Handout 6

Electrons in Periodic Potentials

In this lecture you will learn:

• Bloch’s theorem and Bloch functions

• Electron Bragg scattering and opening of bandgaps

• Free electron bands and zone folding

• Energy bands in 1D, 2D, and 3D lattices

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Reciprocal Lattice and X-Ray Diffraction

 X-rays will scatter in only those directions for which:

 k'      k'  k  G and k'  k

 k

 2   G  k . G  2 Bragg condition for X-ray scattering

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

1 X-Ray Diffraction and Bragg Planes 2D square  Consider x-rays with wavevector k incident reciprocal lattice on a crystal, as shown:

 k

 k

   k'  k  G   k'  k

 What is k ' ?

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

X-Ray Diffraction and Bragg Planes 2D square  Consider x-rays with wavevector k incident reciprocal lattice on a crystal, as shown:

 k

 k

   k'  k  G   k'  k

 What is k ' ?

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

2 X-Ray Diffraction and Bragg Planes 2D square  Consider x-rays with wavevector k incident reciprocal lattice on a crystal, as shown:

 k

 k

   k'  k  G   k'  k

 What is k ' ?

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Atomic Potentials in Crystals V r The potential energy of an electron due 0 to a single isolated atom looks like: Potential of an isolated Energy levels In a crystal, the potential energy due to atom all the atoms in the lattice looks like: x V r 0 Vacuum 0 level

Energy Potential in levels a crystal

x 0 The lowest energy levels and wavefunctions of electrons remain unchanged when going from an isolated atom to a crystal

The higher energy levels (usually corresponding to the outermost atomic shell) get modified, and the corresponding wavefunctions are no longer localized at individual atoms but become spread over the entire crystal

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

3 Properties of Atomic Potentials in Crystals

• The atomic potential is lattice periodic (even for a lattice with a basis):  V r  R  V r  where R is any lattice vector

• Because the atomic potential is lattice periodic, it can be written as a convolution (assuming a lattice in “d ” dimensions)

  d   V r  potential in one primitive cell V r  V r   r  R j   j and expanded in a Fourier series of the type:  V G           j  i G j . r  i G j . r Verify that: V r  R  V r V r   e  V G j e j d j where only the reciprocal lattice vectors appear in the exponential

 The Fourier components of the periodic potential contain only the reciprocal lattice vectors

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Properties of Electron Wavefunctions in Crystals

Electrons in a crystal satisfy the Schrodinger equation: 2   2 r  V r  r  E  r 2m Where:  V r  R  V r Since the potential is periodic, and one lattice site is no different than any other lattice site, the solutions must satisfy:  2  r  R   r 2 This implies that the wavefunction at positions separated by a lattice vector can only differ by a phase factor:    r  R  ei  R  r It follows that both the following relations must hold:        r  R  R'  ei  R  r  R'  ei  R  R'  r      r  R  R'  ei  RR'  r Which implies:      R  R'   R  R'

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

4 Properties of Electron Wavefunctions in Crystals The simplest, and the only way, that the relation:      R  R'   R  R' can hold for all lattice vectors is if the phase is a linear scalar  function of the vector R :     R  k . R  where k is some vector. It follows that our solutions must satisfy:     r  R  ei k . R  r Bloch’s Theorem:

The above is one version of the so called Bloch’s theorem, which says that associated with every solution of the Schrodinger equation in a periodic potential  there is a wavevector k such that:     r  R  ei k . R  r Solutions of the Schrodinger equation for periodic potentials with the above property are called Bloch functions

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Case Study: Electron in a 1D Periodic Potential

Consider the 1D Bravais lattice,

a x

The position vector R n of any lattice point is given by: Rn  n a And the reciprocal lattice and reciprocal lattice vectors are:

2 2 k Gm  m a a Free Electron Approach: We will suppose that the periodic atomic potential V(x) is small, and that the electrons are essentially free, and we will treat the potential as a perturbation and see how it effects the free electrons. We have:

V x  n a  V x

Consequently, the Fourier series expansion of V(x) will be:

i Gm x V k  Gm  V x  V Gm e where : V Gm  m a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

