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

MTLE-6120: Advanced Electronic Properties of Materials Quantum kinetics

Reading:

◮ Kasap: 3.7 ◮ Griffiths QM: 9.1 - 9.2

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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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Example: particle in a box absorption spectrum

◮ States with discrete energies and (normalized) wavefunctions:

En = n2 2π2 2mL2 , ψn(x) =

• 2

L sin nπx L

◮ Start at n = 1, apply EM potential eExe−iωt ◮ Absorb photons and go to higher n ◮ Will excitations occur to all n with equal probability? ◮ Matrix element for transition:

ψn|eEx|ψ1 ≡ eE L dxψ∗

n(x)xψ1(x) =

4eELn π2(n2 − 1)2 [1 − (−1)n−1]

◮ Transition (absorption) rate:

Γ = 2π

• even n

δ(En − E1 − ω)

• 8eELn

π2(n2 − 1)2

• 2

◮ Selection rule: transitions from n = 1 only to even n

L 8

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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