Brillouin Zone Size: Debate Settled

[RadonX]

Hi guys,
Me and a few of my coursemates are revising for a solid state exam but have hit a problem.

Is the area of every 2D brillouin zone (independent of lattice type) (2Pi/a)^2?

For a square lattice in 2D real space with lattice constant, a, the reciprocal lattice vectors can easily be found (with magnitudes of 2Pi/a, right?) and so the brillouin zone in reciprocal space has a area of (2Pi/a)^2.
Half of us are under the impression that this area is true of ALL brillouin zones of varying lattice type (square, oblique square, hexagonal etc), the other half are of the thought that you define the lattice spacing as 'a' and therefore the brillouin zone sizes vary.
Particularly, for a hexagonal lattice this would render an area of 2/sqrt(3)*(2Pi/a)^2.

Obviously the lattice spacing in reciprocal space is important for answering questions on the area of the fermi circle etc.

Hope someone can help us! This is causing a lot of debate!

Thanks,
Will

 

[daveyrocket]

The area of any Brillouin zone is (2Pi)^d/V where d is the dimensionality, and V is the "volume" of the real-space crystal. In 2D, V is the area of the unit cell, a*a fora square lattice, a*b for a rectangular lattice, and something else for other shapes of cells.

You remember how to calculate the primitive vectors of the reciprocal lattice, right? They satisfy
\vec{a}_i \cdot \vec{b}_j = 2\pi \delta_{i,j}.

For three dimensions, you can calculate the volume of the unit cell by (\vec{a}_1 \times \vec{a}_2) \cdot \vec{a}_3. Do the same for the reciprocal cell and with the relationship above you can prove that the product of the volumes of the real-space cell and the reciprocal cell is (2pi)^3. Carry out the same procedure for 2D and you should be able to prove to your friends the correct relationship.

 

[RadonX]

Daveyrocket!
Thank you very much! That's a great help.
Just checking through our stuff to see if we get your (2pi)^3 result when multiplying the reciprocal cell by the real space cell.
We get (2Pi)^2 in our case which makes sense for a 2d lattice area we think?

 

[daveyrocket]

Yes that is correct.