MTLE-6120: Advanced Electronic Properties of Materials Review of - - PowerPoint PPT Presentation

MTLE-6120: Advanced Electronic Properties of Materials Review of basic quantum mechanics

Reading:

◮ Kasap: 3.1 - 3.8, 4.6 ◮ Griffiths QM: 1 - 2, 9.1 - 9.2

1

Blackbody radiation

2 4 6 8 10 12 14 1 2 3 4 5 6 Spectral radiance [kW/(sr m2 nm)] Wavelength [µm] 3000 K 4000 K 5000 K 5000 K classical T ◮ Spectrum of light (EM waves) emitted by a perfect absorber (black body) ◮ Experimental realization of blackbody: pinhole in a closed box ◮ Spectrum peaks at a wavelength inversely proportional to T ◮ Solar spectrum ≈ black body radiation at 5800 K ◮ (All this was known before 1900!)

2

EM modes in a box

◮ Standing EM waves in a box: sin nπx L

in each direction

◮ Overall modes: sin nxπx L

sin nyπy

L

sin nzπz

L ◮ Wavevector

k = (kx, ky, kz) = nxπ

L , nyπ L , nzπ L

• ◮ Between wavevector magnitude k and k + dk:

◮ Volume in

k-space:

4πk2dk 8 ◮ Volume per

k: π

L

3

◮ Number of modes per

k: 2 polarizations

◮ Number of modes per unit volume:

2 · 4πk2dk 8 · L π 3 · 1 L3 = k2dk π2

◮ In terms of wavelength λ = 2π/k:

1 π2 2π λ 2

• d2π

λ

• = 1

π2 2π λ 2 2πdλ λ2 = 8π λ4 dλ L

3

Classical theory: equipartition theorem

◮ Each mode: classical wave with any amplitude A with energy E = c0A2 ◮ At temperature T, probability of energy E is ∝ e−E/(kBT ) ◮ Average energy at temperature T is

E ≡ ∞ dAe−E/(kBT )E ∞ dAe−E/(kBT ) = ∞ d

• E/c0e−E/(kBT )E

∞ d

• E/c0e−E/(kBT )

= ∞ dEe−E/(kBT )E1/2 ∞ dEe−E/(kBT )E−1/2 = (kBT)3/2Γ(3/2) (kBT)1/2Γ(1/2) = kBT 2

◮ Oscillators: kinetic and potential energies ⇒ E = kBT ◮ EM waves: electric and magnetic fields ⇒ E = kBT

4

Rayliegh-Jean’s law

◮ Number of modes per wavelength: 8π λ4 ◮ Energy per mode: kBT ◮ Power radiated per surface area: × c 4 ◮ Spectral power per surface area:

Iλ = 8π λ4 · kBT · c 4 = 2πckBT λ4

2 4 6 8 10 12 14 1 2 3 4 5 6 Spectral radiance [kW/(sr m2 nm)] Wavelength [µm] 3000 K 4000 K 5000 K 5000 K classical

5

Planck hypothesis

◮ Energies for a mode with frequency ν only allowed in increments of hν ◮ In terms of angular frequency ω = 2πν, in increments of ω ◮ With an as yet-undetermined constant h (or = h/(2π)) ◮ At temperature T, n units of energy hν with probability ∝ e−nhν/(kBT ) ◮ Average number of energy units

n ≡

• n e−nhν/(kBT )n
• n e−nhν/(kBT ) =
• n e−nαn
• n e−nα

(α ≡ hν/(kBT)) = − d

dα

• n e−nα
• n e−nα

= − d dα ln

• n

e−nα = − d dα ln 1 1 − e−α = e−α 1 − e−α = 1 ehν/(kBT ) − 1 E = nhν = hν ehν/(kBT ) − 1

6

Modification of equipartition theorem

◮ Average energy per mode changes to

E = hν ehν/(kBT ) − 1

◮ For hν ≪ kBT

E ≈ hν hν/(kBT) = kBT classical regime with n ≫ 1

◮ For hν ≫ kBT

E ≈ hν ehν/(kBT ) = hνe−hν/(kBT ) new regime with n ≪ 1

7

Planck’s law

◮ Number of modes per wavelength: 8π λ4 (as before) ◮ Energy per mode: hc/λ ehc/(λkBT )−1 (new, using ν = c/λ) ◮ Power radiated per surface area: × c 4 (as before) ◮ Spectral power per surface area:

