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1. PHY 140A: Solid State Physics
Solution to Homework #2
TA: Xun Jia1
October 14, 2006
1Email: jiaxun@physics.ucla.edu

2. Fall 2006 Physics 140A c Xun Jia (October 14, 2006)
Problem #1
Prove that the reciprocal lattice for the reciprocal lattice is the original lattice.
(a). Prove that the reciprocal lattice primitive vectors (eqn 2.13 in Kittel) satisfy
b1 · (b2 × b3) =
(2π)3
a1 · (a2 × a3)
. (Hint: Write b1 (but not b2 or b3) in terms of
the ai, and use the orthogonality relations 2.14.)
(b). Suppose primitive vectors ci are constructed from the bi in the same manner
as the bi are constructed from the ai. Prove that these vectors are just the ai
themselves.
(c). Prove that the volume of a Bravais lattice primitive cell is V = |a1 · (a2 × a3)|,
where the ai are the three primitive vectors. (In conjunction with the result of
part (a), this establishes that the volume of the reciprocal lattice primitive cell
is (2π)3
/V .)
Solution:
Two useful identities regarding vector multiplication are:
a · (b × c) = b · (c × a) = c · (a × b)
a · (b × c) = (a · c)b − (a · b)c
(1)
where a, b, and c are three vectors in three dimensional space.
(a). write b1 in terms of the ai, and use the orthogonality relations 2.14, we have:
b1 · (b2 × b3) =
(2π)(a2 × a3) · (b2 × b3)
a1 · (a2 × a3)
=
(2π)(b2 × b3) · (a2 × a3)
a1 · (a2 × a3)
=
(2π)a2 · [a3 × (b2 × b3)]
a1 · (a2 × a3)
=
(2π)a2 · [(a3 · b3)b2 − (a3 · b2)b3]
a1 · (a2 × a3)
=
(2π)2
a2 · b2
a1 · (a2 × a3)
=
(2π)3
a2 · (a3 × a1)
[a1 · (a2 × a3)]2
=
(2π)3
a1 · (a2 × a3)
(2)
(b). Suppose primitive vectors ci are constructed from the bi in the same manner
as the bi are constructed from the ai, then c1 satisfies:
c1 =
(2π)b2 × b3
b1 · (b2 × b3)
=
(2π)3
(a3 × a1) × (a1 × a2)
[a1 · (a2 × a3)]2
(2π)3
a1 · (a2 × a3)
=
(a3 × a1) × (a1 × a2)
a1 · (a2 × a3)
=
[(a3 × a1) · a2]a1 − [(a3 × a1) · a1]a2
a1 · (a2 × a3)
=
[(a3 × a1) · a2]a1
a1 · (a2 × a3)
=
[(a2 × a3) · a1]a1
a1 · (a2 × a3)
= a1
(3)
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3. Fall 2006 Physics 140A c Xun Jia (October 14, 2006)
here we used the fact that (a3 × a1) · a1 = (a1 × a1) · a3 = 0 due to equation
(1). Similarly, we can proof that:
c2 =
(2π)b3 × b1
b1 · (b2 × b3)
= a2 (4)
c3 =
(2π)b1 × b2
b1 · (b2 × b3)
= a3 (5)
Therefore, the reciprocal lattice for the reciprocal lattice is the original lattice.
Figure 1: A general parallelepiped with edges of a, b, and c, where
S = a × b.
(c). As in Fig. 1, consider a general parallelepiped with edges of a, b, and c. The area
of its base parallelogram spanned by vectors a and b is A = |a||b| sin θ = |a×b|,
where θ is the angle between a and b. Note that S = a × b is perpendicular
to this base, the height corresponding to this base is thus h = c ·
S
|S|
, which
is just the projection of c onto the direction of S. therefore the volume of this
parallelepiped is:
V = Ah = c ·
a × b
|a × b|
|a × b| = |c · (a × b)| (6)
Thus, the volume of a Bravais lattice primitive cell spanned by primitive vectors
ai is V = |a1 ·(a2 ×a3)|. Moreover, from part (a), the volume of the reciprocal
lattice primitive cell is then (2π)3
/V .
