ECE606: Solid State Devices Lecture 3 Gerhard Klimeck [email protected]
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Motivation
Periodic Structure E
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Time-independent Schrodinger Equation
Assume ℏ2d 2 Ψ d Ψ − + Ψ= ℏ Ψ= ψ −iEt / ℏ 2 U( x ) i (,)xt () x e 2m0 dx dt
−iEtℏ2d 2 ψ ( x ) − iEt −iE − iEt −eℏ + e ℏ Uxxi()()ψ = ℏ ψ () x e ℏ 2 ℏ 2m0 dx ℏ2 2 ψ −d +ψ = ψ 2 U( x ) E 2m0 dx
d 2ψ 2m +0 (E − U )ψ = 0 dx 2ℏ 2
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Time-independent Schrodinger Equation
d 2ψ 2m +0 (E − U )ψ = 0 dx 2ℏ 2
If E >U, then ….
2m[]E − U d 2ψ 0 +2ψ = ψ ( x) = Asin( kx) + Bcos( kx ) k ≡ 2 k 0 ℏ dx ikx− ikx ≡Ae+ + Ae −
If U>E, then …. 2ψ []− d 2 −αx + α x 2m0 U E −α ψ = 0 ψ ( x) = De + Ee α ≡ 2 ℏ dx
Klimeck – ECE606 Fall 2012 – notes adopted from Alam A Simple Differential Equation
ℏ2 2 ψ −d +ψ = ψ 2 U( x ) E 2m0 dx
• Obtain U(x) and the boundary conditions for a given problem.
• Solve the 2 nd order equation – pretty basic
• Interpret ψ2 = ψ* ψ as the probability of finding an electron at x
• Compute anything else you need, e.g.,
∞ ∞ ℏ ℏ d * d =Ψ* − Ψ p= Ψ Ψ dx E dx ∫ ∫ i dt 0 i dx 0
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Presentation Outline
• Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Additional notes »Discretizing Schroedinger’s equation for numerical implementations
Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Full Problem Difficult: Toy Problems First
Periodic Case 1: Structure Free electron E >> U E
Case 3: Electron in finite well E < U
Case 2: Electron in infinite well E << U
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Case 1: Solution for Particles with E>>U
2ψ []− d 2 2m0 E U + k ψ = 0 k ≡ E dx 2 ℏ
1) Solution ψ (x) = Asin( kx) + Bcos( kx )
ikx− ikx U(x) ~ 0 ≡Ae+ + Ae −
ikx 2) Boundary condition ψ (x) = A+ e positive going wave
−ikx =A− e negative going wave
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Free Particle …
ψ (x) = Asin( kx) + Bcos( kx )
ikx− ikx ≡Ae+ + Ae −
E ikx ψ (x) = A+ e positive going wave
−ikx =A− e negative going wave
U(x) ~ 0 2* 2 2 Probability: ψ= ψψ = A+ or A −
∞ ℏ =Ψ* d Ψ = Momentum: p∫ dxkkℏor - ℏ 0 i dx
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Full Problem Difficult: Toy Problems First
Periodic Case 1: Structure Free electron E >> U E
Case 3: Electron in finite well E < U
Case 2: Electron in infinite well E << U
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Case 2: Bound State Problems
• Mathematical interpretation of Quantum Mechanics(QM) ℏ2 ∂ − ∇ 2Ψ + VΨ = iℏ Ψ 2m ∂t » Only a few number of problems have exact mathematical solutions » They involve specialized functions
1 V = ∞ V = ∞ V = kx 2 V = ∞ V = ax 2
1 V ∝ = r V 0 x = 0 x = L 0 x x = 0 x Coulomb Potential Particle in a box Harmonic Oscillator Triangular Potential Well by nucleus in an atom
2 1 −Cx Ψ ( )= ( ) (θ ϕ) Ψ ( )= ( ) Ψ ()= 2 Ψ ( )= (ξ ) n r AR n r Yn , n x Asin kn x n x A Hn (Bx )e n x NAi n 2n n! Klimeck – ECE606 Fall 2012 – notes adopted from Alam
1-D Particle in a Box – A Solution Guess
• (Step 1) Formulate time independent Schrödinger equation
2 2 < < ℏ d 0 0 x Lx − ψ()x + V() x ψ()x = Eψ(x) where, V() x = 2m dx 2 ∞ elsewhere
V = ∞ V = ∞ • (Step 2) Use your intuition that the particle will never exist outside the energy barriers to guess, 0 0 ≤ x ≤ L ψ()x = x ≠ 0 in the well V = 0 • (Step 3) Think of a solution in the well as: = x = 0 x Lx nπ ψ ()x = Asin x , n =1,2,3,… n Lx
Klimeck – ECE606 Fall 2012 – notes adopted from Alam 1-D Particle in a Box – Visualization
• (Step 4) Plot first few solutions nπ ψ ()x = Asin x , n =1,2,3,… n Lx n = 2 n = 3 n = 4 n = 1
======x 0 x Lx x 0 x Lx x 0 x Lx x 0 x Lx 0 0 ≤ x ≤ L ψ()x = x Matches the condition we guessed at step 2! ≠ 0 in the well But what do the NEGATIVE numbers mean? • (Step 5) Plot corresponding electron densities
2 nπ ψ ()x = A2 sin 2 x , n =1,2,3,… The distribution of SINGLE particle n Lx
n = 1 n = 2 n = 3 n = 4 ======x 0 “s-type”x Lx x 0 “p-type”x Lx x 0 “d-type”x Lx x 0 “f-type” x Lx ONE particle => density is normalized to ONE Klimeck – ECE606 Fall 2012 – notes adopted from Alam
1-D Particle in a Box – Normalization to ONE particle
2 nπ (Step 6) Normalization (determine the constant A) ψ ()x = A2 sin 2 x n L Method 1) Use symmetry property of sinusoidal function x
2 L A ()Area =1= x × A2 2 2 ∴ A = L = = = = x x 0 x Lx x 0 x Lx 2 ψ ()x Method 2) Integrate n over 0 ~ Lx 2nπx − 1 cos 2 Lx 2 Lx nπ Lx L L ∴ A = 1= ∫ ψ ()x dx = ∫ A2 sin 2 x dx = A2 ∫ x dx = A2 x 0 n 0 0 Lx Lx 2 2
2 nπ n =1,2,3,… ∴ψ ()x = sin x, n < < Lx Lx 0 x Lx
Klimeck – ECE606 Fall 2012 – notes adopted from Alam 1-D Particle in a Box – The Solution
(Step 7) Plug the wave function back into the Schrödinger equation
2 nπ ℏ2 d 2 ψ ()= − ψ()+ ()ψ()= ψ( ) n x sin x 2 x V x x E x Lx Lx 2m dx ℏ2 n 2π 2 2 nπ 2 nπ 2 nπ sin + 0 × sin = E sin 2 n 2m Lx Lx Lx Lx Lx Lx Lx
2 nπ ψ ()x = sin x n L L x x n = 4 ℏ2π 2 E = n 2 n 2mL 2 x n = 3 = … < < n 1,2,3, , 0 x Lx n =2 Discrete Energy Levels! n = 1
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
1-D Particle in a Box – Quantum vs. Macroscopic
• Quantum world → Macroscopic world » What will happen with the discretized energy levels if we increase the length of the box? ℏ2π 2 E = n 2 n 2mL 2 = = x = Lx 5nm Lx 50 nm Lx 50 cm
