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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；

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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?