Iλ = 8π λ4 · hc/λ ehc/(λkBT ) − 1 · c 4 = 2πhc2 λ5 ehc/(λkBT ) − 1

• 2

4 6 8 10 12 14 1 2 3 4 5 6 Spectral radiance [kW/(sr m2 nm)] Wavelength [µm] 3000 K 4000 K 5000 K 5000 K classical

8

Planck’s law features

◮ Has a maximum at λ ≈ hc 5kBT (Wein’s displacement law) ◮ Determine h = 6.626 × 10−34 Js, = h/(2π) = 1.055 × 10−34 Js ◮ Total energy per surface area radiated by black body (Stefan’s law)

PS ≡ ∞ dλIλ = ∞ dλ 2πhc2 λ5 ehc/(λkBT ) − 1

• =

∞ d

• hc

xkBT

• 2πhc2

(hc/(xkBT))5 (ex − 1) (x ≡ hc/(λkBT)) = 2πhc2 hc kBT −4 ∞ x−2dx x5 (ex − 1) = T 4 · 2πk4

B

h3c2 π4 15

• σ

σ = 2π5k4

B

15h3c2 = 5.67 × 10−8 W/(m2K4)

◮ Agrees very well with radiated heat measurements.

(What is the classical result for σ?)

9

Photoelectric effect

◮ Light ejects electrons from cathode ⇒ I at V = 0 ◮ V ↑⇒ I ↑ till saturation (all ejected electrons collected) ◮ V ↓⇒ I ↓ till I = 0:

all electrons stopped at V = −V0

◮ Increase intensity I:

higher saturation I but same stopping V

◮ Increase frequency ω:

higher stopping V

◮ Stopping action: eV0 = KEmax ◮ Experiment finds eV0 ∝ (ω − ω0) ◮ In fact eV0 = (ω − ω0) ◮ Different cathodes ⇒ different ω0

but same slope identical to that from Planck’s law!

◮ Light waves with angular frequency ω behave like

particles (photons) with energy ω (Einstein, 1905)

◮ Why does the saturation I ↑ when ω ↑ at constant I?

V I Light I V

10

Compton scattering

◮ X-ray ejects electron with part of its energy and remainder comes out as

secondary X-ray

◮ Energy conservation ω = ω′ + 1 2mv2 (assuming ω0 << ω) ◮ Output X-ray at angle θ has specific frequency ω′. Why? ◮ Photon also has momentum

p = k (magnitude ω/c)

◮ Momentum conservation

ω c = ω′ c cos θ + mv cos θ′ 0 = ω′ c sin θ − mv sin θ′

◮ Eliminate electron unknowns (v, θ′)

cos θ = ω2 + ω′2 − 2mc2

• (ω − ω′)

2ωω′

X-ray source Sample X-ray spectrometer

11

Wave-particle duality

◮ Light is a wave: electric and magnetic fields oscillating ∼ e−iωt

◮ All of classical wave-optics: diffraction etc.

◮ Light is particulate: photons with energy ω and momentum ω/c

◮ Black-body radiation ◮ Photoelectric effect ◮ Compton scattering

12

Electrons

◮ Discovered as ‘cathode rays’ in vacuum tube experiments (1869) ◮ Deflection by magnetic fields to measure charge/mass (1896) ◮ Charge measured in Millikan’s oil drop experiment (1909) ◮ Particle with mass m ≈ 9 × 10−31 kg and charge −e ≈ −1.6 × 10−19 C

known by early days of atomic theory and quantum theory

◮ Is it also a wave?

13

De Broglie wavelength

◮ Yes! With wavelength λ = h/p where p is the momentum ◮ For wavevector

k (of magnitude 2π/λ), this ⇒ p = k (same as photon)

◮ In terms of kinetic energy:

KE [eV] electrons photons λ =

h √ 2mKE [˚

A] λ =

h

√

2mKE+(KE/c)2 [˚

A] λ =

hc KE [˚

A] 1 12.3 12.3 1.24 × 104 10 3.88 3.88 1.24 × 103 100 1.23 1.23 124 103 0.388 0.388 12.4 104 0.123 0.122 1.24 105 0.0388 0.0370 0.124 106 0.0123 0.00872 0.0124

◮ This rule applies to all particles / matter, not just electrons and photons ◮ What is your typical wavelength when walking? (Why don’t you diffract?)

