Elements of Quantum Mechanics Chapter 2 Elements of Quantum Mechanics Classical (Newtonian) mechanics is inaccurate when applied to electrons in crystals or any systems with atomic dimensions. Quantum mechanics is necessary background knowledge for understanding electrons in crystals.

THE QUANTUM CONCEPT Blackbody Radiation 𝐵 λ = 2𝑐𝑘𝑇 λ4 𝐸𝑛=𝐧ℎ𝜈=𝑛ℏ𝜔 An opaque non-reflective ideal body classical model, assuming continuum of allowed energy (valid for long wave length) 𝐵 λ = 2𝑐𝑘𝑇 λ4 real observation explained by Max Planck, assuming discrete energy spectrum 𝐸𝑛=𝐧ℎ𝜈=𝑛ℏ𝜔 𝐧=0, 1, 2, 3,……… 𝐵 λ = 2ℎ𝑐2 λ5 1 𝑒 ℎ𝑐/λ𝑘𝑇 −1 Vibrating atoms in a material can only radiate or absorb energy in discrete packets (energy quantization) λ (𝜇m)

The Bohr Atom 𝐿𝑛=𝑚0ʋ 𝑟 𝑛 =𝒏ℏ For the simple hydrogen atom with Z = 1, Postulations: Electrons exist in certain stable circular orbits. Electrons can shift between orbitals by gaining or losing energy. Angular momentum is quantized. 𝐿𝑛=𝑚0ʋ 𝑟 𝑛 =𝒏ℏ 𝐧=1, 2, 3,……… centripetal force −𝑞 +𝑞 𝑟𝑛 𝐧=1 𝐧=2 𝐧=3 = 𝑚0ʋ2 𝑟𝑛 Continuous energy radiation electromagnetic radiation Coulombl force = 𝑞2 4𝜋𝜖0 𝑟 𝑛 2 Classically, accelerating charge radiates electro magnetic wave.

𝑟𝑛= 4𝜋𝜖0 𝒏ℏ 2 𝑚0𝑞2 𝑚0ʋ2 𝑟𝑛 = 𝑞2 4𝜋𝜖0 𝑟 𝑛 2 P.E = - 𝑞2 4𝜋𝜖0𝑟𝑛 𝑚0ʋ2 𝑟𝑛 = 𝑞2 4𝜋𝜖0 𝑟 𝑛 2 K.E = 1 2 𝑚0ʋ2= 1 2 ( 𝑞2 4𝜋𝜖0𝑟𝑛 ) P.E = - 𝑞2 4𝜋𝜖0𝑟𝑛 E = K.E + P.E 𝐸𝑛=− 𝑚0𝑞4 2 4𝜋𝜖0𝒏ℏ 2 =− 13.6 𝒏2 𝑒𝑉

Wave-Particle Duality Light: wave nature (diffraction, refraction, interference….) particle nature (photoelectric effect, Compton effect) called “photon” Electron: particle nature (m0, q ….) wave nature (SEM, TEM, diffraction, refraction, interference…) De Broglie’s matter wave: All particles have the properties of wave. λ= ℎ 𝑚ʋ p= ℎ λ de Broglie’s hypothesis Low dimensional materials (2D, 1D) in terms of electrical properties: size of reduced dimensions approaches λ of electron

BASIC FORMALISM General Formulation Quantum mechanics In 1926, Schrödinger: wave mechanics Heisenberg: matrix mechanics Five basic postulates of wave mechanics: There exist a wavefunction, Ψ(x, y, z, t), describing the dynamic behavior of the system. mathematically complex quantity (2) The Ψ for a given system and specified system constraint is determined by solving the time dependent Schrödinger wave equation, − ℏ 2 2𝑚 𝛻2Ψ + U(x, y, z) = ℏ 𝑖 𝜕Ψ 𝜕𝑡 (3) Ψ and 𝛻Ψ must be continuous, finite, and single-valued for all values of x, y, z, and t. not single-valued not continuous not finite