5 Electron in a 1D Periodic Potential: Bragg Scattering The key idea is that the electrons will Bragg scatter from the periodic atomic potentials just like X-rays: k

x k' a For Bragg scattering, the difference between the final and initial wavevector must equal a reciprocal lattice vector: k'k  Gm AND the final and initial electron energies must be equal: 2 2 2 2  k'  k  The initial electron 2m 2m wavevector must be Both the above conditions are satisfied if: one-half of a G k'  k & k   m reciprocal lattice 2 vector OR the initial electron wavevector G2 G2 must be on a Bragg plane (or point in 1D) G1 G1

 2 a   a  a 2 a k

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron in a 1D Periodic Potential: Bragg Scattering The Bragg condition can also be thought in terms of interference of waves in scattering: k x a Consider an electron with wavevector k. The electron will “Bragg scatter” from the atoms if the electron wave, with wavelength , reflecting off successive atoms adds in phase in the backward direction This condition gives: 2 a  m  2  2 a  2 m  2  2 k  m  G a m G  k  m 2 G  k   m 2

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

6 Perturbation Theory: A Review

Consider a Hamiltonian with eigenfunctions and energies given by: ˆ Ho n  en n

In the presence of a perturbing potential, the new eigenfunctions and energies are given by: ˆ ˆ Ho  V   n  En  n If the perturbation is small, then the new eigenfunctions are slightly perturbed from the original eigenfunctions and, to first order in the perturbation, can be written as: ˆ m V n  n  n   m  higher order terms mn en  em Thus, the perturbation “mixes” the eigenfunctions of the original Hamiltonian to generate the eigenfunction of the new Hamiltonian.

Note: The effect of the perturbation is not small, and the perturbation theory breaks down, if for: ˆ m V n  0 we have: en  em  0

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron in a 1D Periodic Potential: Perturbation Theory The goal here is to treat the periodic potential as a perturbation to the free electron Hamiltonian. So in the absence of the perturbation we have the free electron case:

ˆ 2 2 2 2 2 ˆ P   1 i k x  k Ho      x  e ek  2m 2m x 2 k L 2m

ˆ Energy Ho k  ek  k

The energy dispersion relation of free electrons 2 2 is parabolic, as shown in the figure  k 2m Now assume that the perturbation is the periodic potential of the atoms: V x  k which can also be expressed in a Fourier series as:

i Gm x V k  Gm  V x  V Gm e where : V Gm  m a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

7 Electron in a 1D Periodic Potential: Perturbation Theory

So we try perturbation theory and write: ˆ ˆ Ho  V x   k  Ek  k

And write the new eigenfunction as: ˆ k' V k  k  k   k'  higher order terms k' ek  e k' 

First evaluate the potential matrix element (L is the size of the entire 1D crystal):

L 2 ˆ 1 i k'x 1 i k x k' V k   dx e V x e L 2 L L L 2 1 i k'x i Gm x i k x  V Gm  dx e e e L m L 2  V G   0 unless k'  k  G  m k'k , Gm m m

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron in a 1D Periodic Potential: Perturbation Theory

The new eigenfunction is: ˆ ˆ k' V k k' V k  k  k   k'  higher order terms  V G  ek  e k'  m k'k , Gm k' m V G    m   higher order terms k  k Gm m ek  ek  Gm  The new eigenfunction corresponding to the wavevector k consists of a superposition of only those plane waves whose wavevectors differ from k by reciprocal lattice vectors The effects of the periodic perturbation will be large for those electron states for which the denominator is zero or is close to zero:

e(k)  e(k  Gm )  0 and V Gm   0 2    G2  2kG   0 2m m m Gm   k    m Bragg condition 2 a  Perturbation theory breaks down for those electron states that Bragg scatter from the periodic potential!

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

8 Electron in a 1D Periodic Potential: Variational Solution We consider a periodic atomic potential of the form: Energy

i G1 x i G1 x 2 2 V x  V G1 e  V G1 e  k 2 2 2m G  G    G 1 a 1 a 1 * Since the potential is always real: V G1  V G1 The potential will strongly couple plane wave eigenstates with wavevectors that differ by  G1 and the strongest coupling will be between states   k with wavevectors,    G G a a ,   1 , 1 a a 2 2 because they have equal energy

Variational Solution (Finite Basis Expansion): For states with wavevectors k near +/a, we assume a variational solution for the perturbed state:   c k   c k  G  k   k  1 k G1

