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 - 3.6, 4.6 ◮ Griffiths QM: 1 - 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

1D wave in a box

◮ Find non-trivial solutions of the wave equation

∂2f ∂x2 = ∂2f v2∂t2 trapped inside a 1D box i.e. boundary conditions f(0) = f(L) = 0

◮ General solution f(x, t) = Aei(kx−ωt) + Bei(−kx−ωt) with ω = vk ◮ Apply boundary conditions:

f(0) = 0 ⇒ A + B = 0 f(L) = 0 ⇒ AeikL + Be−ikL = 0

◮ Solve to get B = −A and sin(kL) = 0 ◮ Not all wavevectors allowed, k = nπ/L for n = 1, 2, . . . ◮ Net solution are standing waves f(x, t) = C sin nπx L e−i(nπv/L)t

3

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

4

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

5

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

6

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

7

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

8

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

9

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?)

10

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

11

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

12

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

13

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?

14

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?)

15

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

16

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

17

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

18

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

19

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

20

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

21

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

22

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

23

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)

24

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

25

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

26

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

27

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)

• 28

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

29

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) 30

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.

31

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

32

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

33