Condensed Matter Physics
Text book: "Fundamentals of Solid Physics", by J.
Richard Christman
Reference: "Introduction to Solid State Physics", by
Charles Kittel
Weight Assessment
Continuous assessment (two tests)
Examination http://ap.polyu.edu.hk/apacwong/
40%
60%
1

Chapter 1 Crystal Lattices and Crystal Structures
1. Crystalline Periodicity

A crystal is a solid in which the atoms are arranged in a periodic pattern.

A basis is a basic replica consisting of a group of atoms, which when being translated successively can construct a crystal.

A lattice point is a geometric point used to represent a basis. It can be located at any position within a basis, but must be consistent with those of others.
 
2


Basis vectors are a group of vectors used for specifying the positions of the atoms in a basis.
Example :

P
1
and

P
2
are the basis vectors which connect the lattice point and the two atoms consisting the basis.

P
2
P
1

A lattice is an array of lattice points reflecting the locations of the basis. Periodicity refers to the periodic feature of the pattern.
lattice point
unit cell
 a
 b

R
B
A
3


Crystal structure

Lattice + Basis

The lattice planes (lines) of a specific orientation are equally spaced and parallel.

A fundamental lattice vector is a vector connecting two lattice points. A 3-D crystal has three independent fundamental lattice vectors denoted as a

and b and c

.

Primitive fundamental lattice vectors :
If a set of fundamental lattice vectors gives a basis which is the smallest in size, the vectors are named as the primitive fundamental lattice vectors. There are more than one set of
and b

' ) can also be non-primitive: b

'
 a '

A lattice vector is a vector presenting the position of a lattice point.

R
R

= n
1 a + n
2 b + n
3 c

,
links two arbitrary lattice points, where n
1
and n
2
are two integers.
4


A b unit cell is a parallelepiped formed by
(2-D case), or a and b and c
 a

and
(3-D case). A crystal is constructed by translating the unit cell along the directions of the lattice vectors. b

 a
B b
 c
 a

A
Exercise 1: Write down the positions of A and B .

A primitive unit cell is a unit cell of the smallest size. It is formed by a set of primitive fundamental lattice vectors
 a and b

. A nonprimitive unit cell is formed by a set of nonprimitive fundamental lattice vectors.
 Primitive basis is a basis of the smallest size.
The positions of the atoms of a primitive basis of the following example are specified P
1
and P
2
, and those of the non-primitive basis are specified by P
1
P
2
P
3 and P
4
.
5

Example : Cesium chloride CsCl
R = n
1 a x
ˆ + n

2 a y
ˆ
Basis vectors : p
Cl
+ n
3
 a z
ˆ
= 0, p
Cs
= a /2 ( x
ˆ + y
ˆ
+ z
ˆ ).
Cl a z
ˆ
Cs a y
ˆ a x
ˆ
Example : Equivalent choices of primitive unit cells.
Parallelepiped
(unit cell)
Unit cell
(parallelogram)
Another possible choice
6


The volume of a unit cell formed by a set of
 fundamental lattice vectors a , b
 and c

is :
= base area

height
= ( a b sin

)

( c cos

)
=

 a

= ( a

 b b

)

 n
 c
 c

=
 c

 b


 a

Exercise 2: (a) Show that a c

= a ( x
ˆ + y
= a x
ˆ , b

= a y and
+ z
ˆ ) can also be selected as the fundamental lattice vectors of
CsCl.
(b) Show that the volume of unit cell is

= a 3 .
Exercise 3: A unit cell of zinc (atomic mass = 65.38), contains two atoms. The rhombus base has a = 0.266 nm and

= 60 o . The height of the unit cell is c = 0.495 nm. Find the density of Zn.
7

2. Crystal Symmetry

Symmetry : If a lattice is not changed after an operation associated with an operator, the lattice is symmetric with respect to that operation.

All crystal has translational symmetry .
Operation : translation
Operator : translation vector T

= n
1 a

+ n
2 b

+ n
3 c

, where n
1
, n
2
and n
3
are integers.

