Outline • Quantum Mechanics Applied to Heterostructures • Quantum Well

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Outline • Quantum Mechanics Applied to Heterostructures • Quantum Well 6.772SMA5111 - Compound Semiconductors Lecture 5 - Quantum effects in heterostructures,I - Outline • Quantum mechanics applied to heterostructures Basic quantum mechanics as applied to carriers in heterostructures: 1. Effective mass theory ("free-space" motion with a different mass) 2. Dingle's "Potential energy landscape" model (particle-in-a-box analogy) Examples: 1. Potential steps (hetero-interfaces) 2. Potential walls (tunnel barriers) 3. Quantum confinement structures (the focus of the next two sections) • Quantum wells, wires, and dots Infinitely deep wells, wires, dots: 1. Energy level systems 2. Densities of states Real-world issues: (the impact of these issues on quantum wells) 1. Finite barriers 2. Effective mass changes 3. Non-parabolicity 4. Multiple bands 5. Strain C. G. Fonstad, 2/03 Lecture 5 - Slide 1 Quantum mechanics for heterostructures Two powerful, simplifying approximations • Effective Mass Theory (the most important concept in semiconductor physics) Electrons in the conduction band of a nearly perfect semi- conductor lattice can be modeled as though they are moving in free space, but with a different mass (their "effective mass"). Their wavefuction, y(x), satisfies the Schrodinger Equation with an appropriate mass and potential energy function: (-h–2/2m*)∂2y(x)/∂x2 = [E - V(x)] y(x) Similarly, empty valence bonds can be modeled as positively charged particles (holes) moving in free space with their own effective mass. • Potential energy landscape model (Ray Dingle, et al, ca. 1974) In heterojunctions electrons and holes can be modeled by the effective mass theory, even though perfect periodicity is destroyed. In this case, the potential energy profile, V(x), seen by electrons is Ec(x) and that seen by holes is Ev(x). C. G. Fonstad, 2/03 Lecture 5 - Slide 2 Quantum mechanics, cont. Schrodinger Equation in three dimensions: (-ћ2/2m*)[∂2/∂x2 +∂2/∂y2 +∂2/∂z2] y(x,y,z) y = [Ex + Ey + Ez - V(x,y,z)] (x,y,z) Corresponding wave function: y y y y (x,y,z) = x(x) y(y) z(z) Where V is a constant, the solutions are y (1) Plane waves if E > V, i.e. (x,y,z) = A exp [±i(kxx+ kyy+kzz)], with ћ2 2 2 2 E-V=[Ex + Ey + Ez - V(x,y,z)] = ( /2m*)(kx +ky+ k z ) y (2) Decaying exponentials if E < V, i.e. (x,y,z) = A exp -(kxx+ kyy+ kzz), with ћ2 2 2 2 V-E=[V-E x - Ey -Ez] = ( /2m*)(kx + ky+ k z ) C. G. Fonstad, 2/03 Lecture 5 - Slide 3 Common 1-d potential energy landscapes A one-dimensional potential step: GaAs AlGaAs V(x) DEc x Classically, electrons with 0 < E < DE cannot pass x = 0, while those D c with E > Ec do not see the step. Quantum mechanically, electrons with 0 < E < DE penetrate the c D barrier with an exponential tail, and those with E > Ec have a finite probability of being reflected by the step. C. G. Fonstad, 2/03 Lecture 5 - Slide 4 Common 1-d potential energy landscapes, cont. A one-dimensional potential barrier (tunnel barrier): GaAs AlGaAs GaAs V(x) DEc x Classically, electrons with 0 < E < DE can again not pass x = 0, while D c those with E > Ec do not see the barrier at all. D Quantum mechanically, electrons with 0 < E < Ec can penetrate the barrier and some fraction can pass right through it, i.e. "tunnel," D while a of fraction of those with E > Ec will be reflected by the step. C. G. Fonstad, 2/03 Lecture 5 - Slide 5 Quantum Tunneling through Single Barriers Transmission probabilities - Ref: Jaspirit Singh, Semiconductor Devices - an introduction, Chap. 1 Rectangular barrier E DEc Eco x d 2 2 T = 4 {4 cosh ad + [(a / k) - (k /a]2sinh ad} 2 = * - h2 a 2 = [D - - ] h2 where k 2m (E E c ) and 2moE c (E E c ) C. G. Fonstad, 2/03 Lecture 5 - Slide 6 Quantum Tunneling through Single Barriers Triangular barrier: E DEc Eco x È- * 1/ 2 ˘ = 4(2m ) {D - - }3/ 2 = T exp Í E c (E E co ) ˙ where q F dE c (x) dx Î 3qFh ˚ Trapezoidal barrier: DEc E Eco x d È- * 1/ 2 ˘ = 4(2m ) {D - - }3/ 2 -D{ - - - }3/ 2 T exp Í E c (E E co ) Ec (E E co qFd) ˙ Î 3qFh ˚ C. G. Fonstad, 2/03 Lecture 5 - Slide 7 Common 1-d potential