Transcript *** 1

2D Symmetry
(1.5 weeks)
From previous lecture, we know that, in 2D,
there are 3 basics symmetry elements:
Translation, mirror (reflection), and rotation.
What would happen to lattices that fulfill the
requirement of more than one symmetry
element (i.e. when these symmetry elements
are combined!).
Combination of translation with rotation:

T
Start with the translation
A
lattice
point
Add a rotation
lattice point


pT
T: scalar
T
A
T



lattice point
A translation vector connecting two
lattice
 points! It must be some integer
of T or we contradicted the basic
Assumption of our construction.

T
p: integer
T
Therefore,  is not arbitrary! The basic constrain has to be met!
A
B’
B
b
T
T
T
A

T
T
tcos
A’

tcos
T
To be consistent with the
original translation t:
b  pT
b  T  2T cos   pT  p  1  2 cos 
2 cos   1  p
p
cos

4
3
2
1
0
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
--

2/3
/2
/3
0
--
n (= 2/)
-2
3
4
6
 (1)
--
b
-3T
2T
T
0
-T
--
p must be integer
p > 4 or
P < -2:
no solution
Allowable rotational
symmetries are 1, 2,
3, 4 and 6.
Look at the
 case
 of p = 2
n = 3; 3-fold
pT  2T

=


T1  T2

T2
 
T1 T2  120o

T1
120o
angle
3-fold lattice.
Look at the
case of p = 1

n = 4; 4-fold
pT  T

 = 90o


T1  T2

T2

T1
4-fold lattice.
 
T1 T2  90o
Look at the
 case
 of p = 0
n = 6; 6-fold
pT  0T

T2

=


T1  T2

T1
60o
 
T1 T2  60o
Exactly the same as 3-fold lattice.
Look at the case of p = 3
n = 2; 2-fold


pT  3T
Look at the case
 of p = -1
pT  1T
1
2
n = 1; 1-fold
1-fold
2-fold
3-fold
4-fold
6-fold
Parallelogram
 


T1  T2
T1 T2  general
Hexagonal
Net
 


T1  T2
T1 T2  120
Square

Net 
T1  T2
o

T1 T2  90o
Can accommodate
1- and 2-fold
rotational symmetries
Can accommodate
3- and 6-fold
rotational symmetries
Can accommodate
4-fold rotational
Symmetry!
Combination of mirror line with translation:

T2

T1
constrain
m
m
Unless
Or
0.5T


T1  T2
 
T1 T2  90o
Primitive cell
centered rectangular
Rectangular
(1) Parallelogram
 


T1  T2
T1 T2  general
(2) Hexagonal
Net
 


T1  T2
T1 T2  120o
(3) Square

Net 
T1  T2

T1 T2  90o
(4) Centered rectangular
 


T1  T2
T1 T2  90o
(5) Rectangular 


T1  T2

T1 T2  90o
Double cell (2 lattice points)
Primitive cell
Lattice + symmetries of motif (point group) = plane group
(5)
(1, 2, 3, 4, 6, m, etc)
The symmetry that the
lattice point can accommodate
Five kinds of lattice
Oblique
+
1, 2
Rectangular
+
m
Centered rectangular
+
m
Square
+
4
Hexagonal
+
3,6
We will show the concept of group!
3D: space group.
Plane group
Group theory
Group theory: set of elements (things) for a law of combination
is defined and satisfies 3 postulates.
(1) the combination of any two elements is also a
member of the group;
(2) “Identity” (doing nothing) is also a member
of the group. “I” aI=Ia=a (a : an element)
(3) for element, an inverse exists. a; a-1
a . a-1 = I a-1. a = I
http://en.wikipedia.org/wiki/Group_(mathematics)
Example: Group {1, -1}; rank 2
rank (order) of the group = number of elements contained in a set.
1
-1
1 1
-1 -1
-1
1
Another Example:
Group {1, -1, i, -i}; rank 4
We will show examples for point groups later!
In a point, there is no translation symmetry!
Therefore, consider 2D point group, we only consider
rotation and mirror!
Put rotation symmetry and mirror together ?
n
1
2
3
4
6
m
[]
[]
[]
[]
[]
Example:
(4)
2+m
L
R
2mm: point group
1  1: 1
1  2: 2
1  3: A
1  4: 1
{1, 1, 2, A} group of rank 4
(1)
m2
(3) R
m1
L (2)
1 1
2
A
1 1
2
A
 1 1 1
A
2
2 2 A
1
1
1
1
1
A
A 2
Abelian group: a.b=b.a
6-fold 1 A / 3 A2 / 3 A A4 / 3 A5 / 3 
1
A  is a subset of
2-fold axis
1
A2 / 3 A4 / 3  subgroup
3-fold axis
1  A2
 
