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1

That is exactly right. I would say "exactly one spin up" rather than "any spin up", but that is just phrasing. Another way of looking at this is that there is an increase in entropy (of $\log n$ times Boltzman's constant) associated with flipping one spin, which will sometimes overcome the energy cost.


3

For the 2D planar Ising Edward-Anderson model, it can be solved in polynomial time. For the 3D case it is NP-hard, moreover NP-Complete.[1] [1] Barry A. Cipra, The Ising Model Is NP-Complete


2

The plot you have for $Z$ looks right. To get some intuition, at low $T$(large $\beta$) the spins are ordered, so the dominant term in the partition function is just $e^{2\beta N J}$, which grows like an exponential with $\beta$. (the Hamiltonian for the Ising model is $H=-\sum_{\langle ij\rangle} \sigma_i\sigma_j$, notice the important minus sign in $H$!) ...


1

I will explain how I measured the spin-spin correlation function for the 2d Ising model. Generalization to more than 2 dimensions should be straightforward as long as you have hypercubic lattices. Just to get the notation straight: Let's use the name $\sigma_{(i,j)}$ for the spin at position $\vec r_{(i,j)} = \begin{pmatrix} i \\ j \end{pmatrix}$. Let's ...


0

This model can be solved exactly by Jordan-Wigner transformation and the expectation value of $\sigma_z$ can be calculated rather straightforwardly(it maps to a local quantity in the fermion model). Physically, when $g>1$ the spins are ordered, so $\sigma_z$ has a finite expectation value at $T=0$ which will be reduced at finite temperature. For $g<1$, ...



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