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- Graphene bands.
(a) The real lattice vectors are
a 1 = a(
p 3 = 2 ; 1 =2); a 2 = a(0; 1)
The formulas for the primitive lattice vectors b 1 and b 2 are written assuming that there are three primitive real lattice vectors. We imagine that there is a third primitive lattice vector a 3 perpendicular to the x-y plane. Then
b 1 =
2 a 2 a 3
a 1 :a 2 a 3
p 3 a
x^
b 2 =
2 a 3 a 1
a 1 :a 2 a 3
p 3 a
x^ +
p 3
2
y^)
where ^x and ^y are unit vectors in the x- and y-directions, respectively. The vectors b 1 and b 2 have the same magnitude (4=
p 3 a) and the angle between them is 60. The reciprocal lattice vectors are G = m 1 b 1 + m 2 b 2 ; m 1 ; m 2 2 Z. To draw the rst Brillouin zone (FBZ), choose one reciprocal lattice point, draw all reciprocal lattice vectors starting from this point and draw the perpendicular bisectors of these vectors. The area enclosed by these perpendicular bisectors, and centered on the chosen point, is the FBZ. For the case of graphene, the FBZ is a regular hexagon. The center of the FBZ is called the -point. The point M has coordinates (2=
p 3 a; 0), the point K has coordinates (2=
p 3 a; 2 = 3 a), and the point K^0 has coordinates (0; 4 =
p 3 a).
(b) A k (r) =^
p N
X
n
e ik:Rn (r Rn); B k (r) =^
p N
X
n
e ik:Rn (r Rn)
These are normalized Bloch functions; they satisfy Bloch's theorem:
A;B k (r^ +^ Rm) =^ e
ik:Rm A;B k (r)
Since we are neglecting the overlap between atomic orbitals on di erent sites, A k (r) and B k (r) are orthogonal: (^) Z
A k (r)^
B k (r)d
3 r = 0
To solve the Schrodinger equation H (^) k(r) = Ek k(r), we consider a solution of the form
k(r) =^ a^
A k (r) +^ b^
B k (r)
Since A k (r) and B k (r) are orthogonal, (^) k(r) is normalized if jaj^2 + jbj^2 = 1. The Schrodinger equation becomes X
n
e ik:Rn [aH (r Rn) + bH (r Rn)]
= En
X
n
e ik:Rn [aH (r Rn) + bH (r Rn)]
We multiply the above equation by (r) and integrate over d^3 r. First, we note that
Z
(r)H (r Rn)d 3 r = 0
This is because if Rn = 0 , the integral is equal to the orbital energy which we set
equal to zer0, and if Rn 6 = 0 , then (r) and (r Rn) are atomic orbitals centered
on atoms of type A, and such atoms are not nearest neighbors. Since we assume that
interactions exist only between nearest neighbors, the integral vanishes. Since we also
ignore the overlap between orbitals on di erent sites, we set
R
(r) (r Rn) equal
to zero.
Taking account of these observations, the Schrodinger equation becomes
X
n
e ik:Rn b
Z
(r)H (r Rn)d 3 r = Ek
X
n
e ik:Rn a
Z
(r) (r Rn)d 3 r
On the RHS, the integral vanishes unless Rn = 0 , in which case the integral is equal to 1
(we are neglecting the overlap between orbitals on di erent atoms and we are assuming
that the atomic orbitals are normalized). Hence, RHS = aEk.