L
L

if  2 ||  1
 2 1  ?
R
1

Chirality not changed:
2

T
 2 1  A?
Rotation is the right choice!
 2 1  A2

(4)
Combination theorem
(3)
2
(1)
(2)
1
 2 1  A
2mm
Show it is a group
1 1
2
A
1 1
2
A
 1 1 1
A
2
2 2 A
1
1
1
1
1
A
A 2
Satisfy 3 postulates?
Rank 4
The number of motif in the pattern is exactly the same as the
rank (order) of the group!
Notation:
Rotation axis
Hermann and Mauguin
International notation
Schonllies notation
n
1, 2, 3, 4, 6
Cn
C1, C2, C3, C4, C6
C: cyclic group – all elements are “powers” of some basic
Operation e.g. A
A  A2
A  A3 1  A  A4

 /2

 /2
3 / 2
 /2
2
 /2

http://en.wikipedia.org/wiki/Group_(mathematics)#Cyclic_groups
Mirror plane
Hermann and Mauguin
International notation
m
Schonllies notation
CS
Cnv : Rotational symmetry with mirror plane vertical to
the rotation axis.
E.g. 2mm – C2v .
(3)
A / 2
R
(1)
1 (2)
A / 2 1  ?
L
A / 2 1   2

4
2
L
m
m
4+m
The rank of this group is ?
m
m
S:C4v
HM: 4mmmm
Only independent symmetry elements.
A / 3 1   2
(3)
2
(1) 1 (2)
A2 / 3
(1) 1 (2)
L
R
(3)
R
/6
The rank of this group is 12!
A2 / 3  1   2
2
S:C6v
HM: 6mm
S:C3v
HM: 3mm (correct?)
The rank of this group is 6!
2 is not independent of 1.
HM (international notation): 3m
So far we have shown 10 point group or specifically 10 2-D
crystallographic point group.
HM notation
1 , 2 , 3 , 4 , 6 , m , 2mm , 3m , 4mm , 6mm ;
Schonllies notation
C1 , C2 , C3 , C4 , C6 , Cs , C2v , C3v , C4v , C6v .
10 2-D crystallographic point group
5 2-D lattices
2-D crystallographic
space group
Oblique
1, 2
Primitive Rectangular
Centered rectangular
Square
Hexagonal
m
Compatible with
m
4
3,6
Compatibility: 2mm, 3m, 4mm, 6mm
2mm
m
Put mirror planes along the edge
of the cell.
m
m
m
Primitive Rectangular
Centered rectangular
Compatible with
Square
m, 2mm
Compatible with
4, 4mm
30o
Hexagonal
Compatible with

3m with m T
3m with m || T

Red ones 3m with m T
Blue ones 3m with m || T
Hexagonal
Compatible with
6mm
Oblique
1 , 2
Primitive Rectangular
m , 2mm
Centered rectangular
Square
Hexagonal
Compatible with
m , 2mm
4 , 4mm
3 , 6 , 3m , 6mm
General oblique net.

T2
atoms

T1
Symbol used to describe the space group
Space group: p1
Type of
lattice
Point
group
P (for primitive) 1
Upper case P for 3D
lower case p for 2D
Primitive oblique net + 2 = p2
A

T2
A

T1
A
(1)
(2)

T2

T1
plane group: p2
B

T
(3)
p2
positions with symmetry  the lattice point!
p2
General relation between new symmetry position generated by
combining rotation with translation
B
/2 /2
(2)
(3) x
(1)
A A

T T A
2
T /2
 tan( / 2)
x

T  A  B
A : (1)  (2);
/2

T : (2)  (3);
Question: what kind of symmetry
operation is required in order for
motif (1) get to motif (3)?
T /2
 tan(  / 2)  x  (T / 2) cot(  / 2)
x
x  (T / 2) cot( / 2)

at along the -bisector of T
2
1
/2 /2
(2)
Could we always rotate /2 respect to the
dashed line  T!
(1)
A

You can always define it that way!
/2
p4
4 + lattice
4  {A / 2 A A3 / 2 A2 }

T2

T1
||
||
A / 2
1
Correct?

