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.

 

[paranormal]

As a scientist, it is important to accurately understand and define concepts such as the Brillouin zone size. After reviewing the discussion between you and your coursemates, it seems that there is some confusion about the definition of the lattice constant, a, and how it relates to the Brillouin zone size.

Firstly, it is important to note that the Brillouin zone is a concept in reciprocal space, which is the Fourier transform of real space. In reciprocal space, the lattice constant, a, represents the spacing between adjacent reciprocal lattice points. This means that for a square lattice, the magnitude of the reciprocal lattice vectors will be 2π/a, as you correctly stated.

However, for other lattice types such as hexagonal or oblique square, the magnitude of the reciprocal lattice vectors will differ from 2π/a due to the different lattice structures. This means that the area of the Brillouin zone will also be different and cannot be generalized as (2π/a)^2 for all lattice types.

In summary, the area of the Brillouin zone is not solely dependent on the lattice constant, but also on the specific lattice structure. It is important to carefully consider the lattice type when calculating the Brillouin zone size for a specific material. I hope this helps to clarify any confusion and good luck on your exam!