# What's the difference between Equivalence and Degeneracy?

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?

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.