3. The Schrödinger Equation

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

z y

x ∂

+ ∂

∂ + ∂

∂

= ∂

∇ ψ ψ ψ ψ

V E

E =

+

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

z y x t

z y

x = ψ ⋅

Ψ ( , , , ) ( , , ) ω ω

ψ i e

i t

t

i

= Ψ

∂ = Ψ

∂

t i

∂ Ψ

⋅ ∂

− Ψ ω =

t E i

∂ Ψ

⋅ ∂

− Ψ

= h

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

∂

− ∂

= h

E

p = − h i ∇ m V

E p E

E

=

+

= +

2

2

= ∇ Ψ + Ψ

∂ Ψ

− ∂ V

m i

i t

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”

2 0

2 2

2

ψ

ψ

m E dx

d 0 h

) 2 (

2

2 + − =

∇

ψ

ψ

x

Ae

x

ψ ( ) =

m E

2

2

= h α

t i x

i

e

Ae

x =

⋅

Ψ ) (

2 2

2 α E = hm

p k m E

=

=

=

= λ

α ² ₂ ²π h

h

2 2

2 k E = hm

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, treat1-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

ψ = +

^{2 m}

^E

= h

α

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”

= 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

1,2,3,....

2 ,

2

2 2 2 2 2

2

=

=

= n

ma n E

^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

sin

λ π r = n 2

n r π

λ

= 2

1 ] cos

2sin [ 1 ) 4

( sin

4 ₀

2

0 2 2

0

* =

∫

∫

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

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 V₀ (> 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

Be

ψ = +

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

De

ψ = +

−

= h β

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 V

m −

= h β

β γ

)

2 (

2

V

E

m −

= h

γ ψ

= Ce

+ De

ψ

= Ce

+ De

⋅ 0 +

∞

⋅

= C D

ψ

→0 C

x

II

De

ψ =

ψ

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

) ( t kx i

x

e

De

⋅

=

Ψ

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 i x

i x

i

Be De

Ae

+

=

^{A + B = D}

(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

) 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

m V E

−

= h γ

e

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

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

)

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

+

=

+

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(

) 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 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

2 0

2 m V

= h

γ m V b b

b 2 ( )

2 0

= h γ

ⅹb →

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, V₀b is large, P is also large and the

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

(b) V₀b and P are small, the allowed band becomes wider.

(c) If V₀b goes 0, thus, P → 0 From 4.67,

cos α a cos = ka

m E k

2

2

h

=

4. Solution of Schrödinger Equation

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

(d) 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 h ⋅

α _{a = n} π

Combining 4.46 and 4.69