The questions Quantum theory of light and matter · 2016-11-15 · Why photons? Shot noise and photon counting Shot noise: origin 0 1 2 3 4 5 0.1 0.2 0.3 0.4 0.5 0.6 0 1 2 3 4 5 6

Quantum theory of light and matter

Paul Eastham

The questions

What is a photon?What is the Hamiltonian for the electromagnetic field?How do we work with this Hamiltonian?What are its experimental consequences?

Course outline

Photons in practiceReview of electromagnetismPlanck distributionPhoton counting and shot noise

The theoretical framework . . .Review of quantum mechanicsHamiltonian for the electromagnetic fieldSecond quantization

. . . and its consequencesUncertainty in the EM fieldInterference experiments and photon statisticsClassical and non-classical states of light

Light-matter interactionsDipole coupling for an atomJaynes-Cummings model and Rabi oscillationsLight emission and lasing

Photons in Practice

1 Revision: classical electromagnetism

2 Why photons?Planck DistributionShot noise and photon counting

Revision: classical electromagnetism

Fields and force laws

Practical form of Coulomb/Ampere/Faraday etc. . .

F = q(E + v × B)

∇× B = µ0J ∇ · E = ρ/ε0

∇ · B = 0 ∇× E = −B.


Displacement current

∇× B = µ0J violates charge continuity –

∇.J = 1µ0∇.∇× B = 0 but need ∂ρ

∂t +∇.J = 0

Fix: add displacement current ∇× B = µ0J + µ0ε0∂E∂t


Displacement current⇒ waves

∇× E = −B

∇×∇× E = −∇2E = −∇× B

= −µ0ε0∂2Edt2

Non-trivial solutions without source terms–lightGeneral solution–superposition of plane waves

E =∑

e,k ,±E±,e,k e exp(ik .r ± ωt)


Light waves and polarization

General solution–superposition of plane waves

E =∑

e,k ,±E±,e,k e exp(ik .r ± ωt)

Except : e must be orthogonal to k = 0Plane wave w/1 e: “linearly polarized light”.Superposed w/e1.e2 = 0 + π/2 shift: “circularly polarized light”


Complex representation, energy and intensity


Gauges

∇× B = µ0J + µ0ε0E ∇ · E = ρ/ε0

∇ · B = 0 ∇× E = −B.

Second set are restrictions on the structure of the fieldsDealt with automatically if we work with potentials

B = ∇× A E = −∇φ− ∂A∂t

Gauge freedom A→ A +∇ξ(r , t), φ→ φ− ξ(r , t)Here use E ,B directlyBooks often use A in Coulomb gauge ∇.A = 0.


Longitudinal and transverse fields

Write any vector field R = RL + RT

∇.RT = 0 ∇× RL = 0

EL is not an independent variable for a field+particles :

∇.E = ρ/ε0

In Coulomb gauge, particle coordinates and A are independentvariables– this is why Coulomb gauge is easiest for quantum optics

Why photons?

Planck Distribution

Planck Distribution

Properties of electromagnetic radiation in thermal equilibrium– in e.g. cubic metal box?

Lx

y

z

Ex (r , t) = Ex (t) cos(kxx) sin(kyy) sin(kzz)

Ey (r , t) = Ey (t) sin(kxx) cos(kyy) sin(kzz)

Ez(r , t) = Ez(t) sin(kxx) sin(kyy) cos(kzz),

Ex ,y ,z(t) = Ex ,y ,z(0)eiωt , ω = c|k |.

(kx , ky , kz) = π(l ,m,n)/L, l = 0,1,2 . . . ,

lmn 6= 0, two independent polarizations for each mode.

Why photons?

Planck Distribution

Counting modes

x

y

z

k

k

k

Why photons?

Planck Distribution

Densities-of-modes

Why photons?

Planck Distribution

Classical energy distribution

Electromagnetic field equivalent to set of harmonic oscillators

Classical equipartition: energy per mode = kT⇒ energy in ω → ω + dω per unit volume

I(ω)dω = kTρ(ω)dω

= kTω2

c3π2 dω.

or in λ→ λ+ dλ

I(ω)dω = I(λ)dλ

⇒ I(λ) =8kTπλ4 (Rayleigh-Jeans)

Why photons?

