NOTE: Because this was a long question I have split it up in two different questions!

For a course on quantum integrability I am reading these notes. (Franchini: Notes on Bethe Ansatz Techniques. Lecture notes (2011))

Some questions have arisen to me, concerning the Heisenberg XXZ model. The general idea is that we will solve several versions of this model in class, using the Bethe Ansatz Approach. However, the basics are not yet clear to me. Consider the Hamiltonian: \begin{equation} \hat{H} = - J \sum_{i=1}^N \left(S^x_jS^x_{j+1} + S^y_jS^y_{j+1} + \Delta S^z_jS^z_{j+1}\right) - 2h\sum_{i=1}^NS^z_j, \end{equation} where we have periodic boundary conditions: $S^{\alpha}_{j+N} = S^{\alpha}_j $. In the following I will set $h=0$.

  1. Letting $J\Delta\rightarrow\infty$ we retrieve an Ising model with also the ground state $|0> = |\uparrow\uparrow\uparrow\dots\uparrow>$. Does this mean that the Ising model and the XXX model are equivalent? I also noticed that the states $|\uparrow\uparrow\uparrow\downarrow\uparrow\uparrow\uparrow>$ and $|\uparrow\uparrow\uparrow\downarrow\downarrow\uparrow\uparrow>$ have the same energy, namely $E=-J\Delta/2$. This means that when a single spin is flipped, it costs no more energy to flip more! Is this the reason that the Ising model in 1D has no phase transition? I know it is related to the fact that in 1D the number of nearest neighbours is equal for a single spin and a domain of spins, but I forgot the exact reasoning I once understood. (Peierls argument?) Finally the notes I mentioned state that domains in this ferromagnetic state have spin $S=1$. For one excited spin I can see this, but for more I don't. How to see this?
  2. Letting $J\Delta\rightarrow-\infty$ we get an anti-ferromagnetic ground state. The low-energy excitations are now domain walls. I don't really see why these carry spin $S=1/2$. Again, how to think about this?

1 Answer 1


1.) The easiest way to count the energy is as that of domain walls. Both of the configurations you have drawn have two domain walls. Each domain wall costs energy $E\Delta/2$, so both configurations have energy $E\Delta$ above the ground state. I don't understand either how one would count the spin of a domain as 1. Perhaps that is the spin/site?

2.) To see that a domain wall between the two antiferromagnetic ground states has spin $ 1/2$, look at a spin configuration with a domain wall and divide the spin at each site evenly between the adjacent links. For example, if there is a spin $S_z=1/2$ at site $i$ and a spin $S_z=-1/2$ at site $i+1$, you'd say there is $0$ spin on the link $<i,i+1>$. Likewise, if there is a spin $S_z=1/2$ at site $i$ and a spin $S_z=1/2$ at site $i+1$, you'd say there is spin $S_z=1/2$ on the link $<i,i+1>$. In this way you end up with 0 everywhere except where there are two spins of the same type in a row (i.e., where there is a domain wall). At the domain wall you get $S_z=\pm 1/2$.


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