Electronic Properties of Materials

재료의 전자기적 성질

Electronic Properties of Materials

Instructor : Prof. Sang-Im Yoo

Office : 33-224, Tel : 880-5720, E-mail : [email protected] Teaching Assistant : Jung-Woo Lee

Website : sm2l.snu.ac.kr

Fall Semester, 2008

Grading

Midterm Exam 30%

Final Exam 40%

Homework & Attendance 30%

(# absence more than 4 lectures = F)

Overall Contents

Part I Fundamentals

Electron Theory : Matter Waves

Electromagnetic Theory : Maxwell Equations

Part II Electrical Properties of Materials Part III Optical Properties of Materials Part IV Magnetic Properties of Materials Part V Thermal Properties of Materials

Lattice Waves

Part I Fundamentals

Electron Theory : Matter Waves

Chap. 1 Introduction

Chap. 2 The Wave-Particle Duality

Chap. 3 The Schördinger Equation

Chap. 4 Solution of the Schördinger Equation for Four Specific Problems

Chap. 5 Energy Bands in Crystals Chap. 6 Electrons in a Crystal

Electromagnetic Theory : Maxwell Equations

Chap. 4 Light Waves

(Electrons in Solids, 3

^rd

Ed., R. H. Bube)

Three approaches to understand electronic properties of materials

- Continuum theory :

consider only macroscopic quantities, interrelate experimental data

ex) Ohm’s law, Maxwell equations, Newton’s law, and Hagen-Rubens equation

- Classical electron theory :

postulate that free electrons in metals drift as a response to an external force and interact with certain lattice atoms

ex) Drude equations

- Quantum theory :

explain important experimental observations

which could not be readily interpreted by classical means ex) Schrödinger Equation

1. Introduction

1. Introduction

¾ Basic equations

ma aw : F

Newton's l =

2

2 : 1

energy E m υ

Kinetic

_kin

= υ m p

Momentum : = m

E

_kin

p

2

=

2

νλ

= c Speed of light :

νλ υ =

: wave of

elocity

V ^{Angular frequency :} ω = ² πν

:

2

e equivalenc energy

- mass s

Einstein' E = mc

2. The Wave-Particle Duality

¾ Light : electromagnetic wave

light quantum (called a photon)

Energy

Planck constant

1924 yr de Broglie

“Wave nature of electrons” “Matter wave”

For a general wave

“ Wave number”

ω h

ν = E = h

π 2

= h h

h p =

λ

νλ υ ⁼

λ π

= 2

k k

υ ⁼ ω

2. The Wave-Particle Duality

¾ Description of electron wave

- The simplest waveform : harmonic wave

- A wave function (time- and space-dependent)

Electron wave : a combination of several wave trains Assuming two waves,

) sin( kx − ω t

= Ψ

]

1

= sin[ kx − ω t Ψ

] ) (

) sin[(

2

= k + Δ k x − ω + Δ ω t

Ψ

2. The Wave-Particle Duality

¾ Description of electron wave Supposition of two waves:

Modulated amplitude sine wave

] 2 )

( 2 )

sin[(

2 ) cos( 2

2

2

1

k x t

k k x

t ω ω

ω Δ

+ Δ −

+ Δ ⋅

Δ −

= Ψ

= Ψ + Ψ

“Wave Packet”

2. The Wave-Particle Duality

The extreme conditions

(a) No variation in angular frequency and wave number :

monochromatic wave

2. The Wave-Particle Duality

The extreme conditions

(b) Very large variation in angular frequency and wave number

' '

2 /

2 /

k k

k t

x ω ω ω

υ =

Δ +

Δ

= +

=

dk d k

t x

g

ω

υ ω =

Δ

= Δ

=

Phase velocity :

velocity of a matter wave

Group velocity:

velocity of a pulse wave

(i.e., a moving particle)

2. The Wave-Particle Duality

The extreme conditions

(b) Very large variation in angular frequency and wave number

h x

p ⋅ Δ ≥ Δ

τ

d

dxdydz

^*

*

= ΨΨ

ΨΨ

Heisenberg’s Uncertainty principle Probability of finding a particle

at a certain location

Q&A

1. Plank constant h = 6.63×10

^-34

J·sec 2. Traveling wave

3. Phase velocity versus Group velocity

4. Prove υ

_g

= υ (velocity of particle)?

