# Heisenberg interaction Hamiltonian for square lattice

by JVanUW
Tags: hamiltonian, heisenberg, interaction, lattice, square
 P: 23 Hi, I just started self studying solid state and I'm having trouble figuring out what the hamiltonian for a square lattice would be when considering the heisenberg interaction. I reformulated the dot product into 1/2( Si+Si+δ+ +Si+δ+S-- ) + SizSi+δz and use Siz = S-ai+ai Si+ = √2S]ai ... Si+δz=-S+ai+δ+ai+δ ... Etc. But I'm getting for the terms of the hamiltonian aiai+δ +ai+δ+ai+ .... but don't these terms violate momentum conservation? What is the real heisenberg interaction hamiltonian for the square lattice?
 P: 20 Firstly, let's correct your terminology a little bit. The Heisenberg interaction is just: $$\mathcal{H}=\mathcal{J}\sum_{i} S_i \cdot S_{i+\delta}$$ You have rewritten it in terms of $S^z, S^+$ and $S^-$ operators which is fine. Your next step is to write it with respect to bosonic operators $a, a^\dagger$ in the Holstein-Primakoff representation, in which case the bosonic operators create and destroy spin waves. It appears you have taken $\mathcal{J}$ to be positive, in which case you have the antiferromagnetic model where spins on neighbouring sites prefer to be antiparallel. This is implicit in your choice of S and -S in the H-P representation. So far your bosonic operators are in the position representation. When you work all this out, you get terms with $a^\dagger_i a^\dagger_{i+\delta}$. These do not violate momentum conservation because they are still in the position representation - if you fourier transform them you'll see there is no problem. You are SUPPOSED to get them. This is what makes a ferromagnet (J<0) different from an antiferromagnet (J>0). In order to diagonalize the Hamiltonian, you must do two steps. 1. Fourier transform it. 2. Use a Bogoliubov transformation to get rid of the $a^\dagger_i a^\dagger_{i+\delta}$ terms. Google this if you don't know what it is.

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