Graphene lattice
real
|a1| = |a2| = a
reciprocal
Graphene lattice
real
|a1| = |a2| = a
reciprocal
| b1 || b2 | b=
2
4

3
3a
a
2
a1
a1  a2
a2
Our result, Eq. (5) is beautiful, because it fully shows the inherent symmetry, by not
using the x-y coordinate.
Compare: Eq.(1) in Bostwick Nature Phys 3, 36 (200);
Eq.(6) in Neto 81, 109 (2009) (not so correct!)
Considering nearest neighbor hopping only
a1
a1  a2
a2
Our result, Eq. (5) is beautiful, because it fully shows the inherent symmetry, by not
using the x-y coordinate.
We now evaluate E at the K points.
Graphene lattice
b1b2
b2
K’
b1
K
b2
real
|a1| = |a2| = a
b1·a2 = 0, b21·a12 = 0
4 3
 2
3a 2
Similarly, b1·a1 = abcos30 = 2
b1·a1 = abcos30 = a
reciprocal
2
4
| b1 || b2 | b 

3
3a
a
2
K’ = (1/3)(b1b2)
K = (1/3)(b2b1)
Graphene lattice
b1b2
b2
b1
K’
K
K’
K’
K
b2
real
|a1| = |a2| = a
reciprocal
K and K’ are two in-equivalent points. The
inequivalence has nothing to do with AB sublattices;
it’s intrinsic to this Bravais lattice.
2/a
Compare:
/a
/a
2
4
| b1 || b2 | b 

3
3a
a
2
K’ = (1/3)(b1b2)
K = (1/3)(b2b1)
F
Again, we often write k where it really
should be q.
In most circumstances, it’s actually okay to
ignore the shift K or K’; what matters is k,
not k. k is not real “momentum” anyway...
v
1 dE
 dk
v
1
E in 2D or 3D

For graphene,
K
K’
k
K’
in 1D
v

q
1
1
E  (vF k )  vF kˆ


K
K
K’