chapter b1: crystal structures and symmetries
TRANSCRIPT
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Chapter B1: Crystal Structures and Symmetries
Georg RothInstitute of CrystallographyRWTH Aachen University
http://www.xtal.rwth-aachen.de http://www.frm2.tum.de
Outstation of the IfK at FRM II in GarchingThe „crystal palace“ in Aachen-Burtscheid
Symmetry Principles are widely used in Solid State Science
Mathematics, Physics, Chemistry, Crystallography, …
Why do Crystallographers use symmetry?
A: To understand the direction dependence of macroscopic physical properties: Anisotropy
B: To write down crystal structures in a concise manner
1 cm3 of matter consists of (roughly) 1023 atoms.Write down 3x1023 atom-coordinates?
Or better use the symmetry concept and write down only very few atoms …one to a few hundred at best…
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Outline:
• Crystal lattices The lattice concept: points, directions, planes
• Crystallographic coordinate systemsUnit cell, origin choice, 7 crystal systems, 14 Bravais-lattices
• Symmetry operations and symmetry elements Translation, rotation, roto-inversion, screw-axes, glide-planes
• Crystallographic point groups and space groups 32 point groups, symmetry directions, Hermann-Mauguin symbols, 230 space groups in 3D
• QuasicrystalsOrdered solids without translational symmetry
• Application of symmetry: Crystal structure of YBa2Cu3O7
Hexagonal symmetry
60°
60°
60°60°
60°
60°
Unit cell
a b
a ba2
b2
a3
b3
a4
b4
Choice of origin: point of highest symmetry (6-fold rotation point)
a b
2-dim. periodic pattern of snowflakes
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120° 180°
60°
2-dim symmetry: 6-fold, 3-fold & 2-fold rotation axes:
a b
a b
Basis vectors, translational symmetry
3a 2b
=3a+2b(+0c)
general translation vector :=ua+vb+wc; u, v, w Z (3dim.)
Motive
MotiveCrystal = Lattice
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Positions of atoms in the unit cell, expressed as fractional coordinates:Positional vector rj = xja + yjb + zjc (0 x, y, z < 1) (3D, atom label: j)
a br1 x1ay1b
r1 = x1a + y1b(0 x, y < 1)
How to describe the contents of the unit cell?…usually: atoms, here: snowflakes…
3 dim. periodicity of crystals crystal lattice
3 non-linear basis vectors a, b and c define a parallelepiped,
called unit cell of the crystal lattice
and the crystallographic coordinate system with its origin
any lattice point (point lattice) is given by a vector
= ua + vb + wc (u, v, w )
is also known as translation vector
angles between basis vectors:
angle (a,b) =
angle (b,c) =
angle (c,a) =
faces of unit cell:
face (a,b) = C
face (b,c) = A
face (c,a) = B
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Lattice points and lattice directions
according to the translation vector = ua + vb + wc (u, v, w )
• lattice points are indicated by the corresponding integers → uvw
• lattice directions or lattice rows by → [uvw] [square brackets]
[010]
[001]
[100]
[231]
[-2-3-1]or [ ]132
Lattice planes (crystal faces are special lattice planes)
3 non-collinear lattice points define a lattice plane:
• interceptions of a lattice plane with the axes X (a), Y (b) and Z (c): ma, nb, oc
• reciprocal values: 1/m, 1/n, 1/o
(with smallest common denominator no/mno, mo/mno, mn/mno)
• Miller indices: h = no, k = mo, l = mn
Miller indices (hkl) describe a set of equally spaced lattice planes.
I: 1 1 11/1 1/1 1/1
h=1 k=1 l=1 (111)
II: 1 2 2
1/1 1/2 1/2
h=2 k=1 l=1
(211)
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Projection of the lattice of graphite (hexagonal) down the Z-axis on the XY-plane to illustrate the decrease of the d(hkl)-spacing
between lattice planes (hk0) as their indices h and k increase:
(100)
d(100)
X
Y
Hexagonal crystal system a=b, c, ==90°, =120°
2
2
2
22
34
1)(
c
l
a
hkkhhkld
+=
++
For a crystal: Interference between waves scattered froma set of lattice planes (hkl)
Bragg equation for the reflection condition
2d(hkl)·sin(hkl) =
d(hkl): interplanar distance of a set of lattice planes (hkl)
(hkl): scattering angle, angle between the incident beamand the lattice plane (hkl)
: wavelength of the radiation
[hkl]
(hkl)
d(hkl)(hkl)
(hkl)
Lattice spacings are directly accessible by experiments:Diffraction of X-rays, neutrons, electrons, ...
