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Handout 5

The Reciprocal

In this lecture you will learn:

• Fourier transforms of lattices

• The reciprocal lattice

• Brillouin Zones

• X-ray

• Fourier transforms of lattice periodic functions

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Fourier Transform (FT) of a 1D Lattice

Consider a 1D :

 a1  a xˆ Now consider a function consisting of a “lattice” of delta functions – in which a delta function is placed at each lattice point: f x

 x a1  a xˆ  f x    x  n a  n

The FT of this function is (as you found in your homework):     i kx x i kx n a 2  2  f k x   dx   x  n a e  e    k x  m   n n a m  a 

The FT of a train of delta functions is also a train of delta functions in k-space

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

1 Reciprocal Lattice as FT of a 1D Lattice f x 1

 x a1  a xˆ FT is: f k x  2 a  2 k b  xˆ x 1 a The reciprocal lattice is defined by the position of the delta-functions in the FT of the actual lattice (also called the direct lattice)

Direct lattice (or the actual lattice):

 x a1  a xˆ Reciprocal lattice:

 2 k x b  xˆ 1 a ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Reciprocal Lattice of a 1D Lattice For the 1D Bravais lattice,

 a1  a xˆ    The position vector R n of any lattice point is given by: Rn  n a1 f x

 x a1  a xˆ    f x    x  Rn n        i kx x i k.Rn The FT of this function is: f k x   dx   x  Rn e  e  n n The reciprocal lattice in k-space is defined by the set of all points for which the k- vector satisfies,   ei k . Rn  1  for ALL R n of the direct lattice    For the points in k-space belonging to the reciprocal lattice the summation  e i k . R n becomes very large! n

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

2 Reciprocal Lattice of a 1D Lattice For the 1D Bravais lattice,

 a1  a xˆ    The position vector R n of any lattice point is given by: Rn  n a1 The reciprocal lattice in k-space is defined by the set of all points for which the k- vector satisfies,   ei k . Rn  1  for ALL R n of the direct lattice    i k . Rn  For k to satisfy e  1 , it must be that for all R n :   k . Rn  2  integer   k x na  2 integer 2  k  m where m is any integer x a

Therefore, the reciprocal lattice is:

 2 k x b  xˆ 1 a ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Reciprocal Lattice of a 2D Lattice y Consider the 2D rectangular Bravais lattice: If we place a 2D delta function at each lattice point we get the function:  a2  c yˆ    x f x,y     x  n a  y  m c  a1  a xˆ nm 

The above notation is too cumbersome, so we write it in a simpler way as:

 2   The summation over “j ” is over all the lattice points f r    r  R j j 2  2     A 2D delta function has the property:  d r  r  ro  gr   gro  and it is just a product of two 1D delta functions corresponding to the x and y components of the vectors in its arguments: 2      r  ro    x  ro.xˆ  y  ro.yˆ  Now we the function f  r   :    2   i k . r 2  2   i k . r f k   d r f r e   d r  r  R j  e j   2   i k . R j 2  2   2   e   k x  n   ky  m  j ac nm   a   c 

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

3 Reciprocal Lattice of a 2D Lattice   2    i k . R j 2   2   2  f k  e   k x  n   ky  m  j ac nm   a   c  ky y  2 b2  yˆ Reciprocal lattice  c a2  c yˆ  x  2 k x a1  a xˆ b  xˆ 1 a Direct lattice

• Note also that the reciprocal lattice in k-space is defined by the set of all points for which the k-vector satisfies,   i k . R e j  1  for all R j of the direct lattice

• Reciprocal lattice as the FT of the direct lattice or as set of all points in k-space for which exp  i k . R j   1 for all R j , are equivalent statements

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Reciprocal Lattice of a 2D Lattice y ky

 2 b2  yˆ Reciprocal lattice  c a2  c yˆ  x  2 k x a1  a xˆ b  xˆ 1 a Direct lattice • The reciprocal lattice of a Bravais lattice is always a Bravais lattice and has its own   primitive lattice vectors, for example, b 1 and b 2 in the above figure  • The position vector G of any point in the reciprocal lattice can be expressed in terms of the primitive lattice vectors:    G  n b1  m b2 For m and n integers So we can write the FT in a better way as: 2   2  2  2   2  2  2   f k   k x  n   ky  m    k  G j  ac nm   a   c  2 j

where 2 = ac is the area of the direct lattice primitive cell

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

4 Reciprocal Lattice of a 3D Lattice

Consider a orthorhombic direct lattice: d      R  n a1  m a2  p a3 where n, m, and p are integers a2 a Then the corresponding delta-function lattice is: c 1 a  3   3 f r    r  R j j a 3  3     A 3D delta function has the property:  d r  r  ro  gr   gro 

