Macroscopic quantum spin tunneling with two interacting spins

Abstract

We study the simple Hamiltonian, $H=\ensuremath{-}K({S}_{1z}^{2}+{S}_{2z}^{2})+\ensuremath{\lambda}{\stackrel{P\vec}{S}}_{1}\ifmmode\cdot\else\textperiodcentered\fi{}{\stackrel{P\vec}{S}}_{2}$, of two large, coupled spins which are taken equal, each of total spin $s$ with $\ensuremath{\lambda}$ the exchange coupling constant. The exact ground state of this simple Hamiltonian is not known for an antiferromagnetic coupling corresponding to the $\ensuremath{\lambda}>0$. In the absence of the exchange interaction, the ground state is fourfold degenerate, corresponding to the states where the individual spins are in their highest weight or lowest weight states, $|\ensuremath{\uparrow},\ensuremath{\uparrow}\ensuremath{\rangle},|\ensuremath{\downarrow},\ensuremath{\downarrow}\ensuremath{\rangle},|\ensuremath{\uparrow},\ensuremath{\downarrow}\ensuremath{\rangle},|\ensuremath{\downarrow},\ensuremath{\uparrow}\ensuremath{\rangle}$, in obvious notation. The first two remain exact eigenstates of the full Hamiltonian. However, we show that the two states $|\ensuremath{\uparrow},\ensuremath{\downarrow}\ensuremath{\rangle},|\ensuremath{\downarrow},\ensuremath{\uparrow}\ensuremath{\rangle}$ organize themselves into the combinations $|\ifmmode\pm\else\textpm\fi{}\ensuremath{\rangle}=\frac{1}{\sqrt{2}}(|\ensuremath{\uparrow},\ensuremath{\downarrow}\ensuremath{\rangle}\ifmmode\pm\else\textpm\fi{}|\ensuremath{\downarrow}\ensuremath{\uparrow}\ensuremath{\rangle})$, up to perturbative corrections. For the antiferromagnetic case, we show that the ground state is nondegenerate, and we find the interesting result that for integer spins the ground state is $|+\ensuremath{\rangle}$ and the first excited state is the antisymmetric combination $|\ensuremath{-}\ensuremath{\rangle}$ while for half odd integer spin, these roles are exactly reversed. The energy splitting, however, is proportional to ${\ensuremath{\lambda}}^{2s}$, as expected by perturbation theory to the $2s$th order. We obtain these results through the spin coherent state path integral.