E = hv = ħ ω and k = p/ ħ → d ω _{= dE/} ħ and dk = dp/ ħ υ

_g

= dω/dk = dE/dp

Since E = mυ

²

/2 and p = mυ, dE/dp = υ

Relativistic expressions: E = mc

²

, E = hv and p = mυ

υ

_g

= dω/dk = dω/dυ/dk/ /dυ = υ )

( 2 sin )

,

( x vt

t

x = −

Ψ π λ

) sin( kx − ω t

= Ψ

' '

2 /

2 /

k k

k t

x ω ω ω

υ =

Δ +

Δ

= +

= dk

d k

t x

g

ω

υ ω =

Δ

= Δ

= / 2

2

/

Q&A

- Why is the wave nature of matter not more apparent to us in our daily observation?

- Can the de Bloglie wavelength of a particle be smaller than a linear

dimension of the particle? Larger? Is there necessarily any relation between such quantities?

- Is the frequency of a de Broglie wave given by E/h? Is the velocity given by

υ ? Is the velocity equal to c? Explain

Q&A

Mathematical description of traveling waves

Consider a string stretched along the x axis whose vibrations are in the y direction Assuming simple harmonic motion,

At t = 0, y = Asin2πvt

where A is the amplitude of the vibrations

If t is replaced by , then y =

Asin2πv( ) : Wave Formula where

υ

is the wave speed

Since the wave speed is given by υ = v λ _{, we have} x − t

υ _υ ^{x − t}

) (

2

sin x vt A

y = −

π λ

Part I Fundamentals

Electron Theory : Matter Waves

Chap. 1 Introduction

Chap. 2 The Wave-Particle Duality

Chap. 3 The Schördinger Equation

Chap. 4 Solution of the Schördinger Equation for Four Specific Problems

Chap. 5 Energy Bands in Crystals Chap. 6 Electrons in a Crystal

Electromagnetic Theory : Maxwell Equations

Chap. 4 Light Waves

(Electrons in Solids, 3

^rd

Ed., R. H. Bube)

3. The Schrödinger Equation

3.1 The Time-Independent Schrödinger Equation

- Time-independent Schrödinger equation: a vibration equation

where, m = the (rest) mass of the electron,

E = the total energy of the system, E_kin

= kinetic energy,

V = the potential energy (or potential barrier)

- Applicable to the calculation of the properties of atomic systems in stationary conditions

0 )

2 (

2

2

+ − =

∇ ψ m E V ψ

h

²

2 2

2 2

2 2

z y

x ∂

+ ∂

∂ + ∂

∂

= ∂

∇ ψ ψ ψ ψ

V E

E =

_kin

+

3. The Schrödinger Equation

3.2 The Time-Dependent Schrödinger Equation

Time-dependent Schrödinger equation: a wave equation

Since

and

Then

Applying differential operators to the wave function (Hamiltonian operators)

ω h ν =

= h E

2 0 2

2

2

=

∂ Ψ

− ∂ Ψ

− Ψ

∇ t

mi mV

h h

t

e

i

z y x t

z y

x = ψ ⋅

^ω

Ψ ( , , , ) ( , , ) ω ω

ψ i e

^ω

i t

t

i

= Ψ

∂ = Ψ

∂

0 )

2 (

2

2 + − =

∇

ψ

m E V

ψ

h

2 2 0

2

2

=

∂ Ψ

− ∂ Ψ

− Ψ

∇

t

mi mV

h h

i t

E ∂

− ∂

= h p = − h i ∇ m V

E p E

E

_total

=

_kin

+

_pot

= +

2

2

= ∇ Ψ + Ψ

∂ Ψ

− ∂ V

m i

i t

²

2 2

2

h h

3. The Schrödinger Equation

3.3 Special Properties of Vibrational Problems

- When boundary conditions are imposed, only certain vibrational forms are possible. ex) a vibrating string

- Vibration problems determined by boundary conditions : Boundary (or eigenvalue) problems

A pecularity of these problems : not all frequency values are possible and therefore, not all values for the energy are allowed because of

The allowed values : eigenvalues

The function belonging to the eigenvalues as a solution of the vibration equation : eigenfunctions

The normalized eigenfunction:

h E = ν

2

1

*

= ∫ =

∫ ^{ψψ d τ ψ d τ}

4. Solution of Schrödinger Equation

4.1 Free Electrons

Suppose electrons propagating freely (i.e., in a potential-free space) to the positive x-direction.