(4) Ψ∗ΨdV = Ψ 2 dV : the probability that the particle will be found in the spatial volume element dV. Ψ∗ΨdV = 1: normalization (5) There is a unique mathematical operator with each dynamic variable such as position or momentum. The expectation value can be obtained by operating on the wavefunction. <𝛼> = 𝑉 Ψ∗𝛼𝑜𝑝ΨdV < 𝑝 𝑥 > = 𝑉 Ψ∗( ℏ 𝑖 𝜕 𝜕𝑥 Ψ)dV =ℏ𝑘 𝑉 Ψ∗ΨdV =ℏ𝑘 <𝑥> = 𝑉 Ψ∗𝑥 ΨdV If Ψ = ej(kx - 𝜔𝑡) <𝐸> = 𝑉 Ψ∗(− ℏ 𝑖 𝜕 𝜕𝑡 Ψ)dV =ℏ𝜔 𝑉 Ψ∗ΨdV =ℏ𝜔

Where does operator come from? For plane wave solution( or harmonic solutions of the wave): Ψ = Aei(kx – 𝜔t) operator 𝜕Ψ 𝜕𝑥 =𝑖𝑘Ψ ℏ 𝑖 𝜕Ψ 𝜕𝑥 =ℏ𝑘Ψ expectation value 𝜕2Ψ 𝜕𝑥2 =− 𝑘2Ψ − ℏ2 2𝑚 𝜕2Ψ 𝜕𝑥2 = ℏ2𝑘2 2𝑚 Ψ 𝜕Ψ 𝜕𝑡 =− 𝑖𝜔Ψ − ℏ 𝑖 𝜕Ψ 𝜕𝑡 =ℏ𝜔Ψ Time-independent Formulation E = K.E + P.E − 𝑃2 2𝑚 +𝑈 𝑥, 𝑦, 𝑧 =𝐸:𝑐𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 In operator form with Ψ − ℏ2 2𝑚 𝛻2+𝑈(𝑥, 𝑦, 𝑧)) Ψ=− ℏ 𝑖 𝜕Ψ 𝜕𝑡 time-dependent Schrödinger equation 𝐻 Ψ=− ℏ 𝑖 𝜕Ψ 𝜕𝑡 =𝐸Ψ Hamiltonian operator

(3’) ψand 𝛻ψmust be continuous, finite, and single-valued − ℏ 𝑖 𝜕Ψ 𝜕𝑡 =𝐸Ψ 𝑉 Ψ∗(− ℏ 𝑖 𝜕 𝜕𝑡 Ψ)dV =𝐸 𝑉 Ψ∗ΨdV =𝐸:𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 The equation has the solution of the form Ψ 𝑥, 𝑦, 𝑧, 𝑡 = ψ 𝑥, 𝑦, 𝑧 𝑒 −𝑖𝐸𝑡/ℏ Substituting this into time dependent equation and by using separation of variables. 𝛻2ψ+− 2𝑚 ℏ2 [ 𝐸 −𝑈 𝑥, 𝑦, 𝑧 ]ψ=0 time-independent Schrödinger equation (3’) ψand 𝛻ψmust be continuous, finite, and single-valued for all values of x, y, and z. (4’) Ψ∗Ψ=ψ∗ψdV = ψ 2 dV : the probability that the particle will be found in the spatial volume element dV. Likewise 𝑉 ψ∗ψdV = 1 (5’) The expectation value of the system variable 𝛼 is given by <𝛼> = 𝑉 ψ∗𝛼𝑜𝑝ψdV

SIMPLE PROBLEM SOLUTIONS The Free Particle (electrons in free space) U(x, y, z) = 0, more generally U(x, y, z) = constant with total energy E and mass m no force acting on the particle In 1-D, time-independent S.E 𝑑 2 ψ 𝑑 𝑥 2 + 2𝑚𝐸 ℏ2 ψ=0 By introducing the constant, k ≡ 2𝑚𝐸/ ℏ 2 or equivalently E= ℏ 2 𝑘 2 2𝑚 = < 𝑝 𝑥 >2 2𝑚 Solution becomes ψ 𝑥 = 𝐴 + 𝑒 𝑖𝑘𝑥 + 𝐴 − 𝑒 −𝑖𝑘𝑥 Recalling the relationship, Ψ 𝑥, 𝑦, 𝑧, 𝑡 = ψ 𝑥, 𝑦, 𝑧 𝑒 −𝑖𝐸𝑡/ℏ Ψ 𝑥, 𝑡 = 𝐴 + 𝑒 𝑖(𝑘𝑥 − 𝐸𝑡 ℏ ) + 𝐴 − 𝑒 −𝑖(𝑘𝑥+ 𝐸𝑡 ℏ ) Traveling wave moving (-) x-direction Traveling wave moving (+) x-direction