14

Electron diffraction

◮ Use gold film as grating (GP Thomson, 1927) ◮ Polycrystalline ⇒ rings (like powder X-ray diffraction) ◮ Modern version: transmission electron microscopy (TEM) ◮ ∼ 100 keV energies ⇒ λ ∼ 0.05 ˚

A ⇒ atomic resolution

Gold film Screen

15

Schrodinger equation

◮ The wave equation for non-relativistic particles with mass m

−2∇2ψ 2m + V ( r, t)ψ = i∂ψ ∂t

◮ Wave function ψ(

r, t): analogous to E( r, t) or B( r, t) for EM waves

◮ For EM wave, intensity is proportional to |E|2 ◮ Intepret intensity as the probability of finding light ◮ Quantum mechanically, |ψ(

r)|2 is the probability density of finding particle at r (normalized as

• d

r|ψ( r)|2 = 1)

◮ Note

E( r) actually is the wavefunction of a photon in the quantum theory

• f EM waves

16

Free particle

−2∇2ψ 2m + V ( r, t)ψ = i∂ψ ∂t

◮ Let potential be constant in space and time i.e. V (

r, t) = V0

◮ Solution of the form ψ(

r, t) = ei(

k· r−ωt)

2k2 2m + V0 = ω p2 2m

• Kinetic

+ V0

• Potential

= E

• Total

◮ Note how De Broglie and Planck hypothesis

connect classical and quantum relations.)

◮ Where is the particle?

Everywhere with a well-defined momentum p = k

17

Time-independent Schrodinger equation

−2∇2ψ 2m + V ( r, t)ψ = i∂ψ ∂t

◮ Let potential be constant in space i.e. V (

r, t) = V ( r)

◮ LHS independent of t: separation of variables ψ(

r, t) = ψ( r)T(t) − 2∇2ψ(

r) 2m

+ V ( r)ψ( r) ψ( r) = i ∂T (t)

∂t

T(t) = const. = E (say)

◮ Then T(t) = e−iEt/ and ψ(

r) is an eigenfunction of −2∇2ψ 2m +V ( r)ψ( r) = Eψ( r) (Time-independent Schrodinger equation)

◮ Note for time dependence e−i(E/)t, angular frequency is E/, energy is

the eigenvalue E

18

Particle in a box

◮ Need to solve time-independent Schrodinger equation

−2∂2

xψ

2m + V (x)ψ(x) = Eψ(x)

◮ When V (x) → ∞, ψ(x) → 0 for finite E ◮ Effectively, with ψ(0) = ψ(L) = 0, solve

∂2

xψ = − 2mE

2

k2

ψ(x)

◮ Solutions cos kx and sin kx (or e±ikx) ◮ Boundary conditions only allow sin nπx L

k = n π L and E = n2 2π2 2mL2

◮ Energy is ‘quantized’: only discrete values allowed

L

19

Particle in a box: ground state

◮ States with discrete energies and (normalized) wavefunctions:

En = n2 2π2 2mL2 , ψn(x) =

• 2

L sin nπx L labeled by ‘quantum number’ n

◮ Lowest energy (n = 1 here) is ground state:

E1 = 2π2 2mL2 , ψ1(x) =

• 2

L sin πx L

◮ What should it have been classically?

• Zero. E1 is confinement energy ≈

0.38 eV (L in nm)2 ◮ Where is the particle?

Distributed between 0 and L with probability 2

L sin2 πx L (Range: L) ◮ What is its momentum?

Since sin k1x = (eik1x − e−ik1x)/2i,

• ne of ±k1 i.e. ± π

L (Range: 2π L )

L

20

Heisenberg’s uncertainty principle

◮ In previous example: range in x was L and range in p was 2π L ◮ More precisely, standard deviation in x is ∆x =

• 1

12 − 1 2π2 L ≈ 0.18L and

standard deviation in p is ∆p = π

L ◮ Narrower well ⇒ reduce ∆x, but increase ∆p ◮ In this case, ∆x · ∆p ≈ 0.57 ◮ Heisenberg’s uncertainty principle

∆x · ∆p ≥ 2

◮ What is the corresponding relation for photons? ◮ Exactly the same: in fact this is purely a wave-mechanics property

∆x · ∆k ≥ 1 2 applicable to all classical waves as well

21

Another example: 1D ‘δ-atom’

◮ Consider the potential V (x) = −V0δ(x) (an infinitely-deep,

infinitenely-narrow well)

◮ Except at x = 0, potential is zero everywhere i.e.

