# Spontaneous symmetry breaking: proving the equivalence of two definitions

This question can be posed for both quantum and classical set-ups. For concreteness, let me consider a local, classical Hamiltonian $$H$$. The expectation values I consider are with respect to the usual classical partition function, i.e., $$\langle \mathcal A \rangle = \frac{1}{Z} \int \mathcal A e^{-\beta H}$$ (suppressing the integration variable).

Let $$\mathcal O(\boldsymbol r)$$ be a local, physical observable which would be zero if the system obeys a particular symmetry. There are two common ways of characterizing the spontaneous breaking of this symmetry:

1. Long-range order: $$\boxed{\lim_{|\boldsymbol{r-r'}| \to \infty} \langle \mathcal O(\boldsymbol r) \mathcal O(\boldsymbol{r'}) \rangle \neq 0}.$$
2. The double limit: $$\boxed{\lim_{h \to 0} \lim_{V \to \infty} \langle \mathcal O (\boldsymbol r) \rangle_h \neq 0},$$ where the average is with respect to the perturbed Hamiltonian $$H_h = H + h \int \mathcal O(\boldsymbol r) \mathrm d^D \boldsymbol r$$ and $$V$$ denotes the system size (i.e., the first limit is the thermodynamic limit).

Can one show that these conditions are equivalent? Clearly the locality of the Hamiltonian is essential (but the partition function is not a local object, so one cannot simply factorize, e.g., the two-point function in the first definition!)

Alternatively, is there a counter-example to the claim that these two definitions are equivalent?

## 2 Answers

Yes, this can be proved.

Let me first discuss this in the classical case, and then give you pointers to where you can learn more about the quantum case.

You first have to be careful, when writing down your expectations, to specify the state with respect to which you are actually computing the expectation. In this respect, the physicists' habit of denoting expectation simply by generic brackets $$\langle \cdot \rangle$$ is very bad. There is no problem for your "double limit" expectation (provided you let $$h\to 0$$ using positive values only), but the expectation in your "long-range order" case is ambiguous and results actually depend on what it is supposed to mean.

Indeed, when a first-order phase transition occurs (which is precisely what you are interested in), there are infinitely-many different Gibbs measures describing your system in the thermodynamic limit. The set of all such measures is always a simplex, that is, a convex set such that each element can be written in a unique way as a convex combination of extremal elements, the latter being those that cannot be decomposed in a non-trivial way. In the Ising model, for instance, the (translation invariant) extremal measures are those corresponding to the usual $$+$$ and $$-$$ phases. In a sense, the only physically relevant measures are the extremal ones, the other describing statistical mixtures and thus not containing any additional interesting physics.

The first statement is then that a measure is extremal if and only if it is mixing. The latter means precisely that the expectation of the product of any pair of local observables $$\langle \mathcal{O}(r) \mathcal{O}(r')\rangle$$ converges, as $$|r'-r|\to\infty$$ to the product of the expectations $$\langle \mathcal{O}(r) \rangle \langle \mathcal{O}(r')\rangle$$. Note that this is an "if and only if" statement, so the claim fails if the state you are considering is not extremal (for example, it fails for the Ising model if you consider the state obtained using free or periodic boundary conditions and zero magnetic field).

As for the second part (with the "double limit"), the point is that by taking this limit (with $$h\downarrow 0$$, that is, $$h\to 0$$ using positive values only) you are reaching (in the limit) precisely the extremal state under which the observable has largest expected value. For example, if you take the limit $$h\downarrow 0$$ in the Ising model, you will get the expected value of the observable under the $$+$$ state.

The equivalence you want thus immediately follows if the state you use for the "long-range order" is the state you get by letting $$h\downarrow 0$$ in the "double limit".

The above statements are standard and can be found in several places, among which our book, Georgii's book or Simon's book.

Similar claims also hold in the quantum case. Precise statements and proofs can be found, for instance, in Chapter IV of Simon's book.

