How does the surface of a material always break inversion symmetry?

I am trying to visualize this for an HCP structure. Take the profile view as such:

just working in 2d.

So my understanding is if we can take a point (x,y) -> (-x,-y) and get the same crystal than inversion symmetry is present.

So take the x axis as the surface here. Clearly we take all points and invert them and get the same crystal. So where exactly is the inversion symmetry broken?

The number of layers and number of atoms is arbitrary here.

Thank you!

• Can you provide a reference that claims surfaces "always break inversion symmetry"? Commented Apr 18, 2016 at 15:04
• psi-k.org/newsletters/News_78/Highlight_78.pdf Section 2.1 for example. But I have read this numerous times from other sources with little justification or further explanation. Commented Apr 18, 2016 at 15:17

1 Answer

When we say that a lattice has a particular symmetry we mean that the lattice is mapped onto itself by the symmetry. So if I have a (2d) material which has inversion symmetry in the bulk and which has an atom at a point $(x,y)$ then inversion symmetry tells me that there is another, identical atom at $(-x, -y)$.

At the surface, however, this is no longer true. In your diagram there is a atom at $(\frac{1}{2}a, -\frac{\sqrt{3}}{2}a)$ but there is nothing at $(-\frac{1}{2}a, \frac{\sqrt{3}}{2}a)$, so the system is not inversion symmetric.

What you are thinking of when you say that if I invert a crystal about a point on its surface we get back another valid structure, only with the positions of atom mirrored about the surface. That is is I have some model for the interactions between the atoms with some Hamiltonian $\mathcal{H}$ then if your original crystal was a solution for this model then there is another corresponding inverted solution. In that case we would say the the Hamiltonian has a symmetry but that this symmetry is broken in our particular solution.