Planck Distribution

Classical energy distribution

Rayleigh-Jeans I(ω) ∝ ω2 fails at high frequency

Why photons?

Planck Distribution

Planck hypothesis

Electromagnetic field equivalent to set of harmonic oscillators

Energy of each field mode not continuous

– only discrete multiples of ~ω

Multiples now called photonsField modes are quantum harmonic oscillators

Why photons?

Planck Distribution

Planck hypothesis⇒ energy distribution

P(n) =e−

n~ωkT

∑∞n=0 e−

n~ωkT

⇒ 〈n〉 =1

e~ω/kT − 1

⇒ I(ω)dω =~ωρ(ω)dωe~ω/kT − 1

=~ω3dω

π2c3(e~ω/kT − 1)

Why photons?

Shot noise and photon counting

Thought experiment

Classicallight source

Inte

nsi

ty

Time

Stable ∼ classical EM wave– fluctuations in “measured” intensity always there

Why photons?


Shot noise calculation

Beam power P = IA, frequency ω

Planck: average of r = IA/(~ω) photons/sec arrive at detector.

n = rt in time interval t.

P(detect 1; t → t + ∆t) = r∆tP(0; t → t + ∆t) = 1− r∆t .

Why photons?


Shot noise calculation

Suppose: photon arrivals are statistically independent– or #s in different lengths of beam are statistically independent

P(n; 0→ t + ∆t) = P(n; 0→ t)P(0; t → t + ∆t)+ P(n − 1; 0→ t)P(1; t → t + ∆t)= Pn(t)(1− r∆t) + Pn−1(t)(r∆t)

⇒ dPn(t)dt

= r [Pn−1(t)− Pn(t)].

Why photons?


Photon count distribution

Coherent light + uncorrelated photonsphoton counts in time t

Pn(t) =(rt)n

n!e−rt

– Poisson distribution

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Why photons?


Shot noise

Classicallight source

Inte

nsi

ty

Time

Current readout is photons counted/sampling timePoisson statistics : 〈n2〉 − 〈n〉2 = 〈n〉 = rtCurrent fluctuations 〈j2〉 − 〈j〉2 ∝ j(〈〉 – average with any time window – white noise)

Why photons?


Shot noise: origin

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Measurements of energy in field modegive different answers– different numbers of photonsIn a classical EM wave E and B bothhave definite (oscillating) valuesIn quantum theory [E ,B] 6= 0 – bestapprox to classical wave is asuperposition of quantum states ofdifferent E (and B)⇒ superposition of states of differentenergy,

H =

∫ [12ε0|E |2 +

12µ0|B|2

]dxdydz

Why photons?


Photon statistics and types of light

Classify light by the mean 〈n〉 and variance σ2n = 〈n2〉 − 〈n〉2

seen when measuring the photon number (or the light intensity)

Classical electromagnetic wave + Planck hypothesis⇒“Poissonian light”, “coherent light”, σ2

n = 〈n〉.Smallest variance consistent with semiclassical theoryCorresponding quantum state is “coherent state”– generated by a laser, or classical oscillating current

“Superpoissonian light” σ2n > n

– above + fluctuating intensity“Subpoissonian light”σ2

n < n– correlated photons e.g. “Fock state”

Why photons?


Super-Poisson light: Thermal Light

Statistical variation in photon number/energy possible fromother sources too . . .

e.g. Single mode in thermal equilibrium,

P(n) =e−β~ωn

∑n e−β~ωn =

zn∑

p zp = zn(1− z).

⇒σ2

n = 〈n〉+ 〈n〉2 > 〈n〉– Superpoissonian light

Why photons?


Thermal and quantum intensity noise

0

0.1

P(n)

300Photon number n

300Photon number n

Poisson statistics Thermal statistics

Why photons?


Super-Poisson light: Classical Chaotic Light

Most light is from many statistically independent emitters

What are the statistical properties of the intensity of such light?

Recall: for a perfect classical source :E = E0 sin(ωt − kx)⇒ cycle-averaged intensity I = 1

2ε0cE20 .

Generalization to complex notation :E = E0ei(ωt−kx) ⇒ I = 1

2ε0c|E0|2.

Why photons?