1 i k x 1 i k G x Or:  x  c k e  ck  G e 1 k L 1 L

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron in a 1D Periodic Potential: Variational Solution   c k   c k  G  Energy k  k 1 k G1 2 2 Plug it into the Schrodinger equation:  k ˆ 2m Ho  V x   k  Ek  k  And then take the bra with  k and then with k  G 1 to get the matrix eigenvalue equation:

 ek V G1  ck   ck      Ek   V G1 e k  G1 ck  G1  ck  G1    k  Solution for the energy eigenvalue is: a a

2 ek  ek  G1  ek  ek  G1  2 for k near   a Ek      V G1 2  2  Now, in a similar way, had we started off by trying to find a solution for k near –/a we would have obtained: 2 ek  ek  G1  ek  ek  G1  2 for k near   a Ek      V G1 2  2 

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

9 Electron in a 1D Periodic Potential: Variational Solution Energy The obtained solutions E(k) are plotted on top of the free electron energy dispersion e(k) so that you can see the difference. An energy gap opens Ek  up! Eg  2V G1  2V G1 ek  Eg i G1 x i G1 x V x  V G1 e  V G1 e

2 2   k G  G    G V G   V * G   1 a 1 a 1 1 1 a a

2 ek  ek  G1  ek  ek  G1  2 for k near   a Ek      V G1 2  2 

2 ek  ek  G1  ek  ek  G1  2 for k near   a Ek      V G1 2  2 

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron in a 1D Periodic Potential: Variational Solution Energy Lets find the wavefunctions for k=/a The matrix equation becomes:

e a V G1  c a   c a  E  a     E a   V G1 e  a c  a  c  a  E  a The two solutions for V(G1) real are: 

E  a  e a  V G1   k   c a  1 1  c a  1  1  a a           c  a  2 1 c  a  2  1

  a  c a  a  c  a  a  2       i x i x   cos x  1  a a   L  a    a x  e  e   2L  2      i sin x   L  a 

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

10 Electron in a 1D Periodic Potential: Origin of the Bandgaps Energy We have: E  a  e a  V G1  2      cos x  1  i x i x    a a   L  a  E  a  a x  e  e   2L  2      i sin x   L  a  E  a Note that (for V(G1) real):  V x  V G ei G1 x  V G  ei G1 x 1 1   k  2    2 V G1 cos x  a a  a  V x  • The solutions are a standing waves (as a result of forward and backward x Bragg scattering) Lower energy Higher energy 2 solution  x solution • The higher energy  a solution has larger probability density in the region of higher potential

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Electron in a 1D Periodic Potential: Summary

Summary of Findings: • For a perturbative periodic potential with the following Fourier Series representation,

i G1 x i G1 x V x  V G1 e V G1 e 2 the plane wave eigenfunctions of the free electron G1  with wavevector k get coupled with the wavevectors a (k+G1) and (k+G-1) as a result of the fact that the 2 G1    G1 potential had wavevectors G1 and G-1 in its Fourier a series. • If the electron wavevector k is such that e(k) and Energy e(k+G1) have the same energy, or if e(k) and e(k+G-1) have the same energy, then a bandgap of magnitude

2|V(G1)| will open up in the free electron dispersion for the wavevector value k

ek  e k  G1  ek   ek  G1 2V G1 G G  k   1  k   1 2 2   k  a a Bandgap will open for these values of k

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

11 Electron in a 1D Periodic Potential: More General Case

Now suppose the potential looks like: Energy

i G1 x i G1 x V x  V G1 e V G1 e i G2 x i G2 x V G2 e V G2 e Bandgaps will open at these k-points: 2V G2  (1) ek  e k  G1  G   k   1  2 a (2) ek  e k  G1  G  2V G   k   1   1 2 a (3) 2   2 k ek  e k  G2    G 2 a a a a  k   2  2 a 2 2 (4) G1  G1    G1 ek  ek  G2  a a G2 2 4 4  k     G2  G2    G2 2 a a a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Bandgaps and Bragg Planes

Bandgaps will open at these k-points: Energy

(1) ek  e k  G1  G   k   1  2 a (2) ek  e k  G1  G   k   1   2 a (3) ek  e k  G2  G 2  k   2  2 a (4) ek  ek  G2  G 2 2   2 k  k   2     2 a a a a a Bandgaps open at Bragg points (1D), lines (2D), planes (3D) in the reciprocal space. Recall that a wavevector is on a Bragg point (1D), line (2D), plane (3D) if the following condition holds:  2   G Gm k . G   and for 1D it becomes: k   2 2

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

12 Bandgaps and Brillouin Zone Boundaries

Energy

Some very important observations: 2nd BZ1st BZ 1st BZ 2nd BZ

• Bandgaps open at Bragg points (1D), lines (2D), planes (3D) in the reciprocal space.

