About basis of the honeycomb lattice

  • Context: Graduate 
  • Thread starter Thread starter KFC
  • Start date Start date
  • Tags Tags
    Basis Lattice
Click For Summary
SUMMARY

The discussion centers on the basis vectors of the honeycomb lattice as described in "Condensed Matter Physics" by Michael P. Marder. The user questions the validity of the basis vectors \vec{v}_1 = a [0 \ 1/(2\sqrt{3})] and \vec{v}_2 = a [0 \ -1/(2\sqrt{3})], suggesting an alternative representation. The conversation clarifies that despite the apparent contradiction of having v1 = -v2, all three sets of basis vectors accurately describe the honeycomb lattice when combined with the Bravais vectors a_1 and a_2. The confusion arises from the interpretation of 'basis' in solid state physics, which can differ from conventional definitions.

PREREQUISITES
  • Understanding of honeycomb lattice structures in solid state physics.
  • Familiarity with Bravais lattice vectors.
  • Knowledge of vector representation in crystallography.
  • Basic concepts of solid state physics as outlined in Marder's "Condensed Matter Physics".
NEXT STEPS
  • Study the derivation of Bravais lattice vectors in solid state physics.
  • Explore the implications of basis vectors in crystallography.
  • Review the geometric interpretation of honeycomb lattices in materials science.
  • Investigate the role of symmetry in lattice structures and basis vector definitions.
USEFUL FOR

Students and researchers in solid state physics, crystallographers, and anyone studying the geometric properties of lattice structures, particularly those interested in the honeycomb lattice and its applications in materials science.

KFC
Messages
477
Reaction score
4
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

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

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

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

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

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

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

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

why (\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1?
 
Physics news on Phys.org
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.
 
sam_bell said:
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.

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.
 
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.
 
sam_bell said:
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.

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 <br /> \vec{v}_2 = a [0 \ -1/(2\sqrt{3})] in the book, right?
 
KFC said:
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 <br /> \vec{v}_2 = a [0 \ -1/(2\sqrt{3})] in the book, right?

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

Similar threads

  • · Replies 11 ·
Replies
11
Views
1K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
1
Views
1K
Replies
1
Views
3K
Replies
15
Views
2K