Then V = 0 and thus

The solution for the above differential equation for an undamped vibration with spatial periodicity, (see Appendix 1)

where Thus

“energy continuum”

0 )

2 (

2

2

+ − =

∇ ψ m E V ψ

h

x

Ae

i

x

^α

ψ ( ) =

m E

2

2

= h α

t i x

i

e

Ae

x =

^α

⋅

^ω

Ψ ) (

2 2

2 α

E

= h

m

p k m E

=

=

=

= λ

α ²

₂

² π h

h

2 2

2 k E = h m

λ π

= 2

k

4. Solution of Schrödinger Equation

4.2 Electron in a Potential Well (Bound Electron)

Consider an electron bound to its atomic nucleus.

Suppose the electron can move freely between two infinitely high potential barriers

At first, treat 1-dim propagation along the x-axis inside the potential well

The solution where

= 0 0 ψ

ψ =

2 0

2 2

2

ψ + ψ =

m E dx

d

h

x i x

i

Be

Ae

^{α α}

ψ = +

⁻

^{2 m}

₂

^E

= h

α

4. Solution of Schrödinger Equation

4.2 Electron in a Potential Well (Bound Electron)

Applying boundary conditions,

x = 0,

x = a

With Euler equation,

Finally,

“energy quantization”

= 0 ψ

= 0 ψ

) 2 (

sin ρ 1 e

^iρ

e

^iρ

i

−

−

=

0 sin

2 ]

[ e − e

⁻

= Ai ⋅ a = A

^{iαa iαa}

α

..

0,1,2,3,..

, =

= n n

a π

α

1,2,3,....

2 ,

2

2 2 2 2 2

2

=

=

= n

ma n E

_n

^h m α ^h π

“energy levels”

z e r p

4. Solution of Schrödinger Equation

4.2 Electron in a Potential Well (Bound Electron)

Now discuss the wave function

z e r p

x Ai α

ψ = 2 ⋅ sin ψ

^*

= 2 ^Ai ⋅ sin α ^x x

A α

ψψ

^*

= 4

²

sin

²

λ π r = n 2

n

r π

λ

= 2

1 ]

cos 2sin

[ 1 ) 4

( sin

4 ₀

2

0

2 2

0

* =

∫

= − + =

∫

^{aψψ dτ A a αx dx} _α^{A αx αx α}_x^{x a}

A a

2

= 1

4. Solution of Schrödinger Equation

4.2 Electron in a Potential Well (Bound Electron)

For a hydrogen atom, Coulombic potential

In 3-dim potential

The same energy but different quantum numbers: “degenerate” states

z e r p

) 1 (

6 . 1 13

) 4

(

2

₀ ^{2 2 2}

4

n eV n

E = me = − ⋅

πε h

) 2 (

2 2

2 2

2 2

z y

x

n

n n n

E = _h ma π + + r

V e

0 2

4 πε

−

=

4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

Suppose electrons propagating in the positive x-direction encounter a potential barrier V

₀

(> total energy of electron, E)

- Region (I) x < 0

2 0

2 2

2

ψ + ψ =

m E dx

d

h

x i x

i

I

Ae

^α

Be

^α

ψ = +

^{− m}2 ^E

2

= h

α

- Region (II) x > 0

0 )

2 (

2 0 2

2

ψ + − ψ =

V m E

dx d

h

The solutions (see Appendix 1)

x i x

i

II

Ce

^β

De

^β

ψ = +

^{− 2}m2 ⁽E V⁰⁾

−

= h

β

4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

Since E – V

₀

is negative, becomes imaginary.

To prevent this, define a new parameter,

Thus, , and

Determination of C or D by B.C. For x → ∞

Since Ψ Ψ

^*

can never be lager than 1,

^{→ ∞}

is no solution, and thus , which reveals Ψ-function decreases in Region II

) 2 (

2

E V

0

m −

= h β

β γ = i )

2 (

2

V

0

E

m −

= h

γ ψ

_II

= Ce

^iβx

+ De

^−iβx

⋅ 0 +

∞

⋅

= C D

ψ

II

→ 0 C

x

II

De

^γ

ψ =

⁻

ψ

II

Using (A.27) + (4.39) in textbook, the damped wave becomes

) ( t kx i

x

e

De

⁻

⋅

⁻

=

Ψ

^{γ ω}

4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

As shown by the dashed curve in Fig 4.7, a potential barrier is penetrated by electron wave : Tunneling

