- #1
KFC
- 488
- 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
[tex]
\vec{v}_1 = a [0 \ 1/(2\sqrt{3})], \qquad
\vec{v}_2 = a [0 \ -1/(2\sqrt{3})]
[/tex]
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
[tex]
\vec{v}_1 = a [0 \ \sqrt{3}/2], \qquad
\vec{v}_2 = a [0 \ -\sqrt{3}/2]
[/tex]
By the way, why the primitive vectors are given as that in 1.6a and 1.6b
[tex]
\vec{v}_1 = (1/6 \ 1/6) , \qquad \vec{v}_2 = (-1/6 \ -1/6)
[/tex]
it said [tex](\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1[/tex]
But
[tex]
\vec{a}_1 = a(1 \ 0), \qquad \vec{a}_2 = a (1/2 \ \sqrt{3}/2)
[/tex]
why [tex](\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1[/tex]?
[tex]
\vec{v}_1 = a [0 \ 1/(2\sqrt{3})], \qquad
\vec{v}_2 = a [0 \ -1/(2\sqrt{3})]
[/tex]
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
[tex]
\vec{v}_1 = a [0 \ \sqrt{3}/2], \qquad
\vec{v}_2 = a [0 \ -\sqrt{3}/2]
[/tex]
By the way, why the primitive vectors are given as that in 1.6a and 1.6b
[tex]
\vec{v}_1 = (1/6 \ 1/6) , \qquad \vec{v}_2 = (-1/6 \ -1/6)
[/tex]
it said [tex](\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1[/tex]
But
[tex]
\vec{a}_1 = a(1 \ 0), \qquad \vec{a}_2 = a (1/2 \ \sqrt{3}/2)
[/tex]
why [tex](\vec{a}_1 + \vec{a}_2)/6 = \vec{v}_1[/tex]?