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Heisenberg interaction Hamiltonian for square lattice

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JVanUW
#1
Nov8-12, 09:28 AM
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?
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OhYoungLions
#2
Nov20-12, 08:27 PM
P: 26
Firstly, let's correct your terminology a little bit. The Heisenberg interaction is just:

[tex]\mathcal{H}=\mathcal{J}\sum_{i} S_i \cdot S_{i+\delta} [/tex]
You have rewritten it in terms of [itex]S^z, S^+[/itex] and [itex]S^-[/itex] operators which is fine.

Your next step is to write it with respect to bosonic operators [itex]a, a^\dagger[/itex] in the Holstein-Primakoff representation, in which case the bosonic operators create and destroy spin waves. It appears you have taken [itex]\mathcal{J}[/itex] 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 [itex]a^\dagger_i a^\dagger_{i+\delta}[/itex]. 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 [itex]a^\dagger_i a^\dagger_{i+\delta}[/itex] terms. Google this if you don't know what it is.


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