Brillouin Zones

Physics 3P41Chris Wiebe

Direct space to reciprocal space

ijji aa πδ2* =•

Reciprocal spaceReal (direct) space

Note: The real space and reciprocal space vectors are not necessarily in the same direction

Laue Equations� Another way of expression the diffraction condition ∆k =

G is by the Laue equations, which can be derived by taking the scalar product of ∆k and G with a1, a2, and a3

� These equations have a simple interpretation: for a Bragg reflection, ∆k must lie on a certain cone about the direction of a1 (for example). Likewise, it must be on a cone for a2 and a3. Thus, a Bragg reflection which satisfies all three conditions must lie at the intersection of these three cones, which is at a point in reciprocal space.

a1 � ∆k = 2πν1 a2 � ∆k = 2πν2 a3 � ∆k = 2πν3

(Laue Equations)

Single crystal diffractionEwald sphere

kk�

Each time ∆k = G, a reciprocal

lattice vector, you get a Bragg reflection. This is a point of intensity at some 2θ angle, and some Φ angle in space

Bragg reflections

Actual single crystal diffraction patterns

Experimental setup

Powder vs. Single crystals� So, Bragg peaks are usually at a single spot in real space for single

crystals. For powder samples, however, which are composed of many tiny single crystals randomly oriented, the ∆G vectors exist on a cone of scattering (think of taking the Ewald circle, and rotating about k).

k

k�∆G

Another picture of powder x-ray diffraction

∆Gk�

2θ

k

Whenever Bragg�s law is satisfied (2dsinθ = nλ), we get a diffraction peak. Since there

is usually some crystallite which has it�s planes orienting in the proper direction, we get a cone of scattering about the

angle 2θ

Brillouin ZonesPoint D in reciprocal

space

½ GD

� A Brillouin Zone is defined as a Wigner-Seitz primitive cell in the reciprocal lattice.

� To find this, draw the reciprocal lattice. Then, use the same algorithm as for finding the Wigner-Seitz primitive cell in real space (draw vectors to all the nearest reciprocal lattice points, then bisect them. The resulting figure is your cell).

� The nice result of this is that it has a direct relation to the diffraction condition:

Wigner-Seitz cell

Therefore, the Brillouin Zone exhibits all wavevectors, k, which

can be Bragg-reflected by a crystalk � (1/2 G) = (1/2 G)2

The First Brillouin ZoneAn example: Rectangular Lattice

� The Zone we have drawn above using the Wigner-Seitz method is called the first Brillouin zone. The zone boundaries are k = +/- π/a (to make the total length to a side 2π/a in reciprocal space).

� The 1st Brillouin zone is the smallest volume entirely enclosed by the planes that are perpendicular bisectors of the reciprocal lattice vectors drawn from the origin.

� Usually, we don�t consider higher zones when we look at diffraction. However, they are of use in energy-band theory

Higher Order Brillouin Zones

Reciprocal lattice to SC lattice� The primitive translation vectors of any simple cubic

lattice are:

� Using the definition of reciprocal lattice vectors:

� We get the following primitive translation vectors of the reciprocal lattice:

321

213

321

132

321

321 2 2 2

aaaaab

aaaaab

aaaaab

ו×

=ו

×=

ו×

= πππ

a1 = a x a2 = a y a3 = a z

b1 = (2π/a)x b2 = (2π/a)y b3 = (2π/a)z

This is another cubic lattice of length 2π/a

Reciprocal lattice to SC lattice

The boundaries of the first Brillouin zone are the planes normal to the six reciprocal lattice vectors +/- b1, +/- b2, +/-b3 at their midpoints:

+/- (π/a)

The length of each side is 2π/a and the volume is (2π/a)3

2π/a

Reciprocal lattice to BCC lattice� The primitive translation vectors for the BCC lattice are:

� The volume of the primitive cell is ½ a3 (2 pts./unit cell)� So, the primitive translation vectors in reciprocal space

are:

� What lattice is this?

a1 = ½ a (x + y - z)a2 = ½ a (-x+y + z)a3 = ½ a (x - y + z)

b1 = 2π/a (x + y)b2 = 2π/a (y + z)b3 = 2π/a (z + x)

Reciprocal lattice to BCC lattice� This is the FCC lattice!� So, the fourier transform of the BCC lattice is the FCC

lattice (what do you expect for the FCC lattice, then?)� The general reciprocal lattice vector, for integrals ν1,ν2,

and ν3 is then:

� The shortest G vectors are the following 12 vectors, where choices of sign are independent:

G = ν1 b1 + ν2 b2 + ν3 b3= (2π/a)[(ν2 + ν3)x +(ν1 + ν3)y + (ν1 + ν2)z)]

(2π/a)(+/-y +/- z) (2π/a)(+/-x +/-z) (2π/a)(+/-x +/-y)

Reciprocal lattice to BCC lattice� This is the first Brillouin zone of

the BCC lattice (which has the same shape as the Wigner-Seitz cell of the FCC lattice). It has 12 sides (rhombic dodecahedron).

� The volume of this cell in reciprocal space is 2(2π/a)3, but it only contains one reciprocal lattice point.

� The vectors from the origin to the center of each face are:

(π/a)(+/-y +/- z) (π/a)(+/-x +/-z) (π/a)(+/-x +/-y)

Reciprocal lattice to the FCC lattice

� The primitive translation vectors for the FCC lattice are:

� The volume of the primitive cell is 1/4 a3 (4 pts./unit cell)� So, the primitive translation vectors in reciprocal space are:

� This is, of course, the BCC lattice� The volume of this cell is 4(2π/a)3 in reciprocal space

a1 = ½ a (x + y)a2 = ½ a (y + z)a3 = ½ a (z + x)

b1 = (2π/a) (x + y - z)b2 = (2π/a) (-x+y + z)b3 = (2π/a) (x - y + z)

Reciprocal space to the FCC lattice

� This is the first Brillouin zone of the FCC cubic lattice. It has 14 sides bound by:

(2π/a)(+/-x +/-y +/- z) 4π/a

(8 of these vectors)

and

(2π/a)(+/-2x)(2π/a)(+/-2y)(2π/a)(+/-2z)

(6 of these vectors)

Summary� So, the reciprocal space for a simple cubic lattice is

simple cubic, but the other cubic lattice (BCC, FCC) are more confusing:

bcc BZfcc WS cell

fcc BZbcc WS cell

Lattice k-space Lattice Real Space

The BCC and FCC

lattices are Fourier

transforms of one

another