PHZ6426: Fall 2013
Problem set # 5: Nearly-free-electron and tight-binding models:
Solutions
due Wednesday, 11/13 at the time of the class
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Office hours: TR 3 pm-4 pm
Please help your instructor by doing your work neatly. Every (algebraic) final result must be sup
December 4, 2013
...
P1 [20 points ] Consider a 1D electron system subject to a weak periodic potential
3
4 πx
−
,
U (x) = U0 cos
a
8
Find and plot the dispersions of energy bands in the first Brillouin zone. Plot the energies in units of
~2 π 2 /2ma2 , the wave numbers in units of 1/a, and assume that U0 = 0.1 in these units.
Solution
Using the trigonometric identity cos2 x = (1 + cos(2x))/2, the potential can be re-written as
2πx 1
4πx
1 4πix/a
1
1 2iπx/a
U (x) = U0
cos
+ cos
e
+ e
+ c.c.
= U0
2
a
8
a
4
16
There are two non-zero harmonics with amplitudes u±1 = U0 /4 and u±2 = U0 /16 with wavenumbers
q = ±2π/a and q = ±4π/a. Scattering at harmonic with q = ±2π/a lifts denegeracy of the freeelectron states with energies ε1 = ~2 k 2 /2m and ε2 = ~2 (k ∓ 2π/a)2 /2m. The degeneracy points are
k = ±π/a. The energy levels near these points are given by
s
2
2
ε1 + ε2
(ε1 − ε2 )
U0
ε=
±
+
.
2
4
4
Likewise, scattering at harmonic with q = ±4π/a lifts denegeracy of the free-electron states with
energies ε1 = ~2 k 2 /2m and ε3 = ~2 (k ∓ 4π/a)2 /2m. The degeneracy points are k = ±2π/a. The
energy levels near these points are given by
s
2
2
ε1 + ε3
U0
(ε1 − ε3 )
ε=
±
+
.
2
4
16
1
P2 [20 points ] Consider 2D electrons subject to a weak periodic potential
2πx
2πy
U (x, y) = U0 cos
+ cos
.
a
a
a) [8 points ] Find the energy bands near the edges of the first Brillouin zone. (Hint: the edges of the
Brillouin zone are the Bragg planes where the eigenstates are doubly degenerate.) Plot the bands
and isoenergetic surfaces.
Solution The non-zero harmonics of the potential have wavenumbers (±b, qy = 0) and (qx =
0, qy = ±b), where b = 2π/a, with amplitudes U0 /2. Consider one of the edges, e.g., kx = π/a; by
symmetry, the energy spectrum
states with
will be the same at other edges. The two free-electron
energies ε0kx ,ky = ~2 kx2 + ky2 /2m and ε0kx −2π/a,ky = ~2 (kx − 2π/a)2 + ky2 /2m are degenerate
at this edge. The energy levels near the edge are given by
v
2
u
u
2
0
0
εkx ,ky + εkx −2π/a,ky t ε0kx ,ky − ε0kx −2π/a,ky
U0
.
±
+
ε=
2
4
2
b) [12 points ] Repeat the analysis of part a) for regions near the corners of the first Brillouin zone, where there
are four degenerate eigenstates.
Solution The basis must now include four degenerate states with energies ε0kx ,ky , ε0kx −b,ky ,
ε0kx −b,ky −qy , and ε0kx −b,ky −b . Using the fact the only non-zero harmonics of the periodic potential are Uqx =±b,0 = Uqx =0,qy =±b = U0 /2, we obtain equations for the weighing factors of the
four degenerate states [cf. Eq. (9.19) in AM]:
(ε − ε0kx ,ky )ckx ,ky
=
(ε − ε0kx −b,ky )ckx −b,ky
=
(ε − ε0kx ,ky −b )ckx ,ky −b
=
(ε − ε0kx −b,ky −b )ckx −b,ky −b
=
U
(ckx −b,ky + ckx ,ky −b )
2
U
(ckx ,ky + ckx −b,ky −b )
2
U
(ckx ,ky + ckx −b,ky −b )
2
U
(ckx −b,ky + ckx ,ky −b ).
2
(1)
Inverting the matrix and measuring the wavenumbers in units of 1/a and energies in units of
~2 /2ma2 , we obtain for the four energy levels
ε1
= E0 + R+
ε2
= E0 − R+
ε3
= E0 + R−
ε4
= E0 − R−
where
E0
R±
= kx2 + ky2 − 2π(kx + ky ) + 4π 2
1/2
q
1
4U02 + 32π 2 (kx2 + ky2 − 2π(kx + ky ) + 2π 2 ) ± 4 (U02 + 16π 2 (π − kx )2 )(U02 + 16π 2 (π − ky )2 )
=
2
P3 [20 points ] The Kronig-Penney model allows one to consider both the nearly-free-electron and tight-binding
limits. Previously, we focused on the latter. Now, consider the (repulsive) Dirac-Kronig-Penney model,
which, as we showed earlier, gives the following implicit equation for the energy bands:
cos ka = cos qa + u
sin qa
,
qa
(2)
√
where q = 2mE/~ and u = mU0 a/~2 > 0. Assuming that u 1, find the solution of Eq. (2) near
the Bragg points k ≈ ±π/a, ±2π/a . . . . (Reminder: the ≈ sign means that k is close yet not equal to
the wavenumber of the Bragg point.) Compare your result with that of the nearly-free-electron model.
Solution
The maximum value of the r.h.s. of Eq. (2) (= 1 + u) occurs at q = 0. The minimum value occurs
near qa = π. To see this, substitute qa = π + δ and expand the r.h.s. to second order in δ:
R ≡ cos qa + u
sin qa
δ2
uδ
= −1 +
−
+ ...
qa
2
π
∂R/∂δ = 0 gives δ = u/π, and the minimum value of R is Rmin = −1−u2 /2π 2 < −1. Since cos k ≥ −1,
Eq. (2) has no soluton in the region around the minimum of R, where Rmin ≤ R ≤ −1. This region
corresponds to a gap in the energy spectrum.
To find the energy levels near the gap, substitute ka = π + p into Eq. (2) and expand cos ka in p as
cos(π + p) = −1 + p2 /2. The resulting equation
δ2 −
2
uδ − p2 = 0
π
has two roots
r
u 2
u
.
δ± = ± p2 +
π
π
Returning to the original variables, we obtain for the energy levels
2 2
E±
=
≈

