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So I realised that the rhombic unit cell is in fact not the same as a hexagonal unit cell. (I thought they both gave hexagonal lattices but the rhomic unit cell with two rods gives a honeycomb lattice wheras hexagonal unit cell with one rod gives a hexagonal lattice). Since photonic band diagrams are normally sweeping along the path of the irreducible Brillouin zone, I was wondering, how do I find the irreducible Brillouin zone of a rhombic unit cell?

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First, the reciprocal lattice vectors $\bar a,\bar b,\bar c$ can be calculated from the direct space lattice vectors $a,b,c$ via (Wikipedia reference): $$ \bar a = 2\pi \frac{b\times c}{a\cdot(b\times c)} $$ (equivalently $\bar b$ and $\bar c$ by cyclic permutation of $a,b,c;~$ i.e. $~\bar b=2\pi (c\times a)/[b\cdot (c\times a)]$.)

The next step is to consider neighboring lattice points in reciprocal space connected by $\bar a,\bar b,\bar c$ and to lay normal planes on the midpoints of the corresponding connecting lines (i.e. a Wigner-Seitz cell [or Voronoi cell] is constructed in reciprocal space).

This Brillouin zone could be reduced further depending on the point group symmetry of the lattice (e.g. parts of the reciprocal space Wigner-Seitz cell could be rotated or mirrored on top of each other).

Below is an example of a hexagonal cell in direct space (left subplot), and the corresponding Wigner-Seitz cell in reciprocal space (right subplot; plots created with XCrySDen).

Hexagonal unit cell in direct spave and corresponding Wigner-Seitz cell in reciprocal space

The Wigner-Seitz cell in reciprocal space corresponding to the hexagonal cell in direct space is also a hexagonal cell.

Here is an example of a rhombohedral cell in direct space and the corresponding Wigner-Seitz cell in reciprocal space:

Rhombohedral unit cell in direct spave and corresponding Wigner-Seitz cell in reciprocal space

Depending on the point group of your system, the irreducible Brillouin zone will only be a wedge of the Wigner-Seitz cell.

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