* For the complete solution,

(1) At x = 0 : continuity of the function

ψ

_I

= ψ

_II

x i x

i x

i

Be De

Ae

^α

+

^{− α}

=

^γ

(2) At x = 0 : continuity of the slope of the function

With x = 0 Consequently,

dx d dx

d ψ

_I

_≡ ψ

_II

x x

i x

i

Bi e De

e

Ai α

^α

− α

^{− α}

= − γ

^−γ

D Bi

Ai α − α = − γ

)

2 ( α

i γ D a

A = +

) 1

2 ( α

i γ

B = D −

4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

The behavior of an electron in a crystal → A motion through periodic repetition of potential well

well length : a barrier height : V₀ barrier width : b

Region (I)

Region (II) 2 0

2 2

2

ψ + ψ =

m E dx

d

h

0 )

2 (

2 0 2

2

ψ + − ψ =

V m E

dx d

h

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

(Continued) For abbreviation

The solution of this type equation (not simple but complicate)

Where, u(x) is a periodic function which possesses the periodicity of the lattice in the x-direction

The final solution of the Schrödinger equations;

where

m E

2

2

2

= h

α ²

2

⁽

0

⁾

2

m V E

−

= h γ

e

ikx

x u

x ) = ( ) ⋅

ψ (

ka a a

P sin a + cos α = cos α

α

(Bloch function)

2 0

h

b

P = maV

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State) Mathematical treatment for the solution : Bloch function

Differentiating the Bloch function twice with respect to x

Insert 4.49 into 4.44 and 4.45 and take into account the abbreviation

e

ikx

x u

x ) = ( ) ⋅ ψ (

e

ikx

u k dx ik

du dx

u d dx

d (

₂

2

²

)

2 2

2

ψ = + −

0 )

(

2

^{2 2}

2

2

+ − k − u =

dx ik du dx

u

d α _{2 (}

^{2 2}

_{) 0}

2

2

+ − k + u =

dx ik du dx

u

d γ

(I) (II)

The solutions of (I) and (II)

) (

^{i x i x}

ikx

Ae Be

e

u =

^{− α}

+

^{− α}

^(I) u = e

^−ikx

( Ce

^−γx

+ De

^γx

) ^(II)

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

(Continued) From continuity of the function

du/dx values for equations (I) & (II) are identical at x = 0

Further, Ψ and u is continuous at x = a + b → Eq. (I) at x = 0 must be equal to Eq. (II) at x = a + b, Similarly, Eq. (I) at x = a is equal to Eq. (II) at x = b

Finally, du/dx is periodic in a + b

limiting conditions : using 4.57- 4.60 in text and eliminating the four constant A-D, and using some Euler eq.(see Appendix 2)

dx and d ψ ψ

D C

B

A + = +

) (

) (

) (

)

( i ik B i ik C ik D ik

A α − + − α − = γ − + γ −

b ik

b ik

a ik i

a ik

i

Be Ce De

Ae

^{( α− )}

+

^{(− α− )}

=

^{( +γ )}

+

^{( −γ )}

b ik b

ik k

ia k

ia

Bi k e C ik e D ik e

e k

Ai ( α − )

^{(α− )}

− ( α + )

^{− (α+ )}

= − ( γ + )

^{( +γ)}

+ ( γ − )

^{( −γ)}

) (

cos )

cos(

) cos(

) sin(

) 2 sin(

2 2

b a

k a

b a

b ⋅ + ⋅ = +

− γ α γ α

αγ

α

γ

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

If V

₀

is very large, then E in 4.47 is very small compared to V

₀

so that

Since V

₀b has to remain finite and b → 0,

γ

b becomes very small.

For a small γ

b, we obtain (see tables of the hyperbolic function)

Finally, neglect α

²

compared to γ

²

and, b compared to a so that 4.61 reads as follow

Let , then

2 0

2 m V

= h

γ ⅹ b

→

b b

and

b γ γ

γ ) ≈ 1 sinh ( ) ≈ cosh(

ka a

a b

m V

cos cos

0

sin

2

α + α =

α _h

2 0

h

b

P = maV ^{a ka}

a

P sin a + cos α = cos α

α

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

“Electron that moves in a periodically varying potential field can only occupy

certain allowed energy zone”

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

The size of the allowed and forbidden energy bands varies with P.

For special cases

(a) If the potential barrier strength, V

₀b

is large, P is also large and the

curve on Fig 4.11 steeper. The allowed band are narrow.

(b) V

₀b and P are small, the allowed

band becomes wider.

