In the 2D Ising model of a ferromagnet, Onsager predicts the absolute magnetization as a function of (unitless) temperature as


Does a similar expression for the absolute magnetization exist for the 1D model? I am able to find a theoretical prediction for the non-absolute magnetization in the presence of an external magnetic field $B$ (in unitless variables),


but this vanishes when $B=0$, due to the symmetry of the system. I have so far been unable to find an analytical expression for $|M_{1D}(T)|$. Does such a thing exist?

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    $\begingroup$ Yes, such an expression exists: it is equal to $0$. (Considering the absolute value does not change anything, once the thermodynamic limit has been taken.) $\endgroup$ – Yvan Velenik Nov 16 '17 at 7:06
  • $\begingroup$ I don't understand... at low T, the system is completely magnetized, with equal probability of being up or down. Taking the absolute value (on a per-microstate basis, before averaging) means |M(0)|=1. Perhaps I should specify that I mean to take the absolute value before taking the average, ie, $\langle|M|\rangle$, not $|\langle M\rangle|$ $\endgroup$ – KBriggs Nov 16 '17 at 15:41
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    $\begingroup$ No, at low temperature the magnetization density vanishes almost surely (and so does, obviously, its absolute value). It is only at zero temperature that the magnetization is non-zero. This can very easily be proved mathematically. $\endgroup$ – Yvan Velenik Nov 16 '17 at 17:52
  • $\begingroup$ I see, so the presence of nonzero magnetization at nonzero temperature in simulation is due to finite size effects? $\endgroup$ – KBriggs Nov 16 '17 at 18:05
  • $\begingroup$ Yes, I added more information in an answer. $\endgroup$ – Yvan Velenik Nov 16 '17 at 18:16

Let $$ m_N = \frac1N \sum_{i=1}^{N} \sigma_i $$ be the (empirical) magnetization density and $\mu_{N;\beta}$ be the Gibbs measure (with periodic boundary condition) at inverse temperature $\beta<\infty$, for a one-dimensional system of $N$ spins $\sigma_1,\ldots,\sigma_N$.

One can then prove the following: For any $\beta<\infty$ and any $\epsilon>0$, $$ \lim_{N\to\infty} \mu_{N;\beta}(|m_N| > \epsilon) = 0. $$ (In fact, this probability decays exponentially fast in $N$.) This is standard material (for a proof, see for example Theorem 3.10 in this book). Of course, it follows that the expected value of $|m_N|$ satisfies $$ \langle |m_N| \rangle_{N;\beta} \leq \epsilon + \mu_{N;\beta}(|m_N| > \epsilon). $$ In view of the above, this immediately implies that $$ \lim_{N\to\infty} \langle |m_N| \rangle_{N;\beta} = 0, $$ for any $\beta<\infty$.


I found a reasonable approximation here, in equation 2.36. Getting the absolute magnetization amounts to solving the implicit equation


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    $\begingroup$ It is not a reasonable approximation: the mean-field approximation fails completely in one dimension. $\endgroup$ – Yvan Velenik Nov 16 '17 at 7:08
  • $\begingroup$ It seems to give reasonable agreement with my simulations using the Metropolis algorithm. It fails above T=1 due to finite size effects, but below the transition temperature it's not too bad, at least qualitatively. $\endgroup$ – KBriggs Nov 16 '17 at 15:39
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    $\begingroup$ Well, again,in dimension $1$, the magnetization density is equal to zero at any positive temperature (in the thermodynamic limit). It is only at $T=0$ that the magnetization is positive (and actually equal to $\pm 1$). (This is an exact mathematical statement, not relying on simulations.) $\endgroup$ – Yvan Velenik Nov 16 '17 at 17:54
  • $\begingroup$ To second Yvan Velenik: In one dimension, the Ising model (or any other model) does not have spontaneous magnetization at non-zero temperature. There is no "below the transition temperature" in 1D. $\endgroup$ – Norbert Schuch Nov 16 '17 at 19:23

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