ECE606: Solid State Devices Lecture 3

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1 ECE66: Solid State Devices Lecture 3 Gerhard Klimeck gekco@purdue.edu Motivation Periodic Structure E

2 Time-independent Schrodinger Equation ħ d Ψ dψ + U ( x) Ψ = iħ m dx dt Assume Ψ( x, t) = ψ( x) e iet/ ħ ħ d ψ ( x) ie e + e = e iet iet iet ħ ħ U ( x) ψ ( x) i ψ ( x) ħ ħ m dx ħ ħ d ψ + U ( x) ψ = Eψ m dx d ψ m + ( E U ) ψ = dx ħ Time-independent Schrodinger Equation d ψ m + ( E U ) ψ = dx ħ If E >U, then. k m [ E U ] ħ d ψ dx + k ψ = ψ ( x) = Asin( kx) + B cos ( kx) A e + A e ikx + ikx If U>E, then. α m U [ E] ħ d ψ dx α ψ = ψ x x ( x) De α + α = + Ee

3 A Simple Differential Equation ħ d ψ + U ( x) ψ = Eψ m dx Obtain U(x) and the boundary conditions for a given problem. Solve the nd order equation pretty basic Interpret ψ * = ψ ψ as the probability of finding an electron at x Compute anything else you need, e.g., p = * ħ d Ψ Ψ dx i dx * ħ d E = Ψ Ψ dx i dt 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. (pages 9-45) piece-wise-constant-potential-barrier tool

4 Full Problem Difficult: Toy Problems First Periodic Structure E Case 1: Free electron E >> U Case 3: Electron in finite well E < U Case : Electron in infinite well E << U Case 1: Solution for Particles with E>>U d ψ dx + k ψ = k [ ] m E U ħ E 1) Solution ψ ( x) = Asin( kx) + B cos ( kx) A e + A e ikx + ikx U(x) ~ ikx ) Boundary condition ( x) = A e positive going wave ψ + ikx = A e negative going wave

5 Free Particle ψ ( x) = Asin( kx) + B cos ( kx) A e + A e ikx + ikx ikx ( x) = A e positive going wave ψ + ikx = A e negative going wave E Probability: * ψ = ψψ = A+ or A U(x) ~ Momentum: * ħ d p = Ψ Ψ dx = k or - k i dx ħ ħ Full Problem Difficult: Toy Problems First Periodic Structure E Case 1: Free electron E >> U Case 3: Electron in finite well E < U Case : Electron in infinite well E << U

6 Case : Bound State Problems Mathematical interpretation of Quantum Mechanics(QM) ħ m Ψ + VΨ = iħ t Ψ» Only a few number of problems have exact mathematical solutions» They involve specialized functions V = V = V = 1 kx V = V = ax V 1 r Coulomb Potential by nucleus in an atom V = x = x = L x x = Particle in a box Harmonic Oscillator Triangular Potential Well x Ψ n Ψ Ψ n ( x)= A 1 n n! H n (Bx)e ( r) = AR n ( r)y n ( θ,ϕ) n ( x) = Asin( k n x) Ψ n ( x) = NAi( ξ n ) Cx 1-D Particle in a Box A Solution Guess (Step 1) Formulate time independent Schrödinger equation ħ m d dx ψ x ( )+ V x ( )ψ x ( ) = Eψ x ( ) where, V x ( )= < x < L x elsewhere (Step ) Use your intuition that the particle will never exist outside the energy barriers to guess, ψ( x)= x L x in the well (Step 3) Think of a solution in the well as: ψ n ( x)= Asin nπ x, n =1,,3, L x V = V = V = x = x = L x