(c) If V

₀b goes 0, thus, P → 0

From 4.67, cos α a cos = ka

m E k

2

2

h

2

=

4. Solution of Schrödinger Equation

4.4 Electron in a Periodic Field of Crystal (the Solid State)

(d) If the V

₀b is very large, P → ∞

sin → 0 a

a α

α

0 sin α a →

1,2,3,....

n for

2 2 2

2

= =

a n π α

2 2

2 2

2 n

E = π ma h ⋅

α _{a = n} π

Combining 4.46 and 4.69

Part I Fundamentals

Electron Theory : Matter Waves

Chap. 1 Introduction

Chap. 2 The Wave-Particle Duality Chap. 3 The Schördinger Equation

Chap. 4 Solution of the Schördinger Equation for Four Specific Problems

Chap. 5 Energy Bands in Crystals

Chap. 6 Electrons in a Crystal

Electromagnetic Theory : Maxwell Equations

Chap. 4 Light Waves

(Electrons in Solids, 3

^rd

Ed., R. H. Bube)

5. Energy Bands in Crystals

5.1 One-Dimensional Zone Schemes

For free electrons, the wave number in 1-dim

2 2

2 k

E = h m k

_x

= const .E

^1/2

ka a a

P sin a + cos α = cos α

α ^{If P = 0,} cos α a cos = ka

2 / 1 2

2 m E

= h α

) 2 cos(

cos

cos α a = k

_x

a ≡ k

_x

a + n π π

α a = k

_x

a + n 2

_1/2

2

2

2 m E

n a k

_x

= h

+ π

: more general form in 1-dim ,....

2 , 1 ,

0 ± ±

= n

}

In a crystal

5. Energy Bands in Crystals

2 / 1 2

2

2 m E

n a k

_x

= h

+ π

,..., 3 , 2 1,

, = ± ± ±

= n n

a

k

_x

π

n a k

_x

_{= ⋅} π

1 cos k

_x

a = ±

If an electron propagates in a periodic potential, discontinuities of the electron energies are observed when cosk

_x

a has a maximum or a minimum, i.e., when

E is a periodic function of with the periodicity of

2 π / a

^kx

5.1 One-Dimensional Zone Schemes

or

At these singularities, a deviation from the parabolic E vs k

_x

occurs and the branches

of the individual parabolas merge into the neighboring ones (see Fig.5.3)

5. Energy Bands in Crystals

The electrons in a crystal behave like free electrons for most k

_x

value except k

_x

→

n

π

/a

periodic zone scheme

(see Fig 5.3)

reduced zone scheme

(see Fig 5.4)

π/a

≤

k_x

≤

π/a

5.1 One-Dimensional Zone Schemes

extended zone scheme

(see Fig 5.5) Deviations from the free electron

parabola at the critical points k

_x = n

∙

_π/a

are particularly easy to identify.

free electron bands

(see Fig 5.6)

Free electrons in a reduced zone scheme from

5. Energy Bands in Crystals

,....

2 , 1 0,

, 2 )

2 (

2

2

+ = ± ± ±

= n

n a m k

E _h

_x

π

5.1 One-Dimensional Zone Schemes

2 / 1 2

2

2 m E

n a k

_x

= h

+ π

5. Energy Bands in Crystals

origin) as

0

( 2

,

0

²

2

with parabola

m k E

n = = h

_x

2 2 2

2 2 2 2 2

1 2

, For

4 2

, 0 For

origin) as

2

( 2 )

2 (

, 1

E ma k a

E ma k

with a parabola

k a E m

n

x x

x

h h h

π π

π

π π

=

=

=

=

−

=

−

=

,....

2 , 1 0,

, 2 )

2 (

2

2

+ = ± ± ±

= n

n a m k

E _h

_x

π

By inserting different n-values, one can calculate the shape of branches of the free electron bands

5.1 One-Dimensional Zone Schemes

5. Energy Bands in Crystals

5.2 One- and Two-Dimensional Brillouin Zones

1-d Brillouin Zone

The first Brillouin Zone (BZ) : π/a ≤ _kx≤ π/a : n-Band

The second Brillouin Zone (BZ):

π/a ≤ _kx≤ 2π/a, -π/a ≤ _kx≤-2π/a : m-band

-

Individual branches in an extended zone

scheme (Fig. 5.5) can be shifted by 2π/a to left or to right.

