EE 606: Solid State Devices Lecture 4: of Schrodinger Equation

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EE‐606: Solid State DevicesEE‐606: Solid State DevicesLecture 4: Solution of Schrodinger Equation

Muhammad Ashraful [email protected]

Alam ECE‐606 S09 1

Outline

1) Time‐independent Schrodinger Equation1) Time independent Schrodinger Equation

2) Analytical solution of toy problems

3) B d t li t t3) Bound vs. tunneling states

4) Conclusions

5) Additional Notes: Numerical solution of Schrodinger Equation

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


Motivation

PeriodicStructure

EE


Time‐independent Schrodinger Equation

2 2

( )d dU x iΨ Ψ− + Ψ =

Assume

/( ) ( ) iEtx xt eΨ ψ −=20

( )2

U x im dx dt

+ Ψ =

2 2 ( )iEt iEt iEtd iE

( , ) ( )x xt eΨ ψ=

2 2

20

( ) ( ) ( ) ( )2

iEt iEt iEtd x U x x i xm

iEe e edxψ ψ ψ

− − −− + =

−

2 2

20

( )2

d U x Em dx

ψ ψ ψ− + =

20

2 2

2 ( ) 0md E Ud

ψ ψ+ − =


2 2 ( )dx

ψ

Time‐independent Schrodinger Equation

20

2 2

2 ( ) 0md E Udx

ψ ψ+ − =

If E >U, then ….

22

2 0ddx

kψ ψ+ =[ ]02m Uk

E −≡ ( ) ( ) ( )x Asin kx B cos kxψ = +

dx ikx ikxA e A e−+ −≡ +

If U E th

( ) x xx De Eeα αψ − += +2

22 0d

dxαψ ψ− =[ ]02m U E

α−

≡

If U>E, then ….


dxα ≡

A Simple Differential Equation

2 2

20

( )2

d U x Em dx

ψ ψ ψ− + =

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

• Solve the 2nd order equation – pretty basic

• Interpret as the probability of finding an electron at x2 *ψ ψ ψ=• Interpret as the probability of finding an electron at x

• Compute anything else you need, e.g.,

ψ ψ ψ=

* dp dxi dx

Ψ Ψ∞ ⎡ ⎤= ⎢ ⎥⎣ ⎦∫

0

* dE dxi dt

Ψ Ψ∞ ⎡ ⎤= −⎢ ⎥⎣ ⎦∫


0 i dx⎣ ⎦∫0 i dt⎣ ⎦

Outline




4) Conclusions



Full Problem Difficult: Toy Problems First

PeriodicPeriodicStructure

E

Case 3:Free electronE >> U

Case 1:Electron in finite well E < U

Case 2:Electron in infinite wellE << U


E << U

Five Steps for Analytical Solution

d 2ψdx 2 + k 2ψ = 01)

2N unknowns for N regions

( ) 0xψ = −∞ =

dx for N regions

2) Reduces 2 unknowns( ) 0xψ = +∞ =

B Bx x x xψ ψ− += =

=

Reduces 2 unknowns

3)Set 2N 2 equations forB B

B Bx x x x

d ddx dxψ ψ

− += =

=

Set 2N‐2 equations for 2N‐2 unknowns (for continuous U)

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

i l l ti

4) 2( , ) 1x E dxψ∞

−∞=∫5)

f f i


or numerical solution for wave function

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

U(x)E

E

2) Boundarysin cosA kx B kxψ = +

1)

( ) 0

2) Boundary conditions

x xα α+

x xDe Neα αψ − ++=

1)

( ) 0( ) 0xx

ψψ

= −∞ == +∞ =

x xMe Ceα αψ − ++=

100 a

Step3: Continuity of wave‐function


= C BC kAα=

= −3)

B Bx x x x

d ddx dxψ ψ

− += =

= sin( ) cos( )cos( ) sin( )

a

a

A ka B ka DekA ka kB ka De

α

αα

−

−

+ =

=

sin cosA kx B kxψ = +

cos( ) sin( )kA ka kB ka Deα− = −

xCeαψ = xDe αψ −=

Alam ECE‐606 S09 110 a

Step 3: Continuity of Wavefunction

C B=C kAα = −

sin( ) cos( ) aA ka B ka De α−+ =

cos( ) sin( ) akA ka kB ka De αα −− = −

0 0 00 01 1 A

k⎛ ⎞ ⎡ ⎤ ⎡ ⎤⎜ ⎟ ⎢ ⎥ ⎢ ⎥

−

sin( ) co0 0 0

0 000

s( )cos( ) sin( ) /

a

a

BCka ka e

ka ka e k D

kα

αα

α−

−

⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥=⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎣ ⎣⎦ ⎦

