# Is there a spontaneous symmetry breaking in finite-size Ising model with + boundary condition?

My question is related to Chapter 3 of Prof. Yvan Velenik's book "Statistical Mechanics of Lattice Systems: a Concrete Mathematical Introduction".

For an Ising model defined on a finite volume $$\Lambda \subset {Z^d}$$, $$Z^d$$ is d-dimensional cubic lattice, the magnetization density $$m_\Lambda ^\# (\beta ,h)$$ is $$m_\Lambda ^\# (\beta ,h)\mathop \equiv \limits^{{\rm{ def }}} \left\langle {\frac{1}{{|\Lambda |}}\sum\limits_{i \in \Lambda } {{\sigma _i}} } \right\rangle _{\Lambda ;\beta ,h}^\#$$

Here '#' denotes the boundary condition, $$h$$ is the external magnetic field, $$|\Lambda |$$ is the number of lattice points in $$\Lambda$$ and $${\sigma _i}$$ is the spin on point $$i$$, $$\left\langle {} \right\rangle _{\Lambda ;\beta ,h}^\#$$ is the ensemble average.

In finite system, $$m_\Lambda ^\# (\beta ,h)$$ should be a real analytic function and it is an odd function with respect to $$h$$. Therefore, no net manetization density at $$h=0$$.

My question is dose the above statement also holds for $$m_\Lambda ^ + (\beta ,h)$$, that is with the + boundary condition. As can be proved (page 106), $$\left\langle {{\sigma _0}} \right\rangle _{\beta ,h}^ + \le m_\Lambda ^ + (\beta ,h)$$
where $$\left\langle {} \right\rangle _{\beta ,h}^ +$$ is an infinite-volume Gibbs state with + boundary conddition. When $$\beta$$ is large, such that we have spontaneous symmetry breaking, $$\left\langle {{\sigma _0}} \right\rangle _{\beta ,0}^ + > 0$$. Does that mean we have $$m_\Lambda ^ + (\beta ,0)>0$$ that is spontaneous magnetization density in finite system with + boundary condition at $$h=0$$?

From another angle, $$m_\Lambda ^\# (\beta ,0)=0$$ can be seen from $$m_\Lambda ^\# (\beta ,h) = - \frac{1}{{|\Lambda |}}\frac{{\partial F_\Lambda ^\# (\beta ,h)}}{{\partial h}}$$, where $$F$$ is the Helmholtz free energy. (In the book, $$\psi$$ page 83 is used instead of $$F$$. I think they give the same result) Since, $$F_\Lambda ^\# (\beta ,h)$$ is an even function of $$h$$, $$m_\Lambda ^\# (\beta ,h)$$ is odd which certainly gives $$m_\Lambda ^\# (\beta ,0)=0$$.

But for $$F_\Lambda ^ + (\beta ,h)$$, what we have is $$F_\Lambda ^ + (\beta ,h) = F_\Lambda ^ - (\beta , - h)$$. $$F_\Lambda ^ + (\beta ,h)$$ is not an even function itself.

In my opinion, does that because the system with + boundary condition does not corresponds to real system. We use + boundary condition just to mimic the effect of an external magnetic field which will reduce the probability of microstates with - spins. And that is the basic idea of symmetry breaking. Actually, in finite system with + bondary condition, we truely have the fact that probability of + spins is larger than that of - spin.

I wonder if I got something wrong.

• To tell the truth I can't understand the 5th equation on page 106. Nor how from the monotonicity of the free energy deduce a monotonicity of its derivative (the magnetization). I haven't thought long about it but I don't think the finite size magnetization with whatsoever BC can be non-analytic. – lcv Jan 30 '20 at 14:22
• @lcv Thanks for your reply! Do you mean the equantion $\left\langle {{\sigma _0}} \right\rangle _{\beta ,h}^ + \le m_\Lambda ^ + (\beta ,h)$, It is a consequce of lemma 3.22 on page 99. Like I said, + BC mimic a external magnetic field which becomes weak as the system grows (when the system goes to infinity, the effect of BC should be eliminishes). Therefore, as systems grows, the probability of having + spins should decreases. That equation is a natural results of this as the magnetization density increases monotonically with the probability of + spin. – FaDA Jan 30 '20 at 15:19
• Yes I mean that equation. I haven't looked too carefully but I couldn't figure out why lemma 3.22 (monotonicity of free energy in the size) implies the same result for the magnetization. – lcv Jan 30 '20 at 15:43
• @lcv: Lemma 3.22 is about expectation of arbitrary nondecreasing local functions. It is not about the free energy. – Yvan Velenik Jan 30 '20 at 15:44
• These results are consequences of correlation inequalities (mainly FKG), but they are also intuitive: a nondecreasing function (say the spin at a particular site) will tend to take larger values on the average, if you force spins in the vicinity of its support to be equal to $+1$. Indeed, $+$ spins "attract" each other and this raises the average value of spins, which in turns increases the average value of the function. – Yvan Velenik Jan 30 '20 at 15:48

