Post on 01-Jan-2016
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 therequirement of more than one symmetryelement (i.e. when these symmetry elementsare combined!).
Start with the translation T
Add a rotation A
A
lattice point
lattice point
latticepoint
A T
TT
T: scalar
Tp
T
A translation vector connecting twolattice points! It must be some integerof or we contradicted the basicAssumption of our construction.
T
p: integer
Therefore, is not arbitrary! The basic constrain has to be met!
Combination of translation with rotation:
T
T T
tcos tcos
b
To be consistent with theoriginal translation t:
pTb p must be integer cos21cos2 ppTTTb
p1cos2 p cos n (= 2/) b
-1.5 -- -- -- -1 2 3T-0.5 2/3 3 2T 0 /2 4 T 0.5 /3 6 0 1 0 (1) -T 1.5 -- -- --
p > 4 orP < -2:no solution
T
T T
A A’
B’B
43210-1-2
Allowable rotationalsymmetries are 1, 2,3, 4 and 6.
Look at the case of p = 2
= 120o
TTp
2
1T
2T
21 TT
o21 120 TT
angle
Look at the case of p = 1
n = 3; 3-fold
n = 4; 4-fold
= 90o
TTp
21 TT
o21 90 TT
1T
2T
3-fold lattice.
4-fold lattice.
Look at the case of p = 0
= 60o
TTp
0
1T
2T 21 TT
o21 60 TT
n = 6; 6-fold
Look at the case of p = 3 n = 2; 2-fold
TTp
3
Look at the case of p = -1 n = 1; 1-fold
1 2
TTp
1
Exactly the same as 3-fold lattice.
1-fold2-fold3-fold4-fold6-fold
Parallelogram21 TT
general21 TT
Hexagonal Net21 TT
o21 201 TT
Can accommodate1- and 2-foldrotational symmetries
Can accommodate3- and 6-foldrotational symmetriesSquare Net
21 TT
o21 90 TT
Can accommodate4-fold rotationalSymmetry!
Combination of mirror line with translation:
m
Unless
0.5T
centered rectangular
constrain
Or21 TT
o21 09 TT
Primitive cell
Rectangular
1T
2T
m
Lattice + symmetries of motif (point group) = plane group(5) (1, 2, 3, 4, 6, m, etc)
Parallelogram21 TT
general21 TT
Hexagonal Net21 TT
o21 201 TT
Square Net21 TT
o21 90 TT
(1)
(2)
(3)
21 TT
o21 09 TT
Double cell (2 lattice points)
Centered rectangular(4)
21 TT
o21 09 TT
Primitive cell
Rectangular(5)
Oblique
Rectangular
Centered rectangular
Square
Hexagonal
1, 2
m
m
4
3,6
Five kinds of latticeThe symmetry that thelattice point can accommodate
+
+
+
+
+
Plane group
3D: space group.
Group theory
We will show the concept of group!
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
Example: Group {1, -1}; rank 2rank (order) of the group = number of elements contained in a set.
1 -1
1 -11-1 1-1
Another Example: Group {1, -1, i, -i}; rank 4
http://en.wikipedia.org/wiki/Group_(mathematics)
We will show examples for point groups later!
n12346
m[ ][ ][ ][ ][ ]
In a point, there is no translation symmetry!
Therefore, consider 2D point group, we only considerrotation and mirror!
Put rotation symmetry and mirror together ?
Example:
m1
m2
R
R L
L
{1, 1, 2, A} group of rank 4
1 1 2 A
1
1
2
A
1 1 2 A
11 2A
1 12 A
112A
Abelian group: a.b=b.a
2mm: point group2 + m
(1)
(2)(3)
(4) 1 1: 11 2: 2
1 3: A
1 4: 1
3/53/43/23/ 1 AAAAA6-fold 21 A
A 1 is a subset of
2-fold axis
3/43/2 1 AA subgroup
3-fold axis
1 2
?12 LL
RChirality not changed:
T
Rotation is the right choice!
12 ||
?12 A
212 A Combination theorem
2
1 A 12
(1) (2)
(3)(4)
if
2mm
Show it is a group
1 1 2 A
1
1
2
A
1 1 2 A
11 2A
1 12 A
112A
Satisfy 3 postulates?
Rank 4
The number of motif in the pattern is exactly the same as therank (order) of the group!
Hermann and MauguinInternational notation
Rotation axis
n 1, 2, 3, 4, 6
Schonllies notation CnC1, C2, C3, C4, C6
C: cyclic group – all elements are “powers” of some basicOperation e.g. 4
2/23
2/2/32
2/2/ 1 AAAAAAA
http://en.wikipedia.org/wiki/Group_(mathematics)#Cyclic_groups
Notation:
Hermann and MauguinInternational notation
Mirror plane
m
Schonllies notation CS
Cnv : Rotational symmetry with mirror plane vertical to the rotation axis. E.g. 2mm – C2v .
