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I was wondering why when we have spin chain Hamiltonians, like the Heisenberg model, we always express the eigenstates in the spin z- eigenbasis. Or maybe, I could pose my question this way - to be specific, consider isotropic Heisenberg model with ferromagnetic couplings, then when I try to 'measure' the system, what do I look for? Magnetisation? if yes, then along with direction?

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  • $\begingroup$ Which magnetization you consider depends on which physical question you ask! For Heisenberg, there is no reason to prefer a basis. For other models, there might be a basis in which the model for some choice of parameters has a simple (classical) solution, like the Ising model, and you might want to expand around that basis. $\endgroup$ – Norbert Schuch Mar 17 '17 at 19:24
  • $\begingroup$ It's just a choice. See also hsm.stackexchange.com/q/5965 $\endgroup$ – Jules Lamers Apr 5 '18 at 16:03
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One measures the magnetization along the direction of magnetic field. For example if the magnetic field is along the $z-$ direction then the operators $S_{z}=$ $\frac{1}{N}\sum_{n=1}^{N}\langle \sigma_{z}\rangle$ will be the average magnetization.

Let me edit my answer. I have given answer to the last part of the question. "Or maybe, I could pose my question this way - to be specific, consider isotropic Heisenberg model with ferromagnetic couplings, then when I try to 'measure' the system, what do I look for? Magnetisation? if yes, then along with direction?"

So the question was which direction. I just gave an example that magnetization along the direction of field is an important quantity. In the Heisenberg case there is a rotation symmetry and therefore average magnetizations along any direction are equally important. Typically when one is interested in symmetry breaking phase transitions, spontaneous magnetization is preferred choice, as mentioned by by Norbert for Ising model given below. \begin{equation} H_{I}=\sum_{i=1}^{N}\sigma^{x}_{i}\sigma^{x}_{i+1}-h\sigma^{z}_{i}. \end{equation}

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  • $\begingroup$ I don't think that's quite right. The isotropic Heisenberg model is 'highly' symmetric and I don't see why the magnetisation should present itself along the z-axis. The z-basis is an artificially chosen basis, it could well have been x or y. I suspect it is degenerate and there is some sort of spontaneous symmetry breaking when one observes. $\endgroup$ – user2578520 Mar 17 '17 at 18:34
  • $\begingroup$ That's not correct. For instance, for the 1D transverse field Ising model one is typically interested in the spontaneous magnetization perpendicular to the field. $\endgroup$ – Norbert Schuch Mar 17 '17 at 19:24
  • $\begingroup$ @Nobert yes I agree with your answer. $\endgroup$ – user123 Mar 18 '17 at 16:23
  • $\begingroup$ @user123 Alright. Your edit makes it clear. Thanks. $\endgroup$ – user2578520 Mar 19 '17 at 7:23

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