• Great answer, this is what makes stackexchange awesome! Found the claimed statement on pages 83, 84 of Simon's book. May 17, 2019 at 9:30
• @RyanThorngren : Thanks! And thanks also for providing the precise pages, as I don't have his book with me at the moment. May 17, 2019 at 9:32
• Thanks for this informative answer and the free pdf book. ( Recently I bought one, since I became a frequent user of this book---) Apr 4, 2020 at 4:19
• @dengaku Thanks a lot for your appreciation! Apr 4, 2020 at 8:06
• @dengaku No, if you have more than one, you automatically have infinitely many, since any convex combination of two Gibbs states is still a Gibbs state (e.g., all measures of the form $\alpha\mu^+ + (1-\alpha)\mu^-$ with $0\leq\alpha\leq 1$ in the Ising model). On the other hand, you can have finitely many extremal Gibbs states (this is the case for the 2d Ising model, in which $\mu^+$ and $\mu^-$ are the only extremal ones; but this is not the case in higher dimensions, where you have infinitely many extremal Gibbs states). Jun 5, 2020 at 13:02

Yvan covered the simplest situation where the state used in 1) about the long-range order is an extremal say $$+$$ state. However for a ferromagnetic scalar valued model like Ising or lattice $$\phi^4$$ the equivalence of 1) and 2) still holds but is much more difficult to prove. Here the state I am talking about in 1) is precisely the one Yvan said one should avoid, namely, zero magnetic field and free boundary conditions so there is nothing to break the symmetry between $$+$$ and $$-$$ spins. If one can prove that there are only two extremal states $$\langle\cdot\rangle_{+}$$ and $$\langle\cdot\rangle_{-}$$ then showing 1) implies 2) is not hard. But if one does not know how to rule out the possibility of more extremal states, then showing 1) implies 2) is more challenging.

You can find a proof in the case of a ferromagnetic (unbounded spin) $$\phi^4$$ model due to Chandra, Guadagni and myself in section 4.7 of Chandra's thesis: Construction and Analysis of a Hierarchical Massless Quantum Field Theory.

and for $$\{-1,1\}$$-valued Ising spins in this article by Aizenman, Duminil-Copin and Sidoravicius: Random Currents and Continuity of Ising Model’s Spontaneous Magnetization.

Note this question is about a general circle of ideas called "sharpness of phase transitions". Namely one may have several competing definitions of the critical inverse temperature $$\beta_c$$. For example one could take $$\beta_{c,\chi}$$ as the infimum of $$\beta$$'s where the susceptibility diverges. One can define $$\beta_{c,SM}$$ as the infimum of $$\beta$$'s where the spontaneous magnetization is nonzero (i.e., item 2) in the question above). One can also define $$\beta_{c,LRO}$$ as the infimum of $$\beta$$'s where one has long-range order, i.e., item 1) above for the symmetric $$h=0$$ state. In general one has $$\beta_{c,\chi}\le \beta_{c,SM}\le \beta_{c,LRO}$$ but one can sometimes show equality as in the literature on sharpness of phase transitions.

• What's an example of a phase transition which is not sharp? May 20, 2019 at 10:02
• @RyanThorngren : For example, in the 2d clock model with a large enough number of states, the 2-point function (i) does not decay to $0$ at low enough temperatures, (ii) decays as a power law at intermediate temperatures, (iii) decays exponentially fast at high enough temperatures. This was proved by Fröhlich and Spencer in the 1980s. May 20, 2019 at 11:52
• The paper I mentioned is this one. The relevant claim is Theorem E. May 20, 2019 at 12:22
• @dengaku Well, Fröhlich and Spencer proved that there is algebraic decay at low enough temperatures and it is easy to show that the decay is exponential at high enough temperatures. This implies the existence of (at least) one temperature at which the behavior changes. The problem is that there is no obvious monotonicity in this case that would guarantee that this transition temperature is well defined (and furthermore that there is no intermediate phase in which the behavior is still different). As far as I know, this is still open. Mar 30, 2021 at 8:15
• @dengaku For the clock model, all these transitions are expected to be sharp. I doubt that this has been proved. There has been substantial progress recently on proving sharpness, thanks to work by Duminil-Copin and collaborators, but this remains a difficult problem in general. Mar 30, 2021 at 8:18