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About basis of the honeycomb lattice 
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#1
Apr2812, 08:25 PM

P: 369

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]? 


#2
Apr3012, 04:12 AM

P: 67

This is confusing. How can v1, v2 be a basis when v1 = v2?? You should scan the page and put it up (doublecheck 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
Apr3012, 12:18 PM

P: 369




#4
Apr3012, 10:10 PM

P: 67

About basis of the honeycomb lattice
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
Apr3012, 10:57 PM

P: 369

\vec{v}_2 = a [0 \ 1/(2\sqrt{3})][/tex] in the book, right? 


#6
Apr3012, 11:07 PM

P: 67




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