∂2

xψ = −2mE

2 ψ(x) with solutions e±ix

√ 2mE/ ◮ For E > 0, oscillatory solutions e±ikx with k =

√ 2mE/

◮ For E < 0, bound solutions e±κx with κ =

√ −2mE/

◮ In general ψ(x) and ψ′(x) ≡ ∂xψ must be continuous ◮ But where V → ∞, ψ′(x) will be discontinuous

22

Derivative discontinuity of wavefunction

◮ Schrodinger equation in a δ-potential:

− 2 2m∂xψ′(x) − V0δ(x)ψ(x) = Eψ(x)

◮ Integrate in a small neighbourhood around x = 0

− 2 2m +ǫ

−ǫ

dx∂xψ′(x) − V0 +ǫ

−ǫ

dxδ(x)ψ(x) = E +ǫ

−ǫ

dxψ(x) − 2 2m [ψ′(+ǫ) − ψ′(−ǫ)] − V0ψ(0) = E +ǫ

−ǫ

dxψ(x)

◮ Take limit ǫ → 0:

− 2 2m

• ψ′(0+) − ψ′(0−)
• − V0ψ(0) = 0

ψ′(0+) − ψ′(0−) = −2mV0 2 ψ(0)

23

δ-atom: bound state

◮ Consider the E < 0 case, where solutions are e±κx for x < 0 and x > 0 ◮ For x < 0, only eκx because e−κx → ∞ as x → −∞ ◮ For x > 0, only e−κx because eκx → ∞ as x → +∞ ◮ With continuity, ψ(x) = Ae−κ|x| ◮ Derivative condition:

A(−κ) − A(κ) = −2mV0 2 A gives κ = mV0

2 ◮ Single bound state (E < 0)

ψ(x) = √κe−κ|x| which is the ground state in this potential

24

δ-atom: free states

◮ Consider the E > 0 case, where solutions are e±ikx for x < 0 and x > 0 ◮ For x < 0, Aeikx + Be−ikx (no restrictions) ◮ For x > 0, Ceikx + De−ikx (no restrictions) ◮ Continuity ⇒ A + B = C + D ◮ Derivative condition:

(ikC − ikD) − (ikA − ikB) = −2mV0 2 (A + B)

◮ Two free variables and two dependent

among A, B, C, D

◮ If D = 0, A incoming wave from −∞,

reflects to B and transmits to C

◮ Solve to get reflection and transmission coefficients

25

Tunneling

◮ Consider particle with energy 0 < E < V0 ◮ Classically, particle cannot cross barrier V0 higher than its energy ◮ ψ ∝ e±ikx with k =

√ 2mE/ in V = 0 regions

◮ ψ ∝ e±κx with κ =

• 2m(V0 − E)/ in V = V0 regions

◮ Match wavefunctions at x = 0 and x = L for wave incoming from left ◮ Probability of ‘tunneling’ to the right

P ∼ e−κL = exp

• −L
• 2m(V0 − E)
• ◮ For more general barrier shape

P ∼ exp

• −
• V (x)>E dx
• 2m(V (x) − E)
• ◮ Responsible for atomic resolution in STM

L

26

How do waves show particulate behaviour?

◮ So far, only option in free space are particles distributed everywhere. ◮ Above true for energy and momentum eigenstates ◮ No longer the case for states which combine many energies and momenta ◮ Example: combine k around k0 with Gaussian distribution of width σk

c(k) = 1

• σk

√ 2π e−(k−k0)2/(2σ2

k)

◮ This Gaussian ‘wave-packet’ has

ψ(x, t) = 1 √ 2π ∞

−∞

dkc(k)ei(kx−ω(k)t)

• ω(k) = 2k2

2m

• =

1

• 2πσk

√ 2π ∞

−∞

dk exp

• −(k − k0)2

2σ2

k

+ i(kx − ω(k)t)

• 27

Gaussian wavepacket

ψ(x, t) = 1

• 2πσk

√ 2π ∞

−∞

dk exp

• −(k − k0)2

2σ2

k

+ i(kx − ω(k)t)

• =

1

• 2πσk

√ 2π ∞

−∞

dk exp

• − (k−k0)2

2σ2

k

+ ikx −i(ω(k0)t + ω′(k0)(k − k0)t + · · · )

