The usual trick to find the spontaneous magnetization for the 1D Ising model is to calculate the partition function $Z$ with the Hamiltonian


Then find the Helmholtz Free energy $F=-kT\ln Z$ and then use the formula

$$M(H=0,T)=-\Big(\frac{\partial F}{\partial H}\Big)_{H=0}.$$

But rather than putting $H=0$ at the end of the calculation to obtain the spontaneous magnetization $M(0,T)$, is it possible to start with $H=0$ in the Ising Hamiltonian (1) from the beginning and then use the definition $M(0,T)=\sum\limits_{i}S_i$ to calculate $M(0,T)$?

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    $\begingroup$ I don't think it's possible because the purpose of the field H is to break the symmetry in the z direction; because up and down states have the same probability there's no way spontaneous magnetization will develop unless there's small perturbation (in this case provided by H) to break the symmetry. $\endgroup$ – rnels12 Jan 21 '19 at 15:43
  • $\begingroup$ @rnels12 But physically you don't apply an external field. The symmetry is broken spontaneously as the temperature is lowered. If $H$ is included in the Hamiltonian from the beginning, it doesn't have the $Z_2$ symmetry to start with. $\endgroup$ – SRS Jan 21 '19 at 15:59
  • $\begingroup$ The same question will be, will a system with its initial state at the top of the mexican hat potential ever evolve to any of the lowest states without a perturbation at all? If so, how? $\endgroup$ – rnels12 Jan 21 '19 at 17:28
  • $\begingroup$ Of course, you can compute directly the expectation of the magnetization: just average it w.r.t. the Gibbs measure. Obviously, if you use periodic or free boundary conditions, then the result will not be very interesting (it will be equal to zero by symmetry). You'll get a nontrivial answer if you use, say, $+$ boundary condition (and keep the system finite). You can do the computation either using the transfer matrix, or (which is simpler when $H=0$) using a high-temperature expansion. $\endgroup$ – Yvan Velenik Jan 21 '19 at 17:45
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    $\begingroup$ No, you compute the expected value of $\sum_{i=1}^N S_i$, not the free energy. If you want to go through the free energy, you need to take $H$ into account in your computations, you cannot set it to $0$ from the start! The expected value of $\sum_i S_i$ will give you $0$ at all $T$ (including $0$) if you use free or periodic b.c.. Moreover, the magnetization density will vanish in the thermodynamic limit for all boundary conditions whenever $T>0$. For $T=0$, you'll find $+1$ if you use $+$ b.c., $-1$ if you use $-$ b.c. and $0$ if you use free or periodic b.c.. $\endgroup$ – Yvan Velenik Jan 21 '19 at 18:17

You can just compute the expected value of $M_N=\sum_{i=1}^N \sigma_i$ with respect to the Gibbs measure: $$ \langle M_N\rangle_{N,T} = \sum_{i=1}^N \langle\sigma_i\rangle_{N,T} , $$ where $\langle\cdot\rangle_{N,T}$ represents expectation with respect to the Gibbs measure for a system of $N$ spins at temperature $T$ (and $H=0$).

Of course, if one uses free or periodic boundary condition, then the expectation is $0$, at all temperatures, by symmetry. In order to get something nontrivial, let us assume $+$ boundary condition (that is, assume that the first and last spins interact each with a boundary spin with fixed value $1$).

The expectation of $\sigma_i$ can be computed easily either through the transfer matrix, or using a high-temperature expansion. The latter is shorter, since only two graphs contribute to both the numerators and the denominator: setting $x=\tanh(J/kT)$, $$ \langle\sigma_i\rangle_{N,T} = \frac{x^i+x^{N+1-i}}{1+x^{N+1}}. $$ Therefore, $$ \langle M_N\rangle_{N,T} = 2 \frac{x^{N+1}-x}{x^{N+2}-x^{N+1}+x-1}. $$ In particular, $$ \lim_{T\to 0} \frac1N\langle M_N\rangle_{N,T} = 1. $$

Note that, above, the $+$ boundary condition does introduce a (weak) symmetry breaking. This is not necessary. One can derive spontaneous symmetry breaking (of course, this is more relevant in higher dimensions) without introducing any symmetry breaking, neither external field, nor boundary conditions. I explained this in this answer.

  • $\begingroup$ How do you get 0 at nonzero T in this approach? @YvanVelenik $\endgroup$ – SRS Jan 21 '19 at 18:50
  • $\begingroup$ I get a positive magnetization at nonzero $T$ because I use $+$ boundary condition. Note, however, that the magnetization density does tend to $0$ as $N\to\infty$ for all $T>0$. $\endgroup$ – Yvan Velenik Jan 21 '19 at 18:56
  • $\begingroup$ Very illuminating indeed! Is this $N\to\infty$ limit also needed in deriving M with a nonzero H from the start? $\endgroup$ – SRS Jan 22 '19 at 12:45
  • $\begingroup$ Computing $M$ by differentiating the free energy w.r.t. $H$ and letting $H\to 0$ gives you exactly the same expression. So, yes, you have to let $N\to\infty$ to get a zero magnetization density for $H>0$ when you use $+$ b.c.. Of course, all this is easier to see and discuss in dimension $2$ or higher, since you then have symmetry breaking at positive temperature. $\endgroup$ – Yvan Velenik Jan 22 '19 at 12:59
  • $\begingroup$ "The latter is shorter, since only two graphs contribute to both the numerators and the denominator:" interesting.... is there a reference that I may read about how graph theory is used in the Ising model like this? $\endgroup$ – Daddy Kropotkin Jan 27 '19 at 16:07

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