Combination of A/2 with T

T  A / 2  B / 2

T2
at x  (T / 2) cot( / 4)  T / 2

T1


T1  T2
 
T1 T2  90o
p + 3 = p3
3  {A2 / 3 A4 / 3  A2 / 3 A2  1}

T2
120o

T1
Combination of A2 /3 with
B 2 / 3

T
T  A2 / 3  B2 / 3 at x  (T / 2) cot( / 3) 

2 3
along the -bisector of T
T
X  T 2  (T / 2) 2  3T / 2
X
mass center
30o
X
3T 1
T


A2 / 3 B2 / 3
X/3
3
2 3 2 3

T2

T1
(2)
60o
60o
(3)
(1)(2): A2/3;

(2)(3): Translation T
(1)(3): B2/3;
(1)

T2
60o

T1
60o


T1  T2
 
T1 T2  120o
p + 6 = p6
p has to be hexagonal net as well!
2-fold
6  {A / 3 A2 / 3 A A4 / 3  A2 / 3 A5 / 3  A / 3 A2  1}
3-fold 
T2
From 3-fold rotation

T1
From 2-fold rotation

Combination of A /3 and A- /3 with T

T  A / 3  B / 3
at x  (T / 2) cot( / 6)   3T / 2
T
3T / 2
T /2
Combination of mirror symmetry with the translation!
m+p
+c
p + m = pm
(2) L
(1) R



 is defined with respect
T    ? to mirror line (plane)


T      @ T / 2
(3) L
Independent mirror plane
c + m = cm
(3) L
(2) L

not an independent mirror plane!
(lattice point!)
(1) R
(1)  (3)

 

T    (T  T|| )       T||

T
Glide plane with glide component 
Two-step operation
cm
m g m g m

T

T||

T||
p + g = pg possible?

T   ?
(3) L
(2) L
(1) R
g
(1)  (3)?


T    @T / 2
g
Remind:

 
T    (T  T|| )     T||
General form:

 

T     (T  T|| )      T|| @T / 2

@T / 2
c + g = cg possible?