7 1-D Particle in a Box Visualization (Step 4) Plot first few solutions n = 1 n = ψ n ( x)= Asin nπ x, n =1,,3, n = 3 L x n = 4 x = x = L x x = (Step 5) Plot corresponding electron densities ψ n ( x) nπ = A sin x, n =1,,3, The distribution of SINGLE particle L x x = L x x = Matches the condition we guessed at step! But what do the NEGATIVE numbers mean? x = L x x = x = L x ψ( x)= x L x in the well x = n = 1 n = n = 3 n = 4 s-type x = L x x = p-type x = L x x = d-type x = L x x = f-type ONE particle => density is normalized to ONE x = L x 1-D Particle in a Box Normalization to ONE particle (Step 6) Normalization (determine the constant A) Method 1) Use symmetry property of sinusoidal function ψ n ( x) nπ = A sin x L x x = x = L x A x = x = L x ( Area) =1= L x A A = L x ψ n ( ) ~ L x Method ) Integrate x over nπx 1 cos L x 1= ψ n ( x) nπ L dx = x A sin x dx = A L x L x L dx = A x L x A = L x ψ n ( x)= sin nπ x, n =1,,3, L x < x < L x L x

8 1-D Particle in a Box The Solution (Step 7) Plug the wave function back into the Schrödinger equation ψ n ( x)= ħ n π m L x sin nπ x L x L x sin nπ + L x L x ħ m d dx ψ x ( )+ V x ( )ψ x ( ) = Eψ x ( ) sin nπ = E n sin nπ L x L x L x L x ψ n ( x)= sin nπ x L x E n = ħ π ml x n L x n = 1,,3,, < x < L x Discrete Energy Levels! n = 4 n = 3 n = n = 1 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? E n = ħ π ml n x L x = 5nm L x = 5nm L x = 5cm 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.

9 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. (pages 9-45) piece-wise-constant-potential-barrier tool Full Problem Difficult: Toy Problems First Periodic Structure E Case 1: Free electron E >> U Case 3: Electron in finite well E < U Case : Electron in infinite well E << U

10 Five Steps for Closed System Analytical Solution 1) d ψ dx + k ψ = Solution Ansatz N unknowns for N regions ψ ψ ikx ikx ( x) = A+ e + A e x x ( x) De α + α = + Ee ) ψ ( x = ) = Boundary Conditions at the edge ψ ( x = + ) = Reduces unknowns 3) ψ = ψ + Boundary Condition at each interface: x= xb x= xb Set N- equations for N- unknowns (for continuous U) dψ dx dψ = dx + x= xb x= xb 4) Det (coefficient matrix)= 5) ψ ( x, E) dx = 1 And find E by graphical or numerical solution Normalization of unity probability for wave function Case : Bound-levels in Finite Well (steps 1,) E U(x) E ψ = Asin kx + B cos kx 1) ψ = Me + Ce α x + α x ψ = De + Ne α x + α x ) Boundary conditions at the edge ψ ( x = ) = ψ ( x = + ) = a

11 Case : Bound-levels in Finite Well (steps3) 3) Boundary at each interface ψ dψ dx = ψ + B x xb x= x = dψ = dx + B x xb x= x = C = B αc = ka Asin( ka) + B cos( ka) = De αa kacos( ka) kb sin( ka) = α De αa ψ = Asin kx + B cos kx ψ = Ce α x a x ψ = De α 1 1 A k α B = αa sin( ka) cos( ka) e C αa cos( ka) sin( ka) αe / k D Case : Bound-levels in Finite Well (step 4) Only unknown is E (i) Use Matlab function (ii) Use graphical method ψ = Asin kx + B cos kx ξ (1 ξ ) tan( αa ξ ) = ξ 1 ξ E U α mu ħ det (Matrix)= ψ = Ce α x a x ψ = De α 1 1 A k α B = αa sin( ka) cos( ka) e C αa cos( ka) sin( ka) αe / k D

12 Case : Bound-levels in Finite Well (steps 4 graphical) y1 x = x + 5 = x y = x + 5 ξ (1 ξ ) ta n( αa ξ ) = ξ 1 y x E 1 ζ=e/u x Case : Bound-levels in Finite Well (steps 4 graphical) ζ=e/u ξ (1 ξ ) ta n( αa ξ ) = ξ 1 E 1 E 1 ζ=e/u a Obtained the eigenvalues => could stop here in many cases

