ece 407 –spring 2009 –farhan rana –cornell university...5 ece 407 –spring 2009 –farhan...

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Chapter 11: Matter-Photon Interactions and Cavity QED

In this lecture you will learn:

• Electron-Photon Interactions• Cavity QED• Light-Matter Entanglement• Noise in Optical Transitions


Interaction between a Charged Particle and Field

In quantum mechanics we have:

kjjk itvmtr ˆ,ˆ

i

vm This implies:

In the presence of an E&M field the canonical momentum is:

ttrAqtvmtp ,ˆˆˆ

kjjk itptr ˆ,ˆ

The canonical momentum satisfies the equal-time commutation relation:

i

p

This implies:


The classical expression for the momentum of a field is:

trHtrErdtP oo ,,3

),(),(),( trEtrEtrE TL

The E-field can be decomposed as:

0),(),(),( trEtrtrE LLo

0),( trETo

Where:

),(),(),( tkEtkEtkE TL

In Fourier-Space:

oL

oL

tkkkitkEtkitkEk

),(ˆ

),(),(),(

0),( tkEk T

Longitudinal component

Transverse component

Classical Field Momentum


),(),( trAtrHo

We also have:

),(),( tkAkitkHo

The vector potential can also be decomposed as:

),(),(),( trAtrAtrA TL

),(.ˆˆ),( tkAkktkAL

),(.ˆˆ),(),(ˆˆ1),( tkAkktkAtkAkktkAT

The field momentum becomes:

trHtrErdtrHtrErdtP TooLoo ,,,, 33

? ?

Since:

( , ) ( , ) ( , )L TE k t E k t E k t

( , ) 0TA r t

,( , )L

o

r tE r t



3

3 3

3 3

3 3

3 3

3

3

3

, ,

, , , ,2 2

( , ) ˆ ˆ ˆ, ( , ) 1 ,2 2

( , ) ,2

,

o o L

o o L o L

oo

T

T

d r E r t H r t

d k d kE k t H k t E k t ik A k t

d k k t d ki k ik A k t k t k k A k tk

d k k t A k t

d r r t A r

,t

Look at the first term in more detail:

trrqtr ),(

For a single particle with position and charge q the charge density is:

ttrAqtrHtrErd TLoo ,,,3

Therefore:

r t



trHtrErdttrAqtP TooT ,,, 3

Therefore the field momentum becomes:

The total momentum of the particle and the field is:

trHtrErdttrAqtvmtP TooTtotal ,,, 3

Kinetic momentum of the particle

Transverse field momentum

Longitudinal field momentum

Coordinate of the particle

Total Momentum of Particle and Field


Total Momentum of Particle and FieldFor the particle interacting with the field, define a canonical momentum as:

,p t mv t qA r t t

The total momentum of the particle and the field is then:

trHtrErdttrAqtvmtP TooTtotal ,,, 3

Kinetic momentum of the particle

Transverse field momentum

Longitudinal field momentum

3

3

, , , ,

, , ,total L o o T

L o o T

P t mv t qA r t t qA r t t d r E r t H r t

p t qA r t t d r E r t H r t


Total Angular Momentum of Particle and Field 3 , ,total o oJ t r t mv t d r r E r t H r t

3

3

3

3

3

3

, ,

, ,

, ,

, ,

, ,

,

o o

o T

o L T

o T T

T

o T

d r r E r t H r t

d r r E r t A r t

d r r E r t A r t

d r r E r t A r t

d r r r t A r t

d r r E r

,Tt A r t

, , ,

, 0L T

L

A r t A r t A r t

A r t

2

, , ,,

. ,

,,

L T

Lo

Lo

E r t E r t E r t

r tE r t

r tE k t i k

k

trrqtr

),(For a single charged particle:

3

,

, ,

T

o T T

r t mv t qA r t t

d r r E r t A r t


Total Angular Momentum of Particle and FieldWe therefore get:

3, , ,total T o o TJ t r t mv t qA r t t d r r E r t H r t

Using the canonical momentum defined as:

,p t mv t qA r t t

We get:

3, , ,total L o o TJ t r t p t qA r t t d r r E r t H r t

3 , ,o T Td r r E r t A r t

3 3, ,, , , ,o T T o s T s T

sd r E r t A r t d r E r t r A r t

Photon spin Photon orbital


Total Angular Momentum of Particle and Field

3

3

3, ,

, , ,

,

, ,

, ,

total L o o T

L

o T T

o s T s Ts

J t r t p t qA r t t d r r E r t H r t

r t p t qA r t t

d r E r t A r t

d r E r t r A r t

Particle angular momentum

Photon spin angular momentum

Photon orbital angular momentum

The above decomposition is gauge invariant


Canonical Momentum: Interpretation in the Coulomb GaugeAssume Coulomb gauge:

0),( trAL

trHtrErdtptP Toototal ,,3

The expression for the total momentum (particle + field) is:

Momentum of photons in the Coulomb gauge (Chapter 5) – momentum associated with the transverse part of the field

v

q ttrAqtvmtp T ,

In the Coulomb gauge, the canonical momentum of a charged particle is seen to consist of two parts: i) The kinetic momentum of the charged particle ii) The momentum contributed by the longitudinal field associated with the charged particle, the source of which is the charged particle itselfiii) In other gauges, the above interpretation does not hold

ttrAqtvmtp T ,

Then canonical momentum of the particle becomes;


Classical Hamiltonian of Particles and Field

Assume several particles with:

Charges: Positions: Velocity:q tr tv

The classical Hamiltonian is:

2 31 1 1( , ). ( , ) ( , ). ( , )2 2 2o oH mv t d r E r t E r t H r t H r t

Or using the definition of the canonical momentum:

),().,(

21),().,(

21

2),( 3

2trHtrHtrEtrErd

mttrAqtpH oo

Note: the scalar electromagnetic field does not get included in the Hamiltonian


Assume several particles with:

Charges: Positions: Velocity:q tr tv

The classical Hamiltonian is:

),().,(

21),().,(

21

2),( 3

2trHtrHtrEtrErd

mttrAqtpH oo

Note that:

),(),(),( trEtrEtrE TL

The longitudinal field energy is:

trtrqq

rrtrtrrd

ktktkkd

tkEtkEkdtrEtrErd

o

oo

LLoLLo

421

'4,',

21),(),(

21

2

),().,(21

2),().,(

21

323

3

3

33

Coulomb interaction energy of the charged particles

,( , )L

o

r tE r t

Classical Hamiltonian of Particles and Field


),().,(

21),().,(

21)(

2, 3

2trHtrHtrEtrErdtrqV

mttrAqtpH oTTo

Consider only a single particle moving in the static Coulomb potential of other particles and interacting with radiation:

Includes the static potential of all other fixed charges (not explicitly included)

Includes the longitudinal and transverse vector potential components

Quantizing Interacting Particle and Field


Postulate the following equal-time commutation rules for the particle coordinates and canonical momenta:

kjjk itptr ˆ,ˆ

Choose Coulomb Gauge:

( , ) 0( , ) ( , )L

T

A r t

A r t A r t

,mv t p t qA r t t

i

p

Kinetic Momentum:



2

3ˆˆ ˆ , 1 1ˆ ˆ ˆ ˆˆˆ ( ) ( , ). ( , ) ( , ). ( , )2 2 2T

o T T o

p t qA r t tH qV r t d r E r t E r t H r t H r t

m

The Hamiltonian operator becomes (in the Heisenberg picture):

)(ˆ).(ˆ

21)(ˆ).(ˆ

21)ˆ(

2ˆˆˆˆ 3

2

rHrHrErErdrqVm

rAqpH oTToT

And in the Schrodinger picture:


The field can be quantized following the steps in Chapter 5


Gauge Transformation

Suppose the electron is localized around or

or

rThe we can approximate the Hamiltonian as:

)(ˆ).(ˆ

21)(ˆ).(ˆ

21)ˆ(

2

ˆˆˆ 32

rHrHrErErdrqVm

rAqpH oTTooT

Now we will perform a gauge transformation to get rid of the vector potential

Suppose a quantum state of the particle-field system is:

Under a gauge transformation the new state is: Tnew ˆ

1ˆˆ TTThe operator has to be unitary:T

Any system operator will transform under as:T ˆ ˆˆ ˆnewO TOT O


Gauge TransformationFor all physical observables, we must have:

TTHTTHH newnewnew ˆˆˆˆˆˆˆ

Hamiltonian is gauge invariant but will turn out to be not form invariant under the gauge transformation

ˆ ˆ ˆˆ ˆ ˆ ˆnew new newO O T TOT T

In particular, for the Hamiltonian we must have:


ˆˆ ˆˆˆˆ

new

T o

p TpT

p qA r

Suppose we choose:

ˆ ˆ( ).ˆ T o oi qA r r r

T e

Then:

3 .3

ˆ ