C q b c q v 2 ? ? q b o v 2 c q b c q v 2 ? q ? q b o

c
q + b
= c
q
V
2( ǫ − ǫ 0
q + b
)
+ O ( V
2
) ,
c
q + b
= c
q
V
2( ǫ 0
q
− ǫ 0
q + b
)
+ O ( V
2
) ,
= − c
q
V 2 ma
4 π planckover2pi1 2 ( q +
π
a
)
+ O ( V
2
);
∴ c
q + b
= c
q
V
2
4 planckover2pi1 2
(
q 2 −
π 2
a 2
)
parenleftBig
1 −
qa
π
parenrightBig
+ O ( V
2
) . (1.b.18)
The story changes, however, for c
q − b
. It is not hard to jump a bit in the calculation and
see
c
q − b
= c
q
V
2( ǫ − ǫ
0
q − b
)
+ O ( V
2
) . (1.b.19)
Now, from our calculation of the eigenenergies at the Bragg plane we know that
ǫ
π/a
− ǫ
0
π/a − b
= ±
V
2
−
V
2
ma
16 π planckover2pi1
2
(
q +
π
a
)
+ O ( V
3
) , (1.b.20)
so we see
c
q − b
= c
q
V
2
parenleftbigg
±
V
2
−
V 2 ma
16 π planckover2pi1 2
(
q +
π
a
)
parenrightbigg
+ O ( V
2
) ,
= c
q
1
±
parenleftbigg
1 ∓ V 2 ma
8 π planckover2pi1 2
(
q +
π
a
)
parenrightbigg
+ O ( V
2
) ,
∴ c
q − b
= ± c
q
parenleftBigg
1 +
Vm
8 planckover2pi1 2
(
q 2 −
π 2
a 2
)
parenleftBig
qa
π
− 1
parenrightBig
parenrightBigg
+ O ( V
2
) . (1.b.21)
Putting all this together, we see
ψ
±
q =
π
a
( r ) = c
q
e
iqr
braceleftBigg
1 ± e
− ibr
+
Vm
4 planckover2pi1 2
(
q 2 −
π 2
a 2
)
parenleftBig
1 −
aq
π
parenrightBig
parenleftbigg
e
ibr
∓
e − ibr
2
parenrightbigg
bracerightBigg
,
so that
ψ
+
( r ) ∝ c
q
e
iqr
braceleftBigg
2 e
− i
br
2
cos
parenleftBig
πr
a
parenrightBig
+ i
Vm
2 planckover2pi1 2
(
q 2 −
π 2
a 2
)
parenleftBig
1 −
aq
π
parenrightBig
bracketleftbigg
sin
parenleftbigg
2 πr
a
parenrightbigg
− i
e
− ibr
4
bracketrightbigg
bracerightBigg
; (1.b.22)
and
ψ
−
( r ) ∝ c
q
e
iqr
braceleftBigg
2 ie
− i
br
2
sin
parenleftBig
πr
a
parenrightBig
+
Vm
2 planckover2pi1 2
(
q 2 −
π 2
a 2
)
parenleftBig
1 −
aq
π
parenrightBig
bracketleftbigg
cos
parenleftbigg
2 πr
a
parenrightbigg
−
e
− ibr
4
bracketrightbigg
bracerightBigg
. (1.b.23)
‘ ´ oπǫρ ’ ´ ǫδǫιπoι tieaccentlowercase ησαι

6 JACOB LEWIS BOURJAILY
-0.4
-0.2
0
0.2
0.4
-0.4
-0.2
0
0.2
0.4
-2
-1
0
1
2
0 4
Figure 2. The first Brillouin zone dispersion for a tight-binding model on a two-
dimensional square lattice.
Problem 2: Tight-Binding Model on a Square Lattice
Consider a tight-binding model on a square, two-dimensional square lattice (lattice spacing a ) with
on-site energy ǫ
0
and nearest-neighbour hopping matrix element t :
H =
summationdisplay
r
braceleftBig
ǫ
0
| r )( r | + t
bracketleftBig
| r )( r + a
ˆ
x | + | r )( r − a
ˆ
x | + | r )( r + a
ˆ
y | + | r )( r − a
ˆ
y |
bracketrightBigbracerightBig
.
a) We are to obtain the dispersion relation for this model.
Just for the sake of clearing up notation, our Bravais lattice here will be generated by
vectora
1
= a (1 , 0) and vectora
2
= a (0 , 1) which has the associated reciprocal lattice generated by
vector
b
1
=
2 π
a
(1 , 0) and
vector
b
2
=
2 π
a
(0 , 1). We will write all momenta in terms of the reciprocal
lattice, so vector q = q
1
vector
b
1
+ q
2
vector
b
2
. Using Bloch’s theorem it is quite easy to see that the
Hamiltonian of this system is given by
Hψ =
braceleftbig
ǫ
0
+ t
(
e
ivector q · vectora
1
+ e
− ivector q · vectora
1
+ e
ivector q · vectora
2
+ e
− ivector q · vectora
2
)bracerightbig
ψ, (2.a.1)
=
braceleftbig
ǫ
0
+ t
(
e
i 2 πq
1
+ e
− i 2 πq
1
+ e
i 2 πq
2
+ e
− i 2 πq
2
)bracerightbig
ψ, (2.a.2)
=
braceleftbig
ǫ
0
+ 2 t