Classical and Semiclassical Oscillator [PDF]

Set 7: Classical & Semi-Classical Oscillators

Dipole Approximation • Radiation field 

Erad Brad

 ˆ q n ˙ = × ((ˆ n − β) × β) c κ3 R ˆ × Erad =n

• A collection of particles Erad

 X qi  n ˆi ˙i ) × ((ˆ n − β ) × β = i i 3 c κ i Ri i

• Take the limit of coherent emission of particles such that the wavelength λ  L, the dimension of the emission region. • Approximation equivalent to non-relativistic velocities u since the scale of the oribits l < L L l L τ ∼ < c u u

→uc

Dipole Approximation • The light travel time across L  time scale of change in charged particle orbits λ L 1 τ∼ ∼  ν c c ˆi = n ˆ and Ri ≈ R0 the mean distance • So κi ≈ 1, n  X qi  n ˆ Erad = × (ˆ n × β˙ i ) c R0 i • Defining the electric dipole moment X d= qi ri i

Erad

1 ¨ = 2 [ˆ n × (ˆ n × d)] c R0

Dipole Approximation • Larmor formula dP d¨2 2 sin Θ = 3 dΩ 4πc

2d¨2 P = 3 3c

• Frequency structure reflects d(t) Z ∞ d(t) = e−iωt d(ω)dω −∞ Z ∞ ¨ =− d(t) ω 2 d(ω)e−iωt dω −∞

in the electric field 1 2 E(ω) = − 2 ω d(ω) sin Θ c R0

Dipole Approximation • Energy spectrum dA = R02 dΩ 1 4 dW 2 2 = c|E(ω)| R0 = 3 ω |d(ω)|2 sin2 Θ dωdΩ c dW 8πω 4 2 = |d(ω)| dω 3c3 • Dipole radiation frequencies reflect frequencies in the dipole moment. Generalize to higher order in the expansion of L/λ or k∆Ri

Thomson Scattering • Simplest example: single free electron system Θ E0e1

θ e electron

Θ =π/2 E0e2

• Incoming wave of frequency ω0 provides oscillatory force for acceleration. For non-relativistic velocities the Lorentz force is dominated by the electric field ˆ = ma F = eE = eE0 sin ω0 t e eE0 ˆ a= sin ω0 t e m

Thomson Scattering • Dipole formula 2 e E0 ¨ ˆ sin ω0 t e d = er , d = ea = m 4 2 e E0 1 dP d¨2 2 2 2 sin Θ = sin Θhsin ω0 ti = 3 2 3 dΩ 4πc m 4πc e4 E02 e4 E02 2 = sin Θ →P = 2 3 8πm c 3m2 c3 • Power is independent of frequency

• Differential Cross Section: P = hSiσT dP dσT outgoing power dΩ = = dΩ hSi incoming flux

=

2 e4 2 E sin 8πm2 c3 0 c 2 E 0 8π

Θ

e4 = 2 4 sin2 Θ mc

Thomson Scattering • Cross section decreases with mass (so electrons not protons dominate), defines a classical size for a point particle r0 = e2 /mc2 . Total cross section Z 8π 2 dσT dΩ = r0 σT = dΩ 3 ˆ and n ˆ . Relate to scattering angle θ and • Θ is angle between e polarization ˆ⊥n ˆ (polarization out of scattering plane) then sin Θ = 1. • For e For polarization in the scattering plane Θ = π/2 − θ • Average over the two incoming polarization states dσT 1 2 1 2 2 π = r0 [1 + sin ( − θ)] = r0 (1 + cos2 θ) dΩ 2 2 2

Classical Line Emission • An electron bound in a central force provides a classical model for spontaneous emission in an atomic line

nucleus e electron

"spontaneous emission" F=-kr restoring force

• Given a restoring force F = −kr there is a natural frequency of oscillation. Neglecting radiation k ma = −kr → x¨ + x = 0 → x¨ + ω02 x = 0 m p where ω0 = (k/m)

Radiation Reaction • But an accelerating electron will radiate causing the oscillator to lose energy as a damped oscillator x¨ + Γx˙ + ω02 x = 0 where Γ is the damping rate due to radiation losses • Larmor’s formula gives power radiated or energy lost to radiation per unit time 2e2 x¨2 P = 3c3

Radiation Reaction • If one assumes that losses are small across one period of the oscillation then (like the Rosseland approximation) one can iterate the solution x¨ ≈ −ω02 x

→

x ≈ A cos(ω0 t + δ)

