Handout 17. Lattice Waves (Phonons) in 1D Crystals: Monoatomic Basis and Diatomic Basis

Handout 17 Lattice Waves (Phonons) in 1D Crystals: Monoatomic Basis and Diatomic Basis In this lecture you will learn: • Equilibrium bond lengths • At...
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Handout 17 Lattice Waves (Phonons) in 1D Crystals: Monoatomic Basis and Diatomic Basis In this lecture you will learn: • Equilibrium bond lengths • Atomic motion in lattices • Lattice waves (phonons) in a 1D crystal with a monoatomic basis • Lattice waves (phonons) in a 1D crystal with a diatomic basis • Dispersion of lattice waves • Acoustic and optical phonons

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

The Hydrogen Molecule: Equilibrium Bond Length  V r 

The equilibrium distance between the two hydrogen atoms in a hydrogen molecule is set by the balance among several different competing factors:

EA EB

• The reduction in electronic energy due to co-valent bonding is 2Vss . If the atoms are too far apart, Vss becomes to small



-d

0

x

d



1s r  dxˆ  Hˆ 1s r  dxˆ   Vss • If the atoms are too close, the positively charged nuclei (protons) will repel each other and this leads to an increase in the system energy

E A  E1s  Vss 2 : E1s

2Vss

1: EA 1 : EB

EB  E1s  Vss

• Electron-electron repulsion also plays a role

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

1

A Mass Attached to a Spring: A Simple Harmonic Oscillator Equilibrium position

Stretched position

M

M u

x 0

x

0

Potential Energy:

1 PE  V u   k u 2 2 spring constant = k (units: Newton/meter)

Kinetic Energy:

PE varies quadratically with the displacement “u ” of the mass from the equilibrium position

KE 

M  du    2  dt 

2

Dynamical Equation (Newton’s Second Law):

M

d 2u dt

2



Restoring force varies linearly with the displacement “u ” of the mass from its equilibrium position

dV  k u du

Solution:

u t   A cosot   B sinot 

o 

where:

k M

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

A 1D Crystal: Potential Energy Atoms can move only in the x-direction

  Rn  n a1

Consider a 1D lattice of N atoms:

 a1  a xˆ

x

• The potential energy of the entire crystal can be expressed in terms of the positions of the atoms. The potential energy will be minimum when all the atoms are at their equilibrium positions.



• Let the displacement of the atom at the lattice site given by Rn from its equilibrium  position be u Rn

 

• One can Taylor expand the potential energy of the entire crystal around its minimum equilibrium value:

     



    V u R1 , u R2 , u R3 .......... .. u RN   VEQ   j



0

  EQ  

V  u R j

 u Rj

 2V 1    2 k j u R j u Rk

    EQ u R j  u Rk  



Potential energy varies quadratically with the displacements of the atoms from their equilibrium positions ECE 407 – Spring 2009 – Farhan Rana – Cornell University

2

A 1D Crystal: Potential and Kinetic Energies   Rn  n a1

A1D lattice of N atoms:

 a1  a xˆ

x

Potential Energy:

V  VEQ   VEQ 

1  2V    2 k j u R j u Rk

    EQ u R j , t  u Rk , t 





 

 



    1   K R j , Rk u R j , t u Rk , t 2k j

Kinetic Energy:

KE   j



 M  du R j , t 2  dt







  K R j , Rk 

 2V   u R j u Rk

    EQ

2  

• The kinetic energy of all the atoms is the sum of their individual kinetic energies

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

A 1D Crystal: Dynamical Equation   Rn  n a1

A1D lattice of N atoms:

 a1  a xˆ

x

 Write Newtons law for the atom sitting at the site Rn : M



 d 2u Rn , t dt

2

 



 

   V     K Rn , R j u R j , t u Rn j

 



Remember that:





  K R j , Rk 

 2V   u R j u Rk

    EQ

The restoring forces on the atoms vary linearly with the displacement of the atoms from their equilibrium positions