2/A
1
?12/ A
(1) (2)
(3)
2
212/ A
4
S:C4v
HM: 4mmmm
m m
m
m
Only independent symmetry elements.
The rank of this group is ?
R
L
L
4 + m
1
213/ A
2
/6S:C6v
HM: 6mm
The rank of this group is 12!
1
(1) (2)
(3)
3/2A
(1)L
(2)R
(3)R
213/2 A
2
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 , , , , , , , , , ; Schonllies notation , , , , , , , , , .
10 2-D crystallographic point group
5 2-D lattices
2-D crystallographicspace group
1 2 3 4 6 m 2mm 3m 4mm 6mm
C1 C2 C3 C4 C6 Cs C2v C3v C4v C6v
Oblique
Primitive Rectangular
Centered rectangular
Square
Hexagonal
1, 2
m
m
4
3,6
Compatible with
Compatibility: 2mm, 3m, 4mm, 6mm
2mm m
m
Put mirror planes along the edgeof the cell.
m
m
Primitive RectangularCentered rectangular
m, 2mmCompatible with
Square 4, 4mmCompatible with
30o
T
m with m3T
||m with m3Hexagonal Compatible with
Red ones Blue ones
T
m with m3T
||m with m3
Hexagonal Compatible with 6mm
Oblique
Primitive Rectangular
Centered rectangular
Square
Hexagonal
,
,
,
,
, , ,
Compatible with
1 2
m 2mm
m 2mm
4 4mm
3 6 3m 6mm
General oblique net.
2T
1T
atoms
Type of lattice
Point group
Symbol used to describe the space group
P (for primitive) 1
Space group: p1
Upper case P for 3Dlower case p for 2D
Primitive oblique net + 2 = p2
2T
1T
A
A TB
(1)
(2) (3)
2T
1T
A
plane group: p2
p2
positions with symmetry the lattice point!
p2
General relation between new symmetry position generated bycombining rotation with translation
)2/tan(2/
x
T)2/cot()2/()2/tan(
2/ Txx
T
BAT
)2/cot()2/( Tx
at along the -bisector of T
/2/2
T
A A
(1)(2)
(3)
A2
T
x
/2
B
Question: what kind of symmetryoperation is required in order formotif (1) get to motif (3)?
A : (1) (2);
T
: (2) (3);
/2/2
A
12
(1)(2)
Could we always rotate /2 respect to thedashed line T!
You can always define it that way!
/2
4 + lattice
2T
1T
4 } { 22/32/ AAAA
2/A 1
|| ||
Correct?
Combination of A/2 with T
p4
2T
1T
21 TT
o21 90 TT
2/2/ BAT
2/)4/cot()2/( TTx at
p + 3 = p3
2T
1T
120o
3 }1 { 23/23/43/2 AAAA
Combination of A2 /3 with
3/23/2 BAT
32)3/cot()2/(
TTx
along the -bisector of
at
T
T
X2/3)2/( 22 TTTX
30o
3/23/2 BA
3/2B
mass center
X/3323
1
2
3
3
TTX
2T
1T
60o 60o
(1)
(2) (3)
(1)(2): A2/3;(2)(3): Translation T
(1)(3): B2/3;
2T
1T
60o 60o
21 TT
o21 201 TT
p + 6 = p6 p has to be hexagonal net as well!
2T
1T
6 }1 { 23/3/53/23/43/23/ AAAAAAAA
3-fold
2-fold
From 2-fold rotation
From 3-fold rotation
Combination of A /3 and A- /3 with T
3/3/ BAT
2/3)6/cot()2/( TTx at
2/3T
2/T
T
Combination of mirror symmetry with the translation!
m + p + c
p + m = pm
? T
(1) R
(2) L
(3) L T
@ 2/T
Independent mirror plane
^ is defined with respectto mirror line (plane)
c + m = cm
not an independent mirror plane!(lattice point!)
(1) (3)
(1) R
(2) L
(3) L
T ||T
T ||T
|||| )( TTTT
Glide plane with glide component
Two-step operation
m m m
cm
g g
p + g = pg possible?
g g
? T
(1) R
(2) L (3) L (1) (3)?
2/@ TT
2/@ )(||
|| TTTTT
General form:
)(||
|| TTTT
Remind: 2/@ T
c + g = cg possible?
g gm
2/)2/(@ )2/2/( 121 TTT
2T
1T
2222 2/2/ ;2/ TTTT
g gm
4/@ )2/2/( 12/2122
TTTTT
4/@ 1T
cg = cmrectangular net:
pm, pg, cm!
p + 2mm = p2mm
c + 2mm = c2mm
p (square) + 4mm
Red: p4.Blue: pm.
m
p4mm
Special case of a rectangular.
p (Hexagonal net) + 3m
p360o 60o
60o 60o
two ways centered rectangular net
m edge m || edge
p3
Cell edge|| Cell edge
p31mp3m1
p31mp3m1
3m3m
3m
3
Not yet done! Glide plane (or line).
p (Hexagonal net) + 6mm = p6 + p3m1 + p31m
Red BlueMirror line
Glide linep6mm
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 rotationaxes maintain undisturbed!
