I would like to express the Heisenberg model using the duality analysis. It is shown here how to express the Ising model using Pauli matrices but I cannot get the relation $ \sigma _{i}^{z}= \prod_{j\leq i} S_{j}^{x}$ or why $ \sigma _{i}^{x}=S_{i}^{z} S_{i+1}^{z} $. Also how can the $ \sigma _{i}^{y} $ then be expressed using the duality transition of Pauli matrices?
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$\begingroup$ I have never heard that the Heisenberg model has this duality property, but I might be wrong. However, may it helps that you can transform the Heisenberg model via the Jordan-Wigner transformation to a chain of spinless fermions with nearest neighbour interactions. $\endgroup$– SGGSCommented Jul 8, 2021 at 10:17
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1$\begingroup$ @CVJM The Heisenberg model is not self-dual. However, there is a dual model to the Heisenberg model. $\endgroup$– GecCommented Jul 8, 2021 at 11:56
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$\begingroup$ @Gec I just have seen our answer. So then one obtains one particular dual model by performing the transformation you named in your answer, right? Do you know anything else, except for this one and the spinless fermonic chain? $\endgroup$– SGGSCommented Jul 8, 2021 at 12:55
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1$\begingroup$ @CVJM The operator change of variables can be applied to any spin-$1/2$ model. The problem is, the relation between $S$ and $\sigma$ is nonlocal, and a hamiltonian dual to a local hamiltonian is not guaranteed to be local. The quantum Ising model is known to be self-dual. A less known fact is that the quantum Ising model is also dual to the XY model. $\endgroup$– GecCommented Jul 8, 2021 at 13:16
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It is not significant how to get to the relations between $\sigma$ and $S$. You can think about the equalities for $\sigma^z$ and $\sigma^x$ as an operator change of variables that someone came up with, perhaps Jordan and Wigner. What is important, this change of variables is consistent with Pauli's matrices properties. Relations $$ \sigma_j^y = i\sigma_j^x\sigma_j^z = -i\sigma_j^z\sigma_j^x, $$ $$ S_j^y = iS_j^xS_j^z = -iS_j^zS_j^x,\quad (S_j^x)^2 = 1,\quad S_j^\alpha S_k^\beta = S_k^\beta S_j^\alpha, \quad\forall\ j\neq k,\ \alpha, \beta $$ together with the equalities for $\sigma^x$ and $\sigma^z$ lead to the following equality for $\sigma^y$: $$ \sigma_j^y = -\prod_{k<j} S_k^x\ S_j^y S_{j+1}^z $$ It is straightforward to check the validity of all the usual Pauli's matrices relations now. It is also easy to obtain formulas for $\sigma_j^\alpha\sigma_{j+1}^\alpha$ for $\alpha = x, y, z$ and to find hamiltonian dual to that of the Heisenberg model.