Problem #2
Interplanar separateion.(Problem 2.1 in Kittel.) Consider a plane hkl in a crystal
lattice.
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4. Fall 2006 Physics 140A c Xun Jia (October 14, 2006)
(a). Prove that the reciprocal lattice vector G = hb1 + kb2 + lb3 is perpendicular
to this plane.
(b). Prove that the distance between two adjacent parallel planes of the lattice is
d(hkl) = 2π/|G|.
(c). Show for a simple cubic lattice that d2
= a2
/(h2
+ k2
+ l2
).
Solution:
(a). To prove that the reciprocal lattice vector G = hb1 +kb2 +lb3 is perpendicular
to this plane, it suffices to show that G is perpendicular to two nonparallel
vectors in this plane. For the plane (hkl), it intercepts axis a1, a2, and a3 at a
ratio 1
h
: 1
k
: 1
l
, thus the two vectors in the plane can be chosen as (1
h
a1 − 1
k
a2)
and (1
h
a1 − 1
l
a3). Obviously, they are not parallel to each other. From direct
calculation:
G · (
1
h
a1 −
1
k
a2) = (hb1 + kb2 + lb3) · (
1
h
a1 −
1
k
a2) = 0 (7)
G · (
1
h
a1 −
1
l
a3) = (hb1 + kb2 + lb3) · (
1
h
a1 −
1
l
a3) = 0 (8)
we know G is perpendicular to those two vectors, and hence perpendicular to
the plane.
(b). Among the indices hkl, at least one of them is nonzero, without lose generality,
h = 0. Since, by definition, this family of planes divide vector a1 into h parts
with equal length, then the vector R1 = 0 ends on one plane, and R2 = 1
h
a1
ends on an adjacent plane. Therefore, the projection of R2 − R1 = 1
h
a1 on to
the direction perpendicular to this family of planes will be the distance between
two neighboring planes. From part (a), G is perpendicular to these planes,
hence:
d(hkl) =
1
h
a1 ·
G
|G|
=
1
h
a1 ·
hb1 + kb2 + lb3
|G|
=
2π
|G|
(9)
(c). For simple cubic, the reciprocal lattice has:
b1 =
2π
a
x, b2 =
2π
a
y, b3 =
2π
a
z (10)
where x, y, and z are unit vectors along x, y, and z directions. Thus,
|G|2
= (hb1 + kb2 + lb3)2
=
(2π)2
a2
(h2
+ k2
+ l2
) (11)
and hence:
d(hkl)2
=
(2π)2
|G|2
=
a2
h2 + k2 + l2
(12)
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5. Fall 2006 Physics 140A c Xun Jia (October 14, 2006)
Problem #3
(a). Show that the density of lattice points (per unit area) in a lattice plane is d/v,
where v is the primitive cell volume and d the spacing between neighboring
planes in the family to which the given plane belongs.
(b). Prove that the lattice planes with the greatest density of points are the {111}
planes in a face-centered cubic Bravais lattice, and the {110} planes in a body-
centered cubic Bravais lattice. (This is most easily done by exploiting the
relation between families of lattice planes and reciprocal lattice vectors.)
Solution:
(a). Within each plane, the lattice is a 2-d lattice. Let the area of the primitive cell
for this 2-d lattice be A, then since v is volume of primitive cell for 3-d lattice,
it follows that A · d = v. Since every primitive cell contains exactly one lattice
points, the number density of lattice points on this plane ρ is:
ρ = 1/A = d/v (13)
(b). Once the type of a lattice is given, the volume of primitive cell v is determined.
Thus from part (a), the larger the interplane distance d is, the greater the
density of points ρ is. From Eqn. (9), the largest d corresponds to the family
of planes for which the length of G, a reciprocal lattice vector perpendicular to
them, is minimized. Thus, if we could find out a G0 in reciprocal space with
least length, then the real space lattice planes perpendicular to this G0 will
have largest density of lattice points.
