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Consider the paper https://arxiv.org/pdf/1008.0352.pdf. At the very last page (the two equations before (60)), John Palmer and Grethe Hystad write down a striking identity for the two-point function of the 2D classical Ising model based on duality with the 1D quantum Ising model, which, in the case that the two spins lie on the zeroth row, specializes to $$\langle \sigma^x_{i0} \sigma^x_{j0}\rangle_\beta= \frac{\text{tr}\,(\sigma^x_i\sigma_j^xT^n)}{\text{tr}(T^n)}$$ Where $T$ is the transfer matrix, the lattice is $n\times m$ with any choice of topology (i.e. toric/rectangular/cylindrical). According to the arguments leading to equation (60), the above correlation function is temperature-independent. To paraprase, letting $\lambda$ be the largest eigenvalue of the transfer matrix, $$\frac{\text{tr}\,(\sigma^x_i\sigma_j^xT^n)}{\text{tr}(T^n)}=\text{tr}\,(\sigma^x_i\sigma_j^x\frac{T^n}{\lambda^n})\xrightarrow[n\to\infty]{}\sum_k\langle\Omega_k|\sigma^x_i\sigma_j^x|\Omega_k\rangle$$ Where the $|\Omega_k\rangle$ can be any orthonormal basis of the highest-eigenvalue subspace (i.e., $T^n/\lambda^n$ approaches the projection onto the highest-eigenvalue subspace of the transfer matrix). However, this is now a temperature-independent quantity. Now the quantity on the left is clearly temperature dependent, so, where is the flaw in the argument given? Am I misunderstanding the paper?

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    $\begingroup$ The transfer operator of the 2D classical Ising model is generally not equal to the exponentiated Hamiltonian of the 1D quantum Ising model. $\endgroup$ Nov 4, 2016 at 7:53
  • $\begingroup$ Ok, but why aren't the two-point functions temperature independent by the above argument? $\endgroup$ Nov 4, 2016 at 14:08
  • $\begingroup$ ??? -- Because you are using said equivalence which is not correct? $\endgroup$ Nov 4, 2016 at 17:07
  • $\begingroup$ I removed all mention of the 1D quantum Ising model. The argument still holds. $\endgroup$ Nov 4, 2016 at 18:17
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    $\begingroup$ Why should the leading eigenvector of $T$ (or its correlations) not depend on the temperature? Can you elaborate that? This is clearly not correct. $\endgroup$ Nov 4, 2016 at 19:53

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The leading eigenvector of $T$, and in particular its correlations, are not independent of the temperature (just as $T$ itself depends on the temperature). This is why your conclusions are incorrect.

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