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The RG is not a group, it's a semi group so you can only go in one direction, the one that actually renormalize. Here you can use the relation K'(K) for your flow. but if you use K(K'), you will have something wrong because the RG procedure that give this relation as no physical meaning. it's as if you add spin in your Ising chain. When you do RG, ,you ...


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I think Adam's answer is excellent, and I'd rather make this a comment but can't as I just signed up in order to answer. While I agree with most of what Adam said, there are cases where large-N works well for $N=1$. The case I'm familiar with is the large-N expansion for the "Spin Ices" Dy$_2$Ti$_2$O$_7$ and Ho$_2$Ti$_2$O$_7$, which have Ising spins on a ...


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For the ferromagnetic type order-disorder phase transition, the correlation length $\xi$ diverge to infinity as the temperature approach the critical point. What happens physically is that there are fractal clusters of the same spin with size of order $\xi$ that grow to infinity as the temperature approach the critical point. The SW algorithm is based on ...


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I have tried to reproduce your problem using my own code, but I couldn't: I got the correct value for the exponent. I can't tell you what went wrong in your calculation, but I can tell you exactly what I did! You can inspect my code on GitHub. It was the first thing I ever wrote in C++ and today I would do many things differently, so please don't judge the ...


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Yes, these equations exist and can be derived from the partition function in JGab's answer. The internal energy per spin is: $$u(\beta) = - \frac{\partial}{\partial \beta} \left( \ln(2) + \frac{1}{8 \pi^2} \int_{0}^{2\pi} \mathrm{d} q_1 \int_{0}^{2\pi} \mathrm{d} q_2 \\\ln \left[ \big( 1 - \sinh(2 \beta J) \big)^2 + \sinh(2 \beta J) \left( 2 - \cos q_1 - ...


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I think you have a problem with double counting. The Ising Hamiltonian is $$H = - J \sum_{\langle i,j \rangle} S_i S_j$$ where this strange sum notation means to sum over all bonds between neighbouring spins. It is the bond that matters, so you should not count it twice (for $ij$ and $ji$). Actually you can get the energy difference of a spin flip as ...



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