2e2 4 2 2e2 2 2 P = 3 (−ω0 x) = 3 ω0 x 3c 3c • Set equal to mechanical work / time? by radiation reaction 2e2 4 2 ω0 x = −Frad x˙ 3 3c • Problem: since x ∝ cos(ω0 t + δ) and x˙ ∝ sin(ω0 t + δ) this formula cannot be satisfied instantaneously representing a breakdown of the classical treatment

Radiation Reaction • Try average over an oscillation period T = 2π/ω0 and match integral of cos2 with sin2 Z 0

2π ω0

2

Z

2π ω0

2e 4 2 dt 3 ω0 A cos2 (ω0 t + δ) = − dtFrad [−Aω0 sin(ω0 t + δ)] 3c 0 2e2 4 2 1 Frad 1 ω0 A = Aω0 3 3c 2 sin(ω0 t + δ) 2

• Solve for the reaction force Frad

2e2 ω03 = A sin(ω0 t + δ) 3 3c 2e2 ω02 =− x˙ 3 3c

Radiation Reaction • So the oscillator equation of motion becomes 2 2 2e ω0 2 x¨ + ω0 x = Frad /m = − 3 x˙ 3c m 2 e2 ω02 Γ= 3 mc3 which quantifies the rate at which the orbit decays due to radiation losses - in a quantum description of the line this is related the Einstein coefficient for spontaneous emission

x¨ + Γx˙ + ω02 x = 0

x = eαt

α2 + Γα + ω02 = 0 1 α = (−Γ ± 2

q Γ2 − 4ω02 )

Lorentz Line Profile • Since Γ  ω0 one can expand s Γ2 α = ±iω0 1 − 2 − Γ/2 ≈ ±iω0 − Γ/2 4ω0 x(t) = x(0)e−Γt/2 cos(ω0 t + δ) • Get frequency content (define zero of time so that δ = 0) −Γt/2 1

 iω0 t  −iω0 t e +e

x(t) = x(0)e 2 Z 1 x(t)eiωt dt x(ω) = 2π   x(0) 1 1 = + 4π Γ/2 − i(ω + ω0 ) Γ/2 − i(ω − ω0 )

Lorentz Line Profile • Frequency content dominated by region around ω = ω0 , i.e. the second term x(ω) ≈ |x(ω)|2 ≈ dW = dω =

1 x(0) 4π Γ/2 − i(ω − ω0 ) 1 x2 (0) (4π)2 (ω − ω0 )2 + (Γ/2)2 8πω 4 2 2 e |x(ω)| 3c3 8πω 4 e2 x2 (0) 1 3c3 (4π)2 (ω − ω0 )2 + (Γ/2)2

• Interpret the amplitude in terms of initial energy in the oscillator 1 2 1 Ei = kx (0) = mω02 x2 (0) 2 2

Lorentz Line Profile .

• All of the initial energy is emitted as radiation

1.0 0.8

0.6

dW 8πω 4 e2 2Ei 1 = dω 3c3 (4π)2 mω02 (ω − ω0 )2 + (Γ/2)2 dW ω 4 e2 Ei 1 = 2 dω ω0 3πmc3 (ω − ω0 )2 + (Γ/2)2   2 2 2 e ω0 Γ= 3 mc3 dW Γ/2π = Ei dω (ω − ω0 )2 + (Γ/2)2

Γ FWHM

0.4

0.2

-4

-2

0

(ν-ν0)/Γ

2

4

Lorentz Line Profile • Lorentz profile: FWHM = Λ with normalization  Z Γ/2π =1 dω 2 2 (ω − ω0 ) + (Γ/2) • So that the decay rate is related to the frequency spread of the line ∆ω = Γ ∆ω 4πe2 −12 ≈ 1.2 × 10 cm ∆λ = 2πc 2 = 2 ω0 3mc

Absorption • Γ must be related to A21 the Einstein spontaneous emission coefficient. However a direct association is impeded in that an atomic state is classically unstable. Establish the relation by considering the absorption coefficient F=-kr restoring force nucleus e electron

E0e1

• Driven oscillator: combination of the Thomson and line calculations with incident radiation at frequency ω x¨ + Γx˙ +

ω02 x

eE0 iωt = e m

Absorption • After transients from the initial conditions are gone due to damping x = x0 eiωt 2

