Lecture 4: Solution of Schrodinger Equation

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1 NCN EE 66: Solid State Devices Lecture 4: Solution of Schrodinger Equation Muhammad Ashraful Alam Alam ECE 66 S9 1

2 Outline 1) Time independent independent Schrodinger Equation ) Analytical solution of toy problems 3) Bound vs. tunneling states tt 4) Conclusions 5) Additional Notes: Numerical solution of Schrodinger Equation Reference: Vol. 6, Ch. (pages 9 45) Alam ECE 66 S9

3 Motivation Periodic Structure E Alam ECE 66 S9 3

4 Time independent Schrodinger Equation d Ψ dψ + U ( x ) Ψ = i m dx dt Assume Ψ ( x, t ) = ( x ) e iet / d ( x ) ie e + e = e iet iet iet U( x) ( x) i ( x) m dx d + U( x) = m dx E d m + ( E U ) = dx Alam ECE 66 S9 4

5 Time independent Schrodinger Equation d m + ( E U) = dx If E >U, then. k m [ E U] d + k = = + dx ikx ikx Ae Ae + + ( x) Asin( kx) Bcos( kx) If UEth U>E, then. α m U [ E] d dx α = x x ( x) De α + α = + Ee Alam ECE 66 S9 5

6 A Simple Differential Equation d + U( x) = m dx E Obtain U(x) and the boundary conditions for a given problem. Solve the nd order equation pretty basic Interpret * = as the probability of finding an electron at x Compute anything else you need, e.g., p = * d * d Ψ Ψ dx E = Ψ Ψ dx idx idt Alam ECE 66 S9 6

7 Outline 1) Time independent independent Schrodinger Equation ) Analytical solution of toy problems 3) Bound vs. tunneling states tt 4) Conclusions 5) Additional Notes: Numerical solution of Schrodinger Equation Alam ECE 66 S9 7

8 Full Problem Difficult: Toy Problems First Periodic Structure E Case 3: Free electron E >> U Case 1: Electron in finite well E < U Case : Electron in infinite well E << U Alam ECE 66 S9 8

9 Five Steps for Analytical Solution 1) d + k = dx N unknowns for N regions ) ( x = ) = ( x =+ ) = Reduces unknowns 3) d dx = + B x x B x= x = = d dx + B x xb x= x = SetN N equations for N unknowns (for continuous U) 4) Det (coefficient matrix)= 5) ( xe, ) dx= 1 And find E by graphical or numerical solution for wave function Alam ECE 66 S9 9

10 Case 1: Bound levels in Finite Well (steps 1,) E U(x) E = = Asin kx + B cos kx 1) Me + Ce α x + α x = De + Ne α x + α x ) Boundary conditions ( x = ) = ( x = + ) = a 1

11 Step3: Continuity of wave function 3) d dx = + B x xb x= x = d dx + B x xb x= x = C = B αc = ka = Asin( ka) + Bcos( ka) = De αa ka cos( ka ) kb sin( ka ) = αα De αa = Asin kx + B cos kx = Ce α x = De α x a Alam ECE 66 S9 11

12 Step 3: Continuity of Wavefunction C = B α C = ka Asin( ka) + Bcos( ka) = De αa kacos( ka) kb sin( ka) = α De αa 1 1 A k α B = αa sin( ka) cos( ka) e C αa cos( ka) sin( ka) αe / k D Alam ECE 66 S9 1

13 Step 4: Bound level in Finite Well det (Matrix)= ξ(1 ξ) E mu tan( αa ξ) = ξ α ξ 1 U U Only unknown is E (i) () Use Matlab function (ii) Use graphical method a Alam ECE 66 S9 13

14 Step 4: Graphical Method for Bound Levels y1 x = x+ 5 = x y tnα = x+ 5 a( a ξ ) = ξ (1 ξ ) ξ 1 y x E 1 ζ=e/u x Alam ECE 66 S9 14

15 Step 4: Graphical Method for Bound Levels E 1 ζ=e/u E 1 ζ=e/u a Alam ECE 66 S9 15

16 Step 5: Wave functions = Asin kx + B cos kx = Ce α x = 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 1 1 B α C ka = αa cos( ka) e D Asin ( ka) B 1 1 C α ka = a D cos( ka ) e α Asin( ka ) 1 Alam ECE 66 S9 16

17 = Ce α x Step 5: Calculating Wave function = Asin kx + B cos kx = De α x B 1 1 = C α ka a D cos( ka) e α Asin( ka) 1 C dx = 1 ( ) ( ) a αx αx e dx + Asin kx + B sin kx dx + D e dx a Alam ECE 66 S9 17