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

3

Dynamical Equation for Nearest-Neighbor Interactions A1D lattice of N atoms:

M



 d 2u Rn , t dt

2

 



 

   V     K Rn , R j u R j , t u Rn j

 



Assume nearest-neighbor interactions:





  K Rn , R j    j ,n  1    j ,n 1  2  j ,n This gives:

M



 d 2u Rn , t dt

2

   u R , t   u R , t    u R , t   u R , t  n n 1 n n 1

The constants “” provide restoring forces as if the atoms were connected together with springs of spring constant “” The constant  is called “force constant” (not spring constant) in solid state physics We have N linear coupled differential equations for N unknowns

 u Rn , t 

 n  0,1,2.......N  1

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Solution of the Dynamical Equation: Lattice Waves (Phonons)  a1  a xˆ

A1D lattice of N atoms:

M



 d 2u Rn , t dt

2

  Rn  n a1

   u R , t   u R , t    u R , t   u R , t  n n 1 n n 1

Assume a solution of the form:



    u Rn , t   Re u q  e i q . Rn e  i  t

Or:

    u Rn , t   u q  e i q . Rn e  i  t



Represents a wave with , wavevector q , frequency  and amplitude u q  Slight abuse of notation

Note that:         u Rn  1, t   u q  e i q . Rn  1 e  i  t  u q  e i q . Rn  a1  e  i  t       e i q . a1u q  e i q . Rn e  i  t         u Rn 1, t   u q  e i q . Rn 1 e  i  t  u q  e i q . Rn  a1  e  i  t       e  i q . a1u q  e i q . Rn e  i  t ECE 407 – Spring 2009 – Farhan Rana – Cornell University

4

Solution of the Dynamical Equation: Lattice Waves (Phonons)

M



 d 2u Rn , t dt

2

  Rn  n a1

 a1  a xˆ

A1D lattice of N atoms:

   u R , t   u R , t    u R , t   u R , t  n n 1 n n 1 











Plug in the assumed solution: u Rn , t  u q  e i q . Rn e  i  t To get:



q  q x xˆ

 



           2 M u q    u q   e  i q . a1 u q    u q   e  i q . a1 u q 

Which simplifies to:

  2 1  cosq . a1 M   4  q . a1   sin2   M  2 

2 

Or:



  4  q . a1  sin  M  2 

Since  is always positive, the negative sign is chosen when the sine term is negative

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Solution of the Dynamical Equation: Lattice Waves (Phonons)













Solution is: u Rn , t  u q  e i q . Rn e  i  t

 u Rn 

  Rn  n a1

 a1  a xˆ

A1D lattice of N atoms:

and



  4  q . a1  sin  M  2 

a x



x

• The lattice waves are like the compressional sound waves in the air

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

5

Solution of the Dynamical Equation: Lattice Waves (Phonons)













Solution is: u Rn , t  u q  e i q . Rn e  i  t

 u Rn 

  Rn  n a1

 a1  a xˆ

A1D lattice of N atoms:



and

  4  q . a1  sin  M  2 

a x

 The relation:





  4 4  q . a1  q a sin sin x   M M  2   2 

represents the dispersion of the lattice waves or phonons







a

a

qx

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Lattice Waves and the First BZ













Solution is: u Rn , t  u q  e i q . Rn e  i  t



and

  4  q . a1  sin  M  2 

Question: What is the shortest wavelength (or the largest wavevector) the lattice waves can have?   = 2a uR

 n

x For the shortest wavelength:

  2a

 First BZ

The largest wavevector is then:

qx 

2





 a

The wavevector values can be restricted to the First BZ • No new solutions are found for values of the wavevector outside the first BZ







a

a

qx

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

6

Phase and Group Velocities 



4 q a sin x  M  2 

First BZ

Phase velocity and group velocity of lattice waves are defined as: 

   q  v p q    qˆ q

   v g q    q  q 



1D

  1D

qx

 





d xˆ dq x





a

a

qx

Case I: For qx ≈ 0 (i.e. qx a