(−ω + Γiω +

ω02 )x0 eiωt

eE0 iωt = e m

• Solution eE0 1 x0 = m ω02 − ω 2 + iωΓ ω 2 − ω02 + iωΓ ≡ |A|e−iδ = |A|(cos δ − i sin δ) |A| cos δ = ω02 − ω 2 ,

|A| sin δ = −Γω

Γω tan δ = 2 ω − ω02 eE0 iδ −1 e x0 = m ω 2 − ω02 − iωΓ

Absorption • Reradiated power 4 4 2 e ω E 1 e2 ω 4 0 2 |x0 | = P = 3 3c 3c3 m2 (ω 2 − ω02 )2 + (Γω)2

• Interaction cross section 1 P 8π e4 ω 4 = 4 2 c 2 2 − ω 2 )2 + (Γω)2 E 3c m (ω 0 8π 0 # "   2 ω4 8π e2 = σT 2 σT = 2 2 2 (ω − ω0 ) + (Γω) 3 mc2

P = σ(ω) = hSi

• If ω  ω0 then σ(ω) → σT and the electron behaves as a free particle

Rayleigh Scattering • If ω  ω0 then  σ(ω) → σT

ω ω0

4

and the steep frequency dependence is the reason the sky is blue and sunsets are red - is the limit where e.o.m is ω02 x = (eE0 /m)eiωt • If ω ≈ ω0 then line absorption and resonance (ω 2 − ω02 ) = (ω − ω0 )(ω + ω0 ) ≈ 2ω0 (ω − ω0 ) ω04 σ(ω) = σT 2 4ω0 (ω − ω0 )2 + (Γω0 )2 σT ω02 σ(ω) = 4 (ω − ω0 )2 + (Γ/2)2

Absorption coefficient • Relate to Lorentz profile σ(ω) =

σT 2πω02 4

Γ/2π Γ (ω − ω0 )2 + (Γ/2)2

2 Γ/2π e 2 = 2π mc (ω − ω0 )2 + (Γ/2)2

2

 Γ= "

2 e ω02 3 mc3

8π σT = 3





e2 mc2

2 #

• Absorption coefficient, Einstein coefficient hν0 αν = nσ(ω) = nB12 φ(ν) 4π hν0 σ(ω) = B12 φ(ν) 4π Z Z dω 2π 2 e2 hν0 4π 2 e2 dνσ(ω) = σ(ω) = = B12 → B12 = 2π 2π mc 4π hν0 mc

Absorption coefficient • Quantum results are stated against this classical result as an oscillator strength f12 B12

4π 2 e2 = f12 hν0 mc

• The Einstein A21 spontaneous emission coefficient is then A21

8π 2 ν02 e2 g1 g1 2h 3 g1 f12 = 3Γ f12 = 2 ν0 B12 = 3 c g2 mc g2 g2

so that the rate Γ defines A21 • We shall see that the relation is corrected in the semiclassical oscillator and A21 = Γ(g1 /g2 )f12

Quantum Oscillator • Schrodinger equation in the absence of radiation field ∂ψ Hψ = i¯ h ∂t • The Hamiltonian including the radiation field is in Coulomb gauge [∇ · A = 0 and ∇2 φ = −4πρ (=0 for radiation)] Hrelativistic = [(cq − eA)2 + m2 c2 ]1/2 + eφ e q2 H≈ + eφ − A · q H = H0 + HI 2m mc • Radiative transitions are approximated through an interaction Hamiltonian in time dependent perturbation theory e ie¯h H =− A·q= A·∇ mc mc which connects the initial and final state I

Quantum Oscillator • Original eigenstates |ni such that H 0 |ni = En |ni • Expand the wave function in the original eigenfunctions X |ψi = cn |nie−iEn t/¯h  0  ∂ I i¯ h |ψi = H + H |ψi ∂t dcm X i¯ h = hm|H I |nicn (t)ei(Em −En )t/¯h dt n • Initially the atom is in the initial state ci = 1 and cn6=i = 0 and the perturbation induces a transition to a final state m = f with strength given by the matrix element HfIi (t) = hf |H I |ii. A short time T later Z T i i I i(Ef −Ei )t/¯ h dtHf i (t)e ≡ − 2πHfIi (ωf i ) cf (T ) = − h ¯ 0 h ¯

