Chapter 2 Reciprocal Lattice

An important concept for analyzing periodic structures

Reasons for introducing reciprocal lattice • Theory of crystal diffraction of x-rays, neutrons, and electrons. Where are the diffraction maximum? What is the intensity?

• Abstract study of functions with the periodicity of a Bravais lattice Fourier transformation. Reciprocal lattice is closely related to Fourier transformation

• Consideration of how momentum conservation will be modified in periodic potential

r r Basic knowledge: e ki ⋅r is a plane wave r r r r ki ⋅r r r e = cos (k ⋅r )+ isin (k ⋅r ) a periodic function r 2π Wave vector k = nˆ λ

1 One dimensional reciprocal lattice

f (x) One dimensional periodic function

f (x + a) = f (x)

Expand f(x) into Fourier series  2πnx 2πnx  f (x) = f0 + ∑ cn cos( ) + sn sin( ) n>0  a a  2πnx 2πnx 2πnx 2πnx  i −i i −i   e a + e a e a − e a  = f + c + s 0 ∑  n n  n>0 2 2i   2πnx 2πnx  i −i  cn sn a cn sn a = f0 + ∑ ( + )e + ( − )e  n>0  2 2i 2 2i 

Redefine coefficients a0 an a−n We have 2πnx i a iGx n runs through all integers f (x) = ∑ ane = ∑ane n n * an = a−n Constraint that ensures f(x) is a real function f(x) is expanded into a series of plane waves, with a set of wave vectors G’s. iGx Each plane wave e preserve the periodicity of f(x). 2πn The set of discrete points G = is one dimensional reciprocal lattice a 2 real space

a a0 a a −1 +1 reciprocal space −2 a+2

• Reciprocal space is Fourier space • Reciprocal lattice points of a Bravais lattice indicate the allowed terms in the Fourier series of a function with the same periodicity as the Bravais lattice

• Reciprocal lattice depends on the Bravais lattice, but does not depend on the particular form of f(x), as long as f(x) has the periodicity of the Bravais lattice

•But reciprocal lattice does not indicate the magnitude of the each Fourier term an, which depends on the exact form of f(x).

2πnx a −i a 1 a 1 −iGx an = ∫ dx f (x)e = ∫ dx f (x)e a 0 a 0 Expand to three dimensional case: r r r r Three dimensional periodic function f (r ) that satisfies f (r) = f (r + R)

r r r r r 1 r −iG⋅r Fourier expansion: f (r) = a r eiG⋅r a r = dV f (r)e ∑r G G ∫ G Vc cell

Volume of a primitive cell 3 General definition of reciprocal lattice

r Bravais lattice R

r a plane wave r eiG⋅r

v r r iG⋅r The set of all wave vectors G that yield rplane waves e with the periodicity of a given Bravais lattice R is known as its reciprocal lattice

r r r r r Analytically, the definition is expressed as eiG⋅(r +R) = eiG⋅r r r v iG⋅R r The set of G satisfying e =1 , for all R in a Bravais lattice

• A reciprocal latticev is defined with rreferencer to a particular Bravais lattice r • A set of vectors G satisfying e iG ⋅ R = 1 is called a reciprocal lattice only if the set of R is a Bravais lattice •Reciprocal lattice is a Bravais lattice (proven in the following)

4 Construction of ar reciprocal lattice and Proof that G is a Bravais lattice

r r r r r r r Are primitive vectors of a Bravais lattice = + + a1 a2 a3 R n1a1 n2a2 n3a3 r r r The reciprocal lattice can be generated by the primitive vectors b3 b1 b2 r r r r r a × a r r r × 3 1 r × a2 a3 b2 = 2π r r r a1 a2 r r r b1 = 2π r r r ⋅( × ) b3 = 2π r r r a1 a2 a3 V = a1 ⋅(a2 × a3 ) a1 ⋅(a2 × a3 ) a1 ⋅(a2 × a3 ) r r volume of primitive cell Apparently bi ⋅a j = 2πδij

An arbitrary vector in reciprocal space can be written as a linear combination of r r r r g = g1b1 + g2b2 + g3b3

r r gi ⋅R r To qualify for a reciprocal lattice, e =1 For all R g has to be integers i r r r r v is a integer Reciprocal lattice vector G = v1b1 + v2b2 + v3b3 i r

Apparently, it is also a Bravais lattice with b i as its primitive vectors 5 The reciprocal of the reciprocal lattice

The reciprocal lattice is a Bravais lattice, one can construct its reciprocal lattice, which turns out to be nothing but the original direct lattice. r R Direct lattice r G Reciprocal lattice r r r eiG⋅R =1 for all R r r r r Ri ⋅G Or, every R satisfies e for all G r r r Ki ⋅G Look for the reciprocal of the reciprocal lattice: K which satisfies e =1 r r r r The set of vectors R is a subset of the set of vectors K R ⊆ K

For any vector that does not belong to the direct lattice R: r r r r r = x1a1 + x2a2 + x3a3 At least one xi is non-integer

r r r r r r r r ri ⋅G ri ⋅bi ix iai ⋅bi i2πxi For G = bi we have e = e = e = e ≠ 1

So r does not belong to K

So the set of vectors R is identical to the set of vectors K 6 Examples of reciprocal lattice

Simple cubic Bravais lattice r r r a1 = xaˆ a2 = yaˆ a3 = zaˆ

Reciprocal lattice r 2π r r π = ˆ 2π 2 b2 y ˆ b = xˆ a b3 = z 1 a a 2π Cubic unit cell with an edge of Also a simple cubic lattice a Face centered cubic

r a r a r a a = (yˆ + zˆ) a2 = (zˆ + xˆ) a = (xˆ + yˆ) 1 2 2 3 2 r r r 1 V = a ⋅(a × a ) = a3 1 2 3 4 Primitive vectors of Reciprocal lattice

r 2π r 2π r b = (xˆ − yˆ + zˆ) 2π b1 = (−xˆ + yˆ + zˆ) 2 b = (xˆ + yˆ − zˆ) a a 3 a