18 Aside: Infinite Quantum Well d + k m = [ E U] dx + k = 1) Solutions: = k A sin kx + ) Boundary conditions B cos( kx) ( x ) Asink ( ) Bcosk ( ) = = = + ( x = a) = = Asin( ka) = Asin( nπ ) k n nπ = = a me n E n n π = ma Alam ECE 66 S9 18

19 Five steps for Analytical Solution: Follow rules 1) d + k = dx N unknowns for N regions ) ( x = ) = ( x =+ ) = Reduces unknowns 3) d dx = + B x x B x= x = = d dx + B x xb x= x = SetN N equations for N unknowns (for continuous U) 4) Det(coefficient matix)= 5) ( xe, ) dx= 1 And find E by graphical or numerical solution for wave function Alam ECE 66 S9 19

20 Practical examples: Si/Ag and Si/SiO Si SiO Si Speer et. al, Science 314, 6. Alam ECE 66 S9

21 Outline 1) Time independent independent Schrodinger Equation ) Analytical solution of toy problems 3) Bound vs. tunneling states tt 4) Conclusions 5) Additional Notes: Numerical solution of Schrodinger Equation Alam ECE 66 S9 1

22 Case : Solution for Particles with E>>U d dx + k = k [ ] m E U E 1) Solution ( x) = Asin( kx) + Bcos( kx) A e + A e ikx + ikx U(x) ~ ikx ) Boundary condition ( x) = A e positive going wave + ikx = Ae negative going wave Alam ECE 66 S9

23 Free Particle ( x ) Asin ( kx ) + Bcos ( kx ) = + Ae + Ae ikx + ikx + ( ) ikx x = A e positive going wave ikx = Ae negative going wave E Probability: * = = A+ or A U(x) ~ Momentum: * d p = Ψ Ψ dx = k or - k idx Alam ECE 66 S9 3

24 Case 3: Bound vs. Tunneling State Ce Ae ik x + Be ik x ik1x + Me ik1x 1 ik x De + Ne ik x Boundary conditions N= 5 unknowns (C,M,A,B,D) 4 equations from x= and x=a interfaces No bound levels Ratios of D/C is of Interest. Alam ECE 66 S9 4

25 Practical Example: Oxide Tunneling Alam ECE 66 S9 5

26 Conclusions 1) We have discussed the analytical solution of Schrodinger equation for simple potentials. Such potential arises in wide variety of practical systems. ) Numerical solution is very powerful, but it is easy to get wrong results if one is not careful. 3) Solving bound dlevel lproblem is different compared to the solution of tunneling problem. The corresponding recipes should be followed carefully. Alam ECE 66 S9 6

27 Outline 1) Time independent independent Schrodinger Equation ) Analytical solution of toy problems 3) Bound vs. tunneling states tt 4) Conclusions 5) Additional Notes: Numerical solution of Schrodinger Equation Alam ECE 66 S9 7

28 Numerical solution of Schrodinger Equation d dx dx + k = [ ] k m E U(x) / U (x) α = ik k > α = ik x Alam ECE 66 S9 8

29 (1) Define a grid d +UU ( x ) = E m dx U(x) U N N+1 a x Alam ECE 66 S9 9

30 Aside: Finite difference Connecting neighbors d a d = + a ( x + a ) ( x ) dx x = a dx x = a d a d = x a +... dx dx ( x a) ( ) x = a x = a ( ) + ( x a) ( x ) x + a = a dx d x = a d dx i = i 1 i + i+1 a Alam ECE 66 S9 3

31 () Express equation in Finite Difference Form d ( t ) a + U ( x ) = dx E t ma d 1 = i 1 i + i+1 a dx i t ( ) i 1+ t + Ui i t i+ 1 = E i () = N unknowns (L) ) = i-1 i i+1 1 N+1 x Alam ECE 66 S9 31

32 (3) Define the matrix [ t i 1 + ( t + E Ci ) i t i+1 ]= E i (i =, 3 N-1) ( ) t + t + ECi 1 t = Ei ( ) (i = 1) t N 1 + t + ECi N t N+ 1 = E i (i = N) H = E N x N N x 1 t (t + +EE Ci C1 )) t t = E Alam ECE 66 S9 3

33 (4) Solve the Eigen value Problem H = E wrong U(x) Eigenvalue problem; easily solved with ih MATLAB & nanohub tools a ε 1 ε ε 3 ε (N-1) N x Alam ECE 66 S9 33

ECE606: Solid State Devices Lecture 3

ECE606: Solid State Devices Lecture 3 ECE66: Solid State Devices Lecture 3 Gerhard Klimeck gekco@purdue.edu Motivation Periodic Structure E Time-independent Schrodinger Equation ħ d Ψ dψ + U ( x) Ψ = iħ m dx dt Assume Ψ( x, t) = ψ( x) e iet/