• = ei(k0x−ω(k0)t)
• 2πσk

√ 2π ∞

−∞

d∆k exp

• −∆k2

2σ2

k

+ i∆k(x − ω′(k0)t)

• = ei(k0x−ω(k0)t)−σ2

k(x−ω′(k0)t)2/2

• 2πσk

√ 2π ∞

−∞

d∆ke

−{∆k−iσ2 k(x−ω′(k0)t)}2 2σ2 k

• √

2πσk

= exp [i(k0x − ω(k0)t)] · exp

• − (x−ω′(k0)t)2

2(σ−1

k

)2

• σ−1

k

√ 2π

28

Group and phase velocities

◮ Gaussian wavepacket

ψ(x, t) = exp [i(k0x − ω(k0)t)] · exp

• − (x−ω′(k0)t)2

2(σ−1

k

)2

• σ−1

k

√ 2π

◮ Localized by Gaussian with width σx = σ−1 k

centered at x0 = ω′(k0)t

◮ Spread ∆k = σk/

√ 2 and ∆x = σx/ √ 2 (Why?)

◮ Minimum uncertainty product: ∆k · ∆x = 1/2 ◮ Packet centered at x0 moves with group velocity

vg = ∂ω ∂k

◮ Underlying waves propagate with phase velocity

vp = ω k

◮ For free particle with ω = k2/(2m)

vg = p m and vp = p 2m (p = k)

x0 = ω’(k0)t ei(k0x - ω(k0)t) 29

Spin

◮ EM wave propagating along x: E0ˆ

yei(kx−ωt) or E0ˆ zei(kx−ωt) (two independent polarizations)

◮ Linearly combine to E0 ˆ y+iˆ z √ 2 ei(kx−ωt) or

E0

ˆ y−iˆ z √ 2 ei(kx−ωt): circular polarizations ◮ Quantize to a photon: circular polarizations

will have (internal) angular momentum ±

◮ Angular momentum quantized in units of ◮ Since maximum magnitude is 1, photon is a spin s = 1 particle ◮ Electrons have internal angular momentum ±/2 ◮ Electrons have spin s = 1/2 ◮ Integer spins: bosons, any number per state

• eg. photons, phonons, He4 etc.

◮ Half-integer spins: fermions, maximum one per state

• eg. electrons, protons, neutrons, He3 etc.

30

Particles in a 3D box

◮ Standing EM waves in a box:

• 2

L sin nπx L

in each direction

◮ Electronic wavefunctions: exactly the same! ◮ Overall modes:

2

L

3/2 sin nxπx

L

sin nyπy

L

sin nzπz

L ◮ Wavevector

k = (kx, ky, kz) = nxπ

L , nyπ L , nzπ L

• ◮ Number of EM modes per

k: 2 polarizations

◮ Number of e− states per

k: 2 spins

◮ Number of modes per unit volume

between k and k + dk: 2 · 4πk2dk 8 · L π 3 · 1 L3 = k2dk π2

◮ Energy per photon ε = ω = c|

k|

◮ Energy per electron ε = ω = 2| k|2 2m

L

31

Average number of particles per mode of energy ε

Probability of n particles ∝ e−nα, where α = ε−µ

kBT and chemical potential µ

controls number (zero for massless particles like photons) Bosons: (eg. photons) n ≡ ∞

n=0 e−nαn

∞

n=0 e−nα

= − d dα ln

• n

e−nα = − d dα ln 1 1 − e−α = 1 exp

• ε−µ

kBT

• − 1

(Bose-Einstein distribution) E = nε = ε exp

• ε−µ

kBT

• − 1

Fermions: (eg. electrons) n ≡ 1

n=0 e−nαn

1

n=0 e−nα

= e0 · 0 + e−α · 1 e0 + e−α = 1 eα + 1 = 1 exp

• ε−µ

kBT

• + 1

(Fermi-Dirac distribution) E = nε = ε exp

• ε−µ

kBT

• + 1

Classical equipartition result: E = kBT

32

Hydrogenic atom

◮ Single electron with a nucleus of charge +Ze, where Z is the atomic

number

◮ Z = 1 is hydrogen, Z = 2 is a He+ ion, Z = 3 is Li2+ etc. ◮ Schrodinger equation

−2∇2ψ( r) 2m − Ze2 4πǫ0rψ( r) = Eψ( r) separable in spherical coordinates resulting in eigenfunctions ψnlm( r) = Rnl(r)Ylm(θ, φ) and eigen-energies Enlm = − me4 32π2ǫ2

