Part I Fundamentals

Part I Fundamentals

Electron Theory : Matter Waves

Chap. 1 Introduction

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

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

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

Chap. 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”

2 0

2 2

2

ψ

ψ

m E dx

d 0 

) 2 (

2

2 + − =

∇

ψ

ψ



x

Ae

x

ψ ( ) =

m E

2

2

=  α

t i x

i

e

Ae

x =

⋅

Ψ ) (

2 2

2 α E = m

p k m E

=

=

=

= λ

α ²₂ ²π





2 2

2 k E = m

λ 2π

= k

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



x i x

i

Be

Ae

ψ = +

^m

^E

2

=  α

0 )

2 (

2

2 + − =

∇

ψ

ψ



Chap. 4. Solution of Schrödinger Equation

4.2 Electron in a Potential Well (Bound Electron)

Applying boundary conditions,

x = 0,

B = – A

x = a

With Euler equation,

Finally,

“energy quantization”

The solution

= 0 ψ

= 0

ψ 0 = Ae

+ Be

= A ( e

− e

)

) 2 (

sin ρ 1 e

e

i

−

=

0 sin

2 ]

[ e − e

= Ai ⋅ a = A

α

..

0,1,2,3,..

, =

= n n

a π

α “energy levels”

z e r

x

Ai α ψ = 2 ⋅ sin

Chap. 4. Solution of Schrödinger Equation

2 2 2

2 2

2 2

1, 2, 3, ...

En n

m ma

n

α π

= =

=

 

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

sin

λ π r = n 2

n

r π

λ

= 2

1 ]

cos 2sin

[ 1 ) 4

( sin

4 ₀

2

0

2 2

0

* =

∫

∫

Chap. 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

4

n eV n

E = me = − ⋅

πε 

) 2 (

2 2

2 2

2 2

z y

x

n

n n n

E = _ ma π + + r

V e

0 2

4 πε

−

=

Chap. 4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

Suppose electrons propagating in the positive x-direction encounter a potential barrier V₀ (> total energy of electron, E)

Region (I) x < 0

2 0

2 2

2

ψ + ψ =

m E dx

d



x i x

i

I

Ae

Be

ψ = +

Region (II) x > 0

0 )

2 (

2 0 2

2

ψ

ψ

V m E

dx d



The solutions (see Appendix 1)

x i x

i

II

Ce

De

ψ = +

2 E V0

m −

=  β

Chap. 4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

Since E − V₀ 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 V0

m

−

= 

β

β γ

)

2 (

2 V0 E

m

−

= 

γ ψ

= Ce

+ De

ψ

= Ce

+ De

⋅ 0 +

∞

⋅

= C D

ψ

→ 0 C

x

II

De

ψ =

ψ

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

) ( t kx i

x

e

De

⋅

=

Ψ

Chap. 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

= ψ

x x

i x

i

Be De

Ae

+

=

+

=

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

With x = 0 Consequently,

dx d dx

d ψ

_≡ ψ

x x

i x

i

Bi e De

e

Ai α

− α

= − γ

D Bi

Ai α − α = − γ

) 1

2 ( α i γ

A = D +

2

α

i

γ

Chap. 4. Solution of Schrödinger Equation

4.3 Finite Potential Barrier (Tunnel Effect)

Chap. 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



0 )

2 (

2 0 2

2

ψ + − ψ =

V m E

dx d



E < V

Chap. 4. Solution of Schrödinger Equation

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

For abbreviation

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

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

In 3-d, u(r) = u(r + R), R = Bravais lattice vector The final solution of the Schrödinger equations;

where m E

2

2 2

= 

α

2 m V E

−

= 

γ

e

x u

x ) = ( ) ⋅ ψ (

ka a a

P sin a + cos α = cos α

α

(Bloch function)

2 0

 b P = maV

Chap. 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

x u

x ) = ( ) ⋅ ψ (

e

u k dx ik

du dx

u d dx

d (

2

)

2 2

2

ψ = + −

0 )

(

2

2

2

+ − k − u =

dx ik du dx

u

d α _{2 (}

_{) 0}

2

2

+ − k + u =

dx ik du dx

u

d γ

(I) (II)

The solutions of (I) and (II)

) (

ikx

Ae Be

e

u =

+

u = e

( Ce

+ De

)

(I) (II)

Chap. 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+ = +

b ik b

ik a

ik i a

ik

i

Be Ce De

Ae

+

=

+

b ik b

ik k

ia k

ia

Bi k e C ik e D ik e

e k

Ai ( α − )

− ( α + )

= − ( γ + )

+ ( γ − )

) (

cos )

cos(

) cosh(

) sin(

) sinh(

2

2 2

b a k a

b a

b ⋅ + ⋅ = +

− γ α γ α

αγ α γ

Chap. 4. Solution of Schrödinger Equation

( ) ( ) ( ) ( )

A i

α −

+

− −

α

=

− − γ

+

γ −

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

If V₀ is very large, then E in 4.47 is very small compared to V₀ so that

Since V₀b has to remain finite and b → 0, γb becomes very small.

For a small γb, we obtain (see tables of the hyperbolic function)

Finally, neglect α² compared to γ² and, b compared to a so that 4.61 reads as follow

Let , then

b b

and

b γ γ

γ ) ≈ 1 sinh ( ) ≈ cosh(

ka a

a b

m V

cos cos

0

sin

2

α + α =

α _

2 0



b

P = maV ^{a ka}

a

P sin a cos cos

=

+ α

α α

0 0

2 2

2 2

( )

m m

V b b V b b

γ = × → γ =

 

Chap. 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”

Chap. 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

• If the potential barrier strength, V₀b is large, P is also large and the curve on Fig 4.11 steeper. The

allowed band are narrow.

• V₀b and P are small, the allowed band becomes wider.

• If V₀b goes 0, thus, P → 0 From 4.67,

cos α a = cos ka

m E k

2

2



=

Chap. 4. Solution of Schrödinger Equation

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

See Fig. 7.2 of Bube

Chap. 4. Solution of Schrödinger Equation

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

• If the V₀b 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  ⋅

αa = nπ

Combining 4.46 and 4.69

Other model:

The tight-binding approximation

Chap. 4. Solution of Schrödinger Equation

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

See Fig. 7.3 of Bube

Chap. 4. Solution of Schrödinger Equation