−

−⎝ ⎠


00cos( ) sin( ) /ka ka e k Dα ⎣ ⎣⎦ ⎦−⎝ ⎠

Step 4: Bound‐level in Finite Well

det (Matrix)=0

02

0

2 (1 ) 2tan( )2 1

mUaUEξ ξ

α ξ ξ αξ

−= ≡ ≡

−

O l k i E

02 1 Uξ

U0 Only unknown is E

(i) Use Matlab function

0

0 a

( )(ii) Use graphical method


0 a

Step 4: Graphical Method for Bound Levels

2 5x x= +2 (1 )ξ ξ

2 5y x= +21y x= 0

2 (1 )2

a ( )1

t n aξ ξξ

α ξ−

=−

E1xy

ζ=E/U

E1x0

x


Step 4: Graphical Method for Bound Levels

E/U

ζ= E1E1

ζ=E/U

0 a


0 a

Step 5: Wave‐functionsxC αψ xD α−

sin cosA kx B kxψ = +

0 0 01 1 A⎛ ⎞ ⎡ ⎤ ⎡ ⎤

xCeαψ = xDe αψ =

sin( ) co

0 0 00 0 0

0 0s( )

1 1

a

AB

ka ka e Ck

α

α−

⎛ ⎞ ⎡ ⎤ ⎡ ⎤⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥=⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟

−

00cos( ) sin( ) /aka DDe kka αα −⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎣ ⎣⎦ ⎦− −⎝ ⎠

0 01 1 B⎛ ⎞ ⎡ ⎤ ⎡ ⎤⎜ ⎟ ⎢ ⎥ ⎢ ⎥

−

cos(0 0

sin )) 0 (a

C kAD kae Aka α

α−

⎜ ⎟ ⎢ ⎥ ⎢ ⎥= −⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎝ ⎠

1

( )

0 00 0

i (

1

)

1

ak

BC kAD A kα

α

−⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= −⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

−


cos( ) sin( )0 akaD A kae α−⎜ ⎟⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎝ ⎠

Step 5: Calculating Wave‐function

xCeαψ = xDe αψ −=sin cosA kx B kxψ = +

11 1 0 00 0

BkAC α

−⎛ ⎞−⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= −⎜ ⎟⎢ ⎥ ⎢ ⎥0 0

sin( )cos( ) 0 a

kAA kakaD

Ce α

α−

⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎝ ⎠

2

1dxψ∞

= ⇒∫( ) ( )

20 2 2 2 2

0

ax x

a

dx

e dx Asin kx B sin kx dx e dC xDα α

ψ−∞

∞ −

−∞

⇒

⎡ ⎤+ + +⎣ ⎦

∫

∫ ∫ ∫


Aside: Infinite Quantum Well

d 2ψ + k 2ψ = 0 [ ]02 −m E Ukdx 2 + k ψ = 0 [ ]

≡k

)cos(sin kxBkxA +=ψ1) Solutions:

( ) ( ) ( )0 0 0 0x Asin k Bcos kψ = = = +

2) Boundary conditions

( ) ( ) ( )0 0 0 0x Asin k Bcos kψ = = = +

( ) ( ) ( )0x a Asin ka Asin nψ π= = = =

2 2 2

202n

nEm a

π=02 n

n

m Enkaπ

= =


02m aa

Five steps for Analytical Solution: Follow rules

d 2ψdx 2 + k 2ψ = 01)

2N unknowns for N regions

( ) 0xψ = −∞ =

dx for N regions

2) Reduces 2 unknowns( ) 0xψ = +∞ =


=

Reduces 2 unknowns

3)Set 2N 2 equations forB B

B Bx x x x

d ddx dxψ ψ

− += =

=

Set 2N‐2 equations for 2N‐2 unknowns (for continuous U)

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

i l l ti

4) 2( , ) 1x E dxψ∞

−∞=∫5)

f f i


or numerical solution for wave function

Practical examples: Si/Ag and Si/SiO2

SiO2 SiSi


Speer et. al, Science 314, 2006.

