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 π ψ ()= n = … n x Asin x , n 1,2,3, Lx Klimeck – ECE606 Fall 2012 – notes adopted from Alam 1-D Particle in a Box – Visualization • (Step 4) Plot first few solutions π ψ = n = … n ()x Asin x , n 1,2,3, 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 π ψ () 2 = 2 2 n = … n x A sin x , n 1,2,3, The distribution of SINGLE particle 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 π n =1,2,3,… ∴ψ = 2 n n ()x sin x, < < 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 2π 2 π π π n 2 n + × 2 n = 2 n 2 sin 0 sin E n sin 2 m 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