Reciprocal Lattice Fcc Bcc Sc

Brillouin Zones Physics 3P41 Chris Wiebe

Direct space to reciprocal space

ai • a j * = 2πδ  ij Real (direct) space

Reciprocal 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 a1 • ∆k = 2πν1

a2 • ∆k = 2πν2

a3 • ∆k = 2πν3

(Laue Equations)

• These equations have a simple interpretation: for a Bragg reflection,  ∆k must lie on a certain cone about the direction of a 1 (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.

Single crystal diffraction Ewald sphere

k k’

Bragg reflections

Each time  ∆k = G, a reciprocal lattice vector, you get a Bragg reflection. This is a point of  intensity at some 2 θ angle, and some Φ angle in space

 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’  ∆G

k

 Another picture of powder x-ray diffraction k’  ∆G

2θ k Whenever Bragg’s law is satisfied (2dsin θ = 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 Zones •

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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: k • (1/2 G) = (1/2 G)2

Point D in reciprocal space ½ GD

Wigner-Seitz cell

Therefore, the Brillouin Zone exhibits all wavevectors, k, which can be Bragg-reflected by a crystal

The First Brillouin Zone  An example: Rectangular Lattice

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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: a1 = a x

a2 = a y

a3 = a z

• Using the definition of reciprocal lattice vectors: b1

=

2π 

a2 × a3 a1 • a2 × a3

b2

=

2π 

a3 × a1 a1 • a2 × a3

b3

=

a1 × a2

2π 

a1 • a2 × a3

• We get the following primitive translation vectors of the reciprocal lattice: b1 = (2π/a)x

b2 = (2π/a)y

b3 = (2π/a)z

This is another cubic lattice of length 2 π/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)

2π/a

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

Reciprocal lattice to BCC lattice • The primitive translation vectors for the BCC lattice are: a1 = ½ a (x + y - z) a2 = ½ a (-x+y + z) a3 = ½ a (x - y + z)

• The volume of the primitive cell is ½ a 3 (2 pts./unit cell) • So, the primitive translation vectors in reciprocal space are: b1 = 2π/a (x + y) b2 = 2π/a (y + z) b3 = 2π/a (z + x)

• What lattice is this?

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: G = ν1 b1 + ν2 b2 + ν3 b3 = (2π/a)[(ν2 + ν3)x +(ν1 + ν3)y + (ν1 + ν2)z)] • The shortest G vectors are the following 12 vectors, where choices of sign are independent: (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 WignerSeitz cell of the FCC lattice). It has 12 sides (rhombic dodecahedron). • The volume of this cell in reciprocal space is 2(2 π/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: a1 = ½ a (x + y) a2 = ½ a (y + z) a3 = ½ a (z + x)

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The volume of the primitive cell is 1/4 a3 (4 pts./unit cell)

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So, the primitive translation vectors in reciprocal space are: b1 = (2π/a) (x + y - z) b2 = (2π/a) (-x+y + z) b3 = (2π/a) (x - y + z)

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This is, of course, the BCC lattice

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The volume of this cell is 4(2 π/a)3 in reciprocal space

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) (8 of these vectors)

and (2π/a)(+/-2x) (2π/a)(+/-2y) (2π/a)(+/-2z) (6 of these vectors)

4π/a

Summary • So, the reciprocal space for a simple cubic lattice is simple cubic, but the other cubic lattice (BCC, FCC) are more confusing:

The BCC and FCC lattices are Fourier  transforms of one another 

Lattice Real Space

Lattice k-space

bcc WS cell

fcc BZ

fcc WS cell

bcc BZ