• Energy level spacing goes smaller and smaller as physical dimension increases. • In macroscopic world, where the energy spacing is too small to resolve, we see continuum of energy values. • Therefore, the quantum phenomena is only observed in nanoscale environment.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Presentation Outline
• Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Additional notes »Discretizing Schroedinger’s equation for numerical implementations
Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Full Problem Difficult: Toy Problems First
Periodic Case 1: Structure Free electron E >> U E
Case 3: Electron in finite well E < U
Case 2: Electron in infinite well E << U
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Five Steps for Closed System Analytical Solution
Solution Ansatz ψ ( ) =ikx + − ikx 2ψ x Ae+ Ae − d + 2ψ = 2N unknowns 1) 2 k 0 −αx + α x dx for N regions ψ ( x) = De + Ee
2) ψ (x = −∞ ) = 0 Boundary Conditions at the edge ψ (x = +∞ ) = 0 Reduces 2 unknowns
ψ= ψ 3) =− = + Boundary Condition at each interface: xxB xx B dψ d ψ Set 2N-2 equations for = 2N-2 unknowns (for continuous U) dx=− dx = + xxB xx B
∞ 4) Det (coefficient matrix)=0 5) ∫ ψ (,)xE2 dx = 1 And find E by graphical −∞ or numerical solution Normalization of unity probability for wave function Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Case 2: Bound-levels in Finite Well (steps 1,2)
E U(x)
E
ψ =Asin kx + B cos kx 1) 2) Boundary conditions −α + α at the edge ψ = Dex+ Ne x ψ = Me−αx+ Ce + α x ψ (x = −∞ ) = 0 ψ (x = +∞ ) = 0
Klimeck – ECE6060 Fall 2012 – notes aadopted from Alam Case 2: Bound-levels in Finite Well (steps3)
3) Boundary at each interface ψ= ψ C= B =− = + xxB xx B αC= − kA dψ d ψ = Asin( ka )+ B cos( ka ) = De −αa − + dxxx= dx xx = B B kAcos( ka )− kB sin( ka ) = − α De −αa
ψ =Asin kx + B cos kx
α x 01 −1 0 ψ = Ce ψ = −α x A 0 De k 0α 0 B 0 = sin(ka ) co s( ka ) 0− e −αa C 0 −αa cos(ka )− sin( ka) 0 α e/ k D 0 0 a
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Case 2: Bound-levels in Finite Well (step 4)
2ξ (1− ξ ) Only unknown is E tan(αa ξ ) = 2ξ − 1 2mU (i) Use Matlab function ξ≡E α ≡ 0 ℏ2 (ii) Use graphical method U0 det (Matrix)=0 ψ =Asin kx + B cos kx
α x 01 −1 0 ψ = Ce ψ = −α x A 0 De k 0α 0 B 0 = sin(ka ) co s( ka ) 0− e −αa C 0 −αa cos(ka )− sin( ka) 0 α e/ k D 0 0 a
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Case 2: Bound-levels in Finite Well (steps 4 graphical)
x2 = x + 5 2ξ (1− ξ ) 2 α ξ = = = + ta n (0a ) y1 x y2 x 5 2ξ −1
E x0 1 y ζ=E/U
x
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Case 2: Bound-levels in Finite Well (steps 4 graphical)
2ξ (1− ξ ) ta n (αa ξ ) = 0 2ξ −1 =E/U ζ E1 E1
ζ=E/U
0 a Obtained the eigenvalues
Klimeck – ECE606=> Fall 2012could – notes adoptedstop from here Alam in many cases Case 2: Bound-levels in Finite Well (steps 5 wavefunction) Need another boundary condition!
