Schrödinger Equation: (Time Independent) Hψ = Eψ This Is a Differential Eigenvalue Equation

Physics and Material Science of Semiconductor Nanostructures

PHYS 570P

Prof. Oana Malis Email: [email protected] Lecture 9

Review of quantum mechanics, statistical physics, and solid state Band structure of materials Semiconductor band structure Semiconductor nanostructures

Ref. Davies Chapter 1 Quantum Mechanics (QM) • The Schrödinger Equation: (time independent) Hψ = Eψ This is a differential eigenvalue equation. H  Hamiltonian operator for the system (energy operator) E  Energy eigenvalue, ψ  wavefunction Particles are QM waves! |ψ|2  probability density; ψ is a function of ALL coordinates of ALL particles in the problem! One Page Elementary Quantum Mechanics & Solid State Physics Review

• Quantum Mechanics of a Free Electron: 2 – The energies are continuous: E = (k) /(2mo) (1d, 2d, or 3d) – The wavefunctions are traveling waves: ikx ikr ψk(x) = A e (1d) ψk(r) = A e (2d or 3d) • Solid State Physics: Quantum Mechanics of an Electron in a Periodic Potential in an infinite crystal :

– The energy bands are (approximately) continuous: E= Enk – At the bottom of the conduction band or the top of the valence band, in the effective mass approximation, the bands can be written: 2 Enk  (k) /(2m*) – The wavefunctions are Bloch Functions = traveling waves: ikr Ψnk(r) = e unk(r); unk(r) = unk(r+R) QM Review: The 1d (infinite) Potential Well (“particle in a box”) In all QM texts!!  Consider the case of a particle in a 1-D potential well, with width L e infinite barriers

 V(x) = 0 for 0  x  L  V(x) =  for x<0, x>L

 Schrödinger equation Inside the well (V=0)  d 2 x 2m  E  x  0 dx2 2 Schrödinger Equation – free particle

Jan 2006 QM Review: The 1d (infinite) Potential Well (“particle in a box”) In all QM texts!! • We want to solve the Schrödinger Equation for: x < 0, V; 0 < x < L, V = 0; x > L, V 2 2 2  -[ /(2mo)](d ψ/dx ) = Eψ • Boundary Conditions: ψ = 0 at x = 0 & x = L (Vthere) •Energies: 2 2 En = (n) /(2moL ), n = 1,2,3 Wavefunctions: ½ ψn(x) = (2/L) sin(nx/L) (a standing wave!) Qualitative Effects of Quantum Confinement: Energies are quantized & ψ changes from a traveling wave to a standing wave. Potential Well

 If we consider that the electrons are confined in a 1D potential well whose energy levels are quantified by a set of discrete values in accordance to En.

 Notice that the in between spacing of consecutive levels is bigger the lesser L is. n2 22 E  n 2mL2 n = 4 (x) ENERGY U= U= n = 3 n=1 n=3

n = 2 n = 1 0 L x n=2 Jan 2006 Graphical w.d.f presentation and probability density function for the three first modes 2  nπ  ψn(x) sin x LL 

2 (x) (x)

n= 2 n= 2

n= 1 n= 1

n= 0 n= 0 0 L 0 L

Jan 2006 In 3Dimensions…  •For the 3D infinite potential well: (x, y, z) ~ sin( n x )sin( m y )sin( qz ), n,m,q  integer Lx Ly Lz R 2 2 n2h2 m2h2 q h Energy levels  2  2  2 8mLx 8mLy 8mLz

Real Quantum Structures aren’t this simple!!

•In Superlattices & Quantum Wells, the potential barrier is obviously not infinite! •In Quantum Dots, there is usually ~ spherical confinement, not rectangular. • The simple problem only considers a single electron. But, in real structures, there are many electrons & also holes! • Also, there is often an effective mass mismatch at the boundaries. That is the boundary conditions we’ve used are too simple! QM Review: The 1d (finite) Rectangular Potential Well In most QM texts!! Analogous to a Quantum Well • We want to solve the Schrödinger Equation for:

We want bound

states: ε < Vo

2 2 2 [-{ħ /(2mo)}(d /dx ) + V]ψ = εψ (ε  E)

V = 0, -(b/2) < x < (b/2); V = Vo otherwise Solve the Schrödinger Equation: 2 2 2 (½)b [-{ħ /(2mo)}(d /dx ) + V]ψ = εψ -(½)b (ε  E) V = 0, -(b/2) < x < (b/2)

V = Vo otherwise V Bound states are in Region II o V= 0 Region II: ψ(x) is oscillatory

Regions I & III: ψ(x) is decaying The 1d (finite) rectangular potential well A brief math summary!

2 2 2 2 Define: α  (2moε)/(ħ ); β  [2mo(ε -Vo)]/(ħ ) The Schrödinger Equation becomes: (d2/dx2) ψ + α2ψ = 0, -(½)b < x < (½)b (d2/dx2) ψ - β2ψ = 0, otherwise.  Solutions: ψ = C exp(iαx) + D exp(-iαx), -(½)b < x < (½)b ψ = A exp(βx), x < -(½)b ψ = A exp(-βx), x > (½)b Boundary Conditions:  ψ & dψ/dx are continuous SO: •Algebra (2 pages!) leads to: 2 2 (ε/Vo) = (ħ α )/(2moVo) ε, α, β are related to each other by transcendental equations. For example: tan(αb) = (2αβ)/(α 2- β2) • Solve graphically or numerically.

• Get: Discrete Energy Levels in the well (a finite number of finite well levels!) • Even eigenfunction solutions (a finite number): Circle, ξ2 + η2 = ρ2, crosses η = ξ tan(ξ)

b • Odd eigenfunction solutions: Circle, ξ2 + η2 = ρ2, crosses η = -ξ cot(ξ)

V o b o

|E2| < |E1|