The reciprocal lattice in k-space is defined by the set of all points for which the k-    vector satisfies: exp  i k . R j   1 for all R j of the direct lattice. The above relation will hold if k equals G :      2  2  2 G  n b  m b  p b and b  xˆ b  yˆ b  zˆ 1 2 3 1 a 2 c 3 d Finally, the FT of the direct lattice is:

   3   i k . r 3  3   i k . r f k   d r f r e   d r  r  R j  e  j b 2    3 3  b1 i k . R j 2 3   2 3   b3  e   k  G j   k  G j  j acd j 3 j

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Direct Lattice Vectors and Reciprocal Lattice Vectors ky y  2 b2  yˆ Reciprocal lattice  c a2  c yˆ  x  2 k x a1  a xˆ b  xˆ 1 a Direct lattice       R  n a1  m a2 G  n b1  m b2 Remember that the reciprocal lattice in k-space is defined by the set of all points for which the k-vector satisfies,   ei k . R  1  for all R of the direct lattice   So for all direct lattice vectors R and all reciprocal lattice vectors G we must have:   ei G . R  1

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

5 Reciprocal Lattice of General Lattices in 1D, 2D, 3D    More often that not, the direct lattice primitive vectors, a 1 , a 2 , and a 3 , are not orthogonal Question: How does one find the reciprocal lattice vectors in the general case? ID lattice:  If the direct lattice primitive vector is: a1  a xˆ and length of primitive cell is:  = a   2 Note: Then the reciprocal lattice primitive vector is: ˆ b1  x  a a . b  2     2   1 1 f r    r  R j  f k   k  G       j  i G . R j 1 j e p m  1 2D lattice: If the direct lattice is in the x-y plane and the primitive vectors are: a and a   1 2 and area of primitive cell is: 2  a1  a2   aˆ2  zˆ  zˆ  a1 Then the reciprocal lattice primitive vectors are: b1  2 b2  2 2 2 2  2    2  2   f r    r  R j f k    k  G j j 2 j     i Gp . Rm Note: a j . bk  2  jk and e  1

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Reciprocal Lattice of General Lattices in 1D, 2D, 3D

3D lattice: If the direct lattice primitive vectors are: a , a , and a  1 2 3 and volume of primitive cell is: 3  a1 . a2  a3 

Then the reciprocal lattice primitive vectors are: Note:       a . b  2   aˆ2  a3  a3  a1  a1  a2 j k jk b1  2 b2  2 b3  2      3 3 3 i G . R e p m  1 3  3    2  3   f r    r  R j f k    k  G j j 3 j Example 2D lattice:  2 b  xˆ  yˆ  1 b a1  b xˆ  4  b b  b2  yˆ a   xˆ  yˆ b2 b 2 2 2   a2 b2 b b a   a  a  1 4 b 1 2 1 2 2

b 4 b

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

6 The

The Wigner-Seitz primitive cell of the reciprocal lattice centered at the origin is called the Brillouin zone (or the first Brillouin zone or FBZ) Wigner-Seitz primitive cell 1D direct lattice:

 x a1  a xˆ Reciprocal lattice: First Brillouin zone

 2 k x b1  xˆ 2D lattice: a y Wigner-Seitz ky primitive cell Reciprocal lattice  2 b2  yˆ First Brillouin zone  c a2  c yˆ  k a1  a xˆ x  2 x b  xˆ 1 a Direct lattice

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Wigner-Seitz The Brillouin Zone 2D lattice: primitive cell First Brillouin zone  2  b b b1   xˆ  yˆ a1  xˆ  yˆ b  2 2  a1 b2  2  b b b2   xˆ  yˆ a2  xˆ  yˆ  b 2 2  b1 a2 b2 b   a  a  4 b 2 1 2 2

b 4 b Direct lattice Reciprocal lattice

Volume/Area/Length of the first Brillouin zone: The volume (3D), area (2D), length (1D) of the first Brillouin zone is given in the same way as the corresponding expressions for the primitive cell of a direct lattice:

 Note that in all dimensions (d) the following 1D   b 1 1 relationship holds between the volumes, areas,   2D   b  b lengths of the direct and reciprocal lattice 2 1 2 primitive cells: 2 d      3D 3  b1 . b2 b3  d d