Chaotic Light: Collison-Broadened Model

Model for statistically independent emitters: collison broadened source

Suppose atoms emit a plane wave but occasionally collidecausing jumps in the phase

E = E0ei(ωt+φ(t))

Timewhere collison rate 1/τc– so phase scrambled after t > τc .

Why photons?


Chaotic Light: Collison-Broadened Model

-

-

E

Time

Why photons?


Chaotic Light: Intensity Distribution

Intensity from many collison-broadened emitters

I =12ε0cE0|

∑

i

eiφi |2,

Suppose τc short and consider average inensity over somelonger time.

〈I〉 =12ε0cE0〈|

∑

i

eiφi |2〉,

=12ε0cE0〈

∑

i,j

ei(φi−φj )〉,

Why photons?


Chaotic Light: Average Intensity

〈I〉 =12ε0cE0〈|

∑

i

eiφi |2〉,

=12ε0cE0〈

∑

i,j

ei(φi−φj )〉,

After time-averaging, only terms with i = j survive, so for Memitters

〈I〉 =12ε0cE0M

Why photons?


Chaotic Light: Intensity Variance

Need

〈I2〉 ∝ 〈|∑

i

eiφi |4〉

= 〈∑

i,j,k ,l

ei(φi+φj−φk−φl )〉.

This time, i = j = k = l survivesbut so does i = k , j = l or i = l , j = k .

〈I2〉 =14ε20c2E4

0 (M + 2M(M − 1))

=

(2− 1

M

)〈I〉2 ∼ 2〈I〉2

and the intensity variance is 〈I〉2

Why photons?


More intensity distributions

0

0.1

P(n)

300Photon number n

300Photon number n

Poisson statistics Thermal statistics

Classical chaotic light

- see Loudon p. 107 for derivation

Why photons?


Summary

Black body spectrum correctly obtained byFinding normal modes of fieldAllowing each mode to have energy n~ω– Or treating these modes as quantum harmonic oscillatos

The excitations of these modes are photons

Why photons?


Summary

Light may be characterized by the probability distribution ofits intensity. . . equivalently, the probability distribution of the photonnumberPlanck approach can be extended to calculate intensityfluctuations in beams of light . . .. . . and predicts that the smallest fluctuations consistentwith uncorrelated photons are σ2

n = 〈n〉.These fluctuations arise from quantum effects [E ,B] 6= 0.Excess fluctuations can also arise from conventionalprobabilistic effects – noise etc.σ2

n < 〈n〉 is possible - but not semiclassically.

Quantum Theory of Light and MatterField quantization

Paul Eastham

Quantization of the EM field

Quantization in an electromagnetic cavity

Quantum theory of an electromagnetic cavity

e.g. planar conducting cavity, E× n = 0,B.n = 0

Ex(r, t) =∑

n

√2ω2

nmV ε0

sin(knz)qn(t).

⇒ By (r, t) =∑

n

εµ

k

√2ω2

nmV ε0

cos(knz)qn(t)



Electromagnetic cavity: mode decomposition

Ex(r, t) =∑

n

√2ω2

nmV ε0

sin(knz)qn(t).

⇒ By (r, t) =∑

n

εµ

k

√2ω2

nmV ε0

cos(knz)qn(t)

Normalization of q arbitrary in classical theoryIntroduce some mass m so q is a lengthAny EM wave consistent with boundary conditions can bedecomposed in this wayFunctions in decomposition are “mode functions”



Electromagnetic cavity: energy in terms of modes

H =12

∫d3r

(ε0E2

x +1µ0

B2y

)

=∑

n

mω2nq2

n2

+mq2

n2

=∑

n

mω2nq2

n2

+p2

n2m

.

Energy of the electromagnetic field = energy of set ofharmonic oscillators (“radiation oscillators”)Similarly Maxwell’s equations becomepn = −mω2

nqn, qn = pm .



Electromagnetic cavity: quantization

H =∑

n

mω2nq2

n2

+p2

n2m

.

Hypothesis: quantum theory by replacing qn,pn → qn, pn, with

[qn, pn′ ] = i~δn,n′

Introduce ladder operators as before, (pn, qn)→ (an, a†n)

H =∑

n

~ωn

(a†nan +

12

).



Eigenstates of the field

H =∑

n

~ωn

(a†nan +

12

).

