Lecture 3 MSE 515

Solutions of Schrodinger equation for special cases

Last time:

−ħ2

2m∇2ψ+Vψ=Eψ=iℏ ∂Ψ

∂t

Ψ is called a wavefunction

∫−∞

+∞

|Ψ (x)|2dx=1

The probability of finding it must be unity. This is a critical point and will come up over and over.

1. Free electrons: (1-D) (Simple but very important one)- We consider electrons that propagate freely in potential

free space in the positive x-direction

- ∇2ψ → d2ψd x2

-

- −ħ2

2md2ψd x2

=Eψ _____ Eq. 1

-- ψ ( x )=e iαx

-

1

- dψdx

=iα eiαx

-

- d2ψd x2

=i2α2 eiαx

-- In Eq. 1-

- ħ2

2mα2 eiαx=Eψ

-ħ2α2

2meiαx=Eψ remember ψ=e iαx

∴E=ħ2α2

2m

-Since no boundary conditions, all values of energy are allowed.

α =√ 2mEħ2 ______Eq. 2

p2

2m=E( K . E .) p (momentum )

p2=2mE∈Eq.1 .

α = √ p2

ħ2

= pħ❑= p ∙2π

h hp

→ λ de Brogli e ' s hypothesis

α = 2πλ

=k

= k (wave number)

2

K is a vector → k x k y k z

λ→wave vector

Note: k – vector describes the wave properties of an electron.

Quantum Mechanics (k)

Classical Mechanism (p)

And remember k = pħ

Back to Energy E, we can find

E = ħ2

2mk2

ψ(x) = e ikx

Separation of variables

ψ(x,t) = e ikx ∙ eiωt

This is an equation of traveling wave. It represents a free particle.

2. Bound electrons [ potential wall]- We consider that the electron can move freely between

infinitely high potential barriers Electrons cannot escape because of the potential wall

3

Note: The potential inside the wall is zero. The difference between this case and last case is the boundary conditions.

−ħ2

2md2ψd x2

=Eψ

Assume the solution is ψ(x) = AeiαL + Be-iαL

α=√ 2mEħ2

Applying Boundary Conditions

at x = zero ψ = 0

at x = L ψ = 0 since ψ = 0 , x ≤ 0 , x ≥ L

0 = Aezero+B e zero

4

A = -B or B = -A

Ae iαa−A e−iαa=0

A (e iαa−e−iαa¿=0 A

EULER'S FORMULA IS THE KEY TO UNLOCKING THE SECRETS OF QUANTUM PHYSICS

From Euler’s Equation

sin δ= 12i

( eiδ−e−iδ )

2iA2i

(eiαL−e−iαL )=0

2A* i sin (α L) = 0

∴ sin α L = 0 → αL = nπ

α = nπL

n= 0, 1, 2, 3... (integer)

from α = √ 2mEħ2

E = ħ2

2mα2

= ħ2

2mn2π 2

aL2

E ¿ ħ2π2

2m aL2n2 n = 1, 2, 3…

5

6

- The second case is similar to an electron bound to its atomic nucleus

- Only certain energy levels are allowed for the electron- This is “Energy quantization”- Probability of finding the electron at inside the well:

ψn=¿A (e iαx−e−iαx ¿ na

¿2 Ai ∙ sinαx

ψ ψ¿=4 A2 sin2αx

7

3. Finite potential barrier: (Tunnel effect) Assume a free electron propagating in the positive x-direction meets a potential barrier V0 (higher than the total energy of electron).

8

Write sch. Eq. For each region:

Region I. V=0

d2ψd x2

+2mħ2

Eψ

Region II.

d2ψd x2

+2mħ2

( E−V o ) ψ=0

Region I solutions:

ψ I=A e iax+B e−iax

a=√ 2mEħ2

9

Region II solution:

ψ II=C eiβx+D e−iβx

Only certain solutions exist (for which n is integer)

Let’s plot the energy solutions for the two cases:

10

The second case is similar to an electron bond to a nucleus.

First three levels of the stationary state and probability of those states.

Region I solutions:

ψ I=A e iax+B e−iax

11

a=√ 2mEħ2

Region II solution:

ψ II=C eiβx+D e−iβx

β = √ 2mħ2

( E−V 0 )

(E – V0) is less than zero

β → imaginary

= if ∴ γ=√ 2mh2

(V 0−E )

ψ II=C eγx +D e−γx

Using boundary conditions

X → ∞

ψ II=C ∙∞+D ∙0

C must be zero

∴ψ II=D e−γx

Ψ decreases in region II exponentially

The decrease is higher for larger , for larger potential barrier

12

13

The electron wave propagates in the finite potential barrier.

Tunneling effect: penetration of a potential barrier

- This is only quantum mechanical effect.- In classical mechanics: If the electron kinetic energy is

smaller than V, the electron will be entirely reflected and “cannot overcome the barrier”

Examples of tunneling :

- Tunneling of electrons from one metal to another through an oxide film.

- Emission of alpha particles from nuclei by tunneling through the binding potential barrier.

14

4. Another Case

- We can find that electron can penetrate region II and propagates in region III.

−ħ2

2m∇2ψ+Vψ=Eψ=iℏ ∂Ψ

∂ t

15

Partial differ eq. in Ψ

- ∇2ψ → d2ψd x2

+ d2ψd y2

+ d2ψd z2

-- ℏ=h /2 π

mass = m

V = potential

|ψ ( x , y , z , t )|2 dx dy dz

gives the probably of finding the electron in volume dx dy dz

−ħ2

2m∇2ψ

ψ+V=iħ 1

ωδωδt

Eq. 1

function r function time

ψ (r ,t )=ψ (r ) ω (t ) Eq 2

Separation of variables substitute Eq 2 in Eq. 1

For each Eq. to be correct it must be equal to a constant

- For each equation to be right, it must equal to constant

16

−ħ2

2m∇2ψ

ψ+V=E E is a constant → Eq. 3

iħ∂ ω∂ t

=Eω E is the same const.

solution: ω = exp[−iEtħ ] from eq. 3

(−ħ2

2m∇2+V )ψ=Eψ

time independent Schrödinger Equation.

It will be applied to be calculations of stationery conditions.

17