13 Case : Bound-levels in Finite Well (steps 5 wavefunction) Need another boundary condition! ψ = Asin kx + B cos kx ψ = Ce α x ψ = De α x 1 1 A k α B = αa sin( ka) cos( ka) e C αa cos( ka) sin( ka) α De / k D det (Matrix)= => Linear dependent E system mu => Only 3 variables ξ α U are unique ħ => One variable is undetermined Let s assume A can be freely ξ (1 ξ ) Chosen tan( => can αa get ξ ) = B,C,D ξ 1 ψ dx = 1 Get A ψ = Asin kx + B cos kx ψ = Ce α x ψ = De α x Another boundary condition! Non-linear => no simple linear algebra expression ( ) ( ) Step 5: Wave-functions a α x α x C e dx A sin kx B sin kx dx D e a dx = A k α B = αa sin( ka) cos( ka) e C αa cos( ka) sin( ka) α De / k D 1 1 B ) α C = ka αa => Linear cos( kadependent ) e system D Asin( ka => Only 3 variables are unique 1 => One Bvariable 1 is 1 undetermined Get Let s assume C = A can α be freely ka B,C,D a Chosen D => cos( can ka) get B,C,D e α Asin( ka )

14 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= In the quantum world:» C and D have a non-zero value! Si SiO Si» Electrons can tunnel inside a barrier ψ = Asin kx + B cos kx ψ = Ce α x ψ = De α x 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. (pages 9-45) piece-wise-constant-potential-barrier tool

15 Transmission through a single barrier Scattering Matrix approach Define our system : Single barrier One matrix each for each interface: S-matrices Incident : A Reflected : B D C Transmitted : E Incident : F No particles lost! Typically A=1 and F=. Tunneling through a single barrier Incident : A Reflected : B D C Transmitted : E Incident : F Wave-function each region,

16 Single barrier case Applying boundary conditions at each interface (x= and x=l) gives, Which in matrix can be written as, Generalization to Transfer Matrix Method The complete transfer matrix In general for any intermediate set of layers, the TMM is expressed as: + A n 1 A = M 11 M + 1 A n n 1 M 1 M A n For multiple layers the overall transfer matrix will be A B D C F E Looks conceptually very simple and analytically pleasing Use it for your homework assignment for a double barrier structure!

Single barrier case Transmission can be found using the relations

Case(γL<<1): Weak barrier Case: E>V o Single barrier : Concepts

Transmission above the barrier is not perfect unity!

17 Single barrier case Transmission can be found using the relations between unknown constants, Case: E<V o Case(γL large): Strong barrier Case(γL<<1): Weak barrier Case: E>V o 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

Effect of barrier thickness below the barrier Increased barrier width reduces tunneling probability Thicker barrier increase the reflection

org/tools/pcpbt Single Barrier Key Summary Quantum wavefunctions can tunnel through barriers Tunneling is reduced with increasing barrier height and

18 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 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» interfaces cause constructive and destructive interference»quasi bound states are formed that result in unity transmission

19 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. (pages 9-45) piece-wise-constant-potential-barrier tool Define our system : Double barrier Double Barrier Transmission: Scattering Matrix approach One matrix each for each interface: 4 S-matrices Left Incident Reflected Transmitted 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

Reminder: Single barrier Transmission is finite under the barrier tunneling!

Quasi-bound state above the barrier. Transmission goes to one.

one / unity for discrete energies (reflection probability of zero) for some

20 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. Double barrier: Concepts Double barriers allow a transmission probability of one / unity for discrete energies (reflection probability of zero) for some energies below the barrier height. This is in sharp contrast to the single barrier case Cannot be predicted by classical physics.

Double barrier: Quasi-bound states In addition to states inside

States above the barrier height are quasi-bound or weakly bound.

than the peak for the state inside the well.

21 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. 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.

22 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. 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.

23 Particle in a box The time independent Schrödinger equation is ħ m d dx ψ x ( )+ V x ( )ψ x ( ) = Eψ x The solution in the well is: ψ n ( x)= Asin nπ x, n =1,,3, L x ( ) V x where, ( )= < x < L x elsewhere V = V = Plugging the normalized wave-functions back into the Schrödinger equation we find that energy levels are quantized. ψ n ( x)= sin nπ x L x E n = h π ml x n L x n = 1,,3,K, < x < L x x = V = x = L x n = 4 n = 3 n = n = 1 Double barrier & particle in a box Green: Particle in a box energies. Red: Double barrier energies Double barrier: Thick Barriers(1nm), Tall Barriers(1eV), Well(nm). First few resonance energies match well with the particle in a box energies. The well region resembles the particle in a box setup.