Shift the branches of 2^nd BZ to the positive side of E- k_x diagram by 2π/a to the left, and likewise the left band by 2π/a to the right → The result is shown in Fig. 5.4

(a reduced zone scheme)

- The same can be done in 3^rd BZ and all BZ (because of the 2π/a periodicity) →

relevant information of all BZ can be contained in the 1^st BZ (a reduced zone scheme)

5. Energy Bands in Crystals

2-d Brillouin Zone

Description for the movement of an electron in the potential of 2-d lattice - Wave vector k = (k_x, k_y) : 2-d reciprocal lattice (Fig 5.7)

- A 2-d field of allowed energy regions which correspond to the allowed energy band → 2-d BZ - 1st zone in 2-d: the area enclosed by four “Bragg planes” having four shortest lattice vectors, G₁:

bisectors on the lattice vectors

- For the following zone, construct the bisectors of the next shortest lattice vectors, G₂, G₃… - For the zone of higher order the extended limiting lines of the zones of lower order are used as

additional limiting lines.

5.2 One- and Two-Dimensional Brillouin Zones

Example: in 2-d lattice, an electron travels at 45

^o

to k

_x

-axis, then the boundary of the BZ is reached, according to Fig 5.8, for

“Usefulness of BZ”

- energy bands of solids (discussed in later section)

- the behavior of electrons which travel in a specific direction in reciprocal space

5. Energy Bands in Crystals

a 2 k

_crit

₌ π

) 2 (

1

2 2 2

max

a m

E ₌ π h

k

_crit

₌ π a E a

₂

m

2 2 max

₌ π h

this yields with (4.8) a maximal attainable energy of If the boundary of a BZ is reached at

the largest energy of electrons moving parallel to k

_x

or k

_y

axis

5.2 One- and Two-Dimensional Brillouin Zones

5. Energy Bands in Crystals

- Once the maximal energy has been reached, the electron waves (those of the incident and the Bragg-reflected electrons) form standing waves (the electrons are reflected back into the BZ.)

- Overlapping of energy bands: bands are drawn in different directions in k-space (Fig 5.9) :

the consequence of

a 2 k

_crit

₌ π

k

_crit

₌ π a

and

5.2 One- and Two-Dimensional Brillouin Zones

5. Energy Bands in Crystals

A different illustration of the occurrence of critical energies at which a reflection of the electron wave takes place :

Bragg relation

Since λ = 2π/k

For a perpendicular incidence, θ = 90

^o

, If θ = 45

^o

,

For increasing electron energies, a critical k-value is finally reached for which

“reflection” of the electron wave at the lattice plane occurs.

At , the transmission of electron beam through the lattice is prevented.

1,2,3,...

, sin

2 a θ = n λ n =

n k a sin θ ² π

2 = θ

π sin n a

k

_crit

=

k

_crit

₌ π a a 2

k

_crit

₌ π

k

crit

5.2 One- and Two-Dimensional Brillouin Zones

5. Energy Bands in Crystals

5.3 Three-Dimensional Brillouin Zones

- In previous section, it was shown that at the boundaries of the zones the electron waves are Bragg-reflected by the crystal.

- The wave vector, |k| = 2π/λ, was seen to have the unit of reciprocal length and thus is defined in the reciprocal lattice.

- The construction of 3-d Brillouin zones for two important crystal

structures of face centered cubic (FCC) and body centered cubic

(BCC) : important features in common with “Wigner- Seitz cells”

5. Energy Bands in Crystals

5.4 Wigner - Seitz Cells

Crystals have symmetrical properties - An accumulation of “unit cell”

- Smallest possible cell “primitive cell”

(consist of 1 atom)

- BCC, FCC : conventional non-primitive unit cells

- Wigner-Seitz cell : a special type of

primitive unit cell that shows the cubic symmetry of cubic cells

- W-S cell construction: bisects the vectors from a given atom to its nearest

neighbors and place a plane perpendicular to these vectors at the bisecting

points. For BCC (Fig 5.11) & FCC (Fig. 5. 13)

5. Energy Bands in Crystals

5.4 Wigner - Seitz Cells

- The atomic arrangement of FCC:

corners and faces of cube,

or center points of the edges and the center of the cell (Fig 5.12)

-The W-S cell for FCC shown

in Fig 5.13

5. Energy Bands in Crystals

5.5 Translation Vectors and the Reciprocal Lattice

Fundamental vectors or primitive vectors

: t

₁

, t

₂

, t

₃

Translation vectors, R :

combination of primitive vectors

where n₁, n₂, and n₃ are integers.