On the LHS, the integral vanishes unless (r Rn) is centered on one of the three
nearest neighbors of atom A. Therefore, in summing over n, only three terms survive:
R 1 = 0 ; R 2 = a(
p 3 = 2 ; 1 =2), and R 3 = a(
p 3 = 2 ; 1 =2). For each of these values of
Rn, the integral on the LHS of the above equation is the matrix element of H between
the pz orbital on A and the pz orbital on one of the nearest neighbors of A; it is thus
t. Hence, the above equation becomes
btgk = aEk
where
gk = 1 + exp
i
p 3
2
kxa +
kya
i
p 3
2
kxa
kya
Next we multiply the Schrodinger equation by (r ) and integrate over d^3 r. On the
RHS we end up with bEk, whereas
LHS =
X
n
e ik:Rn a
Z
(r )H (r Rn)d 3 r
The integral vanishes except for three values of Rn, namely, R 1 = 0 ; R 2 = a(
p 3 = 2 ; 1 =2),
and R 3 = a(
p 3 = 2 ; 1 =2). For each of these values of Rn, the integral is t. These three
Rn vectors are simply the negative of the three Rn vectors encountered earlier. We
thus obtain,
atg k =^ bEk
(d) Let kx = 2=
p 3 a + k x^0 ; ky = 2= 3 a + k y^0 , where k x^0 and k y^0 are small: k^0 x ; k y^0 << =a. Setting k 0 xa^ =^ x^ and^ k
0 ya^ =^ y,
Ek = t
3 + 4cos +
p 3
2
x
cos
y
2
= t
3 4 cos(
p 3 x=2)
cos(y=2)
p 3
2
sin(y=2)
cosy
p 3 siny
where in the last step we used the formula cos(a+b) = cosacosb sinasinb. Expanding:
cos = 1 2 =2! + ; sin = 3 =3! + ;
we obtain
Ek = t
n 3 4(1 3 x^2 =8)(1= 2 y^2 = 16
p 3 y=4) 1 + y^2 = 2
p 3 y
o 1 = 2
= t
h 3 2 + y 2 =4 +
p 3 y + 3x 2 = 4 1 + y 2 = 2
p 3 y
i 1 = 2
= t
x 2
p 3
2
ta
k
(^02) x +^ k
(^02) y
p 3
2
tak 0
There are, of course, terms of higher order in k^0 which we have neglected since they are less important when k 0 is small. Measuring k from point K in the FBZ, we can write
Ek = ~vF k
where vF =
p 3 ta= 2 ~ is the magnitude of the Fermi velocity, and the (+) sign refers to the valence (conduction) band.
We have expanded around point K; it is easily veri ed that the same linear dispersion is obtained near point K 0 .
(e) Consider a shell near point K, bounded by the constant energy surfaces = ~vF k and + d = ~vF k + ~vF dk. The area of the shell is 2kdk. Each k-point occupies an area in k-space given by (2) 2 =A, where A is the area of the graphene crystal. Since there are two states for each k-point (jk "i and jk #i), the number of states in the shell is
N = 2(2kdk)=(2) 2 =A = Akdk=
The density of states is thus
D() =
dN
d
Ak
dk
d
Ak
~vF
Ajj
(~vF )^2
4 Ajj
3 a^2 t^2
=) d()
A
D() =
4 jj
3 a^2 t^2
Since there are two valleys, one near point K and another near point K^0 , the total density of states per unit area is
dtotal() =
8 jj
3 a^2 t^2
- More on graphene.
The pz orbital on each atom is represented by the wave function
(r) = Arcose Zr= 2 a 0
Here, A is a normalization constant, a 0 is the Bohr radius, is the angle between r and the z-axis (the one perpendicular to the graphene plane), and Ze is the e ective charge on the carbon nucleus, as seen by the electron in the pz orbital (Z ' 3). We want to evaluate
I(q) =
Z
(r)e iq:r (r)d 3 r
where q is a two-dimensional vector in the FBZ of graphene.
First, note that since (r) is normalized,
2 A
2
Z 1
0
r 4 e Zr=a 0 dr
Z 2
1
cos 2 dcos = 1
A
2
Z 1
0
r 4 e Zr=a 0 = 1
Expanding e iq:r, we can write
I(q) = A 2
X^1
n=
n!
Z 1
0
r 4 e Zr=a 0 dr
Z 2
0
d
Z 1
1
d(cos) cos 2 ( iq:r) n
If we choose the x-axis to be along the direction of q, then q:r = q rsincos. Therefore,
I(q) = A 2
X^1
n=
( iq) n
n!
Z 1
0
r 4+n e Zr=a 0 dr
Z
0
cos 2 sin n+ d
Z 2
0
cos n d
Note that (^) Z 2
0
cos n d = 0 if n is odd;
hence,
I(q) = A 2
X^1
n=
n q
2 n
(2n)!
Z 1
0
r 4+2n e Zr=a 0 dr
Z
0
(1 sin 2 ) sin 2 n+ d
Z 2
0
cos 2 n d
(ii) Following the same steps as above, it is readily shown that
B k e
iq:r B k+q '^1 ;^
A k e
iq:r B k+q =^
B k e
iq:r A k+q = 0
The second equation results from neglecting overlap between atomic orbitals on dif- ferent sites.
