In the article Strange Metals as Ersatz Fermi Liquids , the authors mentioned in the left bottom paragraph on page 4 that

Furthermore the diverging susceptibility of an order parameter at a quantum critical points does not necessarily imply that one of the proximate phases has static order for the corresponding observable, as is known from a number of theoretical examples.

This statement comes from previous literature, and this article just uses it. However, I do not find useful references about it and I am a bit confused. Here are my questions:

  1. What are the theoretical examples the authors referred to?
  2. The authors say "... has static order for the corresponding observable". So does they mean there can exist divergence of susceptibility at critical points without spontaneous symmetry breaking, or there is still spontaneous symmetry breaking, but in some senses the order is not "static" ( I don't know what it means), or the order are formed for other order parameters instead of the one has divergent susceptibility?

1 Answer 1


I don't know what the authors really meant (in the paragraph you quote, they don't cite anything so it's pretty much impossible to know their true intention without asking them directly), but I have one example that comes to my mind: the (Berzinskii-)Kosterlitz-Thouless transition or KT transition in short.

You can go see the wikipedia link to see the details, but the gist is like this. Consider the following classical Hamiltonian: \begin{equation} H=-\sum_{\langle i,j\rangle} \cos(\theta_i -\theta_j), \end{equation} where $\theta_i\in [0,2\pi)$ represents the angle of a spin pointing in some direction in a plane. The subscript $i$ specifies a point in the two-dimensional square lattice, and the summation only runs over nearest neighboring pairs of $i$ and $j$. The Hamiltonian favors two neighboring spins to be aligned in the same direction, just like the Ising model, but with a countinuous degree of freedom.

This Hamiltonian in 2D goes through a thermal phase transition at inverse temperature $\beta_c\sim1.134...$, but not to a long-range ordered phase! The low-temperature phase is only quasi-long-range ordered, meaning that the correlation function still decays to 0, but only inverse polynomially. To be more precise, \begin{equation} C(r):= \langle \cos(\theta_i - \theta_{i+r})\rangle ~~\sim r^{-a} \end{equation} with some exponent (constant) $a$. I am abusing notation here with $i+r$ to mean the spin that is $r$ distance away from the $i$-th spin.

So in this phase (called the KT phase), there is no "static order" technically, because that needs to have a correlation function that decays to a nonzero constant. Nevertheless, the static susceptibility diverges in this model not only at the critical point $\beta_c$ but through out the KT phase, i.e. for all $\beta>\beta_c$. In many ways you can think that the model is like exactly on top of a critical point through out the entire KT phase. It's so close to forming an order but it barely doesn't. Because of this, some people refer to this phase as a critical phase.

  • $\begingroup$ Thanks. Indeed I've learned KT transition, and I think it's an example I want, but it does not come to my mind until I see your answer. $\endgroup$ Feb 12 at 14:51

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