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What I'm wondering is how to take into account finite temperature in the transverse Ising chain and see how that affects the magnetization. The reason why I find it difficult to consider a finite T is that the Hamiltonian seems to be the same as in the $T=0$ case, in particular:

$H=-J\sum_{i=1}^{N}(\sigma_{i}^{x}\sigma_{i+1}^{x}+g \sigma_{i}^{z})$

and I don't know how to implement the fact that T is finite (non-zero).

So my question is: how does one take into account finite $T$ as opposed to the $T=0$ case, and how does that affect, for example, the calculation of the magnetization?

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    $\begingroup$ You need to compute the partition function $Z=e^{-\beta H}$, where $H$ is your hamiltonian. Usually people only find the ground state of transverse Ising, but it is free Majorana fermions, so you do know all the eigenstates. What is hard is the spin correlators of course. That is hard even at $T=0$. $\endgroup$
    – mike stone
    Commented Mar 10, 2022 at 18:30
  • $\begingroup$ @mikestone Ok but after I calculate Z how do I find the Magnetization? I know that it should be the derivative of free energy (which I can calculate with Z) with respect to magnetic field, but here I don't have a B field. $\endgroup$
    – Mathew
    Commented Mar 11, 2022 at 10:29
  • $\begingroup$ @mikestone Or maybe I can I can calculate it as $\frac{Tr(e^{-\beta H}\sigma^{z})}{Z}$ ? $\endgroup$
    – Mathew
    Commented Mar 11, 2022 at 10:43
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    $\begingroup$ Calculating $\langle \sigma^z\rangle = \mathrm{Tr}(\sigma^z e^{-\beta H})/Z$ is very hard, but what can be calculated is $\langle\sigma^z_{1+j} \sigma^z_{1}\rangle$ in the limit $j \rightarrow \infty$, see arxiv.org/abs/cond-mat/9509147 for the finite-temperature result. One should have $\lim_{j \rightarrow \infty}\langle\sigma^z_{1+j} \sigma^z_{1}\rangle = \langle \sigma^z \rangle^2$. $\endgroup$
    – Seth Whitsitt
    Commented Apr 13, 2022 at 15:42

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