# Does (spontaneous) symmetry breaking imply long-range order and vice-versa?

Crystalline solids have a long-range order (where symmetry is broken) but liquids have only a short-range order (where no symmetry is broken). Ferromagnets have a long-range magnetic order while a paramagnet lacks it. The converse also seems to be true, for example, in the Kosterlitz-Thouless transition there is no symmetry breaking and there is no long-range order (but quasi long-range). By long-range order, I understand that below some critical temperature $T_c$ the two-point correlation function of the order parameter (density) becomes a constant (independent of position).

Is this a generic feature? In other words, does long-range order necessarily imply the symmetry-breaking? And does the symmetry-breaking necessarily imply the long-range order?

• Symmetry breaking implies long-range order (of a suitable kind). But there can be long-range order without any symmetry breaking. – Yvan Velenik Mar 25 '18 at 8:36
• "there can be long-range order without symmetry breaking" Can you give an example what you have in mind? – SRS Mar 25 '18 at 12:18
• Ok, as a trivial example, consider the following variant of the Ising model with an additional 3-body term: $H=-\sum_{i\sim j}\sigma_i\sigma_j - h\sum_i \sigma_i - \epsilon\sum_{i\sim j\sim k}\sigma_i\sigma_j\sigma_k$, where $\sim$ means "nearest neighbors". This Hamiltonian has no internal symmetry when $\epsilon$ is non zero. But one can prove that, for any small $\epsilon>0$ and any temperature large enough, one can find a value of $h$ such that there is long-range order (existence of two Gibbs states with positive, resp. negative, magnetization). – Yvan Velenik Mar 25 '18 at 12:31
• Great, I hope you'll like it, once you see that it is actually not so involved. Of course, I wouldn't advise to start with Chapter 7. – Yvan Velenik Mar 28 '18 at 13:22
• I second @NorbertSchuch : there are different answers possible depending on one's definition of SSB and LRO. The OP should add their desired definition (or explicitly state that (s)he would be happy with whatever definition an answerer might prefer). Relatedly, if one counts LRO of 'string orders', then there are clear counter-examples to the supposed equivalence. – Ruben Verresen Aug 26 '18 at 0:49

There are some subtleties, but the answer is basically "yes" in local, translationally invariant systems, because of the cluster decomposition property. Empirically, virtually any "realistic" physical system satisfies the property that $$\lim_{|x - y| \to \infty} \left[ \langle A(x)\, B(y) \rangle - \langle A(x) \rangle \langle B(y) \rangle \right] = 0$$ for any local operators $A(x)$ and $B(y)$. In other words, correlations decay with distance, so expectation values of faraway observables are uncorrelated. (One can derive this result from various technical locality assumptions.) If $m(x)$ is a local symmetry-breaking order parameter, then $\langle m(x) \rangle \equiv \bar{m}$ is constant by translational invariance, so if we let $A(x) = B(x) = m(x)$ in the identity above then we have $$\lim_{|x - y| \to \infty} \langle m(x)\, m(y) \rangle = \bar{m}^2.$$ The left-hand side being nonzero defines long-range order, and the right-hand side being nonzero defines spontaneous symmetry breaking, so we see that either implies the other.
• @NorbertSchuch (a) The cluster property certainly holds for the non-cat degenerate ground states, which are the ones that are physically observed. (b) I'm defining SSB as "when experimentalists measure the symmetry-breaking order parameter $m(x)$, they find a nonzero value." (c) The second equality follows from the first. – tparker Aug 26 '18 at 0:40
• @NorbertSchuch The existence of a symmetry-breaking local order parameter $m(x)$ such that $$\lim_{|x-y| \to \infty} \langle m(x)\ m(y) \rangle = C \neq 0.$$ Unlike SSB, this can't be determined via local measurements. – tparker Aug 26 '18 at 1:59