Quantum Oscillator • The integral is a Fourier transform that picks out frequency ωf i = (Ef − Ei )/¯h in H I with some width determined by how long (T ) one waits before accumulating significant probability. • Transition rate is the probability per unit time for the transition 4π 2 I wf i = 2 |Hf i (ωf i )|2 h ¯ T • Field carries time dependence A(r, t) = A(t)eik·r and integral picks out ωf i component of field HfIi (ωf i )

ie¯ h = A(ωf i ) hf |eik·r ∇|ii mc Z

hf |eik·r ∇|ii =

d3 xψf∗ eik·r ∇ψi

in dipole approx eik·r ≈ 1 across region ψ has support

Einstein Coefficient • Time reversal symmetry gives wif = wf i , is the quantum origin of the relationship between B12 and B21 the absorption and stimulated emission coefficient • For absorption w12 = B12 Jν and what remains is to relate the A embedded in the interaction Hamiltonian with the specific intensity c|E(ω)|2 dW = dAdωdt T dW 2πc|E(ω)|2 = dAdνdt T

Einstein Coefficient • Given B = ∇ × B and E0 = B0 , in Fourier space field and potential related by 2 ω |E(ω)|2 = 2 |A(ω)|2 c

• A plane wave is a delta function in angle so that Jν = simply divides the result by 4π or 1 ω2 Jν = |A(ω)|2 2 cT

1 4π

R

dΩIν

Einstein Coefficient • Eliminate in favor of Jν 2c 2 |HfIi (ωf i )|2 Jν wf i = 2 4π 2 ωf i |A(ω)| 2 8π 2 e2 = 2 2 hf |eik·r ∇|ii Jν ωf i m c • Determines Einstein coefficient B12

2 8π 2 e2 ik·r = 2 2 hf |e ∇|ii ωf i m c

Einstein Coefficient • Determines the oscillator strength f12 , typically less than unity B12

f12

2 8π 2 e2 = 2 2 hf |eik·r ∇|ii ωf i m c 4π 2 e2 f12 = h ¯ ωf i mc 2 2¯h ik·r = hf |e ∇|ii ωf i m

• Stimulated emission can be similarly handled, the difference being for degenerate levels the result is averaged over initial states and summed over final states – hence the g1 , g2 factors

Einstein Coefficient • Spontaneous emission formally requires second (field) quantization but can be derived semiclassically by the Einstein relation. Key of the quantum derivation is the field behaves as a quantized oscillator and the states are normalized as a† |ni ∝ (n + 1)1/2 |n + 1i where the n  1 returns the semiclassical stimulated emission coefficient B21 and the n = 0 returns the spontaneous emission A21 • When the coefficients cannot be calculated A21 is measured and the others inferred

Line Profile • The natural linewidth is determined by A21 = Γ exactly as in the semiclassical theory (but without the relationship Γ = 2e2 ω02 /3mc3 ) , yielding a Lorentzian profile • Linewidth is broadened by thermal motion. Frequency shifted according to the Doppler shift from the line of sight velocity vk vk ν − ν0 = ν0 c • The velocity distribution is Maxwellian given the atomic mass ma  m 1/2 −ma vk2 /2kT e dvk 2πkT • The net result is a Voigt profile Z ∞  m 1/2 Γ 1 −ma vk2 /2kT dvk e dvk φ(ν) = 2 2 2 4π −∞ (ν − ν0 ) + (Γ/4π) 2πkT

Line Profile • Finally, collisions can also broaden the profile. They introduce a random phase in the electric field. As shown in RL Problem 10.7, collisions of a frequency νcol cause h|E(t)|2 i ∝ e−νcol t (a Poisson process) and comparing this to the e−Γt/2 natural decay implies that the total Lorentzian width of the line Γ → Γ + 2νcol

Electronic, Vibrational, Rotational Lines • Electronic lines tend to have an energy given by the physical scale of the orbital (atom) and tends to be in the few eV energy scale Eelect

¯2 1 p2 1 h ∼ ∼ 2 me 2 a2 m e

h ¯ p∼ a

• Molecules can have vibrations. For vibrations, the atoms execute simple harmonic motion around their equilibrium position with the restoring force associated with the electronic binding energy - so that a displacement of order a must given the electronic energy Eelect Evib

1 h ¯2 1 2 1 2 2 = = ka = m ω a a vib 2 a2 me 2 2  1/2 2 h ¯ me =h ¯ ωvib = ∼ Eelect 1/2 1/2 ma a2 me ma

Electronic, Vibrational, Rotational Lines • Vibrational energies are lower by of order a percent of the electronic energies, i.e. 10−2 − 10−1 eV or infrared • Rotational energy is associated with the moment of inertial I ∼ m a a2 Erot

h ¯2 me h ¯ 2 `(` + 1) ∼ 2 ≈ Eelect ≈ 2I 2a ma ma

or 10−3 eV in the far infrared and radio • Ratio of energies  Eelect : Evib : Erot = 1 :

me ma

1/2

me : ma