T1

T2
g m
g
g m
g



(T1 / 2  T2 / 2)    @(T1 / 2) / 2




  T2 / 2;    T2 / 2  T2 / 2  T2



(T1 / 2  T2 / 2)   T2 / 2   T2 @T1 / 4

  @T1 / 4
cg = cm
rectangular net:
pm, pg, cm!
p + 2mm = p2mm


c + 2mm = c2mm


p (square) + 4mm
p4mm
Red: p4.
Blue: pm.
Special case of a
rectangular.
m
p (Hexagonal net) + 3m
p3
60o
60o
60o
60o
two ways
 centered rectangular net
m  edge
m || edge
p3
 Cell edge
|| Cell edge
p3m1
p31m
Not yet done! Glide plane (or line).
3m
3m
3m
3
p3m1
p31m
p (Hexagonal net) + 6mm
= p6 + p3m1 + p31m
Red
Blue
Mirror line
p6mm
Glide line
2mm
compatible with Rectangular!
mirror plane?
p2mm
What if the mirror line is not passing through the rotation axis?
For example this way? Why not?
How about this way? Why not?
Leave all the two fold rotation
axes maintain undisturbed!
 OK
Center rectangular net (c2mm)?
Two fold rotation symmetries
+ offset mirror line
(m ok? X ) (g ok? X )
p2mg
Two fold rotation symmetries + offset glide line
p2gg
OK? X
Three different ways:
The same results
p4gm
This is not C4gm! Because center position is not a lattice!
System (4)
Lattice (5)
Point group (10)
Plane group (17)
Oblique
ab
 general
Primitive
parallelogram
1
p1
2
p2
Primitive
or centered
rectangular
m
pm pg cm
2mm
p2mm p2mg p2gg
c2mm
4
p4
4mm
p4mm p4gm
3
p3
3m
6
p3m1 p31m
p6
p6mm
Rectangular
a  b,
 = 90o
Square
a = b,
 = 90o
Hexagonal
a = b,
 = 120o
Square
Hexagonal
equilateral
6mm
Old notes
Hermann-Mauguin Notation: pnab or cnab
(1) First letter: p for primitive cell, c for centered cell
(2) n: highest order of of rotational symmetry (1, 2, 3,
4, 6)
a
(3) Next two symbols indicate symmetries relative to
one translation axis.
The first letter (a) is m (mirror), g (glide), or 1
b
(none). The axis of the mirror or glide reflection
 main axis.
1, 2
The second letter (b) is m (mirror), g (glide), or 1
a
(none). The axis of the mirror or glide reflection
b
is either || or tilted 180o/n (when n>2) from the
60o
main axis.
a
a b
45o
3
4
6
b
30o
Old notes  The short notation drops digits or an m that can
p1: p111
p3: p311
p4: p411
p6: p611
p3m1
be deduced, so long as that leaves no confusion
with another group. E.g.
p2 (p211): Primitive cell, 2-fold rotation
symmetry, no mirrors or glide reflections.
p4g (p4gm): Primitive cell, 4-fold rotation, glide
reflection perpendicular to main axis, mirror
axis at 45°.
cmm (c2mm): Centred cell, 2-fold rotation,
mirror axes both perpendicular and parallel
to main axis.
p31m (p31m): Primitive cell, 3-fold rotation,
mirror axis at 60°.
Short pm
pg
cm pmm pmg
full p1m1 p1g1 c1m1 p2mm p2mg
pgg
p2gg
p4m
p6m
p4mm p6mm
The information of the international X-ray table
First line
Symbol for
the plane group
p2
Symbol for the group in 3D
Crystal system
No. 2
# of the particular plane
group in the set
p211
2
oblique
Point group
Diagram
symmetry elements
in the net.
origin at 2
(x y)
(0 0)
y
(x y)
( x y)
(x y)
=
(1  x 1 - y )
(1-x 1-y)
x
Number/cell
(rank of position)
x
Special position
(on a symmetry
Element)
General position
(Unique for every
Plane or space group)
Site symmetry
Always 1 for general position
( x y)
(x y)
2
e
1
1
1
1
1
d
c
b
a
(1 / 2 1/2)
2
(1/2 0)
2
(0 1/2)
2
(0 0)
2
Wyckoff symbol
by
convention
 Notation for asymmetric used to represent point
group symmetry:
(a)
: Asymmetric unit in the plane of the page
(b) + : Asymmetric unit above the plane of the page
(c) : Asymmetric unit below the plane of the page
(d) , : Apostrophe indicating a left-handed
asymmetric unit. Clear circle indicating
right-handedness.
(e)  + : Two asymmetric units on top of each other
(f) , : Two asymmetric units on top of one another,
one left-handed and the other right-handed.
and , are mirror images of each other.
Old notes
Another example
pmm
(x y)
No. 6
,
,
( x y)
(x y)
(x y)
p2mm
4
2
2
2
2
1
1
1
1
Rectangular
,
,
,
,
mm
,
,
Origin at 2mm
i
1
m
h
m
g
m
f
m
e
2mm
d
2mm
c
2mm
b
2mm
a
( x y) ( x y ) ( x y )
(1 / 2 y) (1 / 2 y )
(0 y ) (0 y )
(x 1/2) (x 1/2)
(x 0) (x 0)
(1 / 2 1/2)
(1 / 2 0)
(0 1/2)
(0 0)
(x y)
；
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pmg
No. 7
p2mg
(x y)
1
(  x y)
12
(  x y) ,
2
,
(x y)
,
,
mm
Rectangular
An independent
special position
Not an independent
special position
(mirror)
Origin at 2
1
1
(

x
y
)
(
 x y)
( x y) ( x y )
d
4
1
2
2
1
3
( y) ( y)
c
2
m
4
4
1
11
)
(0 ) (
b
2
2
22
2
1
(
0
0
)
(
0)
2
2
a
2
How about glide plane? Atoms do not coincide!
Glide is never a candidate for a special position!
x  1 x
y  1 y
Old notes
Symmetry
of the
equipoints
rank
designation
Condition limiting
possible reflection
(structure factor)
0, 0
0, 1
0, 0
y, 1-x
x, y
0, 1
0, 1/2
01/2, 1
1/2, 0
1, 0
1-x, 1-y
1-y, x
1, 0
1, 1/2
= 41/2
2 c 2
4 d 1
Old notes
0, 0
0, 1
0, 0
0, 1
41/4 = 1
1/2, 1/2
1, 0
1, 0
1 b 4
1, 1
1 a 4
Supplement
Does the crystallographic group abelian?
Some yes, some no!
m
m
m
m
(3)
(2) m
(3)
m (2)
1 (1)
m
Noncommutative group
A / 2 1  (3)
1  A / 2  (3)
(1)
(3)
(3)
(2)
(1)
(1)
(2)
2
1
Commutative:
a.b=b.a
A 1  (3)
1  A  (3)
a.bb.a
Group: 4mm
1 1
1
1
1 1
4
2
2
3
3
 2 3 4
2
1 1 A/2 A A3/2
2 A3/2 1 A/2 A 3 4
3
3 A A3/2 1
4
4 A/2 A A3/2 1
A/2 A/2 4 1
A
A 3 4
A3/2 A3/2 2 3
3
4 A/2 A A3/2
4 A/2 A A3/2
A/2 4
1
(1) 
2
1
1  2
1 2
3
2 3 A A3/2 1
1 2 A3/2 1 A/2
4 1 1 A/2 A
Ask yourself how to get (1) to the rest of position?