# What does the solid phase in a two-dimensional system with Lennard-Jones potential look like?

Consider a system of two dimensional particles interacting via Lennard-Jones pair potential: $$u(r) = 4[(\frac{1}{r^{12}})-(\frac{1}{r^{6}})]$$ where r is the distance between two particles. What does the solid phase** look like? Is it a lattice of triangles? Of squares? Is there more than one solid phase? At what pressure/density/temperature do we get a solid phase?

** When I say solid phase I mean a soft solid phase, something that resembles a crystal.

• The most stable/highest density structure is hcp. But other structures can arise depending on the temp/pressure, including a simple cubic phase at low enough temperatures. A quick google will turn up the phase diagram for a 2D LJ fluid, e.g. this paper. Mar 16, 2016 at 13:19
• I can't find where they say that there could be a cubic phase. can you point me to the relevant page? Mar 16, 2016 at 15:21
• I don't believe it says so in that paper, but the simple cubic phase is metastable so if you construct it and run at a low enough temperature, it will be stable. I don't know whether it would ever be the thermodynamically preferred structure though. Mar 16, 2016 at 15:23

You cannot have a solid phase in two dimensions, as the latter would break translational symmetry and thus violate the Mermin-Wagner theorem.

Rigorous proofs of this fact can be found in the following papers:

Both papers deal with very general interactions (not only Lennard-Jones); the second one even allow for additional hard-core interactions (which are tricky to deal with, from a Mermin-Wagner perspective).

Of course, you might then inquire what ground-states look like, but the latter would not be stable at positive temperatures. Rigorous results about the ground states of systems with a class of interactions resembling Lennard-Jones can be found in the paper

It is proved there that, under suitable assumptions on the boundary conditions, the ground state indeed forms a triangular lattice. Even for ground-states in two dimensions, the problem seems still not fully understood. There is also the following nice (and recent) review paper on this topic:

• Excellent references for the absence of long-ranged crystalline order in 2D. However, a certain kind of solid phase is possible in 2D with quasi long-ranged order. See arxiv.org/abs/1102.4094
– TLDR
Mar 16, 2016 at 18:50
• @Couchyam : Sure, I agree with you: they are the analogues of the massless phase in $O(N)$ models. However they are not considered to be solid phases (the latter are characterized by the breaking of translation invariance), and I doubt that's what the OP had in mind. Mar 16, 2016 at 19:44
• so how did people get solid in 2D lennard jonse? I believe the mermin wagner work only when the potential is short ranged. maybe this is the problem. an article with solid simulation: ac.els-cdn.com/0370157381900995/1-s2.0-0370157381900995-main.pdf?_tid=add2a556-ef43-11e5-ba12-00000aab0f6c&acdnat=1458551146_88b013c35f7fb17cc656be144e4784cf Mar 21, 2016 at 10:00
• @AdiRo : of course not. The proofs in the papers I link to apply just as well to interactions of infinite range. In particular, they apply to Lennard-Jones interactions. In two dimensions, you do not see true solid phases, in the sense that there cannot be positional long-range order. What is possible, however, is to have some kind of "soft crystal" phases, which look like solid locally; this is due to the very slow deformations of the crystal over long distances. All this is completely analogous to the massless (Kosterlitz-Thouless) phase in the 2d XY model. Mar 21, 2016 at 10:23
• @AdiRo: This is even discussed explicitly in the paper you refer to: just read Sections 2 and 3. What the author does is to argue that a new definition of a solid phase is needed. But there is no discussion (it is a mathematical fact) that in two dimensions translation invariance cannot be broken. Mar 21, 2016 at 10:27