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Are Ising spins scalar or operators? I am not a condensed matter physicist hence having some confusion. I have learnt about Ising models from adiabatic quantum algorithm papers. For example this presentation or this paper encodes their quantum adiabatic algorithm for an Ising model. In an Ising model the spins are $\pm 1$. In his original adiabatic quantum computation paper, Farhi didn't mention about Ising models. Moreover he explicitly used Pauli's spin matrices and not the $\pm 1$ spins.

Can we use $\pm$ spins and Pauli's matrices interchangeability at least in the context of adiabatic quantum algorithms?

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Simultaneous crosspost on TCS. –  Tyson Williams May 29 '13 at 16:39
    
@TysonWilliams, I wanted see if any extra insight is available on that forum. –  Omar Shehab May 29 '13 at 18:30

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As you noted, the Ising model has spins that are $\pm 1$ whereas in a full quantum model such as the Heisenberg model, the spins are represented by Pauli matrices.

This means that they are not interchangeable. The biggest difference is that at zero temperature, there are no spin fluctuations in an Ising model, whereas there are fluctuations in the Heisenberg model.

The Ising Hamiltonian can be written as $$H = J\sum_{\langle i,j\rangle} \sigma_i \sigma_j$$ and all the $\sigma_i \in \{-1,1\}$.

whereas for quantum spins, we'd have $$H = J\sum_{\langle i,j\rangle} \vec{S_i} \cdot \vec{S_j}$$ and the $\vec{S_i}$ are vectors with elements determined by the Pauli matrices. This product can then be expanded as $$\vec{S_i} \cdot \vec{S_j} = S_i^z S_j^z + \frac{1}{2}\left(S_i^+ S_j^- + S_i^- S_j^+\right)$$ where $S_i^+$ and $S_i^-$ are the spin raising and lowering operators. Thus, we get one term that looks just like the Ising term, because $S_i^z$ can be either $-1$ or $1$, but we also get terms that describe how the two spins can flip: They start with opposite spin and then both of them flip.

For a Ferromagnet in the ground state, this is not important, because there the spins are all in parallel and thus the flip terms give zero contribution, but in an antiferromagnet, it makes the ground state highly complicated, whereas in an Ising model the ground state would just have Neel order.

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understood. So, why did the authors in the first two links of my question use Ising model? –  Omar Shehab May 28 '13 at 12:56
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According to the abstract of the arxiv paper you mention, some variants of the Ising model can indeed be experimentally realized. I guess that's why they consider it. –  Lagerbaer May 28 '13 at 14:52

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