Chapter 2_bruillion_Zone.pdf

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Chapter 2Reciprocal Lattice

Phys 175A

Dr. Ray Kwok

SJSU



Crystal Lattice

• Periodic f(r + T) = f(r) for any observable

functions such as electronic density,electric potential….etc. which means theyare all periodic functions because of thetranslational properties of the latticevectors.



Fourier Transform

Fourier series (1D): f(r) = ∑n An ei2πnr/a

where the period is a.

such that f(r + ua) = f(r); u is an integer.

Or write f(r) ≡ ∑G AG eiGr

where G = 2πn/a (for 1D lattice)

and AG = (1/a) ∫cell f(r) e –iGr

dr



1D Translational Invariance

f(r + T) = ∑G AG eiG(r+T) = ∑G AG eiGr eiGT

= f(r) = ∑G AG eiGr

i.e. eiGT = 1

For example, if T = ua (1D),

then GT = 2πun = 2π·(integer).and eiGT = 1.



3D Translational Invariance

For 3D lattice, eiG·T = 1

and T = u1a1 + u2a2 + u3a3 is the translationalvector or the lattice vectorDefine “reciprocal primitive vectors”

b1

= (2π /V) a2

x a3

b2 = (2π /V) a3 x a1

b3 = (2π /V) a1 x a2

where V is the volume of the primitive lattice cell

V = | a1 · a2 x a3|, such that bi· a j = 2πδij

and δij is the kronecker delta function.Note: {a}’s don’t have to be orthogonal.



Reciprocal Lattice Vectors

Define G = v1b1 + v2b2 + v3b3

where the v’s are integersso that

G·T = 2π ( u1v1 + u2v2 + u3v3) = 2π·(integer)

[because bi· a j = 2πδij]and eiG·T = 1

In other words, because of the translational

invariance property of the crystal, there exist aset of vector G such that G·T = 2π·(integer).This set of vector defines another set of latticepoint in the Reciprocal Space.



Reciprocal Vector Space

G has a unit of 1/length. Similar to the

wave-vector k in the plane waveexpression eik·r.

G has a meaning in Fourier transform, k-space, or momentum space.

It defines a set of lattice points in the k-

space.



Direct and Reciprocal Lattice

• For every crystal, there is a set of space lattice

(crystal lattice) – location of lattice points in realspace where atoms and molecules are.

• There is also a set of reciprocal lattice in the

momentum space (k-space) – something we seein diffraction measurement. However, there isno physical object present at the reciprocal

lattice sites.



Light Microscopy vs X-ray Crystallography

λ ~ 500 nm λ ~ 0.1 nm

see real space see reciprocal space



Tunneling Microscope vs X-Ray Diffraction

Reciprocal Lattice

(volume)“see” Direct Lattice(surface only)



d D

Diffraction maximum can be specified by the Miller indices [hkl]

Reciprocal nature of diffraction pattern

Bragg’s law: 2d sinθ = n λ d ∝ 1/sinθ



What do we learn from x-ray?

•Lattice parameters(Space Group)

•Symmetry (Point Group)

•Miller Index (h,k.l) foreach point

•Intensity (square ofstructural factor) of each

reflection



Properties of G – reciprocal lattice

each G is normal to a

lattice plane in realspace

G·T = 2πn. For a

fixed G and n, thereare many T vectorssatisfied this equation.

But they all lie on theplane perpendicular toG.

G

T

Crystal plane



Distance between planes = 2ππππ/G• If G has no common

factor (prime #), then the

distance between crystalplane perpendicular tothe G is 2π /G

d = T cosθ = 2πn/G

d’ = T’cosθ’ = 2π(n+1)/G separation of planes

= d’ – d = 2π /G

and G must be thesmallest reciprocal latticefor a given direction in k-space (prime #).

G

T

Crystal planes

T’

d d’



Small G dominates The larger the G, the closer the crystal plane, and less

atoms on the plane.

For example, in sc, separation of (100) is a,of (110) is a/ √2 = 0.71a

(110)

(100)

a1

a2

(210)

(100)

(110)

a 0.71a



Volume of the reciprocal cell

If the volume of a unit cell of the direct lattice is V, thenthe volume of a unit cell of the reciprocal lattice is (2π)D/Vwhere D is the dimension of the lattice, usually 3.

