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I have a specific Ising model Hamiltonian for a tetrahedral unit cell looks like: $$ H= J1 (S_1^zS^z_2 + S_1^zS^z_3 +S_1^zS^z_4 +S_2^zS^z_3+S_2^zS^z_4+S_3^zS^z_4) -2*J2(S_1^xS_2^x+S_1^yS_2^y+S_1^xS_3^x+S_1^yS_3^y+S_1^xS_4^x+S_1^yS_4^y+S_2^xS_3^x+S_2^yS_3^y+S_2^xS_4^x+S_2^yS_4^y+S_3^xS_4^x+S_3^yS_4^y) + J3[ 2(S_1^xS_2^x-S_1^yS_2^y + S_3^xS_4^x-S_3^yS_4^y)- (S_1^xS_3^x-S_1^yS_3^y + S_1^xS_4^x-S_1^yS_4^y+S_2^xS_3^x-S_2^yS_3^y+S_2^xS_4^x-S_2^yS_4^y)+\sqrt{3} (S_1^xS_4^y+S_1^yS_4^x+S_2^xS_3^y+S_2^yS_3^x-S_2^xS_4^y-S_2^yS_4^x-S_1^xS_3^y-S_1^yS_3^x) ]$$

where J2<0. I am trying to find the ground state for this Hamiltonian, and then I try to add a small perturbation: $$ H_p = J4[-2(S^z_1 S_2^x +S^z_2S_1^x + S^z_3 S_4^x+ S^z_4S_3^x)+(S^z_1S_3^x+S^z_3S_1^x +S^z_1S_4^x+S^z_4S_1^x+S^z_2S_3^x+S^z_3S_2^x+S^z_2S_4^x+S^z_4S_2^x ) + \sqrt{3}(-S^z_1S_3^y- S^z_3S_1^y +S^z_1S_4^y+S^z_4S_1^y+S^z_2S_3^y+S^z_3S_2^y-S^z_2S_4^y-S^z_4S_2^y)] $$

Then I need to find the new ground state after adding this perturbation. We assume the spin behaves classically so on every site, $(S^x)^2+(S^y)^2+(S^z)^2=1$.

I am trying to solve this problem analytically but it seems very hard. I tried to assume spherical coordinates for the spin, but there are too many variables flying around. Can anyone give me some ideas how to do it? Thanks!

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    $\begingroup$ I would definitely not call this an Ising model. There are a variety of reasons, but when you say that spins are classical length-one vector, this is pretty much the opposite situation to the Ising one where one assumes that spin variables are boolean, e.g., $\pm1$. $\endgroup$ – lcv Oct 26 '17 at 5:24

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