Principles of Semiconductor Devices-L4

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Lecture 4: Solution of Schrodinger EquationMuhammad Ashraful [email protected]

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Outline

2) Analytical solution of toy problems

oun vs. unne ng s a es

4) Conclusions5) dditional Notes: Numerical solution of Schrodinger Equation

Reference: Vol. 6, Ch. 2 (pages 29 45)

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Motivation

PeriodicStructure

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Time independent Schrodinger Equation

2 2d d =

Assume

/iEt 2

02m dx dt iEt iEt iEt

,

2

0

( ) ( ) ( )

2

xU x x i x

m

e e e

dx

+ =

2 2

20

( )2

d U x E

m dx

+ =

202 0

md E U

+ =

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dx


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Time independent Schrodinger Equation

20

2 2

2( ) 0

md E U

dx

+ =

If E >U, then .

2

22 0d

k

+ =[ ]02m U

k E

( ) ( ) ( ) x A sin kx B cos kx = +ikx ikx A e A e+ +

( ) x x x De Ee += +2

22 0

d

=[ ]02m U E

> , en .

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x


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A Simple Differential Equation2 2

20

( )2

d U x E

m dx

+ =

Obtain U(x) and the boundary conditions for a given problem.

Solve the 2nd order equation pretty basic

2 *

Compute anything else you need, e.g.,

=

* d p dx

= *d

E dxi dt

=

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0


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Outline



4) Conclusions5) Additional Notes: Numerical solution of Schrodinger Equation

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Full Problem Difficult: Toy Problems First

Structure

E

Case 3:Free electronE >> U

Case 1:Electron in finite well

E < U

Case 2:Electron in infinite well

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Five Steps for Analytical Solution

d 2 2

+k 2 =01)2N unknowns

( ) 0 x = =

2)( ) 0 x = + =

B B x x x x += ==

3)

B B x x x x

d d dx dx

+= ==

2N2 unknowns (for continuous U)

Det (coefficient matrix)=0And find E by graphical

4) 2( , ) 1 x E dx

= 5)

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or numer ca so ut on or wave unction


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Case 1: Bound levels in Finite Well (steps 1,2)

U(x)E

E

sin cos A kx B kx = + oun ary conditions

x x De Ne ++=

( ) 0 x x

= == + = x x

e Ce

+=

100 a


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Step3: Continuity of wave function

B B x x x x += ==

C B

C kA

==

3)

B B x x x x

d d dx dx

+= == sin( ) cos( ) a

a

A ka B ka De

+ =

sin coskx B kx = +

=

xCe = x De =

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Step 3: Continuity of Wavefunction

C B=C kA =

sin( ) cos( ) a A ka B ka De + =cos( ) sin( ) akA ka kB ka De =

0 0 01 1 A

sin( ) co0

0 0s( ) a

a

BC ka ka e

=

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Step 4: Bound level in Finite Well

det (Matrix)=0

02

2 (1 ) 2tan( )

mU a

E =

0

Un y un nown s

i Use Matlab function (ii) Use graphical method

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Step 4: Graphical Method for Bound Levels

2 5 x x= +

2 5 y x= +2

1 y x= 0 2a ( ) 1t n a

=

Ey

=E/U

0

x

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Step 4: Graphical Method for Bound Levels

E / U

=

E1E1

=E/U

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Step 5: Wave functions x

sin coskx B kx = +e = e =

sin( ) co

0

0 0 00 0s( ) a

A

B

ka ka e C

k

=

00cos( ) sin( ) / aka D De k ka

0 01 1 B

cos(

0 0

sin )) 0 (aC kA

D kae Aka

=

10 0

0 0

1 1 B

C kA

=

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cos( ) s n0 aka ae


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Step 5: Calculating Wave function

xCe = x De =sin cos A kx B kx = +

11 1 0 0 B =

sin( )cos( ) 0 a kaka D e

2

1dx

=

( ) ( )20 2 2 2 2

0

a x x

ae dx A sin kx B sin kx dx e d C x D

+ + +

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Aside: Infinite Quantum Well

d 2 2 = [ ]02 m E U

dx2

)cos(sin kx Bkx A += 1) Solutions:

2) Boundary conditions

( ) ( ) ( )0 x a A sin ka A sin n = = = =

2 2 2

2n

n E

=02 nn

m E nk

= =

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Five steps for Analytical Solution: Follow rules

d 2 2

+k 2 =01)2N unknowns

( ) 0 x = =

2)( ) 0 x = + =

B B x x x x += ==

3)

B B x x x x

d d dx dx

+= ==

2N2 unknowns (for continuous U)

Det(coefficient matix)=0And find E by graphical

4) 2( , ) 1 x E dx

= 5)

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or numer ca so ut on or wave unction


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Practical examples: Si/Ag and Si/SiO2

SiO 2 SiSi

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Speer et. al, Science 314, 2006.


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Outline



4) Conclusions5) Additional Notes: Numerical solution of Schrodinger Equation

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Case 2: Solution for Particles with E>>U

d dx 2 +k

2

=0[ ]02m E U

k

E

U x ~ 01) Solution

( ) ( ) ( ) x A sin kx B cos kx = +

x xe A e+ +

2) Boundary condition ( ) positive going wave= negative going wave

ikx

ikx

x A e

A e

+

=

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Free Particle

x A sin kx B cos kx= +ikx ikx A e A e+ +

( ) positive going wave=

ikx

ikx

x A e +

=

2 2 2*= =U(x) ~ 0

+

* d

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0

-i dx


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Case 3: Bound vs. Tunneling State

1 1ik x ik x Ae Be+Boundar conditions

1 1ik x ik xCe Me + 1 1ik x ik x De Ne+N=0

, , , ,

4 equations from x=0 and x=a interfaces

No bound levels

Ratios of D/ C is of

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.


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Practical Example: Oxide Tunneling

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Conclusions

1) We have discussed the analytical solution of Schrodinger equation for simple potentials. Such potential arises in wide variety of practical systems.

2) Numerical solution is very powerful, but it is easy to get wrong results if one is not careful.

o v ng oun eve pro em s eren compare o the solution of tunneling problem. The corresponding recipes should be followed carefully.

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Outline



4) Conclusions

5) Additional Notes: Numerical solution of Schrodinger Equation

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Numerical solution of Schrodinger Equation

d 2 2

+k 2 =0 [ ]02k m E U( x ) / x

U0(x)

k > 0 = ik = ik

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x


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(1) Define a grid

2 2d =2

02m dx1

3

a

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0 1 2 3 N+1N


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Aside: Finite difference Connecting neighbors

2 2d a d a ... x x a

= + + ++0

0

2 x a x adx dx= =

2 2d a d

0 0

00 22 x a x aa ...

dx d x

x x a

= =

= +

( ) ( ) ( )0

220 00 2

x a

xd

a x a adx

x

=

+ + =

d 2 dx 2

= i1 2 i + i+1a 2

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(2) Express equation in Finite Difference Form

22 d =

2

0 2dx

2

0 202m a

dx 2i

= +a 2

=0 1 0 0 1i ii i i +

(0) =0 ( L) =0N unknowns

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0 1 2 N+1- +


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(3) Define the matrix

t 0 i1 + 2 t 0 + E Ci( ) i t 0 i+1[ ]=E i (i = 2, 3N-1)

( )0 0 0 1 0 22 Ci it t E t E + + = (i = 1)0 1 0 0 1 N Ci N N i +

H = E t 0 (2t 0 + E C 1) t 0

t 0 (2t 0 + E Ci) t 0 = E

N x N N x 1

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(4) Solve the Eigen value Problem

wron

4Eigenvalue

problem; easily

12

so ve w t MATLAB & nanohub tools

a

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1 32 4 N(N-1)

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Principles of Semiconductor Devices-L4