Condensed Matter Physics

Text book: "Fundamentals of Solid Physics", by J.

Richard Christman

Reference: "Introduction to Solid State Physics", by

Charles Kittel

Assessment Weight

Continuous assessment (two tests) 40%

Examination 60%

http://ap.polyu.edu.hk/apacwong/

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 : P1 and are the basis vectors
P2
which connect the lattice point and the two
atoms consisting thebasis.
P
1
P2

 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
primitive fundamental lattice vectors  for  a
crystal. Fundamental lattice vectors ( a ' and b ' )
can also be non-primitive: 
 a'
b'

 A lattice vector is a vector presenting the

position
 ofa lattice
 point.

R = n 1 a + n2 b + n3 c ,

R links two arbitrary lattice points, where n1
and n2 are two integers.

4
 
 A unit cell is a parallelepiped
  formed
 by a and
b (2-D case), or a and b and c (3-D case). A
crystal is constructed by translating the unit cell
along the directions of the lattice vectors.
 
a B
b

b  
A
c 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 non-
primitive unit cell is formed by a set of non-
primitive 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 P1 and P2 ,
and those
 of
 the non-primitive
 basis are specified
by P1 P2 P3 and P4 .

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Example : Cesium chloride CsCl
Primitive
 fundamental lattice vectors
R = n1a x̂ + n2 a ŷ + n3 a ẑ
Basis vectors : pCl = 0, pCs = a/2 ( x̂ + ŷ + ẑ
).

Cl

a ẑ
Cs
a ŷ
a x̂

Example : Equivalent choices of primitive unit cells.

Parallelepiped
(unit cell)
Unit cell
(parallelogram)
Another
possible choice

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 The volume of a unit cell formed
  by a set of
fundamental lattice vectors a , b and c is :
= base area  height
  ) (c cos  )
= (a b sin
=  a b ˆ
   n c
= ( a b )  c = 


c
 b


a
 
 that aŷ=aẑx̂ , b =a and
Exercise 2: (a) Show ŷ
c =a ( x̂ + + ) can also be selected
as the fundamental lattice vectors of
CsCl.
(b) Show that the volume of unit cell is 
= a3.

Exercise 3: A unit cell of zinc (atomic mass = 65.38),

contains two atoms. The rhombus base
has a = 0.266 nm and  = 60o. The height
of the unit cell is c = 0.495 nm. Find the
density of Zn.

2. Crystal Symmetry
7
 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 = n1 a + n2 b +n3 c ,
where n1, n2 and n3 are integers.

T1

T2

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 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 = 2r sin(/2) = a, and
A'B' = 2r sin(2  3)/2 = sa.
A'B'/ AB = s = sin(  3/2)/ sin(/2)
= sin (3/2)/sin(/2)
= [3sin(/2)  4sin3(/2)]/sin(/2)
= 3  4 sin 2(/2)
sin 2(/2) = (3  s)/4  1, and must be positive.

A’ sa B’
r sin/2 r sin/2
r X r r
  /2 /2

A a B

Allowed value of s sin2(/2) 

1 1  (2-fold)
0 3/4 2/3 (3-fold)
1 1/2 /2 (4-fold)
2 1/4 /3 (6-fold)
3 0 0 or 2
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 Mirror symmetry
Operator : mirror
Operation : reflection operation

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

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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

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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

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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

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 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 x̂ , a ŷ
and a ẑ being mutually perpendicular
Volume of a BCC unit cell = a3
The number of lattice points/BCC unit cell = 2
Volume  = a3/2
No. of nearest neighbors = 8
No. of second nearest neighbors = 6
Nearest neighbor distance = 3a 2  1/2 = 0.866 a
Packing fraction (not closest)
= 2 (4R3/3)/a3 
= (2 4/3)( 3a2 /4) /a
3 3
 c
= 3 /8 a
= 0.68

b
 
Primitive lattice vectors a = a( x̂ +  ), b = a(
ŷ
1 1
2
ẑ 2

x̂ + ŷ + ẑ ), and c = a( x̂  ŷ + ẑ ) form a
1
2

rhombohedral unit cell.

Reconsider the questions for the primitive lattice.

14
Exercise :
FCC is conventional. A set of primitive fundamental
lattice vectors is a( x̂ + ŷ ), a( ŷ + ẑ ) and a( x̂ + ẑ
1 1 1
2 2 2

). 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
expressed
 as:    
V1 = na a + nb b and V 2 = ma a + mb b ,

where n's and m's are integers.

 An arbitrary position on the plane is expressed as:
   
V = V2 + (V1 V2 )
 
= [ma+ (na ma)] a + [mb+ (nb mb)] b
where  is real.
 
Position I is [ma+ I(na ma)] a + 0 b ,
so mb+ I(nb mb) = 0,
and I = mb/(nb mb).
The intercept is:
x = [ma(nb mb)  mb(na ma)]/(nb mb)
= n/h rational
 Similarly, the intercept at position II is:
y = [ma( nb mb)  mb( na ma)]/( ma  na)
= n/k rational 
a 
b V2
I
-

II

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 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)

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5. Normal of a lattice plane
  
 A lattice plane intercepts the axes at x a , y b and z c
 A normal
 vector of the plane is obtained as:
(y b  x a ) ( z c  x a )
   

= yz ( b c)  yx( 
b a ) xz( a  c)
= xyz( b c/x + c a/y + a b/z)
= A (h b  c + k c  a + l a  b )

  
r1  r2
zc
 
r1 c
  

a b yb
xa

r2

6. Reciprocal lattice vectors

 Fundamental
  
reciprocal lattice vectors are :

A  c /,
 = 2 b
B = 2 c  a/,
 
C = 2 a b /.  
 
 Define a reciprocal lattice vector G = h A +k B +l C
  is 
, which to the (hkl) plane.
   
 a A = b 
 B 
= c  C
 = 2
     
a B = aC = b  A = b C = c  A = c  B
=0

18
Exercise : Find the Miller indices of the lattice planes
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

B.C.C. lattice are

A =2( x̂ + ŷ )/a,

B =2( ŷ + ẑ )/a,

C =2( x̂ + ẑ )/a.

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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 = x a  n̂   
   

= x a (hA +kB +lC )/ h A +k B +l C 
= 2/ hA +k B +l C 
= 2/  G 

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8. Close-Packed Structures
 Metal atoms are very often to show the closest
packing, because metallic bond has weak
preferential orientation.
 The 1st, 2nd and 3rd 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)

C
B
A

 ABCABC stacking gives an F.C.C. lattice (in

fact, they are equivalent).

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Closest packing of spheres has a packing faction of
F = 4  (4R3/3)/
= 4(4/3)( 2 a/4)3/a3
= 0.7405%

2a 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

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.

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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
111
contains two atoms at 000 and 4 4 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
1/4 3/4
3/4 1/4
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

1/2 1/8
c

24