= coherent elastic scattering
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Coordinate systems of crystals = 7 crystal systems in 3 dim.
Name of system Minimum symmetry Conventional unit cell
triclinic 1 or -1 a b c;
monoclinic one diad – 2 or m (‖Y) a b c; ==90°, >90°
orthorhombicthree mutually perpendicular diads – 2 or m (‖X, Y and Z)
a b c; ===90°
tetragonal one tetrad – 4 (‖Z) a = b c; ===90°
trigonal(hexagonal cell)
one triad – 3 or -3 (‖Z) a = b c; ==90°, =120°
hexagonal one hexad – 6 or -6 (‖Z) a = b c; ==90°, =120°
cubicfour triads – 3 or -3
(‖space diagonals of cube)a = b = c; ===90°
|| means: parallel to, diad means: 2-fold rotation, triad: 3-fold etc.
Are the 7 crystal systems in 3D, corresponding to the 7 significantly different, symmetry adapted
coordinate systems all we need?
Not quite:
There are good arguments (again based on symmetry) to define
7 additional lattices with more than one lattice point per unit cell
The 7 non primitive “centered” lattices
Altogether: The 14 Bravais-lattices
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The 14 Bravais lattices (represented by their unit cells)
triclinic P monoclinic Pmonoclinic axis‖c
monoclinic A(0,0,0 + 0, ½, ½)
orthorhombic P
orthorhombic I(0,0,0 + ½, ½, ½)
orthorhombic C(0,0,0 + ½, ½,0)
orthorhombic F(0,0,0 + ½, ½,0
½,0, ½ + 0, ½, ½)
tetragonal P•
tetragonal I Trigonal/hexagonal P hexagonal/rhombohedral
cubic P
cubic I cubic F
The 14 Bravais lattices (cont.)
14 Bravais lattices:
7 primitive lattices P forthe 7 crystal systems with onlyone lattice point per unit cell
+7 centered (multiple) lattices
A, B, C, I, R and F with 2, 3 and 4 lattice points per unit cell
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Diffraction geometry: Concept of the reciprocal lattice:
Reminder: The crystal lattice 'direct lattice' is composed of the set
of all lattice vectors generated by the linear combination of the basis
vectors a1, a2, a3 with coefficients u, v, and w (positive or negative
integers, incl. 0).
a = u a1 + v a2 + w a3.
The Fourier-transform which is occurring during a diffraction
experiment, transforms this direct lattice into the so called
’reciprocal lattice’, with basis vectors 1, 2, 3 and the integer
Miller indices h,k,l as the coefficients.
Diffracted intensity I() is only observed at the nodes of this
reciprocal lattice addressed by the vectors:
= h 1 + k 2 + l 3
of the reciprocal lattice.
*
3
1 a2||2
a3
a1
. d100
.
d001
* = * = = = 90°
* = 180°–
dhkl: lattice spacing for the set of lattice planes (hkl)
Direct and reciprocal basis
vectors satisfy the following
conditions:
1a1 = 2a2 = 3a3 = 1
This means that |ai| = 1 / |i|
and
1a2 = 1a3 = 2a1 = ... = 0
This means that each i is
perpendicular to aj and ak:
i = (aj ak)/Vc
with Vc = a1(a2a3) as the
volume of the direct cell
Example: Monoclinic cell a1, a2, a3, > 90°
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Crystallographic symmetry operationsAll types listed systematically:
1. Translations = ua + vb + wc (u, v, w )properties: no fixed point, shift of entire point lattice
2. Rotations: 1 (identity), 2 (rotation angle 180°), 3 (120°), 4 (90°), 6 (60°)properties: line of fixed points which is called the rotation axis
3. Rotoinversions (combination of n-fold rotations and inversion):(inversion), = m (reflection), , ,
properties: (exactly one) fixed point
4. Screw rotations nm
(combination of n-fold rotations with m/n· translations ‖ to rotation axis)properties: no fixed point
5. Glide reflections a, b, c, n, d(combination of reflection through a plane (glide plane) and translation by glide vectors a/2, b/2, c/2, (a + b)/2, ..., (a b c)/4 ‖ to this plane)properties: no fixed point
Rotations and rotoinversions are called point symmetry operations because they leave at least one point fixed.
