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I'm preparing an exam and in a preparation sheet there is an exercise that I just don't know how to deal with. Could someone please explain it to me?

a) Explain why the 3D Ising model on the cubic box $ \{0,\ldots,n\}^3 $ with $ \beta \to +\infty $ with $+1$ boundary conditions on the three faces such that either $x=0,y=0$ or $z=0$ and $-1$ boundary conditions on the three faces such either $x=n$,$y=n$ or $z=n$, corresponds to a dimer model on the hexagonal domain shown in the previous exercise (image below)

b) Harder: if we define a Markov chain dynamics, on the hexagon dimers by randomly performing moves as above (choosing them uniformly among the possible elementary moves available), do we converge to the uniform distribution on dimers? Hint: look at the equilibrium measure.

c) Much harder: What is the difference between this dynamics and the Metropolis dynamics for a corresponding 3D Ising model with $\beta = + \infty$? Why does one give a uniform distribution on dimer configurations an the other one does not?

enter image description here

enter image description here

a) My idea is that I can assign for example $+1$ spin to the left cube here above and the $-1$ spin to the right cube, and then flipping a spin sign if possible is the same as obtaining another configuration of dimer tiling so every possible domino tiling correspond to exactly one possible configuration of the spin in the 3D Ising model. The boundary condition I'm not very sure why... I mean i think define which cube is $-1$ and which one is $+1$ since we can see this little square as an $n\times n \times n$ square and the one on the left "corresponds" to the condition $x=0,y=0,z=0$, similarly to the other one. But maybe the boundary condition are also important for other reason. Why $\beta \to +\infty $ I don't know.

b) I'm not sure I understand the question. If i understood what the question is yes! Because we are choosing one possible configuration of spin $\sigma = (\sigma_x)$ of the 3D Ising model on the cubic box $ \{0,\ldots,n\}^3 $ and since this correspond to a dimer model, i.e. they can be mapped one to one with the hexagonal lattice dimer configurations so $$ \mathbb{P}(\sigma) = \frac{ e^{- \beta H(\sigma)}}{Z} $$ where $ Z = \sum_{ \tilde{\sigma} } e^{- \beta H(\tilde{\sigma})}$, converge as $ \beta \to \infty $ to $ \frac{1}{\text{number of tillings on the hexagonal grid}} $

c) No idea!

3m0o
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  • For b) and c) you need to include what "performing moves as above" refers to. For a), the role of the boundary conditions and $\beta\to+\infty$ is to ensure that there's a minimal stepped transition surface between $-1$ and $+1$. At finite $\beta$, there could be multiple switches between $-1$ and $+1$ along a coordinate line, so the transitions couldn't necessarily be described by a stepped surface in bijection with the dimer model. – joriki Jan 05 '23 at 07:10
  • The things is that "above" there is no description of the moves... I posted everything related (the image is above and there it is asked to prove that the number of the hexagonal dimer tilling is the determinat of the adjacency matrix), maybe he want to say to choose a single cube and adding to the 3D $n^3$ cube. For a) can you please put more details I don't understand very well the problem of possible multiple switches and why the minimal stepped transition is important – 3m0o Jan 05 '23 at 11:44
  • For c) the metropolis algorithm: we start from an arbitrary configuration (with the boundary values as in the Ising model) and we make random flips: 1) We compute the energy of the current configuration $H_{\sigma}$ 2) We pick a vertex $x$ at random (which is not in the boundary) and we consider the configuration $\rho$ obtained by flipping the spin of $x$ in $\sigma$ and we compute its energy $H_{\rho}$. If $H_{\rho}\leq H_{\sigma}$ we replace $\sigma $ by $\rho$ otherwise we reject $\rho$ and we continue with $\sigma$ with probability $e^{- \beta H_{\rho}}/e^{-\beta H_{\sigma}} $. – 3m0o Jan 05 '23 at 12:20
  • Maybe I understood, for c) we have that the Metropolis algorithm converges to the measure of the Ising model, and since $\beta \to \infty$ we have that $ e^{- \beta H(\sigma)} \to \mathbb{1}{H(\sigma)=0} $, since we may replace $H $ with $H- \min{\sigma}(H\sigma)$. We have that the measure converges to the uniform distribution on the ground states, i.e. the two states for which $\sigma_x = + 1 $ or $\sigma_x= -1$ for all $x$. So the metropolis dynamics does not converges to the uniform distribution on the hexagonal dimer tilling since there are more than two possible configuration. – 3m0o Jan 05 '23 at 14:53

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