_1\cdot\vec S_2$, of two, large, coupled spins which are taken equal, each of total spin $s$. The exact ground state of this simple Hamiltonian is not known for an antiferromagnetic coupling, $\lambda>0$. In the absence of the exchange interaction, the ground state is four fold degenerate, corresponding to the states where the individual spins are in their highest weight or lowest weight states, $|-1 mm\uparrow, \uparrow>, |-1 mm\downarrow, \downarrow>, |-1 mm\uparrow, \downarrow>, |-1 mm\downarrow, \uparrow>$, in obvious notation. The first two remain exact eigenstates of the full Hamiltonian. However, we show the that the two states $ |-1 mm\uparrow, \downarrow>, |-1 mm\downarrow, \uparrow>$ organize themselves into the combinations $|\pm>=\frac{1}{\sqrt 2} (|-1 mm\uparrow, \downarrow> \pm |-1 mm\downarrow \uparrow>)$, up to perturbative corrections. For $\lambda>0$, the ground state is non-degenerate, and we find the interesting result that for integer spins the ground state is $|+>$, and the first excited state is the anti-symmetric combination $|->$ while for half odd integer spin, these roles are exactly reversed. The energy splitting however, is proportional to $\lambda^{2s}$, as expected by perturbation theory to the $2s^{\rm th}$ order. We obtain these results through the spin coherent state path integral.
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http://arxiv.org/abs/1304.3734
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