• The Bragg points (1D), lines (2D), planes (3D) define the boundary between Brillouin zones

 Bandgaps open at the Brillouin zone boundaries

2   2 k   a a a a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Restricted k-Space Convention and Energy Bands Consider any value of the wavevector outside the Energy FBZ, as shown

The unperturbed solution would be plane wave of wavevector k: 1  x  ei k x k L

The periodic potential perturbation would couple this plane wave state with all other states that are separated from it in k-space by reciprocal lattice 2   2 k   vectors. Therefore the actual solution would look a a a a something like: label original k value k value 1  x  c k  G ei k Gm x k  m  label original m L k value k value The above is a superposition of plane waves with wavevectors that differ from the unperturbed wavevector by reciprocal lattice vectors

The convention is to label the actual solutions  k  x  not by the k-value of the unperturbed wavefunction but by that wavevector in the superposition solution that falls in the FBZ, as shown

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

13 The Restricted k-Space Convention and Energy Bands

1 i k Gm x In the actual solution:  k x  c k  Gm  e m L The k-value used for labeling is always understood to be in the first BZ

Consequently, the energy-vs-k dispersion relation is always drawn only for the first BZ by translating the energy-vs-k curves lying in higher BZs to the the first BZ by appropriate reciprocal lattice vectors, as shown below:

The resulting different “bands” of energy in the first BZ are called “energy bands” and are labeled as n=1,2,3,….

Energy Energy

n=3

n=2

n=1 2   2 k 2   2 k     a a a a a a a a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Restricted k-Space Convention and Energy Bands Energy Since now we have multiple energy values for the same k-label, we use an additional label “n” to indicate the energy band. The final solutions n=3 and energy values are then written as follows:

 n,k x and En k  n=2 where k-value is understood to be in the first BZ. And the solution can be expanded as: n=1 1 i kGm x 2   2 k  n,k x  cn k  Gm  e   m L a a a a

Bloch’s theorem check: We know that solutions of the Schrodinger equation in periodic potentials (Bloch functions) need to satisfy the Bloch’s theorem:     r  R  ei k . R  r

1 i k Gm xR i k R 1 i k Gm x  n,k x  R  cn k  Gm e  e cn k  Gm e m L m L i k R  e  n,k x

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

14 From Free Electron Dispersion to Energy Bands – 1D One can always get an approximate idea of how the bands will look: Energy Fold the free- Energy electron band Start from into the first free electron BZ dispersion

Open up bandgaps

2   2 k 2   2 k     a a a a a a a a Energy Open up Energy bandgaps at edges of BZs

Fold into the first BZ

2   2 k   2   2 k a a a a   a a a a ECE 407 – Spring 2009 – Farhan Rana – Cornell University

From Free Electron Dispersion to Energy Bands: Zone Folding

Energy Fold the free- Energy electron band into the first BZ

2   2 k 2   2 k     a a a a a a a a Energy An easy trick to construct the free electron bands folded into the first BZ is to assume that a replica of the free electron band is sitting at each reciprocal lattice point

2   2 k   a a a a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

15 Generalization to Higher Dimensions - I Consider a 2D or a 3D crystal with the periodic potential given as:     i G j . r V r  V G j e j  • The potential will couple the free-electron state with wavevector k to all other   states with wavevectors k  G j 2   2 2  2 • The strongest coupling will be with states whose energy  k  G j equals  k  2m 2m • Therefore, strong coupling will occur if the wavevector k satisfies: 2   2 2  2  k  G j  k  2m 2m    2  2G j .k  G j  0  2   G j  k.G   j 2  • Since, the reciprocal lattice vector G j is arbitrary, one can also write the above condition as:  2   G k . G   Bragg condition 2 ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Generalization to Higher Dimensions - II In a 1D lattice, bandgaps opened up at k-values at the Bragg points (edges of BZs):