Traveling wave (plane wave) Electron in free space Traveling wave (plane wave) particle λ 𝑘= 2𝜋 λ : Expectation value of momentum for free particle wave number <𝑝𝑥> = −∞ ∞ ψ∗ ℏ 𝑖 𝑑 𝑑𝑥 ψ dx =ℏ𝑘 −∞ ∞ ψ∗ψdx =ℏ𝑘 = ℎ λ The phase of the wave, kx – Et/ℏ = constant 𝑑 𝑑𝑡 𝑘𝑥 − 𝐸𝑡 ℏ =0 de Broglie’s hypothesis 𝑘 𝑑𝑥 𝑑𝑡 =𝑘ʋ= 2𝜋 λ 𝜈λ=2𝜋𝜈=𝜔= 𝐸 ℏ Expectation value of the momentum for a free particle is exact , because ∆x = ∞ and hence ∆ p = 0. From classical mechanics From wave mechanics E= 𝑝 2 2𝑚 = 𝑚𝑣2 2 E= ℏ 2 𝑘 2 2𝑚 = < 𝑝 𝑥 >2 2𝑚 continuous energy spectrum Identical because ∆ p = 0 for free particle.

k = 𝑛𝜋 𝑎 , Particle in a 1-D Box ∴ ψ𝑛 𝑥 = 𝐴 𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑎 i) Outside the box, U = ∞, ψ 𝑥 =0 ii) Inside the box, U = 0. 𝑑 2 ψ 𝑑 𝑥 2 +− 2𝑚 ℏ2 (𝐸−𝑈)ψ=0 𝑑 2 ψ 𝑑 𝑥 2 + 𝑘 2 ψ=0 0 < x < a where k ≡ 2𝑚𝐸/ ℏ 2 or E= ℏ 2 𝑘 2 2𝑚 Solution becomes ψ 0 =0 or ψ 𝑥 =𝐴 𝑒 𝑖𝑘𝑥 + 𝐵 𝑒 −𝑖𝑘𝑥 Boundary conditions ψ 𝑥 =𝐴𝑠𝑖𝑛𝑘𝑥 + 𝐵𝑐𝑜𝑠𝑘𝑥 ψ 𝑎 =0 k = 𝑛𝜋 𝑎 , ψ 0 =𝐵=0, ψ 𝑎 =𝐴𝑠𝑖𝑛𝑘𝑎=0 ∴ n = ±1,±2, ±3,……. Two waves with +, - direction form standing wave. ψ𝑛 𝑥 = 𝐴 𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑎 Particle was restricted to a finite range of coordinate value.

E= ℏ 2 𝑘 2 2𝑚 En= 𝑛 2 ℏ 2 𝜋 2 2𝑚 𝑎 2 discrete energy spectrum standing wave Particle’s momentum is zero for all energy states, since the particle periodically changes direction. < 𝑝 𝑥 > = −∞ ∞ ψ∗ ℏ 𝑖 𝑑 𝑑𝑥 ψ dx =0

ψ∗ ψ= ψ 2 is called spatial density (probability density when it is normalized). ψ 2 ψ∗ ψ𝑑𝑥= ψ 2 𝑑𝑥: probability of finding the particle between x and x + dx x0 x −𝑞ψ∗ ψ=−𝑞 ψ 2 : spatial distribution of charge corresponding to a single electron The electron is no longer considered to be identifiable as a point with particular position, the whole density distribution is the “particle”. From normalization, −∞ ∞ ψ∗ψdx = 0 𝑎 ψ∗ψdx = 0 𝑎 𝐴 𝑛 2 𝑠𝑖𝑛2 𝑛𝜋𝑥 𝑎 dx = 1 ∴ 𝐴 𝑛 = ( 2 𝑎 )1/2 −∞ ∞ −𝑞ψ∗ψdx =−𝑞 0 𝑎 ψ∗ψdx =−𝑞