01 −1 0 A 0 k 0α 0 B 0 = sin(ka ) co s( ka ) 0 e −αa C 0 −αa ψ =Asin kx + B cos kx cos(ka )− sin( ka ) 0 − α De/ k D 0 ψ = Ce α x ψ = De −α x det (Matrix)=0 => Linear dependent system2mU ξ≡E α ≡ 0 => Only 3 variablesU are uniqueℏ2 => One variable0 is undetermined ξ− ξ Let’s assumeα A ξcan= be2 freely (1 ) Chosen tan(=> cana get ) B,C,D Klimeck – ECE606 Fall 2012 – notes adopted from Alam 2ξ − 1
Step 5: Wave-functions
∞ 2 Another boundary condition! ψ dx =1 ⇒ ∫−∞ Non-linear => no simple linear algebra expression
0 a 2 ∞ C 2e 2α x dx+ Asinkx()() + Bsinkx dx +D2 e − 2 α x dx = 1 ∫−∞ ∫0 ∫ a 01 −1 0 A 0 Get “A” k 0α 0 B 0 = sin(ka ) co s( ka ) 0 e −αa C 0 −αa ψ =Asin kx + B cos kx cos(ka )− sin( ka ) 0 − α De/ k D 0 ψ = Ce α x ψ = De −α x 1 −1 0 B 0 α = − 0 0 C kA −αa => Linearcos( ka dependent) 0 e systemD−Asin ( ka ) => Only 3 variables are unique −1 => OneB variable1 −is1 undetermined0 0 Get =α − Let’s assumeC0 A can be 0 freely kA “B,C,D” −αa Dcos(ka ) 0 e − Asin( ka ) Klimeck – ECE606 Fall 2012 – notes adoptedChosen from Alam => can get B,C,D Key Summary of a Finite Quantum Well
• Problem is analytically solvable • Electron energy is quantized and wavefunction is localized • In the classical world: » Particles are not allowed inside the barriers / walls => C=D=0 • In the quantum world: » C and D have a non-zero value! Si SiO 2 Si » Electrons can tunnel inside a barrier ψ =Asin kx + B cos kx ψ = Ce α x ψ = De −α x
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Presentation Outline
• Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Additional notes »Discretizing Schroedinger’s equation for numerical implementations
Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Transmission through a single barrier Scattering Matrix approach
Define our system : Single barrier
One matrix each for each interface: 2 S-matrices
D Incident : A Transmitted : E
C Reflected : B Incident : F
No particles lost! Typically A=1 and F=0.
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Tunneling through a single barrier
D Incident : A Transmitted : E Reflected : B C Incident : F
Wave-function each region,
Klimeck – ECE606 Fall 2012 – notes adopted from Alam Single barrier case
Applying boundary conditions at each interface (x=0 and x=L) gives,
Which in matrix can be written as,
Klimeck – ECE606 Fall 2012 – notes adopted from Alam
Generalization to Transfer Matrix Method
• The complete transfer matrix
A D E BFC • In general for any intermediate set of layers, the TMM is expressed as: + + A M M A n−1 = 11 12 n − M M − An−1 21 22 An • For multiple layers the overall transfer matrix will be
• Looks conceptually very simple and analytically pleasing • Use it for your homework assignment for a double barrier structure! Klimeck – ECE606 Fall 2012 – notes adopted from Alam Single barrier case
Transmission can be found using the relations between unknown constants,
γ Case: E γ Case( L<<1): Weak barrier Case: E>V o Klimeck – ECE606 Fall 2012 – notes adopted from Alam Single barrier : Concepts •Transmission is finite under the barrier – tunneling! •Transmission above the barrier is not perfect unity! •Quasi-bound state above the barrier. Case: E>V o Transmission goes to one. •Computed with – http://nanohub.org/tools/pcpbt Klimeck – ECE606 Fall 2012 – notes adopted from Alam Effect of barrier thickness below the barrier •Increased barrier width reduces tunneling probability •Thicker barrier increase the reflection probability below the barrier height. Case: E>V o •Quasi-bound states occur for the thicker barrier too. •Computed with – http://nanohub.org/tools/pcpbt Klimeck – ECE606 Fall 2012 – notes adopted from Alam Single Barrier – Key Summary • Quantum wavefunctions can tunnel through barriers • Tunneling is reduced with increasing barrier height and width • Transmission above the barrier is not unity »2 interfaces cause constructive and destructive interference »Quasi bound states are formed that result in unity transmission Klimeck – ECE606 Fall 2012 – notes adopted from Alam Presentation Outline • Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Additional notes »Discretizing Schroedinger’s equation for numerical implementations Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt Klimeck – ECE606 Fall 2012 – notes adopted from Alam Double Barrier Transmission: Scattering Matrix approach Define our system : Double barrier One matrix each for each interface: 4 S-matrices Left Incident Transmitted Reflected Right Incident No particles lost! Typically Left Incident wave is normalized to one. Right incident is assumed to be zero. Also this problem is analytically solvable! => Homework assignment Klimeck – ECE606 Fall 2012 – notes adopted from Alam Reminder: Single barrier •Transmission is finite under the barrier – tunneling! •Transmission above the barrier is not perfect unity! •Quasi-bound state above the barrier. Transmission goes to one. Klimeck – ECE606 Fall 2012 – notes adopted from Alam • Double barriers allow a transmission probability of one / unity for discrete energies • (reflection probability of zero) for some energiesDouble below the barrier: barrier height. Concepts • This is in sharp contrast to the single barrier case • Cannot be predicted by classical physics. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Double barrier: Quasi-bound states • In addition to states inside the well, there could be states above the barrier height. • States above the barrier height are quasi-bound or weakly bound. • How strongly bound a state is can be seen by the width of the transmission peak. • The transmission peak of the quasi-bound state is much broader than the peak for the state inside the well. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Effect of barrier height •Increasing the barrier height makes the resonance sharper. •By increasing the barrier height, the confinement in the well is made stronger, increasing the lifetime of the resonance. •A longer lifetime corresponds to a sharper resonance. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Effect of barrier thickness •Increasing the barrier thickness makes the resonance sharper. •By increasing the barrier thickness, the confinement in the well is made stronger, increasing the lifetime of the resonance. •A longer lifetime corresponds to a sharper resonance. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Double barrier energy levels Vs Closed system The well region in the double barrier case can be thought of as a particle in a box. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Particle in a box • The time independent Schrödinger equation is ℏ2 d 2 < < − ψ()x + V() x ψ()x = Eψ(x) 0 0 x Lx 2 where, V() x = 2m dx ∞ elsewhere • The solution in the well is: nπ = ∞ = ∞ ψ ()x = Asin x , n =1,2,3,… V V n Lx • Plugging the normalized wave-functions back into = the Schrödinger equation we find that energy n 4 levels are quantized. 2 nπ n = 3 ψ ()x = sin x n L L x x = 2π 2 n 2 = h 2 = E n 2 n n 1 2mL x = = x 0 x Lx n =1,2,3,K , 0 < x < L = Klimeck – ECE606 Fall 2012 – notes adopted fromx Alam V 0 Double barrier & particle in a box • Green: Particle in a box energies. • Red: Double barrier energies • Double barrier: Thick Barriers(10nm), Tall Barriers(1eV), Well(20nm). • First few resonance energies match well with the particle in a box energies. • The well region resembles the particle in a box setup. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Open systems Vs closed systems • Green: Particle in a box energies. • Red: Double barrier energies • Double barrier: Thinner Barriers(8nm), Shorter Barriers(0.25eV), Well(10nm). • Even the first resonance energy does not match with the particle in a box energy. • The well region does not resemble a particle in a box. • A double barrier structure is an OPEN system, particle in a box is a CLOSED system. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Reason for deviation? Potential profile • Wave-function and resonance penetrates into the energies using barrier region. tight-binding. • The effective length of the well region is modified. • The effective length of the well is crucial First excited state in determining the wave-function energy levels in the amplitude using closed system. tight binding. h2π 2 E = n 2 n 2mL 2 Ground state well = < < wave-function n 1,2,3,K , 0 x Lwell amplitude using tight binding. Klimeck – ECE606 Fall 2012 – notes adopted from Alam Double Barrier Structures - Key Summary • Double barrier structures can