3D:

Vg = | b1 · b2 x b3|= (2π /V)3 {(a2 x a3) · (a3 x a1) x (a1 x a2)}

= (2π /V)3 {(a2 x a3) · [a3·(a1 x a2)] a1 - [a1·(a1 x a2)] a3}

= (2π /V)3 {(a2

x a3) · [a

3·(a

1x a

2)] a

1}

= (2π /V)3 {(a2 x a3) · V a1 }

= (2π)3 /V2 {(a2 x a3) · a1 }

= (2π)3 /V





Primitive Reciprocal Lattice Cell

The Wigner-Seitz cell of the reciprocal lattice

is called the Brillouin Zone

Reciprocal Lattice



Cubic – reciprocal lattice

=

=

=

za

ya

xa

3

2

1

a

a

a

r

r

r

( )( )

( )

+=

+=+=

xza

zyayˆ

xˆ

a

21

3

21

2

2

1

1

a

a

a

r

r

r

( )( )

( )

+−=

++−=

−+=

zyxa

zyxa

zyxa

21

3

21

2

21

1

a

a

a

r

r

r

( )

( )

( )

=

=

=

z / 2b

y / 2b

x / 2b

3

2

1

a

a

a

π

π

π

r

r

r

( )( )

( )

+=

+=+=

xzb

zyb

yxb

23

22

21

a

a

a

π

π

π

r

r

r

( )( )

( )

+−=

++−=−+=

zyxb

zyxbzyxb

23

22

2

1

a

a

a

π

π

π

r

r

r

( )3 / 2 aπ

( )3 / 24 aπ

( )3 / 22 aπ

SC

FCC

BCC

Direct Lattice Reciprocal Lattice Volume (k-space)

SC

BCC

FCC



bcc

Direct LatticePrimitive Cell

Wigner Cell

Reciprocal Lattice(fcc)

Brillouin Zone



fcc

Direct LatticePrimitive Cell

Wigner Cell

Reciprocal Lattice(bcc)

Brillouin Zone





1D - Reciprocal Lattice

a

Direct Lattice

bi·a j = 2πδij means b·a = 2π

so b = 2π /a along the same direction

2ππππ/a

First BZ

b=2π /a

Reciprocal Lattice





Diffraction in quantum mechanic Diffraction in quantum mechanic description: (Born

approx)

The transitional amplitude to scatter from k-state to k’-state = Mkk’ = <k’|V(r)|k> = ∫Ψ*(r) V(r) Ψ(r) dr

Ψ(r) = eik·r is the plane wave (think of the Bragg

scattering plane incident & diffracted waves) In Fourier series, V(r) = ∑ VG eiG·r (with the

translational property of the lattice)

Mkk’= ∑ VG ∫ ei(G + k - k’)·r dr = ∑ VG δ(G – ∆k)

= VG if G = ∆k , Mkk’= 0 otherwise. So the condition for Bragg scattering becomes

G = ∆k = k’ - k



Bragg Scattering For elastic scattering, frequency ω = ck is

unchanged. Speed of light is the same before &

after scattering off the target, so |k| = |k’|.write k + G = k’ and square both sides gives

2 k·G + G2 = 0,

k

k’

θ

∆k=G

θ

θ

k αβ

2k·G = 2kG cosα = -G2

-2k cosβ = -G2k sinθ = G

2(2π/λ) sinθ = 2πn/d2d sinθ = nλwhich is the Bragg condition.



X-ray data



Bragg Conditions

2 / G|Gk |

0GGk 2 2

=⋅

=+⋅r

r r

k

k’ ∆k=G

B r a g g P l a

n e

k’

ko

Can use different k to probethe same G. Different circles,

different k and angles.

Or similarly, all k end on theperpendicular bisecting plane

of G would give Braggdiffraction. (BZ !!)

oGG/2

k

G/2

k’

BZ





Ewald Sphere

k

k’2θ

G

A sphere with radius k. Any point lands on the surface ofthe sphere is an allowed G value, i.e. diffraction occurs atthat particular angle & k.In principle, it can map out the whole reciprocal latticeusing different k.



(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

REAL LATTICE

a1

a2

b1

b2

ExampleNote:a1·b2 = 0

a2·b1 = 0here {a} are the directlattice vectors & {b}are the reciprocallattice vectors



(0,1) planes(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d0,1

REAL LATTICE

a1

a2

b1

b2



(1,1) planes(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d1,1

Note:

length is longerthan (0,1) sincespacing between(1,1) planes issmaller.

REAL LATTICE

a1

a2

b1

b2



(0,0)

(2,1) planes

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d2,1

REAL LATTICE

a1

a2

b1

b2



(0,0)

(3,1) planes

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d3,1

REAL LATTICE

a1

a2

b1

b2



(0,1) planes

(1,1) planes(0,0)

(2,1) planes

(3,1) planes

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

REAL LATTICE

a1

a2

b1

b2



(0,2) planes(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d0,2

(0,2)

REAL LATTICE

a1

a2

b1

b2

with basis,for example



(1,2) planes(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d1,2

(1,2)(0,2)

REAL LATTICE

a1

a2

b1

b2



(2,2) planes(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d2,2

(1,2)(0,2)

(2,2)

REAL LATTICE

a1

a2

b1

b2



(0,0)

(3,2) planes

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

(1,2)(0,2)

(2,2)(3,2)

length=1/d3,2

REAL LATTICE

a1

a2

b1

b2



(0,1) planes

(1,1) planes(0,0)

(2,1) planes

(3,1) planes

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

(0,0)(0,0)(0,0)(0,0)

(1,2)

(2,2)(3,2)

(0,2)

(0,2) planes

(1,2) planes

(2,2) planes

(3,2) planes

REAL LATTICE

a1

a2

b1

b2



(0,1) planes(0,0)

RECIPROCAL LATTICE

(0,1)(1,1)

(2,1)(3,1)

length=1/d0,1

How do weorient thecrystal to

observediffraction fromthe (0,1)reflection?