Outline




4) Conclusions



Case 2: Solution for Particles with E>>U

2d 2ψdx 2 + k 2ψ = 0 [ ]02m E U

k−

≡ E

U(x) ~ 01) Solution ( ) ( ) ( )

ik ik

x Asin kx Bcos kxψ = +U(x) 0ikx ikxA e A e−

+ −≡ +

2) Boundary condition ( ) positive going wave

= negative going wave

ikx

ikx

x A e

A e

ψ +

−−

=


Free Particle …

( ) ( ) ( )x Asin kx Bcos kxψ = +

E

( ) ( ) ( )ikx ikx

x Asin kx Bcos kx

A e A e

ψ−

+ −

+

≡ +

E

( ) positive going wave

= negative going wave

ikx

ikx

x A e

A e

ψ +

−

=

2 2 2* A or Aψ ψψ= =U(x) ~ 0

Probability:

negative going waveA e−

A or Aψ ψψ + −Probability:

or -* dp dx k kΨ Ψ∞ ⎡ ⎤= =⎢ ⎥∫Momentum:


0

or -p dx k ki dx

Ψ Ψ= =⎢ ⎥⎣ ⎦∫Momentum:

Case 3: Bound vs. Tunneling State

1 1ik x ik xAe Be−+Boundary conditions

1 1ik x ik xCe Me−+ 1 1ik x ik xDe Ne−+y

N=0

5 unknowns (C M A B D)5 unknowns (C,M,A,B,D)

4 equations from x=0 and x=a interfaces

No bound levels

Ratios of D/C is ofInterest


Interest.

Practical Example: Oxide Tunneling


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.

3) S l i b d l l bl i diff t d t3) Solving bound level problem is different compared to the solution of tunneling problem. The corresponding recipes should be followed carefully.


Outline




4) Conclusions



Numerical solution of Schrodinger Equation

d 2ψdx 2 + k 2ψ = 0 [ ]02k m E U( x ) /≡ −dx 2

U0(x)

k > 0α = ik α = ik


x

(1) Define a grid …

U(x)2 2

( )d EU xψ ψ ψ− + = U(x)20

(2

) Em dx

U x ψ ψ+

UU1

ψ3

x

a


x0 1 2 3 N+1 N

Aside: Finite difference – Connecting neighbors

( ) ( )2 2

00 2

d a da ...xx aψ ψψ ψ= + + ++( ) ( )

0 0

00 22x a x adx dxψ ψ

= =

( ) ( )2 2d a d ψψ( ) ( )

0 0

00 22x a x a

d a da ...dx d

xx

x aψ ψψ ψ

= =

= − + −−

2

( ) ( ) ( )0

22

20 00 2x a

x dax a adx

xψψ ψψ=

+ −+ − =

d 2ψdx 2

i

= ψ i−1 − 2ψ i +ψ i+1

a2


i

(2) Express equation in Finite Difference Form

( )2

2 ( )Udt xa Eψ ψ ψ− + =2

t ≡( )0 2 ( )Ut xa Edx

ψ ψ+ =

d 2ψ ψ i 1 − 2ψ i +ψ i+1

0 202

tm a

≡

d ψdx 2

i

= ψ i−1 2ψ i +ψ i+1

a2

( )2 EUt t tψ ψ ψ ψ⎡ ⎤− + + − =⎣ ⎦( )0 1 0 0 12i ii i iEUt t tψ ψ ψ ψ− +⎡ ⎤+ + =⎣ ⎦

ψ(0) = 0 ψ(L) = 0N unknowns

x

ψ( )

i 1 i i 1


0 x1 2 N+1 i-1 i i+1

(3) Define the matrix …

−t0ψ i−1 + 2t0 + ECi( )ψ i − t0ψ i+1[ ]= Eψ i (i = 2, 3…N-1)

( )0 0 0 1 0 22 Ci it t E t Eψ ψ ψ ψ− + + − =⎡ ⎤⎣ ⎦ (i = 1)

( )2t t E t Eψ ψ ψ ψ− + + − =⎡ ⎤⎣ ⎦ (i = N)( )0 1 0 0 12N Ci N N it t E t Eψ ψ ψ ψ− +− + + − =⎡ ⎤⎣ ⎦ (i = N)

Hψ = Eψ−t0 (2t0 + EC1) −t0

⎡ ⎢ ⎢ ⎢ ⎢ ⎢

⎤ ⎥ ⎥ ⎥ ⎥ ⎥

−t0 (2t0 +ECi) −t0 = E

N x N N x 1⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎦

⎥ ⎥ ⎥ ⎥ ⎥


(4) Solve the Eigen‐value Problem

U(x)Hψ = Eψ wrong U(x)Hψ = Eψ wrong

ε3

ε4Eigenvalueproblem; easily l d i h

ε1

ε2

3solved with MATLAB & nanohub tools

x

a


1 x32 4 N (N-1)

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EE 606: Solid State Devices Lecture 4: of Schrodinger Equation