Three vectors for the reciprocal lattice: b₁

, b

₂

, b

₃

a translation vector for the reciprocal lattice, G

where h₁,h₂, and h₃ is integer

3 3 2

2

1

t t t

R = n

₁

+ n + n

) (

2

₁

b

₂

b

_{2 3}

b

₃

G = π h

₁

+ h + h

) 2 (

1

i j l

t = a − + +

m n

m

n

_nm

nm nm m

n

≠

=

=

=

=

for 0

and

for 1 where

,

δ δ

δ t

b

5. Energy Bands in Crystals

The relation between real and reciprocal lattices

By definition,

. 0

, 0

, 1

3 2

=

•

=

•

=

•

t b

t b

t b

1 1

1

1 Kronecker-Delta symbol

}

3 2

1

. t t

b = const × b

₁

• t

₁

= const . t

₁

• t

₂

× t

₃

= 1

3 2

1

1 t t

t • ×

= const

3 2

1

3 1 2

t t

t

t b t

×

•

= ×

3 2 1

1 3

2

t t t

t b t

×

•

= ×

3 2

1

2 1

3

t t t

t b t

×

•

= ×

5.5 Translation Vectors and the Reciprocal Lattice

5. Energy Bands in Crystals

Calculation for the reciprocal lattice of a BCC crystal Real crystal

a: lattice constant , t₁

, t

₂

, t

₃

: primitive lattice vectors

_,

i, j, l : unit vectors in the x, y, z coordinate system (see Fig. 5.14(b))

Abbreviated,

) 2 (

1

i j l

t = a − + +

) 11 1 2 (

1

= a

t ( 1 1 1 )

2

2

= a

t ( 11 1 )

3

2

= a t

) 2 (

) 2 2

4 (

) 4 (

1 1

1

1 1

4 1

2 2

2 2

3 2

l j l

j

j i l l j i k

j i

t t

+

= +

=

+

− + + +

=

−

−

=

×

a a

a a

5.5 Translation Vectors and the Reciprocal Lattice

5. Energy Bands in Crystals

(continued)

) 2 1 1 0 4 ( )

0 ( ) 4 (

3 3

3 3

2 1

a a

a − + + • + + = + + =

=

×

• t t i j l j l

t

3 2 1

3 1 2

t t t

t b t

×

•

= × 1 ( ),

2 ) 2 (

3 2

1

j l

l j

b + = +

= a a

a

) 011 1 (

1

= a

b 1 ( 101 )

2

= a

b 1 ( 110 )

3

= a b

BCC (reciprocal lattice) FCC (real lattice)

1st

Brillouin zone for BCC Wigner-Seitz cell for FCC Vice versa

5.5 Translation Vectors and the Reciprocal Lattice

Periodicity of E(k) → all information of electron contained in the 1st Brillouin Zone (BZ)

E_k'

for k' for outside 1

^st

BZ → E

_k

with in 1

^st

BZ with a suitable translation vector G

“Energy bands are not alike in different directions in k-space”

for the demonstration, “free electron band” is used (Fig 5.6 ).

In 3-D, from (5.7)

5. Energy Bands in Crystals

5.6 Free electron Bands

G k

k

^'

= +

G

2

k )

2 (

2

'

= +

E

_k

h m

,....

2 , 1 0,

, 2 )

2 (

2 2

±

±

±

= +

= n

n a m k

E _h

_x

π

(5.7)

5. Energy Bands in Crystals

In Fig 5.17, three important directions [100] from (origin) to point H : [110] from to N :

[111] from to P :

Fig 5.18 calculated by using the following eqn.

Γ Γ Γ

Δ Σ

Λ

G

2

k )

2 (

2

'

= +

E

_k

h m

5.6 Free electron Bands

5. Energy Bands in Crystals

band calculation for BCC Γ − H [ 100 ] direction

k

x

k

_Γ−Η

≡

For this direction (5.35) becomes

2 2

2 )

2 ( i + G

= x

a E _h m π

Where x may take values between 0 and 1. to start with, let G = 0, then where

this curve is labeled (000) in Fig 5. 18 since h

₁

,h

₂

,h

₃

= 0,0,0 for G=0 between 0 and 2 π

/a (boundary of BZ)

2 2

2 2

) ( 2 )

2 ( x Cx

a

E = _h m π i ≡

2 2 2

2

2

2 )