T
1

T
2
8


Rotational symmetry
Operator : rotation axis
Operation : rotation by 2

/ n (rad) about the rotation axis. If the crystal is not changed after the rotation, it is said to have an n -fold symmetry. n can only be
2, 3, 4 and 6, but no more.
Proof : A and B are two nearest lattice points, separated by a . If the crystal is rotational symmetric about X , A moves to A ' after rotating +

. B moves to B' by rotating -

. A ' B ' // AB .
Translational symmetry ensures A ' B ' = sa , with s = an integer.
AB = 2 r sin(

/2) = a , and
A ' B ' = 2 r sin(2
 
3

)/2 = sa .
A ' B '/ AB = s = sin(
 
3

/2)/ sin(

/2)
= sin (3

/2)/sin(

/2)
= [3sin(

/2)

4sin 3 (

/2)]/sin(

/2)
= 3

4 sin 2 (

/2) sin 2 (

/2) = (3
 s )/4

1, and must be positiv e.
A’ sa B ’ r sin

/2 r sin

/2
r X r r
 

/2

/2

A a B
Allowed value of s

1
0
1
2
3 sin 2 (

/2)
1
3/4
1/2
1/4
0


(2-fold)

2

/3 (3-fold)

/2 (4-fold)

/3 (6-fold)

0 or 2

9


Mirror symmetry
Operator : mirror
Operation : reflection operation

Inversion symmetry
Operation : inversion ( r to
 r

)

180 o rotation + reflection by a mirror perpendicular to the rotation axis
10

3. Bravais Lattices (3-D crystals)

There are 14 lattice types (Bravais lattices), which are grouped into seven lattice systems.

P

primitive points at the corners

F

with face centers

I

with body center

C

with face centers at the top and bottom surfaces c
  b a

Cubic a= b = c

=

=

=

/2
P I F
Tegragonal a= b

c

=

=

=

/2
11

Orthorhombic a

b

c P C C

=

=

=

/2
I F
Monoclinic a

b

c

=

=

/2
 
P I
Triclinic a

b

c
    
/2
 
Hexagonal a = b

c

=

=

/2

= 2

/3
Trigonal
(Rhombohedral) a = b = c

=

=
  
/2
12

Two different lattices belonging to the same lattice system would have the same set of symmetry elements.
Example : Cubic system (S.C., F.C.C., B.C.C.)
3 tetrads 4 triads
6 diads 3 mirrors 6 mirrors
13


Conventional unit cell is a non-primitive unit cell of a lattice. It is more symmetric than the primitive one, and is more convenient to be used in practice.
Example :
BCC is conventional, with the lattice vectors a x
ˆ , a y
ˆ and a z
ˆ being mutually perpendicular
Volume of a BCC unit cell = a 3
The number of lattice points/BCC unit cell = 2
Volume

= a 3 /2
No. of nearest neighbors = 8
No. of second nearest neighbors = 6
Nearest neighbor distance = 3 a 2

1/2 = 0.866 a
Packing fraction (not closest)
= 2

(4

R 3 /3)/ a 3
= (2

4

/3)( 3 a 2 /4)
= 3

/8
3 / a 3
= 0.68
 a c
 b


Primitive lattice vectors a a (
 x
ˆ + y + z
ˆ ), and c

= 1
2
= 1 a ( x
ˆ + y
ˆ  z
ˆ ), b

= 1
2 2
a ( x
ˆ  y
ˆ
+ z
ˆ ) form a rhombohedral unit cell.
Reconsider the questions for the primitive lattice.
14

Exercise :
FCC is conventional. A set of primitive fundamental lattice vectors is 1
2 a ( x
ˆ + y
ˆ
), 1
2
a ( y + z
ˆ ) and 1
2
a ( x
ˆ + z
ˆ ).
Find:
(a) volume of a conventional unit cell,
(b) no. of lattice points in a conventional unit cell,
(c) volume of a primitive unit cell,
(d) no. of lattice points in a primitive unit cell,
(e) no. of nearest and second nearest neighbors,
(f) nearest and second nearest neighbor distances,
(g) packing fraction. c
 a
 b

15

4. Lattice planes and Miller Indices

Two lattice points on a 2-D lattice planes are
V
1
= n a a + n b b

and V

2

= m a a + m b b

,
where n 's and m 's are integers.

V

= V

2
+

( V

1
= [ m a
+

( n a

V

2

m
) where

is real.
Position I is [ m a
+
 so m and
 b
+

 a
)] a
I
( n b

m
I
=
 m
I
( n
+ [ m b
+

( n b

m a

m a

)] a b
) = 0, b
/( n b

m b
).
+ 0 b
 b
,
)] b

The intercept is: x = [ m a
( n b

m b
)

m b
( n a

m a
)]/( n b

m b
)
= n / h rational

Similarly, the intercept at position II is: y = [ m a
( n b

m b
)

m b
( n a

m a
)]/( m a

n a
)
= n / k rational a
 b
 
V
2

V
1
-

V
2

V
1
I
II
16


For 3-D crystal, if x , y and z are the intercepts of a lattice plane at the three crystal axes (in the units of the lengths of the fundamental lattice vectors)
x = n / h , y = n / k and z = n / l , n is integer
1/ x : 1/ y : 1/ z = h : k : l