(iii)
v k e
iq:r v k+q =
g kgk+q jgkgk+qj
A k je
iq:r j A k+q +^
B k je
iq:r j B k+q
g k
jgkj
A k je
iq:rj B k+q +^
gk
jgkj
B k je
iq:rj A k+q
Using the results of (i) and (ii),
v k e
iq:r v k+q '^
g k gk+q
jgk gk+qj
(iv) Similarly, it is readily shown that
c k e
iq:r c k+q =^
v k e
iq:r v k+q
c k e
iq:r v k+q =^
v k e
iq:r c k+q =
g k gk+q
jgk gk+qj
(v)
gk = 1 + e
i
p 3 2 kxa+^
1 2 ky^ a
i
(^) p 3 2 kxa+^
1 2 ky^ a
We assume that k is near the point K or K' (gk vanishes at these points). Expanding about point K,
gk =
@gk
@kx (^) K
kx
p 3 a
@gk
@ky (^) K
ky
3 a
i
p 3 a
2
kx
p 3 a
p 3 a
2
ky
3 a
Measuring k from point K,
gk =
i
p 3 a
2
(kx iky) = i
p 3
2
kae ik
where k is the angle between k and the x-axis. Therefore,
jgkgk+qj =
3 a 2
kjk + qj
and
g kgk+q^ =
3 a^2
4
kjk + qje
i(k k+q)
Hence, v k e
iq:r v k+q =
h 1 + e i(k k+q)
i
and
v k e
iq:r v k+q
2
h 1 + e i(k k+q)
i h 1 + e i(k k+q)
i
(1 + cos)
where is the angle between k and k + q:
cos =
k:(k + q)
kjk + qj
k 2
kjk + qj
k + q cos
jk + qj
where is the angle between k and q.
Following the same steps as above, we nd
c k e
iq:r v k+q
2
k + q cos
jk + qj
Since c k e
iq:r c k+q =^
v k e
iq:r v k+q
and v k e
iq:r c k+q =^
c k e
iq:r v k+q
we can write
Fss^0 (k; q) =
1 + ss 0 k^ +^ q cos jk + qj
where s; s^0 = 1 (+1) if s; s^0 = v (c).
- Density of states.
D()d = 2
(2)^3
V
Z
dk?dS
In the above, the function dk? is integrated over the surface in k-space on which the energy is a constant equal to . dk? is the perpendicular distance in k-space between the inner and outer surfaces of the shell.
The point to note here is that for the constant energy surface (k) = , the gradient rk(k) is perpendicular to the constant energy surface. Therefore,
jrkj dk? = d
Hence,
D()d = 2
(2)^3
V
Z
dS
jrkj
d
Problems on nearly free electron approximation:
Problem no.2: Nearly free electrons in 1D
The periodic potential is given as the following series:
V (x) = V 0 + V 1 cos
2 πx a
4 πx a
The Schr¨odinger equation in Fourier space is given by:
ℏ^2 k^2 2 m
− ϵk
αk(G⃗ ) +
X
G⃗′
V (G⃗ − G⃗ ′ )αk(G⃗ ′) = 0
We would like to express the potential V (x) in the form:
V (x) =
X
G′
e−iG
′x V (G′) (1)
V (x) = V 0 +
V 2 2
e
2 πix a (^) + e−^ 2 πix a
V 2 2
e
4 πix a (^) + e−^ 4 πix a
Rewriting in a more compact form:
V (x) = V 0 +