show unity transmission for energies BELOW the barrier height »Resonant Tunneling • Resonance can be associated with a quasi bound state »Can relate the bound state to a particle in a box »State has a finite lifetime / resonance width • Increasing barrier heights and widths: »Increases resonance lifetime / electron residence time »Sharpens the resonance width Klimeck – ECE606 Fall 2012 – notes adopted from Alam Presentation Outline • Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Additional notes »Discretizing Schroedinger’s equation for numerical implementations Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt Klimeck – ECE606 Fall 2012 – notes adopted from Alam 1 Well => 1 Transmission Peak • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 2 Wells => 2 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Bonding/Anti-bonding State Klimeck – ECE606 Fall 2012 – notes adopted from Alam 3 Wells => 3 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 4 Wells => 4 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 5 Wells => 5 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 6 Wells => 6 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 7 Wells => 7 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 8 Wells => 8 Transmission Peaks • Vb=110meV, W=6nm, B=2nm Klimeck – ECE606 Fall 2012 – notes adopted from Alam 9 Wells => 9 Transmission Peaks • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 19 Wells => 19 Transmission Peaks • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 29 Wells => 29 Transmission Peaks • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 39 Wells => 39 Transmission Peaks • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 49 Wells => 49 Transmission Peaks • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam N Wells => N Transmission Peaks • Bandpass filter formed • Bandpass sharpens with • Band transmission not symmetric increasing number of wells Klimeck – ECE606 Fall 2012 – notes adopted from Alam 1 Well => 1 Transmission Peak => 1 State • Bandpass filter formed • Bandpass sharpens with • Band transmission not symmetric increasing number of wells Klimeck – ECE606 Fall 2012 – notes adopted from Alam 2 Wells => 2 Transmission Peaks => 2 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 3 Wells => 3 Transmission Peaks => 3 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 4 Wells => 4 Transmission Peaks => 4 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 5 Wells => 5 Transmission Peaks => 5 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 6 Wells => 6 Transmission Peaks => 6 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 7 Wells => 7 Transmission Peaks => 7 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 8 Wells => 8 Transmission Peaks => 8 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 9 Wells => 9 Transmission Peaks => 9 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 19 Wells => 19 Transmission Peaks => 19 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 29 Wells => 29 Transmission Peaks => 29 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 39 Wells => 39 Transmission Peaks => 39 States • Bandpass filter formed • Band transmission not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam 49 Wells => 49 Transmission Peaks => 49 States • Bandpass filter formed • Cosine-like band formed • Band transmission not symmetric • Band is not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam N Wells => N Transmission Peaks => N States • Bandpass filter formed • Cosine-like band formed • Band transmission not symmetric • Band is not symmetric Klimeck – ECE606 Fall 2012 – notes adopted from Alam N Wells => N States => 1 Band • Vb=110meV, W=6nm, B=2nm => ground state in each well => what if there were excited states in each well => Vb=400meV Klimeck – ECE606 Fall 2012 – notes adopted from Alam N Wells => 2N States => 2 Bands Vb=110meV, W=6nm, B=2nm Vb=400meV,Klimeck – ECE606 W=6nm, Fall 2012 – notesB=2nm adopted from Alam N Wells => 2N States => 2 Bands Vb=110meV, W=6nm, B=2nm 1 state/well => 1band Can we get more states/well? => Increase well width 2 states/well => 2bands Vb=400meV,Klimeck – ECE606 W=6nm, Fall 2012 – notesB=2nm adopted from Alam X States/Well => X Bands Vb=110meV, W=6nm, B=2nm 1 state/well => 1 band Vb=400meV W=6nm, B=2nm 2 states/well => 2 bands Vb=400meV W=10nm, B=2nm 3 states/well => 3 bands Klimeck – ECE606 Fall 2012 – notes adopted from Alam X States/Well => X Bands Vb=110meV, W=6nm, B=2nm 1 state/well => 1 band Vb=400meV W=6nm, B=2nm 2 states/well => 2 bands Vb=400meV W=10nm, B=2nm 3 states/well => 3 bands Klimeck – ECE606 Fall 2012 – notes adopted from Alam Formation of energy bands •Each quasi-bond state will give rise to a resonance in a well. (No. of barriers -1) •Degeneracy is lifted because of interaction between these states. •Cosine-like bands are formed as the number of wells/barriers is increased •Each state per well forms a band •Lower bands have smaller slope = > heavier mass Klimeck – ECE606 Fall 2012 – notes adopted from Alam Presentation Outline • Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Procedure Summary • Additional notes »Discretizing Schroedinger’s equation for numerical implementations Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt Klimeck – ECE606 Fall 2012 – notes adopted from Alam Five Steps for Closed System Analytical Solution Solution Ansatz ψ ( ) =ikx + − ikx 2ψ x Ae+ Ae − d + 2ψ = 2N unknowns 1) 2 k 0 −αx + α x dx for N regions ψ ( x) = De + Ee 2) ψ (x = −∞ ) = 0 Boundary Conditions at the edge ψ (x = +∞ ) = 0 Reduces 2 unknowns ψ= ψ 3) =− = + Boundary Condition at each interface: xxB xx B dψ d ψ Set 2N-2 equations for = 2N-2 unknowns (for continuous U) dx=− dx = + xxB xx B ∞ 4) Det (coefficient matrix)=0 5) ∫ ψ (,)xE2 dx = 1 And find E by graphical −∞ or numerical solution Normalization of unity probability for wave function Klimeck – ECE606 Fall 2012 – notes adopted from Alam Open System: Generalization to Transfer Matrix Method • The complete transfer matrix A D E BFC • In general for any intermediate set of layers, the TMM is expressed as: + + A M M A n−1 = 11 12 n − M M − An−1 21 22 An • For multiple layers the overall transfer matrix will be • Looks conceptually very simple and analytically pleasing • Use it for your homework assignment for a double barrier structure! Klimeck – ECE606 Fall 2012 – notes adopted from Alam Presentation Outline • Time Independent Schroedinger Equation • Analytical solutions of Toy Problems »(Almost) Free Electrons »Tightly bound electrons – infinite potential well »Electrons in a finite potential well »Tunneling through a single barrier • Numerical Solutions to Toy Problems »Tunneling through a double barrier structure »Tunneling through N barriers • Procedure Summary • Additional notes »Discretizing Schroedinger’s equation for numerical implementations Reference: Vol. 6, Ch. 2 (pages 29-45) • piece-wise-constant-potential-barrier tool http://nanohub.org/tools/pcpbt Klimeck – ECE606 Fall 2012 – notes adopted from Alam Numerical solution of Schrodinger Equation 2ψ d 2 ≡[ − ] ℏ + k ψ = 0 k2 m0 E U(x)/ dx 2 U0(x) k > 0 α = i k α = i k x Klimeck – ECE606 Fall 2012 – notes adopted from Alam (1) Define a grid … ℏ2d 2 ψ − +ψ = ψ U(x) 2 U( x ) E 2m0 dx U1 ψ 3 a x 0 1 2 3 N N+1 Klimeck – ECE606 Fall 2012 – notes adopted from Alam Second Derivative on a Finite Mesh ψ 2 2 ψ ψ ()+ =+ψ () d + a d + x0 a x0 a2 ... dxx= a 2 d x = 0 x0 a ψ 2 2 ψ ψ ()− =−ψ () d + a d − x0 a x0 a2 ... dxx= a 2 d x = 0 x0 a 2ψ ψ ()+ +ψ ()− − ψ () = 2 d x0 a x0 a 2 x0 a 2 dx = x0 a 2ψ ψ − ψ +ψ d = i−1 2 i i+1 dx 2 a 2 i Klimeck – ECE606 Fall 2012 – notes adopted from Alam (2) Express equation in Finite Difference Form d 2ψ ℏ2 −()2 +ψ = ψ ≡ t0a2 U (x ) E t0 2 dx 2m0 a 2ψ ψ − ψ +ψ d = i−1 2 i i+1 2 2 dx i a −ψ ++( ) ψψ − = ψ t01i−2 t 0Ui i t 01i + E i ψ = (0) 0 N unknowns ψ(L) = 0 x 0 1 2 i-1 i i+1 N+1 Klimeck – ECE606 Fall 2012 – notes adopted from Alam (3) Define the matrix … − ψ + + ψ − ψ = ψ [ t0 i−1 (2t0 ECi ) i t0 i+1] E i (i = 2, 3…N-1) −++ψ( ) ψψψ − = (i = 1) t002 tE 0Ci 102 t E i −++ψ( ) ψψψ − = t010N− 2 tE Ci N t 01 N+ E i (i = N) ψψψ = ψψψ H E −t (2t + E ) −−t t00 (2t00 ECiC1) 0t0 = E N x N N x 1 Klimeck – ECE606 Fall 2012 – notes adopted from Alam (4) Solve the Eigen-value Problem Hψψψ = Eψψψ wrong U(x) ε Eigenvalue 4 problem; ε easily solved 3 ε with MATLAB 2 ε & nanohub 1 tools a x 1 23 4 (N-1) N Klimeck – ECE606 Fall 2012 – notes adopted from Alam