π ~ 
1+
2ma2

2 2
π ~ 
1+
2ma2
u
π2
2u
π2
2
2
ka
u 
±
−1 +
.
π
π2

s
2 2
ka
u 
±2
−1 +
,
π
π2
s
2
(3)
where the last line is valid to first order in u.
Recall the result of the NFE model (with diagonal matrix elements U11 and U22 equal to zero)
s
εk + εk−2π/a + U12
εk − εk−2π/a 2
2 ,
E± =
±
+ U12
2
2
(4)
where εk = ~2 k 2 /2m. The difference of the free-electron energies can be re-written as
2π 2 ~2 ka
εk − εk−2π/a =
−1 ,
ma2
π
while in their sum we can replace k by π/a:
εk + εk−2π/a =
~2 π 2
ma2
Now it is obvious that the NFE result is equivalent to Eq. (4).
P4 [15 points ] Find the dispersions of energy bands for fcc and bcc lattices taking into account hopping between the
nearest and next-to-nearest neighbors.
Solution:
The tight binding dispersion relation is:
Ek = E0 − t
X
~ ~
eik·Rnn − t
0
~ nn
R
(a) case of bcc
8 n.n.:
a
a
a
2 (±1, +1, +1) , 2 (±1, +1, −1) , 2 (±1, −1, +1) ,
6 n.n.n.:
a(±1, 0, 0) , a(0, ±1, 0) , a(0, 0, ±1)
X
X
~ ~
eik·Rnnn
(5)
~ nnn
R
a
2 (±1, −1, −1)
~ ~ bcc
eik·Rnn = e(1/2)ia(+kx +ky +kz ) + e(1/2)ia(−kx +ky +kz )
~ bcc
R
nn
+e(1/2)ia(+kx +ky −kz ) + e(1/2)ia(−kx +ky −kz )
+e(1/2)ia(+kx −ky +kz ) + e(1/2)ia(−kx −ky +kz )
+e(1/2)ia(+kx −ky −kz ) + e(1/2)ia(−kx −ky −kz )
(6)
= 2 cos[(a/2)(kx + ky + kz )] + 2 cos[(a/2)(kx − ky + kz )]
+2 cos[(a/2)(kx + ky − kz )] + 2 cos[(a/2)(kx − ky − kz )]
= 8 cos(kx a/2) cos(ky a/2) cos(ky a/2)
X
~ ~ bcc
eik·Rnnn = eiakx + e−iakx + eiaky + e−iaky + eiakz + e−iakz
(7)
~ bcc
R
nnn
= 2 cos(kx a) + 2 cos(ky a) + 2 cos(kz a)
(b) case of f cc
12 n.n.:
a
a
a
a
2 (+1, 0, ±1) , 2 (0, +1, ±1) , 2 (−1, 0, ±1) , 2 (0, −1, ±1) ,
a
a
a
a
2 (+1, +1, 0) , 2 (+1, −1, −) , 2 (−1, +1, 0) , 2 (−1, −1, 0) ;
6 n.n.n.:
a(±1, 0, 0) , a(0, ±1, 0) , a(0, 0, ±1); (same as bcc)
X
~ ~ f cc
eik·Rnn = 2 cos[(a/2)(kx + ky )] + 2 cos[(a/2)(kx − ky )]
f cc
~ nn
R
+2 cos[(a/2)(kx + kz )] + 2 cos[(a/2)(kx − kz )]
(8)
+2 cos[(a/2)(ky + kz )] + 2 cos[(a/2)(ky − kz )]
= 4[cos(kx a/2) cos(ky a/2) + cos(ky a/2) cos(kz a/2) + cos(kz a/2) cos(kx a/2)]
X
f cc
~ nnn
R
~ ~ f cc
eik·Rnnn =
X
~ bcc
R
nnn
~ ~ bcc
eik·Rnnn
(9)
P5 [25 points ] Bilayer graphene consists of two graphene layers stacked on top of each other in a Bernal manner,