2 m ( a ma

C _{= h} π _{= h} π

5.6 Free electron Bands

5. Energy Bands in Crystals

For the case of h

₁

,h

₂

,h

₃

= 0,-1,0

combined (5.36) and (5.38)

) 2 2

( ]

1 )

1 [(

] )

1 (

[ )]

2 ( [ 2

2

2 2

2 2

2

+

−

= +

−

=

−

−

= +

−

=

x x

C x

C

x a C

a x

E _h m π i π i l i l

C E

x

C E

x

1 1

for and

2 0

=

→

=

=

→

=

) 0 1 0 (

The band labeled in Fig 5.18 obtained.

Similarly, For FCC, see Figs. 5.19 & 5. 20

) 2 (

l i

G = − +

a π

For

5.6 Free electron Bands

Band structure of actual solids:

Figs. 5.21-24

(results of extensive, computer-aided

calculations)

Directions in k-space [100] :

[110] : [111]:

5. Energy Bands in Crystals

5.7 Band Structures for Some Metals and Semiconductors

− X Γ

− K Γ

− L Γ

Band diagram for aluminum

- parabola-shaped band: free- electron like

5. Energy Bands in Crystals

Band diagram for copper

- Lower half of the diagram closely spaced and flat running bands (due to 3d-bands of Cu)

Band diagram for silicon - Band gap : near 0~ 1eV →

“semiconductor properties”

5.7 Band Structures for Some Metals and Semiconductors

5. Energy Bands in Crystals

Band diagram gallium arsenide:

so called III – IV semiconductor

Important for “optoelectronic devices”

5.7 Band Structures for Some Metals and Semiconductors

5. Energy Bands in Crystals

5.8 Curves and Planes of Equal Energy

Energy vs. wave vector, k

Fig 5.25: curves of equal energy for free electrons Fig 5.26: near boundary of BZ- deviation from a circular form (2-d)

Fig 5.27: 3-d BZ for Cu

Q&A2

Brillouin Zone in 2-d

From Solid State Physics, N.W. Aschcroft & N.

D. Mermin, Holt, Rinehart and Winston

Brillouin

Zone

in 3-d

Part I Fundamentals

Electron Theory : Matter Waves

Chap. 1 Introduction

Chap. 2 The Wave-Particle Duality Chap. 3 The Schördinger Equation

Chap. 4 Solution of the Schördinger Equation for Four Specific Problems

Chap. 5 Energy Bands in Crystals

Chap. 6 Electrons in a Crystal

Electromagnetic Theory : Maxwell Equations

Chap. 4 Light Waves

(Electrons in Solids, 3

^rd

Ed., R. H. Bube)

6. Electrons in a Crystal

6.1 Fermi Energy and Fermi Surface

The Fermi energy, E_F:

- An important part of an electron band diagram

- Defined as “the highest energy that the electrons assume at T = 0 K”

- Fermi energy for Al and Cu : see Fig 5.21, 5.22 Fermi energy for semiconductor:

- The above definition can occasionally be misleading, particularly when dealing with semiconductors

- Fermi function at E_F, F(E_F) = ½ : see Section 6.2 for more accurate definition

Fermi surface (in 3-d k-space) for Cu : see Fig 5.27

6. Electrons in a Crystal

6.2 Fermi Distribution Function

Fermi function, F ( E ) :

The probability that a certain energy level is occupied by

electrons

Fermi distribution for T = 0 K (Fig 6.1) and for higher T (T^≠ 0 K) (Fig 6.2) At high energy (E >> E_F), F(E) is

approximated by classical Boltzmann distribution

“B lt t il”

1 exp

) 1 (

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

= ⎛ −

T k

E E E

F

B F

⎥ ⎦

⎢ ⎤

⎣

⎡ ⎟⎟

⎠

⎜⎜ ⎞

⎝

− ⎛ −

≈ k T

E E E

F

B

exp

F

) (

~ 1% E_F

at room temp.

6. Electrons in a Crystal

6.3 Density of States

“How energy levels are distributed over a band?”

Assume free electrons are confined in a square potential well of crystal.

Similar to the case in Sec. 4.2, by using B.C., the solution of the Schrödinger equation

where n_x.n_y,n_z are principal quantum numbers, a is the length of the crystal

) 2 (

2 2

2 2

2 2

z y

x

n

n n n

E = π ma h + +

2 2

2 2

z y

x

n n

n

n = + +