3 2231

1D reciprocal lattice 2D square reciprocal lattice

Same thing happens in higher dimensions: bandgaps open up for wavevectors that lie on the Bragg lines (2D), planes (3D). 3 3 2 • Recall that a wavevector will like on a Bragg 3 3 line/plane if it satisfies: 1 2  2 2   G k . G   3 3 2  for some reciprocal lattice vector G 2 3 3 • Bragg lines/planes in k-space are perpendicular bisectors of some reciprocal lattice vector

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

16 Generalization to Higher Dimensions - III • Bandgaps will open up at the edges of the 2D square reciprocal lattice Brillouin zones

• Wavevector is restricted to the first BZ, and electron energy-vs-k dispersion curves in higher BZs can be translated by appropriate 3 3 reciprocal lattice vectors to be in the first BZ 2 3 3 to obtain energy bands 1 2 • Electron energies and solutions are written 2 as: 3 3     n,k r  and En k  2 3 3

• The solutions satisfy the Bloch’s theorem:     r  R  ei k . R  r and can be written as a superposition of plane waves, as shown below for 3D:

    1 i kG . r   j  n,k r  cn k  G j  e j V

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Energy Bands of a 2D Square Lattice - I

2D square direct lattice 3 3 2 a 3 3 1 2 2 a 2 3 3

a 2 Question: How to draw the free electron 3 3 bands?

Answer: Assume a free electron band sitting at each reciprocal lattice point and 2 then consider its contribution to the bands in the first BZ a Reciprocal lattice  2 2 G  m xˆ  n yˆ a a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

17 Energy Bands of a 2D Square Lattice - II • It is obviously difficult to draw bands for 2D or 3D lattices

• The bands are usually drawn along some 3 3 specific high-symmetry directions in k-space. 2 X The figure below shows the bands from  to 3 M 3 the X point (the numbers indicate the 1 degeneracy of each energy band) 2 2  Energy 2 3 3 a 2 3 3 1

2 2 a 1 Reciprocal lattice

1  2 2 G  m xˆ  n yˆ a a (0,0) (0,/a)  X ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Energy Bands of a 2D Square Lattice - II

Energy Energy

1 1

2 2

1 1

1 1

(0,0) (0,/a) (0,0) (0,/a)  X  X

• Once the free electron energy bands have been drawn in the first BZ then the locations where bandgaps are likely to be opened are identified

• A rough sketch of the actual bands can then be made, as shown above

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

18 Energy Bands of a 2D Square Lattice - III

   0,   a     0,   a         ,0 0,   a   a  X

 2 FBZ a 2 a

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Appendix: Obtaining the 2x2 Matrix Equation (On Slide 14)

Remember the matrix element of the periodic potential between the plane wave states:  Vˆ   V G  k' k   m  k'k , Gm m

Trial solution for values of k near G1:   c k   c k  G  k  k  1 k G1 Plug it into the Schrodinger equation: ˆ Ho  V x   k  Ek  k

And then take the bra with  k to get: ˆ ˆ k Ho  V x   k  Ek  k  k   Hˆ  Vˆ x c k   c k  G  k o k 1 k G1  E k  c k   c k  G  k k 1 k G1  e k c k   Vˆ x  c k  G  E k c k  k k G1 1 

 ek c k  V G1 c k  G1  Ek c k First result

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

19 Appendix: Obtaining the 2x2 Matrix Equation

  c k   c k  G  k  k  1 k G1 Plug it into the Schrodinger equation: ˆ Ho  V x   k  Ek  k  And then take the bra with k  G  1 to get:

 Hˆ  Vˆ x   E k   k G1  o  k   k G1 k   Hˆ  Vˆ x c k   c k  G  k G1 o k 1 k G1  E k  c k   c k  G  k G1 k 1 k G1  e k  G c k  G   Vˆ x  c k  E k c k  G 1 1 k G1 k  1

 ek  G1 c k  G1  V G1 ck  E k ck  G1 Second result

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Appendix: Obtaining the 2x2 Matrix Equation

We have the two equations:

1 e k c k  V G1 ck  G1  Ek  ck

2 ek  G1 ck  G1  V G1 ck  Ek  ck  G1

which can be written in the matrix form:

 ek V G1  ck   ck      Ek   V G1 e k  G1 ck  G1  ck  G1 

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

20