02 · Z2

n2 = −Z2 2n2 Eh = −Z2 n2 Ryd ≈ −Z2 n2 (13.6 eV)

33

Atomic quantum numbers

◮ For the box, we has nx, ny and nz ◮ Now in spherical coordinates, so correspond to r, θ and φ ◮ Principal quantum number n = 1, 2, . . . is for the radial r direction ◮ Angular quantum number l = 0, 1, 2, . . . , n − 1 is for the θ direction ◮ Azimuthal quantum number ml = −l, −l + 1, . . . , +l is for the φ direction ◮ But energy Enlml ∝ n−2 only depends on n ◮ States of various l and ml at same n are ‘degenerate’ i.e. have same energy

34

Radial wavefunctions

◮ Radial functions of the form

Rnl(r) ∝ exp −2Zr na0

• · rl · p(l)

n−l−1(r)

where a0 = 4πǫ02/(me2) ≈ 0.529 ˚ A is the Bohr radius

◮ Typical radial extent ∼ na0/Z ◮ Polynomial degree n − l − 1: first n of given l

has no nodes, next has one node etc.

◮ Remember l = 0, 1, 2, 3 denoted by s, p, d, f ◮ 1s has no nodes, 2s has 1 node etc. ◮ 2p has no nodes, 3p has 2 nodes etc.

1s 2s 3s 2p 3p

35

Angular wavefunctions

◮ Spherical harmonics Ylml(θ, φ) = P ml l

(cos θ)eimlφ

◮ Characteristic orbital shapes used in

chemistry (typically ReYlm and ImYlm)

◮ l controls number of lobes ◮ ml controls number in xy-plane ◮ All ml related by spherical symmetry

36

Electronic configuration of atoms

◮ Pauli exclusion principle: one electron per state (Fermi-Dirac statistics) ◮ Spin: ms = ±1/2 (2 states) ◮ Azimuthal: m = −l, −l + 1, . . . , +l (2l + 1 states) ◮ Per n and l: 2(2l + 1) states ◮ Periodic table by orbital being filled (Z range):

1s (1-2) 2s (3-4) 2p (5-10) 3s (11-12) 3p (13-18) 4s (19-20) 3d (21-30) 4p (31-36) 5s (37-38) 4d (39-48) 5p (49-54) 6s (55-56) 4f (57-70) 5d (71-80) 7p (81-86) 7s (87-88) 5f (89-102)

37

The size of atoms

◮ Orbital size ∼ na0/Z ◮ Hydrogen atom Z = 1, n = 1: size ∼ a0 ≈ 0.53 ˚

A

◮ Sodium atom Z = 11, n = 3: size ∼ 3a0/11 ≈ 0.14 ˚

A

◮ Platinum atom Z = 78, n = 6: size ∼ 6a0/78 ≈ 0.04 ˚

A

◮ What’s wrong? ◮ Hydrogenic orbitals are for one electron systems only! ◮ When more than one electron, electron-electron repulsion matters ◮ Effective charge seen by outer electrons is approximately that of nucleus +

inner electrons

38

Many-electron Schrodinger equation

◮ So far, we discussed wavefunction ψ(

r) satisfying − 2 2m∇2ψ + V ( r)ψ = Eψ which is strictly a one-electron theory only.

◮ For N electrons, need to keep track of all N electronic coordinates with a

wavefunction ψ( r1, r2, . . . , rN)

◮ Corresponding Schrodinger equation with e-e interactions:

− 2 2m

• i

∇2

• riψ
• Kinetic

+

• i

V ( ri)ψ

• e-nuc

+

• i=j

e2 4πǫ0| ri − rj|ψ

• e-e

= Eψ which is impossible to solve exactly beyond special N = 2 cases

39

Many-electron non-interacting case

◮ Without e-e interactions:

− 2 2m

• i

∇2

• riψ
• Kinetic

+

• i

V ( ri)ψ

• e-nuc

= Eψ which is separable in each ri.