If a primitive lattice is selected, h , k and l are integers with no common factor. ( hkl ) is defined as the Miller indices of a set of lattice planes of a specific orientation.
Example
(i) If the primitive lattice of a crystal is a simple cubic lattice, a (100) plane would contain real lattice points, but a (200) plane would not.
(ii) If the FCC lattice of a crystal is considered, since it is non-primitive, both (100) and (200) planes would contain lattice points.
(100)
17

5. Normal of a lattice plane

A lattice plane intercepts the axes at
 xa , y b

and zc


A normal vector of the plane is obtained as:
( y b

= yz ( b

xa

 c

)

( zc

)
= xyz (
 b

= A ( h b
 c

 c

/ x + c
+ k
 c




)
 yx ( b
xa
 a
 a
 a



)
)
 xz (
 a

/ y + a

+ l a

 b

 b

/ z )
) c
 z c

 r
1 a x a
 c

 b
 r
2

 yb
 r
1

 r
2
6. Reciprocal lattice vectors


A

B
C

= 2
 b
= 2
 c

= 2
 a



 a c

 b

/

,
/

,
/

.


Define a reciprocal lattice vector G = are :
 h A +
 k B + l C

,
 which is
 a
 a


A
B
= b

= a


to the (


C
B =

= b c


 hkl

A

C
) plane.
= 2
= b



C = c



A = c



B = 0
18

in parallel with both the lattice vectors a and c of:
(a) the conventional lattice of F.C.C.,
(b) the primitive lattice of F.C.C.
Exercise : Find the Miller indices of the lattice planes in parallel with both 3 a

+ c


and b of an arbitrary lattice.
Exercise : Show that the reciprocal lattice vectors of a
=2

( x
ˆ + y
ˆ )/ a ,
B
C
=2

( y
ˆ
+ z
ˆ )/ a ,
=2

( x
ˆ + z
ˆ )/ a .
19

7. Separation between lattice planes

A ( hkl ) plain closest to the origin (which is a lattice point) cuts the axes at x = 1/ h , y = 1/ k and z = 1/ l .
  
  
  
  
  
  

The shortest distance from the origin to the plane is d =
=
 xa xa
= 2


/


 ( h A
 n
ˆ h A
= 2

/

G



+ k B

+ k B

+ l C

+ l C


)/
 h A +
 k B + l

C

20

8. Close-Packed Structures

Metal atoms are very often to show the closest packing, because metallic bond has weak preferential orientation.

The 1 st , 2 nd and 3 rd layers of atoms are located to fill up the A, B and C sites.

Two stacking sequence are possible:
(i) ABCABC … 
F.C.C.
(ii) ABAB
… 
hexagonal close-packed (HCP)
B
C
A

ABCABC stacking gives an F.C.C. lattice (in fact, they are equivalent).
21

Closest packing of spheres has a packing faction of
F = 4

(4

R
3
/3)/

= 4(4

/3)( 2 a /4) 3 / a 3
= 0.7405%
2 a a

12 nearest neighbors for each atom

ABAB stacking sequence

HCP. Two atoms in one unit cell. 12 nearest neighbors. F = 0.7405
A D
B
A A C B a
Exercise : In an ideal HCP, a regular pyramid formed by four atoms is found. Show that c / a =
1.633.
22

9. Body-Centered Cubic Structures

All alkali elements (Li, Na, K, Rb, Cs, Fr) at room temperature show B.C.C. structures.

F = 0.68. Not the closest packing.
10. Covalent Structures

Mainly in solids with bonds of strong preferential orientation.

Smaller F value.
Example: Diamond structure including diamond, Si and Ge.
Conventional lattice structure is F.C.C.
Each atom is associated with and separated from another atom with ¼ of a body diagonal.
Primitive unit cell is rhombohedral. A basis contains two atoms at 000 and
1
4
1
4
1
4
.
Each atom is tetrahedrally bonded to other four, forming a regular pyramid.
0 1/2
3/4
1/4
1/2 0 1/2
3/4 1/4
1/4 3/4
1/2
23


Zinc blende structure is analogous to the diamond structure, where one F.C.C. is formed by one type of atoms, and the other F.C.C. is formed by another type of atoms.
Examples: CdS, InAs, GaAs, AlP, cubic BN.

Wurtzite structure
It consists of two types of atoms.
A primitive cell contains four atoms, two of each type.
Atoms of one type form a hexagonal lattice.
The two hexagonal lattices are separated by c/8 along the c-axis.
Examples: ZnO, BeO, MgTe, SiC.
0 5/8 c
1/2
1/8
24