– set of quantum Harmonic oscillators– one for each mode of the field

Eigenstates– first Harmonic oscillator in eigenstate |n1〉,– second in |n2〉 . . .– denote |n1〉 ⊗ |n2〉 ⊗ |n3〉 . . . ≡ |n1,n2,n3 . . .〉Energy is ∑

i

~ωi(ni + 1/2)



Field operators

Ex(r, t) =∑

n

√2ω2

nmV ε0

sin(knz)qn(t).

⇒ Ex(r) =∑

n

√~ωn

V ε0(an + a†n) sin(knz)

=∑

n

En(an + a†n) sin(knz).

also By (r, t) =∑

n

−iEn

c(an − a†n) cos(knz)



Field operators

Ex(r) =∑

n

√~ωn

V ε0(an + a†n) sin(knz)

=∑

n

En(an + a†n) sin(knz).

Mass disappearsNormalization En– natural quantum scale of electric field En made from ~, ω– “electric field of a photon”Ex and By ∼ the position and momentum of a quantumharmonic oscillator


Other geometries: quantization in general

Schematic: generalization to other geometries

Can generalize this procedure to other geometries.We expect to get

HEM =∑

modes,i

~ωi a†i ai ,

and field operators like e.g.

E(r) =∑

modes,i

Ei(r)a + h.c.

(Normalization of mode function Ei(r)– EM energy is ~ωi when one photon in mode)


Other geometries: quantization in general

Example: free-space quantization

Introduce periodic boundary conditions in box of side L→∞.

⇒ mode functions are plane waves ek,seik.r.

– labelled by wavevector k = (lπ,mπ,nπ)/L

+ polarization ek,s (two for every k)

Giving

E(r) =∑

k,s

√~ωk

2ε0V

(ak,seik.r + a†k,se−ik.r

)

Physical consequences

Electric field distribution in a Fock state

Electric field in Fock states

What would we find if we measure the electric field of a singlemode of the cavity?Depends on state – suppose we had an energy eigenstate |n〉Result Ex with probability

|〈Ex |n〉|2

– basically the position-space wavefunction of the harmonicoscillator

〈Ex |n〉 ∝ e−E2x /(2E2 sin2 kz)Hn

(Ex

E sin(kz)

).

Hn are “Hermite” polynomials :H0 = 1,H1(x) = 2x ,H2(x) = 4x2 − 2 . . .



Electric field in Fock states



Distribution in Fock states

Single-mode cavity field in a Fock state

Ex(r) = E(a + a†) sin(kz) E =

√~ωV ε0

⇒ 〈n|Ex |n〉 ∝ 〈n|(a + a†)|n〉 = 0

and

〈n|E2x |n〉 = E2 sin2(kz)〈n|(a + a†)2|n〉

= E2 sin2(kz)〈n|aa† + a†a|n〉= E2 sin2(kz)(2n + 1).



Distribution in Fock states

〈n|Ex |n〉 = 0

〈n|E2x |n〉 = E2 sin2(kz)(2n + 1) ≡ σ2

E

So –

Expected field is always zeroField fluctuates, variance σ2

E follows mode profileMore photons⇒ stronger fluctuations – σ2

E ∝ nSpatial average of σ2

E associated with one photon is E2

Range of electric fields ∼ E even when n = 0– “vacuum fluctuations”



Distribution in Fock states: origins

Fluctuations because [Ex ,By ] 6= 0(like position and momentum of harmonic oscillator)

H involves Ex and By ⇒ [H,Ex ] 6= 0

So an eigenstate of energy is not an eigenstate of Ex

Summary

Summary

Quantum theory of light is constructed byIdentifying the normal modes of the EM fieldTreating these normal modes as quantum harmonicoscillators

This reproduces the Planck ansatz . . .. . . and gives us expressions we can work with for theelectric and magnetic field operatorsThe eigenstates of the field involve Fock states |n〉 . . .. . . in which the electric field is zero on average but will notalways be zero in a measurementEven in the lowest energy state |0〉 we will measure Ex 6= 0often – vacuum fluctuations

Coherent states and the classical limit

Paul Eastham

. . . and its consequencesUncertainty in the EM fieldCasimir effectInterference experiments and photon statisticsClassical and non-classical states of light← this lecture

Light-matter interactionsDipole coupling for an atomJaynes-Cummings model and Rabi oscillations

Motivation: electric fields in number states

Number states become classical waves?