1 2 3 4 6
Point symmetry operations
rotations rotoinversions
1=identity
2-fold = 180°-rotation 2-fold rotation combinedwith inversion = reflection
inversion
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1
2
3
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How many distinct combinations of point symmetry operations (rotation & roto-inversion) are possible in 3 D?
The answer is: 32
32 crystallographic point groups (“crystal classes”)
The point group symmetries determine the anisotropic (macroscopic) physical properties of crystals:
Mechanical, Electrical, Optical, Thermal, ...Tensorial Crystal Physics
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Hermann-Mauguin symbols of point groupse.g. m m 2
and symmetry directions
Basis vectors are conventionally chosen parallel to important symmetry directions of the crystal system.
Example: In the cubic lattice, a, b and c are parallel to the 4-fold rotation axes.
A maximum of 3 independent main symmetry directions(“Blickrichtungen”) are sufficient to describe the complete
point group symmetry of a crystal. These symmetry directions are specific for each of the 7 crystal
systems and are essential to understand theHermann-Mauguin symbol
symmetry directions in the orthorhombic lattice
a b c; = = = 90°
x
y
z
[100]
m2
[010]
m2
[001]
m2
x
y
z
x
y
z
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symmetry directions in the tetragonal lattice
a = b c; = = = 90°
x
y
z
[001]
m4
[100]
m2
[110]
m2
x
y
z
x
y
z
symmetry directions in the cubic lattice
a = b = c; = = = 90°
y
x
[100]
m4
[111]
3
[110]
m2
x
y
z
x
y
zz
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All possible combinations of point symmetry operations in 3 dim. lead to 32 crystallographic point groups (crystal classes)
Nomenclature:
Left: Schoenfliess-Symbol
Right: Hermann-Mauguin-Symbol
Plotted:„stereographicprojections“ : Point on upperhemisphere : Point on lower hemisphere
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Example: Orthorhombic system
Crystallographic point group: mm2
a
b
−
−
=
−
−
z
y
x
z
y
x
100
010
001symmetry operation
represented by
a rotation matrix
2 ‖ to [001]:
a
b
x
y x,y,z
-x,-y,z
ma
ma ⊥ a:
−
=
−
z
y
x
z
y
x
100
010
001
mb
mb ⊥ b:
−=
−
z
y
x
z
y
x
100
010
001
Crystallographic point groups which have a centre of symmetry 11 Laue classes
Crystal systems (7) Laue classes (11)
triclinic -1
monoclinic 1 2/m 1 = 2/m
orthorhombic 2/m 2/m 2/m = 2/m m m
tetragonal4/m
4/m 2/m 2/m = 4/m m m
trigonal-3
-3 2/m = -3 m
hexagonal6/m
6/m 2/m 2/m = 6/m m m
cubic2/m -3 = m -3
4/m -3 2/m = m -3 m
By diffraction methods, only the 11 Laue classes can be distinguishedand not all the 32 crystal classes.
The diffraction experiment – by its nature - always adds a centre of symmetry!