• For fcc, choose:
a1 =
a
2
(y + z), a2 =
a
2
(z + x), a3 =
a
2
(x + y) (14)
then the reciprocal lattice has:
b1 =
2π
a
(y + z − x), b2 =
2π
a
(z + x − y), b3 =
2π
a
(x + y − z) (15)
Above equations indicate that the reciprocal lattice of fcc is bcc. Since the
shortest vector in bcc is the one from the corner of the cube to the body
center, without lose generality, take:
G0 = b1 =
2π
a
(y + z − x) (16)
then in the real space lattice, the lattice planes perpendicular to this G0
belong to family of {111}, i.e. the lattice planes with the greatest density
of points are the {111} planes in a face-centered cubic Bravais lattice.
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6. Fall 2006 Physics 140A c Xun Jia (October 14, 2006)
• For bcc, choose:
a1 =
a
2
(y + z − x), a2 =
a
2
(z + x − y), a3 =
a
2
(x + y − z) (17)
then the reciprocal lattice has:
b1 =
2π
a
(y + z), b2 =
2π
a
(z + x), b3 =
2π
a
(x + y) (18)
Above equations indicate that the reciprocal lattice of bcc is fcc. Since the
shortest vector in fcc is the one from the corner of the cube to the face
center, without lose generality, take:
G0 = b1 =
2π
a
(y + z) (19)
then in the real space lattice, the lattice planes perpendicular to this G0
belong to family of {110}, i.e. the lattice planes with the greatest density
of points are the {110} planes in a body-centered cubic Bravais lattice.
Remark: You can proof these in a mathematical way, say, write explicitly the
expression of |G| and minimize it.
Problem #4
Hexagonal space lattice.(Problem 2.2 in Kittel.) The primitive translation vectors
of the hexagonal space lattice may be taken as:
a1 =
√
3a
2
x +
a
2
y, a2 = −
√
3a
2
x +
a
2
y, a3 = cz (20)
(a). Show that the volume of the primitive cell is
√
3a2
c/2.
(b). Show that the primitive translations of the reciprocal lattice are:
b1 =
2π
√
3a
x +
2π
a
y, b2 = −
2π
√
3a
x +
2π
a
y, b3 =
2π
c
z (21)
so that the lattice is its own reciprocal, but with a rotation of axes.
(c). Describe and sketch the first Brillouin zone of the hexagonal space lattice.
Solution:
(a). By direct calculation, the volume of the primitive cell is:
V = |a1 · (a2 × a3)| = (
√
3a
2
x +
a
2
y) · [(−
√
3a
2
x +
a
2
y) × (
2π
c
z)] =
√
3a2
c
2
(22)
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7. Fall 2006 Physics 140A c Xun Jia (October 14, 2006)
(b). From Eqn. (2.13) in Kittel, we can obtain that:
b1 =
(2π)a2 × a3
a1 · (a2 × a3)
=
2π
√
3a
x +
2π
a
y
b2 =
(2π)a3 × a1
a1 · (a2 × a3)
= −
2π
√
3a
x +
2π
a
y
b3 =
(2π)a1 × a2
a1 · (a2 × a3)
=
2π
c
z
(23)
a3 and b3 are both along z axis; in the xy plane, the lattice and the reciprocal
lattice are as in Fig. 2, obviously, the reciprocal lattice is also a hexagonal
lattice, but with a rotation about z axis by an angle of 30◦
.
(a) (b)
Figure 2: (a) xy plane of hexagonal lattice. (b)xy plane of the recip-
rocal lattice for hexagonal lattice. Shaded area indicates the
Brillouin zone.
(c). The first Brillion zone of the hexagonal lattice is also a hexagonal structure.
The cross section of the Brillouin zone in xy plane is illustrated by the shaded
area in Fig. 2(b). Sweep this shaded area along z axis from z = −
π
c
to z =
π
c
,
we get the whole first Brillion zone.
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