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I am considering the Heisenberg XXZ Hamiltonian: $$ H(\Delta, J) = J\sum_{i=1}^L\left(\sigma^x_i\sigma^x_{i+1} + \sigma^y_i\sigma^y_{i+1} + \Delta \sigma^z_i\sigma^z_{i+1} \right) $$ Apparently, one can show: $$-H(-\Delta) = UH(\Delta)U^{-1}$$, where $U=\Pi_{m=1}^{M/2}\sigma_z^{2m}$ is a unitary transformation. I am having some trouble showing this. Of course the transformation commutes with the z-Pauli matrices and using $(\sigma^z)^{\dagger} = \sigma_z$ and $(\sigma_z)^2 = \textrm{id}$ this term is unchanged. However, I am left with mix terms for the first and second term.

Furthermore, my reference claims that because of this simularity transformation, $J$ can be omitted if $-\infty<\Delta<\infty$. I don't really see how this follows from the simularity transform.

Any help is much appreciated!

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1 Answer 1

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I don't see how you get mixed terms. Your $U$ acts on every other site. For instance (as operators acting on different sites commute) $$\ldots+ U \sigma^{2i}_x \sigma^{2i+1}_x U^{-1} +\ldots= \ldots \sigma^{2i}_z \sigma^{2i}_x \sigma^{2i}_z \sigma^{2i+1}_x \ldots=\ldots i \sigma^{2i}_y \sigma^{2i}_z \sigma^{2i+1}_x \ldots=\ldots-\sigma^{2i}_x \sigma^{2i+1}_x\ldots,$$ and similarly for the Pauli $y$-matrices.

EDIT: Maybe you made a mistake multiplying the Pauli matrices? Here is the full multiplication table for the Pauli matrices: http://www.its.caltech.edu/~hmabuchi/Ph125b/HW7-solutions.pdf

$J$ can be omitted because it is not zero compared to $\Delta$ and you can divide (rescale) the Hamiltonian by $J$ and absorb it into $\Delta/J \rightarrow \Delta'$. If $\Delta \rightarrow \infty$ then you just recover the Ising model. Remember, this transformation is unitary, which means it's just a change of basis in Hilbert space. It can't change anything "physically" and that tells us that it's only the relative $\pm$ - sign between $J$ and $\Delta$ that is physical.

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