OK
Center rectangular net (c2mm)?
(m ok? ) (g ok? )
p2mg
X XTwo fold rotation symmetries+ offset mirror line
p2gg
Two fold rotation symmetries + offset glide line
Three different ways:
OK? X
p4gm
The same results
This is not C4gm! Because center position is not a lattice!
System (4) Lattice (5) Point group (10) Plane group (17)
Obliquea b
general
Rectangulara b, = 90o
Squarea = b, = 90o
Hexagonala = b,
= 120o
Primitiveparallelogram
Primitiveor centeredrectangular
Square
Hexagonalequilateral
2
pm pg cm
3
3m6
6mm
p3
p3m1 p31mp6
p6mm
4
4mm
p4
p4mm p4gm
m
2mmp2mm p2mg p2gg
c2mm
1
p2
p1
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) (3) Next two symbols indicate symmetries relative to one translation axis. The first letter (a) is m (mirror), g (glide), or 1 (none). The axis of the mirror or glide reflection main axis. The second letter (b) is m (mirror), g (glide), or 1 (none). The axis of the mirror or glide reflection is either || or tilted 180o/n (when n>2) from the main axis.
a
b
1, 2
a
b
3
60o
b
445o
a b
630o
a
Old notes
The short notation drops digits or an m that can 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°.
Shortfull
pmp1m1
pgp1g1
cmc1m1
pmmp2mm
pmgp2mg
pggp2gg
p4mp4mm
p6mp6mm
p1: p111p3: p311p4: p411p6: p611
p3m1
Old notes
Symbol forthe plane group
# of the particular planegroup in the set
Symbol for the group in 3D
Point group
Crystal system
p2 No. 2 p211 2 oblique
The information of the international X-ray table
Diagram
symmetry elementsin the net.
First line
origin at 2
(0 0)
x
y(x y)
(1-x 1-y)
) ( yx
) ( yx) ( yx General position(Unique for every
Plane or space group))-1 1( yx
=
Number/cell(rank of position)
2 e 1
Site symmetryAlways 1 for general position
) ( yx) ( yx
Special position(on a symmetry
Element)
1 d 2
1 c 2
1 b 2
1 a 2
)/21 2/1(
)1/2 0(
)0 /21(
)0 0(
x
Wyckoff symbol
byconvention
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 No. 6 p2mm mm Rectangular
,
,
,
,
,
,
,
,
Origin at 2mm
(x y)
) ( yx ) ( yx
) ( yx
2 m ) 2/1( y) 2/1( y2 m ) 0( y) 0( y2 m )/21 (x)/21 (x2 m )0 (x1 2mm )/21 2/1(1 2mm )0 2/1(1 2mm )/21 0(1 2mm )0 0(
)0 (x
4 1 ) ( yx) ( yx ) ( yx) ( yxihgfedcba
;
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pmg No. 7 p2mg mm Rectangular
,
,
(x y)
) ( yx
) 2
1( yx
) 2
1( yx
Origin at 2
,
,
4 1 ) ( yx) ( yx ) 2
1( yx)
2
1( yx
2 2 )0 0( )0 2
1(
Not an independentspecial position
(mirror)
2 m ) 4
1( y )
4
3( y
2 2 )2
1 0( )
2
1
2
1(
An independentspecial position
a
b
c
d
How about glide plane? Atoms do not coincide!Glide is never a candidate for a special position!
rank
Symmetryof the
equipoints
designation
yy
xx
1
1
Condition limitingpossible reflection(structure factor)
Old notes
0, 0
1, 0
0, 1
x, y
1-y, x1-x, 1-y
y, 1-x
4 d 1
0, 0
1, 0
0, 1
1/2, 0
2 c 2
0, 1/2
1, 1/2
01/2, 1
= 41/2
0, 0
1, 0
0, 1
1/2, 1/2
1 b 4
0, 0
1, 0
0, 1
1, 1
1 a 4
Old notes
41/4 = 1
Supplement
Does the crystallographic group abelian?Some yes, some no!
m
m
mm
m
m
m
(1)1
12/ A
2/1 A(2)
(3) (3)
(1)
(2)(3)
(3)
2
1(1) (2)
(3)
1A
A1
(3)
(3)(1)
(2)(3)
Commutative: a.b=b.a
Noncommutative group a.bb.a
1 1 2 A/2
1
1
2
A
Group: 4mm
A A3/2
A/2
A3/2
1
2
3
4
3 4
3
4
1
1
2
A
A/2
A3/2
3
4
1 2 A/2 A A3/23 4
1
24
A/2 A A3/2 2 3 4
A3/2 1 A/2 A 3 4 1
A A3/2 1 A/2 1
A/2 A A3/2 1 1 2 3
4 1 2 3 1A A3/2
3 4 1 2 A3/2 1 A/2
2 3 4 1 1 A/2 A
(1)
Ask yourself how to get (1) to the rest of position?