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I am seeking recommendations on how to compute the Binder ratio numerically accurate when doing Monte Carlo simulation on spin models. Binder ratio is defined as:

$$ B = \frac{\langle M^4\rangle}{\langle M^2\rangle^2}. $$

Given a safe method to compute $M$ per metropolis sweep, one can get directly $M^2$ and $M^4$. If we take $N$ samples of these values, we can then get the average ones; $\langle M\rangle$, $\langle M^2\rangle$ and $\langle M^4\rangle$.

Then one can compute the Binder ratio. But near the critical temp, the results are not so precise. Am I having too much floating point error in $\langle M^2\rangle$ and $\langle M^4\rangle$?

An example of how the Binder ratio looks, it was a very short simulation. A proper simulation generates the very smooth curve, with little standard error, but the spike remains.

enter image description here

Edit: It is a parallel tempering simulation, it might be possible that the cause for the unexpected Binder Ratio could be related to measuring too early when replicas are not thermalized properly.

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  • $\begingroup$ Why would rounding error be a suspect here? I think it's more likely that the variance of sample averages becomes too high. $\endgroup$
    – Kirill
    Commented Mar 23, 2015 at 17:39
  • $\begingroup$ It came to my mind after I saw that starting from a certain lattice size the curve got somehow different to what it ussually is. Now that you mention, It is possible that the variance might be too high, making $\langle M^4 \rangle$ be inferior or superior to the denominator, in cases where it should be the opposite. In that case, is there something one can do to redice it for these kind of Simulations? $\endgroup$
    – Cristobal Navarro
    Commented Mar 23, 2015 at 17:50
  • $\begingroup$ Can you post more information (plots, data, etc.)? I think that would make the question easier to understand and answer, as well as give it more context. $\endgroup$
    – Kirill
    Commented Mar 23, 2015 at 17:53
  • $\begingroup$ Sure, I will look for some data and pictures, in the meanwhile I added that the simulation is through a parallel tempering method. By the way, thanks for improving the format of the mathematical expressions. $\endgroup$
    – Cristobal Navarro
    Commented Mar 23, 2015 at 18:02

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The solution to get better estimates near the critical temperature is to get better statistics. In other words, you need more samples (or at least more independent samples, for example by using better sampling techniques) -- that's precisely what makes computations near the critical temperature so difficult.

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  • $\begingroup$ Yes, I have to consider this too. At this moment I am investigating if maybe I am measuring at a stage of the algorithm that might be not so convenient. Since parallel tempering makes swaps, maybe i have to wait more simulation time for them to stabilize, before taking measure samples. I need to test with and without PT. $\endgroup$
    – Cristobal Navarro
    Commented Mar 26, 2015 at 13:35
  • $\begingroup$ I agree. If you can, you should work up from small volumes and beat them to death with statistics and calculate your integrated-auto-correlation times and keep track near the critical point. you should expect it to get nasty. There are cluster methods if your spins take on a finite number of values. These can eliminate critical slowing down. $\endgroup$
    – kηives
    Commented Mar 31, 2015 at 17:47

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