# Bravais lattices and lattices with a basis

• Niles
In summary, a Bravais lattice can be defined as a collection of lattice points that can be reached by a linear combination of three basis vectors. However, this is not the case for lattices with a basis. Despite this, a non-Bravais lattice can still have primitive vectors, as it can be viewed as a Bravais lattice with the basis sitting at the lattice points. This is demonstrated in the example of the CuO2 planes in a cuprate superconductor. It is important to note that a crystal consists of both a basis and a lattice, and the primitive vectors refer to the lattice and are independent of the basis. Additionally, any Bravais lattice has three basis vectors in 3D,

#### Niles

Hi guys

Ok, so one way to define a Bravais lattice is to say that each lattice point can be reached by R = la1+ma2+na3 for some integer m, l and n. Obviously, this cannot be the case when we have a lattice with a basis.

But does that also mean that a lattice with a basis does not have primitive vectors?

A non-Bravais lattice can be viewed as some Bravais lattice with the basis sitting at the lattice points, and so it will have primitive vectors, as long as we realize our lattice points are now a collection of atoms.

For example, think of the CuO2 planes in a cuprate superconductor. We can think of a basis consisting of a copper atom with two neighbouring oxygen atoms. We can think of the primitive vectors of this lattice as those of the square lattice. A picture would explain this far more clearly, but unfortunately I don't have one to hand.

A lattice is a lattice and a basis is a basis. The primitive vectors refer to the lattice and are independent of whether one considers a basis or not.

I was quite in a hurry when I wrote the last message.
So what I wanted to say is that an ideal crystal can be thought off to be composed of a basis and a lattice, the lattice being just the collection of all vectors R you defined. So in mathematical terminology, the lattice is just a vector space (of the vectors R). Physicist often mingle this up by meaning with lattice the complete crystal structure including the basis. A crystal is never only a lattice. Even if the basis contains only one atom, the electron density within the basis is not constant and you need both the position inside the basis and a lattice vector to uniquely specify a point of the crystal.

DrDu said:
I was quite in a hurry when I wrote the last message.
So what I wanted to say is that an ideal crystal can be thought off to be composed of a basis and a lattice, the lattice being just the collection of all vectors R you defined. So in mathematical terminology, the lattice is just a vector space (of the vectors R). Physicist often mingle this up by meaning with lattice the complete crystal structure including the basis. A crystal is never only a lattice. Even if the basis contains only one atom, the electron density within the basis is not constant and you need both the position inside the basis and a lattice vector to uniquely specify a point of the crystal.

I see, thanks. But can I always "decompose" a crystal into its basis and the underlying lattice, which is always Bravais?

Best,
Niles

Niles said:
I see, thanks. But can I always "decompose" a crystal into its basis and the underlying lattice, which is always Bravais?

Best,
Niles

Any bravias lattice has 3 basis vectors in 3D. There are some Bravias lattices with centering translations, for instance bcc (Im3m), but this is done for convenience to have more right angles and so on. Real Bravias lattice on which electrons are scattered are _always_ primitive. Simply if you choose a bigger direct space unit cell you get a smaller reciprocal cell, but some nodes will be extinct.

Niles said:
But can I always "decompose" a crystal into its basis and the underlying lattice, which is always Bravais?

Yes, as long as there is periodicity, which is implied by "crystal" (as opposed to amorphous).