So far : number/Fock states |n〉 of a single-mode field.

– “n photons”.

Do we recover classical wave as n gets large?

No : E = E(a + a†)⇒ 〈E〉 is always zero.


Number states: fields and uncertainties

Suppose we have a state |n〉 at t=0 – fields at later time?

Use Heisenberg equation, e.g.

i~dadt

= [a, H] and H = ~ω(a†a + 1/2)

⇒ a(t) = e−iωt a(0) ≡ e−iωt a

⇒ E(t) = E(ae−iωt + a†eiωt )

So

〈n|E(t)|n〉 = E〈n|(ae−iωt + a†eiωt )|n〉= 0.



〈n|E(t)|n〉 = E〈n|(ae−iωt + a†eiωt )|n〉= 0.

Also:

〈n|E2(t)|n〉 = E2〈n|(ae−iωt + a†eiωt )2|n〉= E2〈n|(aa† + a†a|n〉= E2(2n + 1).



Time

Electric Field

Coherent states as eigenstates of the annihilation operator

Coherent states: motivation and definition

Approx. to classical wave should have

〈E〉 6= 0 ∼ cos(ωt),

Since E ∝ a + a†, try finding eigenstate of a.

Eigenvalue equation a|λ〉 = λ|λ〉⇒ eigenstates

|λ〉 = e−λ∗λ/2eλa† |n = 0〉

= e−λ∗λ/2

∞∑

n=0

λn√

n!|n〉.

Properties of coherent states

Useful maths

Coherent states: formal properties

Right eigenstates of non-Hermitian operator a.

Continuous complex eigenvalue λ.

State 〈λ| is the left eigenstate of a† (with eigenvalue λ∗)

Normalised 〈λ|λ〉 = 1.

Not orthogonal 〈λ1|λ2〉 = e−|λ1|2/2−|λ2|2/2+λ∗1λ2 .

Complete 1π

∫|λ〉〈λ|d(<(λ))d(=(λ)) = 1.


Photon number distribution

Coherent states: physical properties

Average photon number 〈λ|a†a|λ〉 = |λ|2.Probability of n photons

P(n) = |〈n|λ〉|2

= e−|λ|2 |λ|2n

n!.

– Poisson distribution (as before) – with average n=|λ|2.

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Fields and uncertainties

Coherent states: fields and uncertainties

Suppose we have a mode in a coherent state |λ〉 at time zero,and field operator

E(t) = E(ae−iωt + a†eiωt )

Then

〈λ|E(t)|λ〉 = E(2<(λ) cos(ωt) + 2=(λ) sin(ωt)).

〈λ|E(t)2|λ〉 − 〈λ|E |λ〉2 = E2.


Fields and uncertainties

Coherent states: electric fields

Time

Electric Field

Coherent states are minimum uncertainty states

Visualising states in quadratures

Quadrature operators

A related visualisation of single-mode states by considering the“quadrature operators”

X1 =12

(a + a†) ∝ Ex ,

X2 =−i2

(a− a†) ∝ By .

– For a single mode field, dimensionless Ex and By .– For a harmonic oscillator, dimensionless position andmomentum operators.– ∼ real and imaginary parts of a



Quadrature operators: time dependence

Time evolution from Heisenberg equation

X1(t) =12

(ae−iωt + a†eiωt )

X2(t) =−i2

(ae−iωt − a†eiωt ),

⇒

X1(t) = X1(0) cos(ωt) + X2(0) sin(ωt)

X2(t) = −X1(0) sin(ωt) + X2(0) cos(ωt).



Quadrature visualisations - classical oscillators


Quadrature picture of coherent states

Coherent states: quadrature representation

For the coherent state |λ〉,

〈X1〉 =12

(λ+ λ∗) = <(λ)

〈X2〉 =−i2

(λ− λ∗) = =(λ)

And the uncertainties

∆X 21,2 = 〈X 2

1,2〉 − 〈X1,2〉2 =14



Coherent states: quadrature visualisation



Coherent states: uncertainty minimisation

Remember uncertainty ∆A∆B ≥ |c|/2 if [A,B] = c.For quadrature operators [X1,X2] = i/2, so ∆X1∆X2 ≥ 1/4.For coherent states –

This bound is saturated – “minimum uncertainty state”The uncertainty is symmetrically distributed ∆X1 = ∆X2.

