PHY 140A: Solid State Physics. Solution to Homework #2

Solution to Homework #2

TA: 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_a 2· b2 a1· (a2× a3) = (2π) 3_a 2· (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)

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

(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 _{= a}2_/(h2_{+ k}2_{+ l}2_).

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 : k1 : 1l, thus the two vectors in the plane can be chosen as (h1a1− 1ka2)

and (1

ha1− 1la3). Obviously, they are not parallel to each other. From direct

calculation: G · (1 ha1− 1 ka2) = (hb1+ kb2+ lb3) · ( 1 ha1 − 1 ka2) = 0 (7) G · (1 ha1− 1 la3) = (hb1+ kb2+ lb3) · ( 1 ha1 − 1 la3) = 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 6= 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 = _h1a1

ends on an adjacent plane. Therefore, the projection of R2− R1 = 1_ha1 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 ha1· G |G| ¯ ¯ ¯ ¯ ¯= ¯ ¯ ¯ ¯ ¯ 1 ha1· 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 az (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 (h 2_{+ k}2_{+ l}2_{) (11)} and hence: d(hkl)2 ₌ (2π)2 |G|2 = a2 h2_{+ k}2 _{+ l}2 (12)

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.

• 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 2y, a2 = − √ 3a 2 x + a 2y, 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 √ 3ax + 2π a y, b2 = − 2π √ 3ax + 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 2y) · [(− √ 3a 2 x + a 2y) × ( 2π cz)] ¯ ¯ ¯ ¯ ¯= √ 3a2_c 2 (22)

(b). From Eqn. (2.13) in Kittel, we can obtain that: b1 = (2π)a2× a3 a1· (a2× a3) = √2π 3ax + 2π ay b2 = (2π)a3× a1 a1· (a2× a3) = −√2π 3ax + 2π ay b3 = (2π)a1× a2 a1· (a2× a3) = 2π cz (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,