Open systems Vs closed systems Green: Particle in a box energies.

Even the first resonance energy does not match with the particle in a box energy.

A double barrier structure is an OPEN system, particle in a box is a CLOSED system. Reason for deviation?

First excited state wave-function amplitude using tight binding.

24 Open systems Vs closed systems Green: Particle in a box energies. Red: Double barrier energies Double barrier: Thinner Barriers(8nm), Shorter Barriers(.5eV), Well(1nm). 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. Reason for deviation? Potential profile and resonance energies using tight-binding. First excited state wave-function amplitude using tight binding. Ground state wave-function amplitude using tight binding. Wave-function penetrates into the barrier region. The effective length of the well region is modified. The effective length of the well is crucial in determining the energy levels in the closed system. E n = h π ml well n n =1,,3,K, < x < L well

25 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 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. (pages 9-45) piece-wise-constant-potential-barrier tool

26 1 Well => 1 Transmission Peak Vb=11meV, W=6nm, B=nm Wells => Transmission Peaks Vb=11meV, W=6nm, B=nm Bonding/Anti-bonding State

27 3 Wells => 3 Transmission Peaks Vb=11meV, W=6nm, B=nm 4 Wells => 4 Transmission Peaks Vb=11meV, W=6nm, B=nm

28 5 Wells => 5 Transmission Peaks Vb=11meV, W=6nm, B=nm 6 Wells => 6 Transmission Peaks Vb=11meV, W=6nm, B=nm

29 7 Wells => 7 Transmission Peaks Vb=11meV, W=6nm, B=nm 8 Wells => 8 Transmission Peaks Vb=11meV, W=6nm, B=nm

30 9 Wells => 9 Transmission Peaks Bandpass filter formed Band transmission not symmetric 19 Wells => 19 Transmission Peaks Bandpass filter formed Band transmission not symmetric

31 9 Wells => 9 Transmission Peaks Bandpass filter formed Band transmission not symmetric 39 Wells => 39 Transmission Peaks Bandpass filter formed Band transmission not symmetric

32 49 Wells => 49 Transmission Peaks Bandpass filter formed Band transmission not symmetric N Wells => N Transmission Peaks Bandpass filter formed Band transmission not symmetric Bandpass sharpens with increasing number of wells

33 1 Well => 1 Transmission Peak => 1 State Bandpass filter formed Band transmission not symmetric Bandpass sharpens with increasing number of wells Wells => Transmission Peaks => States Bandpass filter formed Band transmission not symmetric

34 3 Wells => 3 Transmission Peaks => 3 States Bandpass filter formed Band transmission not symmetric 4 Wells => 4 Transmission Peaks => 4 States Bandpass filter formed Band transmission not symmetric

35 5 Wells => 5 Transmission Peaks => 5 States Bandpass filter formed Band transmission not symmetric 6 Wells => 6 Transmission Peaks => 6 States Bandpass filter formed Band transmission not symmetric

36 7 Wells => 7 Transmission Peaks => 7 States Bandpass filter formed Band transmission not symmetric 8 Wells => 8 Transmission Peaks => 8 States Bandpass filter formed Band transmission not symmetric

37 9 Wells => 9 Transmission Peaks => 9 States Bandpass filter formed Band transmission not symmetric 19 Wells => 19 Transmission Peaks => 19 States Bandpass filter formed Band transmission not symmetric

38 9 Wells => 9 Transmission Peaks => 9 States Bandpass filter formed Band transmission not symmetric 39 Wells => 39 Transmission Peaks => 39 States Bandpass filter formed Band transmission not symmetric

39 49 Wells => 49 Transmission Peaks => 49 States Bandpass filter formed Cosine-like band formed Band transmission not symmetric Band is not symmetric N Wells => N Transmission Peaks => N States Bandpass filter formed Cosine-like band formed Band transmission not symmetric Band is not symmetric

40 N Wells => N States => 1 Band Vb=11meV, W=6nm, B=nm => ground state in each well => what if there were excited states in each well => Vb=4meV Vb=11meV, W=6nm, B=nm N Wells => N States => Bands Vb=4meV, W=6nm, B=nm

