1
$\begingroup$

If I have a 2D ladder of atoms, can I desribe it with two primitive vectors $a_1=a(1,0)$ , $a_2=a(0,1)$, or do I need to desribe it using only one primitve vector $a_1=a(1,0)$ and basis: $b_1=(0,0)$, $b_2=a(0,1)$? The problem is that the crystal is of limited size in the vertical direction and all definitions of primitve vectors, that I'm familiar with, pertain to unlimited crystals.

To summarize: Can I regard this kind of lattice as a 2D crystal or a 1D crystal with a basis?

Example (Ashcroft) : A Bravais lattice consist of all points with position vectors $\textbf R$ of the form $\textbf R=n_1 \textbf a_1 + n_2 \textbf a_2 + n_3 \textbf a_3$, where $\textbf a_1, \textbf a_2 , \textbf a_3$ are any three vectors not all in the same plane and $n_1,n_2,n_3$ rangle through all integral values.

enter image description here

$\endgroup$
1
  • 1
    $\begingroup$ If I was dealing with a structure like this where there was the potential for ambiguity, I would avoid saying that it was either a 1D crystal or a 2D crystal. I would say something along the lines of it is a 2D structure with the symmetry of a 1D lattice $\endgroup$ – By Symmetry Sep 3 '17 at 21:09
1
$\begingroup$

I would treat it as a 1D lattice with a basis. As mentioned, a Bravais lattice is infinitely periodic in whatever dimensions you're using, which is only appropriate for the wide one. In any case, this crystal would certainly behave 1D (i.e. you would have strong state quantization in the narrow dimension, not anything resembling a continuous band). But if you're looking for the reciprocal lattice, I suspect you'll need to do a 2D Fourier transform.

For similar real-world examples, MoS$_2$ is a three-atom thick "2D" material, and carbon nanotubes are "1D" structures several atoms wide. And also you have graphene nanoribbons. Of course, if you want to calculate electronic band structure etc. of the carbon allotropes, you start with the graphene 2D band structure and apply appropriate boundary conditions on one of the dimensions.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.