They are the primitive vectors of a body centered cubic lattice 4π Size of the cubic cell: ( )3 a 7 Body centered cubic

Primitive vectors r a r a = (−xˆ + yˆ + zˆ) r a a 1 a = (xˆ − yˆ + zˆ) a3 = (xˆ + yˆ − zˆ) 2 2 2 2

Primitive vectors for reciprocal lattice

r 2π r r 2π 2π ˆ ˆ b1 = (yˆ + zˆ) b = (xˆ + zˆ) b3 = (x + y) a 2 a a

They are the primitive vectors of a fcc lattice

Fcc and bcc are reciprocal to each other

Brillouin Zones

The Wigner Seitz primitive cell of the reciprocal lattice is known as the first Brillouin zone.

8 First Brillouin zone for bcc structure

The first Brillouin zone is enclosed by perpendicular bisectors between the central lattice point and its 12 nearest neighbors

It is a regular 12 faced solid

9 First Brillouin zone of fcc structure

Primitive vectors of fcc structure Brillouin zones of fcc structure The reciprocal lattice is bcc lattice

10 Lattice planes and reciprocal lattice vectors

Lattice planes: any plane that contains at least three noncollinear Bravais latice points

Due to trnaslational symmetry, any lattice plane will contain infinitely many lattice points, which form a two dimensional Bravais lattice

Family of lattice planes: A set of parallel, equally spaced lattice planes which together contains all the points of the three dimensional Bravais lattice.

Theorem: For any family of lattice planes separated by a distance d, there are reciprocal lattice vectors 2π perpendicular to the planes, the shortest of which have a length of . Conversely, for any r d r reciprocal lattice vector G , there is a family of lattice planes normal to G and separated by a r distance d, where 2 π is the length of the shortest reciprocal lattice vector parallel to G d

The theorem is a direct consequence of r r (1) The definition of reciprocal lattice eiG⋅R =1 (2) the fact that a plane wave has the same value at all points lying in a family of planes that are perpendicular to its wave vector and separated by an integral number of wavelength. 11 Proof of the first part of the theorem:

Given a family of lattice planes with a separation of d , let n ˆ be a unit vector normal to the planes. Let’s make up a wave vector: r 2π K = nˆ d r We need to prove that K is a reciprocal lattice vector, and it is the shortest one in that direction r r r ⋅ To qualify for a reciprocal vector, e Ki R = 1 has to be satisfied for all R of the Bravais lattice r r Ki ⋅0 We know this condition is satisfied for the origin of the Bravais lattice r = 0 because e =1 r r r Ki ⋅r r Then e = 1 must be true for all the lattice points r in the lattice plane that contains the origin r = 0 r r r Because K is perpendicular to that plane, and K ⋅ r should be a constant in the plane

r r Ki ⋅r 2π e is a periodic function with a wavelength λ = r = d in the direction of its wave vector K

The wavelength λ = d happens to be the separation of the planes r r r ⋅ Therefore for all the lattice points R in the family of planes, e Ki R = 1 is satisfied r r 2π So is a reciprocal vector, and we can write it as G = nˆ K d 12 r r r r If there is another vector G ′ shorter than G G′ < G r 2π The wavelength of G ′ is λ′ = r > d G′ r r ′ It is impossible for e i G ⋅ r = 1 to have the same value on two planes with a distance shorter than the wavelength r So G ′ can not be a reciprocal vector Proof of the converse of the theorem r Given a reciprocal lattice vector, let G be the shortest parallel reciprocal lattice vector

r r iG⋅r Let’s construct a set of real space planes (not necessarily lattice planes) on which e =1

Now we need to prove that this set of planes are lattice planes and contains all the Bravais lattice points r r 2π These planes must contain r = 0 . They must be all perpendicular to G and separated by a distance d = λ = G r r r For all Bravais lattice vector R , e i G ⋅ R = 1 must be true for any reciprocal vector

So the set of planes must contain all the Bravais lattice points.

Lastly, we need to prove that each of these planes contain lattice points instead of every nth of them Suppose only every nth of the planes contain lattice points. According to the first part of the theorem, the shortest 2π G reciprocal vector perpendicular to the planes will be, = , which contradict with assmuption that G is the nd n shortest reciprocal vector in that direction 13 Reciprocal lattice vectors and Miller indices of lattice planes

The miller indices of a lattice plane are the coordinates of the shortest reciprocal lattice vector normal to that plane, with respect to a specified set of rprimitiver reciprr ocalr lattice vectors. Thus a plane with Miller indices h, k, l, is normal to the reciprocal lattice vector G = hb1 + bk 2 + bl 3 Proof r r r r Consider a plane with miller indices h, k, l. G = h ′ b 1 + k′b2 + l′b3 is the shortest reciprocal vector normal to that plane. Let’s prove that h′ = h k′ = k l′ = l r a r r r 3 For any point r on plane, G ⋅r = A r A r x a ′ a1 3 3 r h = r 2πx r G ⋅(x a ) = A 1 r 1 1 x1a1 So r r G A G ⋅(x2a2 ) = A k′ = r r 2πx2 G ⋅(x a ) = A 3 3 A l′ = 2πx3 r x2a2 1 1 1 r h′: k′ :l′ = : : a Then it follows that 2 x1 x2 x3

Since G is the shortest reciprocal vector perpendicular to the plane. There should be no common factors between h ′ k ′ l ′ . These parameters satisfy the original definition of miller indices

Therefore h′ = h k′ = k l′ = l 14