– this is the sense in which coherent states are the closestquantum state to a classical wave

Generation of coherent states (non-exam)

Physical origin of coherent states

Laser light is a close approximation to a coherent state

When we have matter (atoms) etc., add

Hint = −∫

d3r E.P

– to the Hamiltonian of the EM field (see later in course).

Suppose classical electric dipoles P→ (−P0/2) cos(iωt)

but quantum E→ Ee(a + a†).



For one mode of the field we now have something like

H0 + (Ee.P0)(a + a†)(eiωt + e−iωt )

Suppose at time t = 0 field is in the ground state |0〉 and switch

on the oscillator.

What happens to the field?



Simplify: field mode of frequency ω

Switch to “interaction picture” operators a→ ae−iωt .

State vector obeys

i~d |ψ〉dt

= Hint|ψ〉,

whereHint ≈ (Ee.P0)(a + a†).

Integrating

|ψ〉 = e−iEe.P0

~ t(a+a†)|0〉

≡ e−α∗a+αa† |0〉 α = − itEe.P0

~



|ψ〉 = e−iEe.P0

~ t(a+a†)|0〉

≡ e−α∗a+αa† |0〉 α = − itEe.P0

~

Gives :

⇒ |ψ〉 = e−|α|2

2 eαa† |0〉.

= a coherent state (of amplitude α ∝ t - why?)

Summary

Summary

Number states do not approach the expected classical limitat large n.Important class of states which do are the coherent states

– defined as eigenstates of the annihilation operator a.In these states -

Poisson distributon of photon numberExpected 〈E〉 and 〈B〉 follow a classical wave withThe smallest fluctuations consistent with the uncertainty[X1,X2] = i/2

Coherent states are generated by classical oscillatingcurrents or dipoles← lasers

Light-Matter Interactions

Paul Eastham

The model

= Single atom in an electromagnetic cavity

MirrorsSingleatom

Realised experimentallyTheory:“Jaynes Cummings Model”⇒ Rabi oscillations

– energy levels sensitive to single atom and photon

– get inside the mechanics of “emission” and “absorption”

Where we’re going

= One field mode, two atomic states

Energy of photon in field mode

H = (∆/2) (|e〉〈e| − |g〉〈g|) + ~ω a†a +~Ω

2(a|e〉〈g|+ a†|g〉〈e|).

Dipole coupling energy

Energy difference between atomic levels

How to get there

Show that ∼ H = Hatom + Hfield − E .(er)

Write electron position operator r in basis– eigenstates of Hatom == atomic orbitalsApproximate to one mode of field and two atomic levelsNeglect non-resonant “wrong-way” terms(like electron drops down orbital and emits photon)

How to get there

Show that ∼ H = Hatom + Hfield − E .(er)←−Write electron position operator r in basis– eigenstates of Hatom == atomic orbitalsApproximate to one mode of field and two atomic levelsNeglect non-resonant “wrong-way” terms(like electron drops down orbital and emits photon)

Atom-Field Interaction Energy

Hamiltonian for Atom+Field

Field∑

modes

~ωi a†i ai

H = HEM + Hatom + Hint

Atom[− ~2

2me∇2

e −e2

4πε0|re − R0|

]

(nucleus fixed @ R0)Interaction energy ?


Interaction energy: classical field

Hatom =1

2mp2 + V (r).

With vector potential A, p→ (p−eA) (minimal coupling)and scalar potential φ, Hatom → Hatom + eφ

Hatom−field =1

2m(p− eA)2 + eφ+ V (r).


Interaction energy: A.p form

Choose the Coulomb gauge where ∇.A = 0, φ = 0

Hatom−field =p2

2m− e

mA.p +

e2

2mA2 + V (r)

= Hatom + Hint

(Can use this form directly – not in this course)


Interaction energy: dipole approximation

Interested in interaction with light waves

A = A0ei(k.r−ωt) + c.c.