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

7 Direct Lattice Planes and Reciprocal Lattice Vectors

There is an intimate relationship between reciprocal lattice vectors and planes of points in the direct lattice captured by this theorem and its converse

Theorem: 3D lattice If there is a family of parallel lattice planes separated by distance “d ” and n ˆ is a unit vector normal to the planes  then the vector given by, G  2 G  nˆ d is a reciprocal lattice vector and so is: 2 m nˆ  m  integer d d d Converse: If G is any reciprocal lattice vector, 1  2D lattice and G is the reciprocal lattice vector d of the smallest magnitude parallel to G 1 , then there exist a family of lattice planes    G perpendicular to G 1 and G , and separated by distance “d ” where: 2 d   G

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Example: Direct Lattice Planes and Reciprocal Lattice Vectors ky y  2 b2  yˆ Reciprocal lattice  c a2  c yˆ  x  2 k x a1  a xˆ b  xˆ 1 a Direct lattice

Consider:     xˆ yˆ  G  b1  b2  2     a c   2 ac There must be a family of lattice planes normal to G and separated by:   G a2  c2 Now consider:     2xˆ yˆ  G  2b1  b2  2     a c   2 ac There must be a family of lattice planes normal to G and separated by:   G a2  4c2

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

8 The BCC Direct Lattice

a 4 a

y a 4 a x z

a 4 a Direct lattice: BCC Reciprocal lattice: FCC

The direct and the reciprocal lattices are not necessarily always the same!

a ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The FCC Direct Lattice a 4 a

y a z x 4 a

a 4 a Direct lattice: FCC Reciprocal lattice: BCC

First Brillouin zone of the BCC reciprocal lattice for an FCC direct lattice

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

9 The Reciprocal Lattice and FTs of Periodic Functions The relationship between delta-functions on a “d ” dimensional lattice and its Fourier transform is: d  d    2  d   f r    r  R j  f k    k  G j j d j Supper W  r  is a periodic function with the periodicity of the direct lattice then by definition:    W r  R j   W r   for all R j of the direct lattice One can always write a periodic function as a convolution of its value in the primitive cell and a lattice of delta functions, as shown for 1D below: W x 

a  a 2 a 2 x W x

a  a 2 a 2 x

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Reciprocal Lattice and FTs of Periodic Functions W x 

a  a 2 a 2 x W x

a  a 2 a 2 x Mathematically:  W x  W x   x  n a  n And more generally in “d ” dimensions for a lattice periodic function W  r   we have:

  d   W r  W r   r  R j  j

Value of the function Lattice of delta in one primitive cell functions

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

10 The Reciprocal Lattice and FTs of Periodic Functions For a periodic function we have:   d   W r   W r    r  R j  j Its FT is now easy given that we know the FT of a lattice of delta functions: d  d    2  d   f r     r  R j  f k    k  G j j d j We get: d d   2 d   2  d    W k  W k   k  G j    k  G j W G j d j d j The FT looks like reciprocal lattice of delta-functions with If we now take the inverse FT we get: unequal weights d   d  d   d k  i k . r d k 2  d    i k . r W r   d W k e   d   k  G j W G j e 2 2 d j   W G j i G . r A lattice periodic function can always   e j  be written as a that only j d has wavevectors belonging to the reciprocal lattice

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Reciprocal Lattice and X-Ray Diffraction X-ray diffraction is the most commonly used method to study crystal structures  In this scheme, X-rays of wavevector k are sent into a crystal, and the scattered  X-rays in the direction of a different wavevector, say k ' , are measured  k' If the position dependent dielectric constant of the medium is given by  r then the diffraction theory tells us that the    amplitude of the scattered X-rays in the direction of k ' is proportional to the integral:    k   3  i k' . r  i k . r Sk  k'   d r e  r e

For X-ray frequencies, the dielectric constant is a periodic function with the periodicity of the lattice. Therefore, one can write:     i G j . r  r    G j e j

   3    Plug this into the integral above to get: Sk  k'    G j  2   k  G j  k' j  X-rays will scatter in only those directions for which:     k'  k  G where G is some reciprocal lattice vector     Because is also a reciprocal vector Or: k'  k  G  G whenever G is a reciprocal vector ECE 407 – Spring 2009 – Farhan Rana – Cornell University

11 The Reciprocal Lattice and X-Ray Diffraction  k'  X-rays will scatter in only those directions for which:    k'  k  G (1)  Also, the frequency of the incident and diffracted X-rays is the k same so: '      k'c  k c    k'  k

