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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. For $\Delta = 1$ we recover the Heisenberg XXX model. At first I thought that a ground state would be all spins making an angle of 45 degrees with the z-axis and the projected part an angle of 45 degrees with both the y and the x axis. Equivalent ground states would then follow by performing rotations of 90 degrees around the z-axis. However, it occurred to me that the model is solved by introducing the spin flip operators: $S^{\pm}_n := S^x_n \pm iS^y_n$. I think this effectively means that you are quantizing along the z-direction, yielding a ground state $|0> = |\uparrow\uparrow\uparrow\dots\uparrow>$, with all spins in the z-direction. Is this reasoning correct? Of course I have done spin in my quantum mechanics course, but I fail to make the connection with this case and have lost my handiness with it.

  2. $\Delta=0$: the XX or XY model. Apparently "the model can be exactly mapped into free lattice fermions". I have no clue what this means and how it works. References?

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Minor comment to the question (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. –  Qmechanic Sep 24 '13 at 8:52
    
@Qmechanic Done. –  Funzies Sep 24 '13 at 8:58
    
It would probably be clearer for readers that $h=0$ if you don't write the $-2h\sum_j S_j^z$ term in the first place. –  Mark Mitchison Sep 24 '13 at 10:16
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1 Answer 1

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  1. Yes, by introducing a homogenous field in the $z$ direction you break full $SU(2)$ rotational symmetry. Otherwise, you would have more degeneracies, but since the Hamiltonian conserves magnetization you can easily add or remove a homogenous field. If you do remove the field then any spin axis is fine and the ground state will be further degenerate. By introducing creation and annihilation operators for the spin projection in $z$ direction you are essentially hiding the full symmetry of the model, but the ground states will still be states which have all spins pointing in some direction. You will see that clearly for those which point in the $z$ direction because that is the basis you chose.
    For, finite $\Delta \neq 1$ the symmetry is actually a q-deformation of the universal enveloping algebra of $su(2)$, $U_q[SU(2)]$. See, for instance, L. D. Faddeev, "How Algebraic Bethe Ansatz works for integrable model" [1]

  2. This is called the Wigner-Jordan transformation. Basically you show that these models are essentially not interacting, that is to say that they are just like a gas of spinless "free" fermions.

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Aha, I see now. Sorry. –  Bubble Sep 24 '13 at 10:18
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Cheers, nice answer, +1. –  Mark Mitchison Sep 24 '13 at 14:39
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