Finite Potential Well U(x) U0 U(x) = 0, 0 <x < a 𝑑 2 ψ 𝑑 𝑥 2 +− 2𝑚 ℏ2 (𝐸−𝑈)ψ=0, U(x) = U0, otherwise x a For 0 < x < a, U = 0, 𝑑 2 ψ0 𝑑 𝑥 2 + 𝑘 2 ψ0=0 where k ≡ 2𝑚𝐸/ ℏ 2 𝑑 2 ψ ± 𝑑 𝑥 2 + 𝛼 2 ψ ± =0 For x > a, x < 0, U = U0, where α≡ 2𝑚(𝑈0 −𝐸)/ ℏ 2 (0 < E < U0) The general solutions, ψ− −∞ =0 𝐵-, A+ = 0 ψ+ ∞ =0 ψ− 𝑥 =𝐴− 𝑒 𝛼𝑥 + 𝐵 −𝑒 −𝛼𝑥 …x < 0 ψ− 0 =ψ0 0 ψ0 𝑥 =𝐴0𝑠𝑖𝑛𝑘𝑥 + 𝐵0𝑐𝑜𝑠𝑘𝑥 …0 <x < a continuity of ψ B.C’s ψ0 𝑎 =ψ+ 𝑎 ψ+ 𝑥 =𝐴+ 𝑒 𝛼𝑥 + 𝐵 +𝑒 −𝛼𝑥 …x > a 𝑑ψ−(0) 𝑑𝑥 = 𝑑ψ0(0) 𝑑𝑥 continuity of 𝑑ψ 𝑑𝑥 𝑑ψ0(𝑎) 𝑑𝑥 = 𝑑ψ+(𝑎) 𝑑𝑥 Four equations four unknowns

𝐴0𝑠𝑖𝑛𝑘𝑎 + 𝐵0𝑐𝑜𝑠𝑘𝑎=𝐵 +𝑒 −𝛼𝑎 𝐴0 𝑘 2 − 𝛼 2 𝑠𝑖𝑛𝑘𝑎 −2𝛼𝑘𝑐𝑜𝑠𝑘𝑎 =0 𝛼𝐴−= 𝑘𝐴0 𝐴− = 𝐵0 𝐴0𝑠𝑖𝑛𝑘𝑎 + 𝐵0𝑐𝑜𝑠𝑘𝑎=𝐵 +𝑒 −𝛼𝑎 𝐴0 𝑘 2 − 𝛼 2 𝑠𝑖𝑛𝑘𝑎 −2𝛼𝑘𝑐𝑜𝑠𝑘𝑎 =0 𝛼𝐴−= 𝑘𝐴0 𝐴0 = 0; trivial solution 𝑘𝐴0𝑐𝑜𝑠𝑘𝑎 − 𝑘𝐵0𝑠𝑖𝑛𝑘𝑎=− 𝛼𝐵 +𝑒 −𝛼𝑎 Non-trivial solution, 𝑘 2 − 𝛼 2 𝑠𝑖𝑛𝑘𝑎 −2𝛼𝑘𝑐𝑜𝑠𝑘𝑎=0 or 𝑡𝑎𝑛𝑘𝑎 = 2𝛼𝑘 𝑘 2 − 𝛼 2 Introducing, α0≡ 2𝑚𝑈0 /ℏ2 (α0 =constant) and ξ≡ 𝐸/𝑈0 ( 0 <ξ < 1) Then, α=α0 1 −ξ and 𝑘=α0 ξ and therefore 𝑡𝑎𝑛⁡(α0𝑎 ξ )= 2 ξ(1 −ξ) 2ξ −1 = 𝑓(ξ) α0𝑎= 𝜋 4 assuming ξ = 0.87 𝐸=0.87𝑈0

there is one and only one allowed energy level. α0𝑎< 𝜋 𝑜𝑟 𝑈0< ℏ 2 𝜋 2 2𝑚 𝑎 2 , For very shallow wells, there is one and only one allowed energy level. When 𝜋< α0𝑎<2𝜋, there is two allowed energy levels. When 2𝜋< α0𝑎<3𝜋, there is three allowed energy levels. Visualization of quantum mechanical reflection finite well e-1 a infinite well penetration depth Visualization of tunneling through a thin barrier