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Crystallographic symmetry operationsare isometric movements in crystals:
1. Translations = ua + vb + wc (u, v, w )properties: no fixed point, shift of entire point lattice
2. Rotations: 1 (identity), 2 (rotation angle 180°), 3 (120°), 4 (90°), 6 (60°)properties: line of fixed points which is called the rotation axis
3. Rotoinversions (combination of n-fold rotations and inversion):(inversion), = m (reflection), , ,
properties: exacly one fixed point
4. Screw rotations nm
(combination of n-fold rotations with m/n· translations ‖ to rotation axis)properties: no fixed point
5. Glide reflections a, b, c, n, d(combination of reflection through a plane (glide plane) and translation by glide vectors a/2, b/2, c/2, (a + b)/2, ..., (a b c)/4 ‖ to this plane)properties: no fixed point
1 2 3 4 6
120°
1/3
31 = 3 + 1/3
Screw rotations nm = n + m/n·
+ 42, 43 and 65
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Crystallographic symmetry operationsare isometric movements in crystals:
1. Translations = ua + vb + wc (u, v, w )properties: no fixed point, shift of entire point lattice
2. Rotations: 1 (identity), 2 (rotation angle 180°), 3 (120°), 4 (90°), 6 (60°)properties: line of fixed points which is called the rotation axis
3. Rotoinversions (combination of n-fold rotations and inversion):(inversion), = m (reflection), , ,
properties: exacly one fixed point
4. Screw rotations nm
(combination of n-fold rotations with m/n· translations ‖ to rotation axis)properties: no fixed point
5. Glide reflections a, b, c, n, d(combination of reflection through a plane (glide plane) and translation by glide vectors a/2, b/2, c/2, (a + b)/2, ..., (a b c)/4 ‖ to this plane)properties: no fixed point
1 2 3 4 6
m
reflection: mirror plane m ⊥ image plane
a
a/2
glide reflection: glide plane a ⊥ with glide vector a/2
a
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In 3 dimensions:
All possible combinations of point symmetries
of the 32 crystallographic point groups
with lattice translations (the 14 Bravais lattices)
and symmetry elements with a translational component (glide planes, screw axes) lead to exactly
230 crystallographic space groups
International Tables for Crystallography Vol. A(Theo Hahn, Ed.)
Conventional graphic symbols for symmetry elements:
• symmetry axes (a) perpendicular, (b) parallel, and (c) inclined to the image plane
• symmetry planes (d) perpendicular and (e) parallel to the image plane
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Summary (intermediate):
To describe the anisotropy of macroscopic physical properties, we need:• The point group symmetry (1 out of 32)
To describe a crystal structure, we need:• Choice of unit cell with basis vectors• The space group symmetry (1 out of 230)• The atomic positions in the unit cell
…see example (YBa2Cu3O7) below…
…just as a reminder:
The “magic” crystallographic numbers in 3D space:
7: Crystal systems (triclinic, monoclinic…)
14: Bravais lattices (P, C, A, B, I, R, F)
32: Crystal classes (point groups)
11: Laue classes (point groups with inversion center)
230: Space groups (all useful combinations of point group symmetry with translational symmetry)
…is this system closed and final…?...…not really!
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Quasicrystals:
Nobel-Prize in physics 2011 awarded to Dan Shechtman
“for the discovery of quasi crystals”
HoMgZn: Icosahedral quasi crystal
HoMgZn: Electron diffraction pattern taken along the -5 axis
Symmetry of quasi crystals? No translation lattice!
A new ordered ground state of solid matter:
Crystal: Quasi crystal:
• long range order
• (3D) periodic
• unit cell (repeat unit)
• translational-symmetry
• long range order
• aperiodic
• no unit cell (in 3D)
• no translational symmetry
2D-analog:
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Diffraction experiment Crystal structure
• Ceramic high-TC superconductor with TC = 92 K
• Technical application with liquid N2 cooling is possible
,
TCExample: YBa2Cu3O7-
Atom positions in YBa2Cu3O6.96
orthorhombic, space group P 2/m 2/m 2/m
a = 3.858 Å, b = 3.846 Å, c = 11.680 Å (at room temperature)
atom/ion multiplicity site symmetry x y z
Cu1/Cu2+ 1 2/m 2/m 2/m 0 0 0
Cu2/Cu2+ 2 m m 2 0 0 0.35513(4)
Y/Y3+ 1 2/m 2/m 2/m ½ ½ ½
Ba/Ba2+ 2 m m 2 ½ ½ 0.18420(6)
O1/O2- 2 m m 2 0 0 0.15863(5)
O2/O2- 2 m m 2 0 ½ 0.37831(2)
O3/O2- 2 m m 2 ½ 0 0.37631(2)
O4/O2- 1 2/m 2/m 2/m 0 ½ 0
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a
b
a
c
c
b
crystal system
P Bravais lattice, symmetry directions: 2/m‖[100], 2/m‖[010], 2/m‖[001]
3 different projectionsof the symmetry
Space group symmetry P m m moperating on a general point x,y,z
gives a total of 8 symmetryequivalent points
From: International Tables for Crystallography
Th. Hahn (ed.)
YBa2Cu3O7-
Ba
O3
O2
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YBa2Cu3O7- O1
O4
Cu2
Cu1
Y1
Thank you for your attention!
Chapter B1: Crystal Structures and Symmetries
Georg RothInstitute of CrystallographyRWTH Aachen University