energy landscapes, cont. A one-dimensional resonant tunneling barrier: GaAs AlGaAs GaAs AlGaAs GaAs V(x) DEc x - xB - xW/2 - xW/2 xW/2 xB + xW/2 Classically, electrons with 0 < E < DE can again not pass from one side Dc to the other, while those with E > Ec do not see the barriers at all. D Quantum mechanically, electrons with 0 < E < Ec with energies that equal energy levels of the quantum well can pass through the struc- D ture unattenuated; while a fraction of those with E > Ec will be reflected by the steps. C. G. Fonstad, 2/03 Lecture 5 - Slide 8 Common 1-d potential energy landscapes, cont. A one-dimensional potential well: AlGaAs GaAs AlGaAs V(x) DEc x Classically, electrons with 0 < E < DE are confined to the well, while D c those with E > Ec do not see the well at all. Quantum mechanically, electrons can only have certain discrete values of 0 < E < DE and have exponential tails extending into the barriers, c D while a fraction of those with E > Ec have a higher probability of being found over the well than elsewhere. C. G. Fonstad, 2/03 Lecture 5 - Slide 9 Three-dimensional quantum heterostructures - quantum wells, wires, and dots The quantities of interest to us are 1. The wave function 2. The energy levels 3. The density of states: As a point for comparison we recall the expressions for these quantities for carriers moving in bulk material: y Wave function: (x,y,z) = A exp [±i(kxx+kyy+kzz)] ћ2 2 2 2 Energy: E - Ec = ( /2m*)(kx +ky + kz ) r 2 ћ 3/2 1/2 Density of states: (E) = (1/2π ) (2m*/ ) (E - Ec) C. G. Fonstad, 2/03 Lecture 5 - Slide 10 Three-dimensional quantum heterostructures - quantum wells, wires, and dots The 3-d quantum well: InGaAs InP InP d x In an infinitely deep well, i.e. ∆Ec =∞,dx wide: y Wave function: (x,y,z) = An sin (nπx/dx) exp [±i(kyy+ kzz)] for 0 ≤ x ≤ dx = 0 outside well ћ2 2 2 Energy:E-E c = En + ( /2m*)(ky +kz ) 2 2 2 2 with En =πh n /2m*dx r 2ћ Density of states: (E) = (m*/π )for E≥ En for each n C. G. Fonstad, 2/03 Lecture 5 - Slide 11 Three-dimensional quantum heterostructures - quantum wells, wires, and dots d The 3-d quantum wire: x d y InGaAs InP In an infinitely deep wire, i.e. ∆Ec =∞,dx by dy: y Wave function: (x,y,z) = Anm sin (nπx/dx) sin (mπx/dy)exp[±i k zz)] for 0 ≤ x ≤ dx, 0 ≤ y ≤ dy = 0 outside wire ћ2 2 Energy:E-E c = En,m +( /2m*) kz 2ћ2 2 2 2 2 with En,m =(π /2m*)(n /dx + m /dy ) r ћ2 2 1/2 Density of states: (E) = {m*/[2 π (E - Ec -En,m)]} for each n,m Note: some combinations of n and m may give the same energies C. G. Fonstad, 2/03 Lecture 5 - Slide 12 Three-dimensional quantum heterostructures - quantum wells, wires, and dots d The 3-d quantum box: x d y InGaAs InP d z In an infinitely deep box, i.e. ∆Ec = ∞, dx by dy by dz: y Wave function: (x,y,z) = Anm sin (nπx/dx) sin (mπx/dy) sin (pπx/dz) for 0 ≤ x ≤ dx, 0 ≤ y ≤ dy , 0 ≤ z ≤ dz = 0 outside box Energy: E - Ec = En,m,p 2ћ2 2 2 2 2 2 2 with En,m,p = (π /2m*)(n /dx + m /dy + p /dz ) Density of states: r(E) = one per box for each combination of n, m, and p Note: some combinations of n and m may give the same energies C. G. Fonstad, 2/03 Lecture 5 - Slide 13 Quantum Wells, Wires, Boxes: Density of States 1. Bulk material Volume in k-space per state: (2p /L)3 Volume in k-space occupied by states with energy less than E: 2m V = 4p k 3 /3 where k = E - E k h2 c Number of electron states in this volume: p 3 3 Ê ˆ3/2 = ⋅4 k /3 = L Á2m ˜ - 3/2 N(E ) 2 ( E E c ) (2p /L)3 3p 2 Ë h ¯ Density of states with energies between E and E+dE per unit volume: 3/2 - - dN E Ê mˆ 1/2 r 1 ⋅ 3 = 1 ( ) = 1 Á2 ˜ ( - ) (E ) [eV m ] 3 2 E E c L dE 2p Ë h ¯ C. G. Fonstad, 2/03 Lecture 5 - Slide 14 Quantum Wells, Wires, Boxes: Density of States 2. Quantum well 2 Area in 2-d k-space (i.e., ky,kz) per state: (2p / L) Area in k-space occupied by staes with energy less than En: 2m p 2h2n 2 A = p k 2 where k = E - E - E , and E = k h2 n c n * 2 2m dx Number of electrons in this area in band n: p 2 2 * = ⋅ k = L m - - Nn (E ) 2 ( E E n E c ) (2p / L)2 p 2h 2 Density of states in well band n between E and E+dE per unit area: * - - dN E m r 1 ⋅ 2 = 1 n ( ) = n (E ) [eV m ] 2 2 2 L dE p h C.
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