Solid State Physics Lecture 4 – Reciprocal lattices Professor Stephen Sweeney

Advanced Technology Institute and Department of Physics University of Surrey, Guildford, GU2 7XH, UK

[email protected]

Solid State Physics - Lecture 4 Recap from Lecture 3

• The structure of crystals can be probed with X-rays using diffraction techniques

• There is a relationship between the lattice spacing and the Miller indices (hkl) which is straightforward for cubic structures

• There are various techniques for X-ray diffraction: single crystal, powder method, Laue (energy dependent) technique

• Applications go well beyond condensed matter physics, e.g. archaeology etc.

Solid State Physics - Lecture 4 More on diffraction

• Diffraction from a crystal has two contributions:

• Diffraction due to the underlying lattice

• Diffraction due to the basis

Solid State Physics - Lecture 4 Bragg revisited

2D Case

n  2d sin 

Solid State Physics - Lecture 4 Revision of Waves



A

Definition of wave: Aexpik r t (complex form)

amplitude time dependence (often ignore)

observation point wave-vector 2 k  

Solid State Physics - Lecture 4 Bragg revisited

General 3D Case Path difference = AB+BC:

AB  d cos 

A BC  d cos'

AB  BC  d cos' d cos 

B C  n  dcos' d cos  k is wave-vector of incident wave k’ is wave-vector of diffracted wave d is the distance between two lattice points (d is the displacement vector)

Solid State Physics - Lecture 4 Laue condition

Since d is the displacement vector General 3D Case between the scattered waves: 2 k d  d cos   k 2 A k'd   d cos'  Combine with our previous result (previous slide): d  k’ n  d cos  d cos'  k k'd 2 B C K  k k' is the scattering vector k is wave-vector of incident wave k’ is wave-vector of diffracted wave  2n  Kd or expiK d 1 d is the displacement vector Laue (diffraction) condition

Solid State Physics - Lecture 4 The Reciprocal lattice

• The diffraction pattern that is generated in X-ray diffraction is a representation of a “reciprocal lattice”

• The reciprocal lattice is of fundamental importance when considering periodic structures and processes in a crystal lattice where momentum is transferred (e.g. diffraction)

• It is a set of imaginary points in which the direction of a vector from one point to another corresponds to a direction normal to a plane in a real lattice • The separation of the reciprocal lattice points (magnitude of the vector) is G proportional to the reciprocal of the real separation between planes, i.e. 2 G  dhkl

Solid State Physics - Lecture 4 The Reciprocal lattice

Things to remember:

• Convention: the reciprocal lattice vector is 2 times the reciprocal of the interplanar distance (NB: crystallographers often don’t bother with the 2 but we always use it in Solid-State physics)

• Real lattice points correspond to the location of real objects (e.g. atoms, ions etc.) and have dimensions [L]

• Reciprocal lattice points are rather more abstract (e.g. magnitude/direction of momentum) with dimensions [L-1]

• We can relate the real lattice to the reciprocal lattice using Fourier Transforms (but we won’t in this course)

Solid State Physics - Lecture 4 2D Reciprocal lattice

• The basis set of the reciprocal lattice vectors are defined from:

real lattice reciprocal lattice (where i,j are the directions) vector vector

Therefore: b1 must be perpendicular to a2 and a1 must be perpendicular to b2

Solid State Physics - Lecture 4 2D Reciprocal lattice

Real Lattice Reciprocal Lattice

Solid State Physics - Lecture 4 2D Reciprocal lattice

General 2D lattice:

Real Lattice Reciprocal Lattice

General 2D reciprocal lattice vector

Solid State Physics - Lecture 4 3D Reciprocal lattice

General Case (3D) lattice:

area of plane in unit cell x vector perpendicular to plane

a1 a2

a3 volume of real unit cell

Thus, b1 is perpendicular to a2 and a3 with magnitude 2/a1

Solid State Physics - Lecture 4 3D Reciprocal lattice

General Case (3D) lattice:

a1 b2  0

as before

General 3D reciprocal lattice vector

Solid State Physics - Lecture 4 Cubic Lattices

For simple cubic Bravais Lattice

a

a a

Q. What are the reciprocal lattice vectors? z a2 a3  b1  2 y a1 a2 a3  x x y z 2 a a  0 a 0  a2x  b  x 2 3 1 a 0 0 a

Solid State Physics - Lecture 4 Cubic Lattices – bcc structure

bcc cell primitive vectors:

Q. What are reciprocal lattice primitive vectors?

Solid State Physics - Lecture 4 Cubic Lattices – bcc structure

Primitive reciprocal lattice vectors for bcc (real space):

The reciprocal lattice is a fcc structure with cube unit cell length of 2/a

(similarly a bcc real space lattice has a reciprocal lattice fcc structure)

Solid State Physics - Lecture 4 Laue condition (again)

From before: d is the real space vector K is the scattering vector which separating the lattice points of depends on the wavelength and interest, i.e. the geometry of the scattering d  n1a1  n2a2  n3a3

 2n  Kd or expiK d 1

Laue (diffraction) condition

We will see that if K=Ghkl (where Ghkl is a reciprocal lattice vector) that the Laue condition is satisfied

Solid State Physics - Lecture 4 Laue condition (again)

(where h,k & l are the Miller Indices…) We may write that: Ghkl  hb1  kb2 lb3

k’ If d  a1

K=k’-k Ghkl d  hb1  kb2 lb3 a1 =Ghkl k a2 a3  But b1  2 and b2 a1  b3 a1  0 a1 a2 a3 

a2 a3   Ghkl d  2h a1  2h a1 a2 a3 

Satisfies Laue condition

Solid State Physics - Lecture 4 Reciprocal lattice and Miller Indices

Since the scattering vector K for each diffracted beam corresponds to a point on the reciprocal lattice, we can use (hkl) to label that beam

Ghkl  hb1  kb2 lb3 is orthogonal to the planes with indices (hkl)

2 2 2 is inversely proportional to the spacing (dhkl) Ghkl  h  k  l of the (hkl) planes, as can be proven…

Solid State Physics - Lecture 4 Reciprocal lattice and Miller Indices

Real Space Reciprocal Space

The vector normal to lattice plane (hkl) is parallel to

Solid State Physics - Lecture 4 Reciprocal lattice and Miller Indices

The vector normal to lattice plane (hkl) is parallel to

Remember:

 a a   a a    2  1  3  1   k h   l h   a a   a a   a a    2 3    1 2    3 1   kl   kh   hl 

1 volume  ha a  la a  ka a  hkl 2 3 1 2 3 1 1 volume volume G  hb  kb  lb   hkl hkl 2 1 2 3 2 hkl

Solid State Physics - Lecture 4 Reciprocal lattice and Miller Indices

(since ai.bj=2ij)

Solid State Physics - Lecture 4 Ewald Construction

An elegant construction to show the condition for diffraction

Procedure (2D):

1. Construct the reciprocal lattice 2. Draw a vector AO (k) of length 2/ ending on a lattice point 3. Draw a circle of radius 2/ A 4. Draw a vector AB (k’) to a point of intersection with another lattice point  k’ 5. Draw vector OB (G) joining the k intersecting lattice points 6. Draw a line perpendicular to OB back to B A ( is the angle between the incident or G E scattered wave and the plane AE (i.e. it O is the scattering angle) 7. Repeat for all points of intersection

Solid State Physics - Lecture 4 Ewald Construction

An elegant construction to show the condition for diffraction

Vector OB (G) joins two lattice points, is normal to a set of real planes and has

length 2/dhkl 2 OE has length sin  A 2 OB has length 2 sin   k’ k 2 2  2 sin     2dhkl sin B  dhkl E G Braggs Law O

Solid State Physics - Lecture 4 Ewald Construction

An elegant construction to show the condition for diffraction

Other points of note:

If length of AO<½ 2/a then diffraction is not possible (i.e. if >2a)

Shorter wavelength (larger circle) leads to more intersections (higher probability of A diffraction)  k’ k In 3D this is a sphere (Ewald sphere) B E G O

Solid State Physics - Lecture 4