◮ Therefore solution must be consist of products

ψ( r1, r2, . . . , rN) ∼ φ1( r1)φ2( r2) · · · φN( rN)

◮ Each ‘orbital’ φi: N = 1 Schrodinger equation with orbital energy εi ◮ Total energy E = i εi ◮ Strictly, fermionic wavefunctions need to be antisymmetric

⇒ ψ = det[φi( rj)] (Slater determinant)

40

Kohn-Sham density functional theory (DFT)

◮ A single-particle theory

− 2 2m∇2φi( r) + VKS( r)ψ = εiφi( r) in an effective potential VKS( r)

◮ VKS(

r) = V ( r) + contribution from electron density n( r)

◮ Total energy E = i εi + contribution from electron density n(

r)

◮ Electron density n(

r) =

i |φi(

r)|2 made self-consistent

◮ DFT works surpisingly well even for strongly interacting electrons ◮ When DFT does not work, material called ‘strongly-correlated’!

41

Atoms revisited

◮ Orbital energies are those from effective potential (not hydrogenic) ◮ For spherical atoms, still degenerate in m and ms, but not in l ◮ At same n, energy increases with l ◮ In particular, energy of (n + 1)s < (n − 1)f < nd < (n + 1)p ⇒

1s (1-2) 2s (3-4) 2p (5-10) 3s (11-12) 3p (13-18) 4s (19-20) 3d (21-30) 4p (31-36) 5s (37-38) 4d (39-48) 5p (49-54) 6s (55-56) 4f (57-70) 5d (71-80) 7p (81-86) 7s (87-88) 5f (89-102)

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Operators and expectation values

◮ |ψ(

r)|2 probability distribution of r

◮ Average value of

r:

• r ≡
• d

r|ψ( r)|2 r Expectation value of operator r in state with wavefunction ψ ψ| r|ψ ≡

• d

r ψ∗( r) r ψ( r)

◮ Expectation value of r2:

r2 ≡ ψ|r2|ψ ≡

• d

r ψ∗( r) r2 ψ( r)

◮ Uncertainty in x, ∆x ≡

• x2 − x2

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Momentum operator

◮ For a free particle with momentum

p = k, ψ = ei

k· x/

√ L

◮ Consider expectation value of gradient

ψ|∇|ψ ≡

• d

r ψ∗( r) ∇ ψ( r) = 1 L

• d

r e−i

k· x ∇ei k· x

= 1 L

• d

r e−i

k· x i

kei

k· x

= i k

◮ Momentum operator ˆ

• p defined by

p = ψ|ˆ

• p|ψ

◮ So the momentum operator is

ˆ

• p = −i∇

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Hamiltonian operator

◮ If we define the Hamiltonian operator as

ˆ H ≡ 2∇2 2m + V ( r) = ˆ

• p2

2m + V ( r) the Schrodinger equation becomes ˆ Hψ = Eψ

◮ Expectation value of the Hamiltonian is ψ| ˆ

H|ψ = E

◮ What about time dependence? Remember ψ(

r, t) = ψ( r)e−iEt/ ψ| ˆ H|ψ ≡

• d

rψ∗( r, t) ˆ Hψ( r, t) =

• d

rψ∗( r)eiEt/ ˆ Hψ( r)e−iEt/ =

• d

rψ∗( r) ˆ Hψ( r) =

• d

rψ∗( r)Eψ( r) = E

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Time dependence due to perturbations

◮ Let Hamiltonian ˆ

H have two eigenstates Hψ1 = E1ψ1 and Hψ2 = E2ψ2

◮ Eigenstates are orthogonal

• ψ∗

1ψ2 = 0 and complete: any

ψ = c1ψ1 + c2ψ2

◮ Say apply electric field

Ee−iωt, changes Hamiltonian to ˆ H + ˆ H′e−iωt with ˆ H′ =−e E · r

◮ Time-dependent Schrodinger equation ( ˆ

H + ˆ H′e−iωt)ψ = i ∂ψ

∂t ◮ Substitute expansion ψ(t) = c1(t)e−iE1t/ψ1 + c2(t)e−iE2t/ψ2

( ˆ H + ˆ H′e−iωt)(c1e−iE1t/ψ1 + c2e−iE2t/ψ2) = (i˙ c1 + E1c1)e−iE1t/ψ1 + (i˙ c2 + E2c2)e−iE2t/ψ2

◮ Rewrite using eigenvalues of ˆ

H c1e−i(E1+ω)t/ ˆ H′ψ1 + c2e−i(E2+ω)t/ ˆ H′ψ2 = i˙ c1e−iE1t/ψ1 + i˙ c2e−iE12/ψ2

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Time dependence due to perturbations (contd.)

