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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)
Alam ECE‐606 S09 2
Motivation
PeriodicStructure
EE
Alam ECE‐606 S09 3
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+ − =
Alam ECE‐606 S09 4
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 ….
Alam ECE‐606 S09 5
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∞ ⎡ ⎤= −⎢ ⎥⎣ ⎦∫
Alam ECE‐606 S09 6
0 i dx⎣ ⎦∫0 i dt⎣ ⎦
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
Alam ECE‐606 S09 7
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
Alam ECE‐606 S09 8
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
Alam ECE‐606 S09 9
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
B Bx x x xψ ψ− += =
= 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−
−
⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥=⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎣ ⎣⎦ ⎦
−
−⎝ ⎠
Alam ECE‐606 S09 12
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
Alam ECE‐606 S09 13
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
Alam ECE‐606 S09 14
Step 4: Graphical Method for Bound Levels
E/U
ζ= E1E1
ζ=E/U
0 a
Alam ECE‐606 S09 15
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
−⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= −⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
−
Alam ECE‐606 S09 16
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−∞
∞ −
−∞
⇒
⎡ ⎤+ + +⎣ ⎦
∫
∫ ∫ ∫
Alam ECE‐606 S09 17
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
= =
Alam ECE‐606 S09 18
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 = +∞ =
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 matix)=0And find E by graphical
i l l ti
4) 2( , ) 1x E dxψ∞
−∞=∫5)
f f i
Alam ECE‐606 S09 19
or numerical solution for wave function
Practical examples: Si/Ag and Si/SiO2
SiO2 SiSi
Alam ECE‐606 S09 20
Speer et. al, Science 314, 2006.
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
Alam ECE‐606 S09 21
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 +
−−
=
Alam ECE‐606 S09 22
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:
Alam ECE‐606 S09 23
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
Alam ECE‐606 S09 24
Interest.
Practical Example: Oxide Tunneling
Alam ECE‐606 S09 25
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.
Alam ECE‐606 S09 26
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
Alam ECE‐606 S09 27
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
Alam ECE‐606 S09 28
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
Alam ECE‐606 S09 29
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
Alam ECE‐606 S09 30
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
Alam ECE‐606 S09 31
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⎣
⎢ ⎢ ⎢ ⎢ ⎢ ⎦
⎥ ⎥ ⎥ ⎥ ⎥
Alam ECE‐606 S09 32
(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
Alam ECE‐606 S09 33
1 x32 4 N (N-1)
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