i.e., such that only one of the two inequivalent lattice sites, e.g., A, has the nearest neighbor along the
vertical (see Fig. 1). The tight-binding Hamiltonian in the nearest-neighbor-hopping approximation
reads
X †
X †
Ĥ = −γ0
âm,R b̂m,R0 − γ1
â1,R â2,R0 + H.c.
(10)
m,R,R0
R,R0
where âm and b̂m with m = 1, 2 are the electron annihilation operators for sites A and B in the bottom
(1) and top (2) planes, correspondingly, sum over R and R0 includes only the nearest neighbors, and
parameters γ0 and γ1 are real (FYI, γ0 ≈ 3 eV and γ1 ≈ 0.4 eV). The planes are separated by distance
c (≈ 3Å). Find the energy spectrum for arbitrary k in the Brillouin zone. Plot the isonergetic contours.
Analyze the spectrum near the K and K 0 points.
Solution
The unit cell of a graphene bilayer consists of four atoms. The electron state is a 4-spinor


a1
 b1 

|ψi = 
 a2  ,
b2
where a1 . . . b2 are the amplitudes of finding an electron
Hamiltonian can be written as a 4 × 4 matrix

0
S −γ1
 S∗ 0
0
Ĥ = 
 −γ1 0
0
0
0 S∗
on site A1 . . . B2 . In basis of |ψi states , the

0
0 
,
S 
0
(11)
where S is the matrix element connecting A and B sites in the same plane. The top left and bottom
right 2 × 2 blocks are the same as for a monolayer. Diagonalizing Hamiltonian in Eq. (11), we obtain
for the energy levels
s
r
γ 2 |S|2
3
γ12 + |S|2
γ14
±
+ 1
− |S|4 .
(12)
ε=±
2
4
2
4
Recall that |S| = (3/2)γ0 ak ≡ ~vF k near the K and K 0 points, where k is measured from either of
these two points. In the low-energy limit |S| γ1 , two out of the four branches in Eq. (12) approach
finite value (equal to γ1 ), whereas the other two vanish. Expanding these last two dispersions in Taylor
series, we find
ε = ±|S|2 /γ1 = ±(~2 vF2 /γ1 )k 2 ≡ ±~2 k 2 /2m∗ ,
where m∗ ≡ γ1 /2vF2 . Therefore, the low energy excitations in bilayer graphene are massive electrons
and holes. Degeneracy of the spectrum at K and K 0 points is not lifted by inter-layer coupling but the
functional form of the dispersion around the degeneracy point is modified from linear to quadratic.
Figure 1: Crystal structure of bilayer graphene, from A. H. Castro Neto, F. Guinea, N. M. R. Peres, K. S.
Novoselov, A. K. Geim, Rev. Mod. Phys. 81, 109 (2009).