Some remarks that may clarify some of your misconceptions:

• The finite-volume magnetization density $$m_\Lambda^\#(\beta,h)$$ is not odd in $$h$$ in general. Only the limiting quantity (as $$\Lambda\uparrow\mathbb{Z}^d$$) is odd. Exceptions for which the result does hold for finite systems are the cases of free and periodic boundary conditions.
• In particular, it is not true, in general, that $$m_\Lambda^\#(\beta,0) = 0$$.
• It is true that $$m_\Lambda^+(\beta,0)>0$$ uniformly in $$\Lambda$$ when $$\beta>\beta_{\rm c}$$. This follows, as you say, from a combination of the Peierls argument and the fact that $$m_\Lambda^+(\beta,0) \geq \langle\sigma_0\rangle_{\beta,0}^+$$ (first inequality in the proof of Proposition 3.29).
• As above, $$F_\Lambda^\#$$ is not an even function of $$h$$ in general. This is only true in the thermodynamic limit.

Given the above, I wouldn't say that there is spontaneous symmetry breaking under $$\mu_{\Lambda;\beta,0}^+$$, since there is no symmetry to be broken there (the spin-flip symmetry being explicitly broken by the boundary condition). Spontaneous symmetry breaking, just like phase transitions in general, only strictly makes sense in infinite systems. This does not prevent finite systems to display ordered phases, of course (below the critical temperature, a typical configuration of the Ising model in a finite box with $$+$$ boundary condition will be composed of a density $$>1/2$$ of $$+$$ spins with very high probability).

Concerning boundary conditions, the latter can indeed be seen as a mathematical trick to break the symmetry (instead of, or in addition to, using a magnetic field).

But this is not the only motivation for introducing them. You can use boundary conditions to model the interaction between the microscopic constituents of your system and the boundary of the vessel that contains them. This is particularly important in the lattice gas interpretation of the model, since this point of view allows one to discuss very interesting surface phase transitions, such as the wetting transition.

Here is a picture (taken from this review paper) of the type of things you can prove for the Ising lattice gas: fix the total number of particles (i.e., $$+$$ spins) in the box. If the density in the box is between the density of the dense and the dilute phases, then the system spontaneously creates a macroscopic bubble of dilute phase inside the dense phase, of deterministic shape (minimizer of the surface tension). Moreover, playing with the attraction between the bottom wall of your vessel and the particles (i.e., the adsorption energy), you can prove that the droplet will start to attach itself to the wall (spreading more or less depending on the parameters). To model such phenomena, you need to take boundary conditions seriously. • I am not sure I answered all your questions. Do not hesitate to ask for complements if needed. – Yvan Velenik Jan 30 '20 at 15:42
• I think the short question was if there is spontaneous symmetry breaking for the + BC at finite size. – lcv Jan 30 '20 at 15:52
• @lcv: Then the answer is yes. A typical configuration in a finite square box $\Lambda$ of "radius" $n$ with $+$ boundary condition consists in a "sea" of $+$ spins with small (of diameter at most $K\log n$, for some $K$ depending on $\beta$) islands of $-$ spins. And the magnetization density you measure in this finite box is very close (if $n$ is large) to the infinite-volume magnetization density. In particular, it is positive (it is in fact even larger than the limiting quantity!) – Yvan Velenik Jan 30 '20 at 15:55
• Can the given magnetization (with + BC) be non-analytic as a function of $h$ for finite size? – lcv Jan 30 '20 at 16:09
• @lcv: No, the magnetization density in finite volume is analytic in all the parameters. But this does not prevent a positive magnetization density in finite volumes (again: it is easier to have a positive magnetization density in a finite box with $+$ boundary condition than in an inifinite-volume system!).. – Yvan Velenik Jan 30 '20 at 16:09