V 2 2
eib^1 x^ + e−ib^1 x
V 2 2
ei(2b^1 )x^ + e−i(2b^1 )x
where the reciprocal lattice vectors are:
G⃗ = {±b⃗ 1 , ± 2 ⃗b 1... }
Given:
b⃗n =^2 πn a
ˆi
When G⃗ = 0, we get: (^) ℏ^2 k^2 2 m
− ϵk
αk(⃗0) +
X
G⃗′
αk( G⃗ ′)V (⃗ 0 − G⃗′) = 0
Expanding the summation:
ℏ^2 k^2 2 m
− ϵk
αk(⃗ 0) + αk(⃗0)V (⃗ 0) + αk(b⃗ 1 )V (−⃗b 1 ) = 0
=⇒
ℏ^2 k^2 2 m
− ϵk
αk(⃗ 0) + αk(⃗0)V 0 + αk(b⃗ 1 )
V 1 2
= 0
=⇒
ℏ^2 k^2 2 m
− ϵk + V 0
αk(⃗ 0) + αk(b⃗ 1 )
V 1 2
= 0
When G⃗ 1 = b⃗ 1 : (^) ℏ^2 k^2 2 m
− ϵk
αk(b⃗ 1 ) + αk(⃗0)V (b⃗ 1 ) + αk(b⃗ 1 )V (⃗0) = 0
=⇒
ℏ^2 k^2 2 m
− ϵk
αk( b⃗ 1 ) + αk(⃗0)
V 1 2
=⇒
ℏ^2 k^2 2 m
− ϵk + V 0
αk( b⃗ 1 ) + αk(⃗ 0)
V 1 2
= 0
For nontrivial solutions of αk(⃗0) and α( b⃗ 1 ) from equations (i) and (ii) we must satisfy
ℏ^2 k^2 2 m −^ ϵk^ +^ V^0
V 1 2 V 1 2
ℏ^2 k^2 2 m −^ ϵk^ +^ V^0
= 0
1
Expanding the determinant: ℏ^2 k^2 2 m
− ϵk + V 0
ℏ^2 k^2 2 m
− ϵk + V 0
−
V 1 2
2 = 0
=⇒
ℏ^2 k^2 2 m
− ϵk + V 0
2
V 1 2
2
Taking the square root on both sides:
ℏ^2 k^2 2 m
− ϵk + V 0 = ±
V 1 2 Solving for ϵk:
ϵk =
ℏ^2 k^2 2 m
∓
V 1 2
Problem no.3: Nearly free electrons in 2D ℏ^2 k^2 2 m
− ϵk
αk(G⃗ ) +
X
G⃗′
V (G⃗ − G⃗ ′)αk(G⃗ ′) = 0
Unlike the 1D case there is a degeneracy since the energy is same at ⃗k = b⃗ 1 and ⃗k = b⃗ 2. In this case we
consider only perturbations that effect the band structure at the boundaries of the 1st BZ.
Points of degeneracy include G⃗ = ⃗0 ; G⃗ = b⃗ 1 ; G⃗ = b⃗ 2 ; G⃗ = b⃗ 1 + b⃗ 2
The potential at points are given V ( (^2) aπ , (^2) aπ ) = 0 and V (0, (^2) aπ ) = V 0. Considering symmetry of square lattice
it is obvious that V ( (^2) aπ , 0) = V 0 and V (0, 0) = 0
When G⃗ = ⃗0: ℏ^2 k^2 2 m
− ϵk
αk(⃗ 0) +
X
G⃗ 1
V ( G⃗ 1 )αk( G⃗ 1 ) = 0
ℏ^2 k^2 2 m
− ϵk
αk(⃗ 0) + V (⃗ 0)αk(⃗ 0) + V (⃗ 0 − b⃗ 1 )αk(b⃗ 1 ) + V (⃗ 0 − b⃗ 2 )αk(b⃗ 2 ) + V (⃗ 0 − b⃗ 1 − b⃗ 2 )αk(b⃗ 1 + b⃗ 2 ) = 0
ℏ^2 k^2 2 m
− ϵk
αk(⃗0) + V 0 αk( b⃗ 1 ) + V 0 αk( b⃗ 2 ) = 0 .....(i)
When G⃗ = b⃗ 1 : ℏ^2 k^2 2 m
− ϵk
αk(b⃗ 1 ) +
X
G⃗ 1
V ( b⃗ 1 − G⃗ 1 )αk(G⃗ 1 ) = 0
ℏ^2 k^2 2 m
− ϵk
αk( b⃗ 1 ) + V (b⃗ 1 )αk(⃗0) + V (⃗0)αk(b⃗ 1 ) + V ( b⃗ 1 − b⃗ 2 )αk( b⃗ 2 ) + V (⃗ 0 − b⃗ 2 )αk(b⃗ 1 + b⃗ 2 ) = 0
ℏ^2 k^2 2 m
− ϵk
αk( b⃗ 1 ) + V 0 αk(⃗ 0) + V 0 αk(b⃗ 1 + b⃗ 2 ) .....(ii)
When G⃗ = b⃗ 2 : ℏ^2 k^2 2 m
− ϵk
αk(b⃗ 2 ) +
X
G⃗ 1
V ( b⃗ 2 − G⃗ 1 )αk(G⃗ 1 ) = 0
2