For an atom the wavefunction extends over about 1Å

For light |k| = 2π/λ ≈ (500nm)−1

∴ A approximately constant in space over the atom,

– A(r, t)→ A(r = 0, t)


Interaction energy: E.r form

Coulomb gauge form:

Hatom−field =1

2m(p− eA)2 + eφ(= 0) + V (r).

Now change gauge :

A→ A +∇χ(r, t)

φ→ φ− ∂χ(r, t)∂t

,

so

Hatom−field =1

2m(p− e(A +∇χ))2 − e

∂χ

∂t+ V (r).


Interaction energy: E.r form

Hatom−field =1

2m(p− e(A +∇χ))2 − e

∂χ

∂t+ V (r).

Choose χ(r, t) = −A.r

so that ∇χ = −A

and∂χ

∂t= −∂A

∂t.r = E.r

[E = −∇φ− ∂A

∂t= −∂A

∂t

]

– (A, φ-Coulomb gauge)

Hatom−field =1

2m(p2 + V (r))− er.E(t).


Hamiltonian: E.r form

Quantum form: E(r)→ E(r0) at the position of the atomSo for our cavity quantization + one electron atom:

H = Hatom +∑

n

~ωna†nan

+∑

n,s

√~ωn

ε0Vsin(knzat)(an + a†n)es.(−er).


How to get there


Write electron position operator r in basis– eigenstates of Hatom == atomic orbitals←−Approximate to one mode of field and two atomic levelsNeglect non-resonant “wrong-way” terms(like electron drops down orbital and emits photon)


Second quantization: general

Generally have Z indistinguishable electrons

⇒ Atomic eigenstates labelled by occupation of orbitals(1s22s1 etc)

These states – |i〉 – form a complete set (for Z electrons)

This allows us to formally write atomic operators– in terms of “transition operators” |i〉〈j |


Second quantization: Hamiltonian

1 =∑

i

|i〉〈i |.

Formal representation of H :

H = 1H1

=∑

i

|i〉〈i |H∑

j

|j〉〈j |

=∑

i,j

|i〉〈i |Ej |j〉〈j | (H|j〉 = Ej |j〉, 〈i |j〉 = δij)

=∑

i

Ei |i〉〈i |.


Second quantization: One-body operators

Eigenstates of H – |i〉 – form a complete set for Z electrons, so

1 =∑

i

|i〉〈i |.

Formal representation of D =∑

i

eri :

D = 1D1

=∑

i

|i〉〈i |D∑

j

|j〉〈j |

=∑

i,j

〈i |D|j〉|i〉〈j |


Second quantization: One-body operators

D = 1D1

=∑

i

|i〉〈i |D∑

j

|j〉〈j |

=∑

i,j

〈i |D|j〉|i〉〈j |

If we know the orbitals in real-space, can calculate

Dij = 〈i |D|j〉 =

∫dxdydzψ∗i (r)(er)ψj(r).


Dipole matrix elements

Dij = 〈i |D|j〉 =

∫dxdydzψ∗i (r)(er)ψj(r).

Dij only non-zero between some states⇒ selection rulesi.j different parity.∆l = ±1 (if l good quantum number)

Magnitude |D| ≈ e× 1 Å


Atom-field Hamiltonian

So for our cavity problem

H =∑

n

~ωna†nan

+∑

i

Ei |i〉〈i |

+∑

n,s

∑

ij

En sin(knzat)(an + a†n)es.Dij |i〉〈j |.


How to get there


Write electron position operator r in basis– eigenstates of Hatom == atomic orbitalsApproximate to one mode of field and two atomic levelsNeglect non-resonant “wrong-way” terms(like electron drops down orbital and emits photon)


Why two levels?

Light-matter interactions weak(cf. energy differences in uncoupled problem)

⇒ small effects which can be treated in perturbation theory

except: if ~ω ≈ ∆, degeneracy between |n,g〉, |n − 1,e〉∴ can focus on the physics of one mode

+ nearly resonant atomic transition


Rotating-Wave approximation

Interaction is

~Ω

2[ a|e〉〈g|+ a†|g〉〈e| + a|g〉〈e|+ a†|g〉〈e| ].

Energy changes ≈ 0 ≈ ±2∆

∴ drop these terms

Jaynes-Cummings Model in Rotating-Wave Approx

H = (∆/2)(|e〉〈e| − |g〉〈g|) + ~ωa†a

+~Ω

2(a|e〉〈g|+ a†|g〉〈e|).