◮ Equation in terms of expansion ψ(t) = c1(t)e−iE1t/ψ1 + c2(t)e−iE2t/ψ2

c1e−i(E1+ω)t/ ˆ H′ψ1 + c2e−i(E2+ω)t/ ˆ H′ψ2 = i˙ c1e−iE1t/ψ1 + i˙ c2e−iE12/ψ2

◮ Now integrate equation

• ψ2(

r, t)∗ to get c1ψ2| ˆ H′|ψ1ei(E2−E1−ω)t/ + c2ψ2| ˆ H′|ψ2e−iωt = i˙ c2

◮ If we start at t = 0 in state ψ1 i.e. c1(0) = 1, c2(0) = 0, then at t = 0

i˙ c2 = ψ2| ˆ H′|ψ1ei(E2−E1−ω)t/ which is the rate at which state ψ2 starts appearing

◮ c2(t) oscillates in time with zero average value as long as E2 = E1 + ω

(Energy conservation)

◮ If E2 = E1 + ω, then c2(t) grows in time

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Fermi’s Golden rule

◮ Upon applying a perturbation Hamiltonian H′eiωt,

Γ1→2 = 2π |ψ2| ˆ H′|ψ1|2δ (E2 − (E1 + ω)) is the rate of transitioning from ψ1 to ψ2

◮ More generally,

Γi = 2π

• f

|ψf| ˆ H′|ψi|2δ (Ef − (Ei + ω)) is the rate of transitioning out of initial state ψi

◮ Fundamental equation of ‘quantum kinetics’

(say analogous to Arrhenius equation)

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Orbital angular momentum

◮ Classical picture: electrons revolving around nuclei with

L = r × p

◮ In quantum picture, ˆ

• L =

r × ˆ

• p = −i

r × ∇

◮ In particular ˆ

Lz = −i(x∂y − y∂x) = −i∂φ

◮ In atomic orbitals, angular dependence Ylml(θ, φ) = P ml l

(cos θ)eimlφ

◮ Azimuthal angular momentum ˆ

Lz = ml

◮ Account for all directions, magnitude of angular momentum

ˆ L2 = l(l + 1)2

◮ Number of projections quantized to 2l + 1 (number of allowed ml)

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Spin angular momentum

◮ Electrons have spin s = 1/2 ◮ Corresponding ms = ±1/2 (2 = 2s + 1 values) ◮ Projected angular momentum Sz = ms ◮ Angular momentum magnitude S2 = s(s + 1)2 ◮ Both orbital and spin angular momentum for electron ◮ Total angular momentum

J = L + S

◮ Also quantized, with quantum numbers j, mj ◮ j = |l − s| to l + s in increments of 1 ◮ mj = −j, −j + 1, . . . , +j ◮ Projected angular momentum Jz = mj ◮ Angular momentum magnitude J2 = j(j + 1)2

z y x

S L S L = +

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Angular momentum consequence: magnetic moments

◮ Consider particle with charge q and mass m moving with speed v in circle

• f radius r

◮ Angular momentum L = mvr ◮ Current I = qv 2πr ◮ Magnetic moment µ = 1 2

• r × d

lI = 1

2r(2πr) qv 2πr = qvr/2 ◮ Classical particle µ = q 2mL ◮ Exactly true for orbital angular momentum

µz = −e 2mml = −mlµB where µB ≡ e

2m is the Bohr magneton ◮ What about spin?

µz = −gemsµB where ge ≈ 2.0023 = 2 +

e2 4πǫ0hc + · · · is called the gyromagnetic ratio

(Relativity ⇒ ge = 2, rest quantum correction)

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Angular momentum conservation: selection rules

◮ Light absorption: photon excites electron from lower state nlml to higher

state n′l′m′

l ◮ Dominant electron-photon interaction through electric field ⇒ involves

• nly L of electron (not S)

◮ Initial angular momentum s = 1 in photon and l of electron

⇒ j = l − 1, . . . , l + 1

◮ Projection mj = ms + ml = ml − 1, . . . , ml + 1 ◮ Angular momentum conservation (l′, m′ l) must equal (j, mj) ◮ Process allowed only if ∆l = 0, ±1 and ∆ml = 0, ±1 ◮ More careful analysis ∆l = 0 disallowed (because ψ2| ˆ

H′|ψ1 = 0), ⇒ ∆l = ±1 and ∆ml = 0, ±1

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