REAL LATTICE

a1

a2

b1

b2



(0,0 )

(0,1) planes

nλ=2dsinθ

θ

Bragg condition-- upper beamhas to be an integral number ofwavelengths from the lower

beam forconstructive interference.



(0,0 )

( 0, 1 ) ( 1, 1 )

( 2, 1 )

( 3, 1 )

( 0, 0 )

(0,1) planes

nλ=2dsinθ

θ 2θ



(1,1) planes

( 0, 0 )

( 0 , 1 ) ( 1 ,

1 )

( 2 , 1 )

( 3 , 1 )

( 0 , 0 )

3D



( 0 , 0 )

(2,1) planes

( 0 , 1 )

( 1 , 1 )

( 2 , 1 )

( 3 , 1 )

( 0 , 0 )





Diffraction by a monatomic lattice

Bragg condition:In phase to get constructive interference (+ + + +)

Zero reflection otherwise (+ - + - + -)monatomic lattice

k k’

+

+

primitive cell

d

From here we got:

G = ∆k // d

G=2π /dBragg diffraction

Laude Equations

Construction of reciprocal lattice from EACH diffraction observedi.e. each reciprocal lattice point gives a spot on the screen



Diffraction by a lattice with a Basis

diatomic lattice

k k’+

+

d

conventional cell

d1

+

+

For diatomic lattice (assume identical atoms)ALL in phase to get max amplitude (++ ++)

Zero amplitude if the reflection from the basis

cancel (−+ −+).Likewise, if the basis reflections are in phase,

but the lattice reflections are out-of-phase,

amplitude still cancels (++ −−)i.e. to get max Bragg scattering, both latticereflections and basis reflections should be inphase. Amplitude is less if they are partially inphase or out-of-phase.

With multiple atoms, or non-identical atoms, scattering amplitudewould not be uniform. There are partial cancellations.

The resulting reciprocal lattice of the direct lattice would be modified.



Fourier Analysis of a Basis Scattering amplitude

Mkk’ = <k’|V(r)|k> =∫Ψ*(r) V(r) Ψ(r) dr

where Ψ(r) = eik·r

V(r) = ∑ j V(r-r j) = ∑G,j VG eiG·(r-r j)

r j is the position of the jth atom in the unit cell. Mkk’= ∑ VG ∫ ei(G + k - k’)·(r-r

j) dr

= ∑G ∑ j e-iG·r j [VG ∫ ei(G + k - k’)·r dr]

Define SG = ∑ j e-iG·r j f j the Structure Factor and f jis the Atomic Form Factor such that Mkk’∝ SG



Now if we specify G and rj as

we get:

Note that S G can be complex, because the scatteringintensity involves the magnitude squared of S G.

The evaluation of atomic form factor is complicated but forspherically-symmetric electron density, it can be written as

(Kittel)

aaar bbbG 321332211 and j j j j z y xvvv ++=++=

( )3212

1

v zv yv xis

j jG

j j je f S ++π−

=∑=

( )∫π=∞

0

2 sin)(4 dr

Gr

Gr r r n f j j





For a BCC lattice, r j = (000) and (½ ½ ½). The structurefactor is then: ( )

f eS

vvvi

G321

1

++π−

+=

bcc structure factor

The structure factor is maximum SG = 2f when the sumof the indices is even, i.e. v1+v2+v3=2n .

The structure factor is SG = 0 when the sum of theindices is odd, i.e. v1+v2+v3=2n+1 .

(0,0,0)

The only G show up in diffraction

pattern are the ones fit the fcc latticepoints.

Same physical measurementregardless how the cell was defined.



Example – fcc

primitive cellfccV=a3 /4

reciprocal cell

bccV =(4π /a)3 /2

* see primitive vectors

•lattice cell sc•w/ 4-pts basis

•V=a3

(0,0,0)

•reciprocal lattice sc

•V= (2π /a)3 [8x smaller

than the cube (4π /a)3

]•extra lattice points•extra diffraction spots?



For a FCC lattice, we have four atoms per init cell located at(000), (0 ½ ½), (½ 0 ½) and (½ ½ 0). The structure factor isthen:

fcc structure factor

(0,0,0)

The only G show up in diffractionpattern are the ones fit the bcc latticepoints.

Same physical measurement

regardless how the cell was defined.

( ) ( ) ( ) f eeeS

vvivvivvi

G

3231211+π−+π−+π− +++=

When all indices are even or odd, then SG = 4f.When the indices are partially even and partially odd, thenSG = 0.



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