Summary



H = (∆/2) (|e〉〈e| − |g〉〈g|) + ~ω a†a +~Ω

2(a†|e〉〈g|+ a|g〉〈e|).



Physics of the Jaynes-Cummings Model

Paul Eastham

Outline

1 The model

2 Solution

3 Experimental ConsequencesVacuum Rabi splittingRabi oscillations

4 Summary

The model

= Single atom in an electromagnetic cavity

MirrorsSingleatom

Realised experimentallyTheory:“Jaynes Cummings Model”⇒ Rabi oscillations

– energy levels sensitive to single atom and photon

– get inside the mechanics of “emission” and “absorption”

The model

Atom-field Hamiltonian

Last lecture –

H =∑

n

~ωna†nan

+∑

i

Ei |i〉〈i |

+∑

n,s

∑

ij

En sin(knzat)(an + a†n)es.Dij |i〉〈j |.

The model

→ Jaynes-Cummings Model



H = (∆/2) (|e〉〈e| − |g〉〈g|) + ~ω a†a +~Ω

2(a|e〉〈g|+ a†|g〉〈e|).



Solution

Solving the JCM

H only connects within disjoint pairs |n,g〉 and |n − 1,e〉∴ eigenstates are

un,±|n,g〉+ vn,±|n − 1,e〉.

⇒ En,± = ~ω(n − 12

)± 12

√(∆− ~ω)2 + ~2Ω2n

and at resonance states are

1√2

(|n,g〉 ± |n − 1,e〉).

Solution

Jaynes-Cummings Spectrum

Solution

Jaynes-Cummings Spectrum

Experimental Consequences

Vacuum Rabi splitting

Transmission experiments: idea

Laser Detector

Transmission

Frequency/(Resonance frequency)

With no atom

(Fabry-Perot resonator -- SF Optics?)


Vacuum Rabi splitting

Transmission experiments

Transmission

Frequency/(Resonance frequency)

2

4

-40 0 40Probe Detuning ωp (MHz)

-40 0 40

2

4

⟨(

nω

p )×

⟩0

12-

2

4

0.3

0.2

0.1

0.0

0.3

0.2

0.1

0.0

T1( ω

p)

0.3

0.2

0.1

0.0

A. Boca et al., Physical Review Letters 93, 233603 (2004)


Rabi oscillations

Rabi oscillations

Different way to observe the Jaynes-Cummings physics

Suppose we start with no light, add atom in |e〉What happens?

Photon number oscillates – “Rabi oscillations”


Rabi oscillations

Rabi oscillations

Easiest for resonant case ∆ = ~ω.

Eigenstates with one “excitation” are |±〉 =1√2

(|0,e〉 ± |1,g〉)

Energies E± and E+ − E− = ~Ω


Rabi oscillations

Rabi oscillations

Eigenstates with one “excitation” are |±〉 =1√2

(|0,e〉 ± |1,g〉)

∴ initial state is |0,e〉 =1√2

(|+〉+ |−〉) .

⇒ state at time t is

1√2

( |+〉eiE+t/~ + |−〉eiE−t/~)

= ei(E++E−)t/~ [cos (Ωt/2) |e,0〉+ i sin (Ωt/2) |g,1〉] .


Rabi oscillations

Rabi oscillations

Expected photon number is 〈n〉 = sin2(Ωt/2)

<n>

Time


Rabi oscillations

Rabi oscillations

Rempe et al., Physical Review Letters 58, 393 (1987)


Rabi oscillations

Photon statistics in Rabi oscillations

What about the intensity fluctuations?

〈n〉 = sin2(Ωt/2). 〈n2〉 − 〈n〉2 = sin2(Ωt/2)− sin4(Ωt/2).

⇒ strongly sub-Poissonian light

Summary

Summary: light-matter coupling

Interaction between light and matter is the dipole couplingP.E.Seen how to write this in terms of a, |i〉〈j |Single mode+two-level atom+Rotating-waveapproximation=Jaynes-Cummings modelEigenstates of JCM are superpositions like|n,g〉+ |n − 1,e〉Coupling splits the energy levelsSeen experimentally in optical cavities in transmissionand Rabi oscillations

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