2D Ising model simulation using Metropolis algorithm.

There is one thing which I don't understand. The difference in energy $\Delta E$ between initial state and new state is:

$\Delta E = 2Js\sum_rs_r$ (btw can someone confirm that pls) where J is constant, s-initial spin and sum is equal to the sum of spins of the nearest neighbours.

The new state is always accepted when $\Delta E<0$ or if $\Delta E>0$ accepted with probability $p=\exp(-\Delta E/k_bT)$.

The question is what happens when $\Delta E = 0$?

  • $\begingroup$ If you apply either of the $\Delta E < 0$ and $\Delta E > 0$ rules to the $\Delta E = 0$ case, you get the same result (accept with probability $p=1$). $\endgroup$ – Noiralef Jan 28 '17 at 11:29

The answer comes from the implementation. The simplest way to code this is to generalize the process of acceptance by always calculating the acceptance probability $p=\exp(-\Delta E / k_b T)$, which, all other quantities assumed constant:

  • is between 0 and 1 for $\Delta E > 0$
  • is larger than 1 for $\Delta E < 0$
  • is exactly 0 for $\Delta E = 0$

Then you pick a real number between 0 and 1 at random and if it's less than $p$, accept the spin flip. And here's the kicker - since you're picking a random float, the chance of hitting precisely 1 is basically zero. So it's not something that's going to impact your simulation either way.

This, at least, is how it was implemented in the amazing MOOC "Statistical Mechanics: Algorithms and Computations" on Coursera, so I'm basing the answer on that.

Your $\Delta E$ expression looks okay, by the way.

  • $\begingroup$ For ΔE = 0 p =1 and spin flip is always accepted? $\endgroup$ – ad1v7 Jan 27 '17 at 22:05
  • $\begingroup$ Yeah, because your random number for checking acceptance of the flip is always below zero. $\endgroup$ – Perfi Jan 27 '17 at 22:07

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