41 N Wells => N States => Bands Vb=11meV, W=6nm, B=nm 1 state/well => 1band Can we get more states/well? => Increase well width Vb=4meV, W=6nm, B=nm states/well => bands X States/Well => X Bands Vb=11meV, W=6nm, B=nm 1 state/well => 1 band Vb=4meV W=6nm, B=nm states/well => bands Vb=4meV W=1nm, B=nm 3 states/well => 3 bands

X States/Well => X Bands Vb=11meV, W=6nm, B=nm 1 state/well => 1 band Vb=4meV W=6nm, B=nm states/well => bands Vb=4meV W=1nm, B=nm 3 states/well => 3 bands Formation of energy bands Each quasi-bond

42 X States/Well => X Bands Vb=11meV, W=6nm, B=nm 1 state/well => 1 band Vb=4meV W=6nm, B=nm states/well => bands Vb=4meV W=1nm, B=nm 3 states/well => 3 bands 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

43 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. (pages 9-45) piece-wise-constant-potential-barrier tool Five Steps for Closed System Analytical Solution 1) d ψ dx + k ψ = Solution Ansatz N unknowns for N regions ψ ψ ikx ikx ( x) = A+ e + A e x x ( x) De α + α = + Ee ) ψ ( x = ) = Boundary Conditions at the edge ψ ( x = + ) = Reduces unknowns 3) ψ = ψ + Boundary Condition at each interface: x= xb x= xb Set N- equations for N- unknowns (for continuous U) dψ dx dψ = dx + x= xb x= xb 4) Det (coefficient matrix)= 5) ψ ( x, E) dx = 1 And find E by graphical or numerical solution Normalization of unity probability for wave function

44 Open System: Generalization to Transfer Matrix Method The complete transfer matrix In general for any intermediate set of layers, the TMM is expressed as: + A n 1 A = M 11 M + 1 A n n 1 M 1 M A n For multiple layers the overall transfer matrix will be A B D C F E Looks conceptually very simple and analytically pleasing Use it for your homework assignment for a double barrier structure! 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. (pages 9-45) piece-wise-constant-potential-barrier tool

45 Numerical solution of Schrodinger Equation d ψ dx + k ψ = [ ] k m E U( x ) / ħ U (x) α = ik k > α = ik x (1) Define a grid ħ d ψ + U ( x) ψ = Eψ m dx U(x) U 1 ψ3 1 3 N N+1 a x

46 Second Derivative on a Finite Mesh ψ ψ ψ dψ a d ψ = + a ( x + a) ψ ( x ) dx x = a dx x = a ( x a) ψ ( x ) dψ a d ψ = a +... dx x = a dx x = a ( ) + ψ ( x a) ψ ( x ) d ψ x + a = a dx d ψ = ψ i 1 ψ i +ψ i+1 dx a i x = a () Express equation in Finite Difference Form d ψ ( ta ) + U ( x) ψ = Eψ dx t ħ ma d ψ dx i = ψ i 1 ψ i +ψ i+1 a ( U ) tψ i 1 + t + i ψ i tψ i+ 1 = Eψ i ψ() = N unknowns ψ(l) = i-1 i i+1 1 N+1 x

47 (3) Define the matrix [ t ψ i 1 + ( t + E Ci )ψ i t ψ i+1 ]= Eψ i ( ) tψ + t + ECi ψ1 tψ = Eψ i ( ) (i =, 3 N-1) (i = 1) tψ N 1 + t + ECi ψ N tψ N + 1 = Eψ i (i = N) Hψ = Eψ N x N N x 1 t (t + +EE Ci C1 )) t t = E (4) Solve the Eigen-value Problem Hψ = Eψ wrong U(x) Eigenvalue problem; easily solved with MATLAB & nanohub tools a ε 1 ε ε 3 ε (N-1) N x

Lecture 4: Solution of Schrodinger Equation

NCN www.nanohub.org EE 66: Solid State Devices Lecture 4: Solution of Schrodinger Equation Muhammad Ashraful Alam alam@purdue.edu Alam ECE 66 S9 1 Outline 1) Time independent independent Schrodinger Equation