 2  2  2   (1) gives: k'  k  G  2 k . G

 2  2  2    k'  k  G  2 k . G  2  2  2    k  k  G  2 k . G  2   G   k . G  Condition for X-ray diffraction 2

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Reciprocal Lattice and X-Ray Diffraction  k' The condition,  2   G k . G   2  is called the Bragg condition for diffraction k Incident X-rays will diffract efficiently provided the incident wavevector satisfies the Bragg condition for some  reciprocal lattice vector G

A graphical way to see the Bragg condition is that the incident wavevector lies on a plane in k-space (called the Bragg plane) that is the perpendicular bisector of some  reciprocal lattice vector G

        k k'  k  G k k'  k  G

  G G

Bragg plane Bragg plane

k-space k-space

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

12 The Reciprocal Lattice and X-Ray Diffraction  k' The condition,  2   G k . G   2  k can also be interpreted the following way:

Incident X-rays will diffract efficiently when the reflected waves from successive atomic planes add in phase

**Recall that there are always a family of lattice planes in real space perpendicular to any reciprocal lattice vector

Condition for in-phase reflection from successive lattice planes:    2d cos m      G 2 2  2  1  2    m  cos   m    d  2  d   d k  2  2   G Real space G  m nˆ  k . G  d 2

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Bragg Planes 2D square reciprocal lattice Corresponding to every reciprocal lattice vector there is a Bragg plane in k-space that is a perpendicular bisector of that reciprocal lattice vector

Lets draw few of the Bragg planes for the square 2D reciprocal lattice corresponding to the reciprocal lattice vectors of the smallest magnitude

1D square reciprocal lattice

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

13 Bragg Planes and Higher Order Brillouin Zones 2D square reciprocal lattice Bragg planes are shown for the square 2D reciprocal lattice corresponding to the reciprocal lattice vectors of the smallest magnitude 3 3 2 3 3 1 2 2 Higher Order Brillouin Zones 3 3

2 The nth BZ can be defined as the region 3 3 in k-space that is not in the (n-1)th BZ and can be reached from the origin by crossing at the minimum (n-1) Bragg planes

The length (1D), area (2D), volume (3D) of BZ of any order is the same 1st BZ 2nd BZ 3rd BZ

3 2231

1D square reciprocal lattice

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Appendix: Proof of the General Lattice FT Relation in 3D This appendix gives proof of the FT relation: 3  3    2  3   f r    r  R j f k    k  G j j 3 j for the general case when the direct lattice primitive vectors are not orthogonal     Let: R  n1 a1  n2 a2  n3 a3 Define the reciprocal lattice primitive vectors as:       aˆ2  a3  a3  a1  a1  a2 b1  2 b2  2 b3  2 3 3 3   Note: a j . bk  2  jk

Now we take FT:    3   i k . r 3  3   i k . r f k   d r f r e   d r  r  R j  e j   i k . R   e j j

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

14 Appendix: Proof  One can expand k in any suitable basis. Instead of choosing the usual basis:  k  kx xˆ  ky yˆ  kzzˆ I choose the basis defined by the reciprocal lattice primitive vectors:     k  k1 b1  k2 b2  k3 b3   Given that: a j . bk  2  jk

I get:        i k . R j i k . n a n a n a  f k   e   e 1 1 2 2 3 3 j n1 n2 n3    k1  m1  k2  m2  k3  m3 m1 m2 m3 Now: 3    k1  m1 k2  m2  k3  m3    k  G     where: G  m1 b1  m2 b2  m3 b3   But we don’t know the exact weight of the delta function  3 k  G

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Appendix: Proof      Since:k  kx xˆ  ky yˆ  kzzˆ and k  k1 b1  k2 b2  k3 b3

This implies: kx  b1x b2x b3x k1   k   b b b k  (1)  y   2x 2y 2z  2    kz  b3x b3y b3z k3     Any integral over k-space in the form:  dk1  dk2  dk3       can be converted into an integral in the form:  dkx  dky  dkz    by the Jacobian of the transformation:       kx , ky , kz   dkx  dky  dkz   dk1  dk2  dk3    k1 , k2 , k3   

Therefore: kx ,ky ,kz  3    k1  m1  k2  m2  k3  m3   k  G k1,k2,k3

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

15 Appendix: Proof

From (1) on previous slide: 3 kx ,ky ,kz     2   b1 . b2  b3  3  k1,k2,k3 3

Therefore:    i k . R j f k   e    k1  m1  k2  m2  k3  m3 j m1 m2 m3 3 2 3     k  G j 3 j

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

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