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I'm in a middle of an advanced stat-mech course, and we are now learning about liquid crystals (LC). In the last tutorial, we developed the Landau Free Energy of the LC system near the transition between the isotropic liquid and the nematic phase.

As a first try for the order parameter, we thought about using an Ising-Like OD:

$$\mathbf m = \frac{1}{N} \left< \sum_\alpha \mathbf{n}_\alpha\right>$$

Which since we want the same energy for $\mathbf n$ and $-\mathbf n$, the energy will have to be a polynomial in ${|\mathbf m|}^2$. This all seemed reasonable to me, and I was surpried to see it won't work. The reason was:

It is not enough for the states $\mathbf n_\alpha$ , $-\mathbf n_\alpha$ to have the same energy, but we rather want them to represent the same state. [...] In other words, we want equivalence rather than degeneracy.

This claim motivated us to use a rank-2 tensor as our order parameter, which is the known and correct OD for this problem.

My question is, what is the difference between equivalence and degeneracy, and why does a rank 2 tensor represents equivalence in this case?

Qmechanic
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Yoav Zack
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1 Answers1

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OK, I think I got it, and the solution was a lot simpler then I thought at the beginning. The trick was to think about the meaning of "degeneracy" in this case, since it is the easier to parse.

When we said "degeneracy", we meant "degeneracy of the Landau Free Energy in the order parameter", which gave us the quadratic dependency.

Since the Free Energy is identical for $\mathbf m$ and $-\mathbf m$, the "equivalence" part cannot relate to it. What is not identical? the OD itself! So, "Equivalence" means "The OD itself is identical for $\mathbf m$ and $-\mathbf m$", and the simplest object that satisfies that is the rank 2 traceless tensor, which is the correct OD for the problem.

We want the OD itself to be identical, since in contract to the Ising model (for example), the particles do not have a preferred vector - just a preferred axis. Flipping a single particle won't change anything, while flipping a single one in the Ising model would.

My misunderstanding came from a lack of words in the explanation, not some deep understanding of mathematical equivalence or something like that. I'm keeping this question up in case it will be relevant to anyone else, even though it sounds a lot more specific and shallow than I thought.

Yoav Zack
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