# About basis of the honeycomb lattice

1. Apr 28, 2012

### KFC

Hi there, I am reading the book "Condensed Matter Physics" second edition by Michael P. Marder. It stated in page 9 that one basis of the the honeycomb lattice is

$$\vec{v}_1 = a [0 \ 1/(2\sqrt{3})], \qquad \vec{v}_2 = a [0 \ -1/(2\sqrt{3})]$$

which is based on figure 1.5(B) in page 10. But in that case when two (vertical) atoms are bind together, so should this basis be

$$\vec{v}_1 = a [0 \ \sqrt{3}/2], \qquad \vec{v}_2 = a [0 \ -\sqrt{3}/2]$$

By the way, why the primitive vectors are given as that in 1.6a and 1.6b

$$\vec{v}_1 = (1/6 \ 1/6) , \qquad \vec{v}_2 = (-1/6 \ -1/6)$$

it said $$(\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1$$

But
$$\vec{a}_1 = a(1 \ 0), \qquad \vec{a}_2 = a (1/2 \ \sqrt{3}/2)$$

why $$(\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1$$?

2. Apr 30, 2012

### sam_bell

This is confusing. How can v1, v2 be a basis when v1 = -v2?? You should scan the page and put it up (double-check the Forum rules first .. i'm not an expert). Few people are so eager to help that they would go to the library and check out the book. You have to make the helpers' life easy.

3. Apr 30, 2012

### KFC

Sorry for the confusing ... and sorry also the book has been returned to the library and I don't have one now. But one thing I could explain here, in solid state physics, in some book 'basis' mean the combination of atoms only, nothing to do with the basis vector, so it is possible to have v1=-v2 in that case.

4. Apr 30, 2012

### sam_bell

All three bases describe a honeycomb lattice, when combined with Bravais vectors a1, a2. The second set (v1 = a[0, sqrt(3)/2] and v2 = a[0, -sqrt(3)/2]) is translated by a[1/2,0] relative to the first. The third set (v1 = a1/6 + a2/6 and v2 = -a1/6 -a2/6) is rotated by 60 degrees relative to the first.

5. Apr 30, 2012

### KFC

Thanks for your reply. I get the point now. So, there is a mistake to write $$\vec{v}_1 = a [0 \ 1/(2\sqrt{3})], \qquad \vec{v}_2 = a [0 \ -1/(2\sqrt{3})]$$ in the book, right?

6. Apr 30, 2012

### sam_bell

Err, no. That's what I was referring to as the "1st" set.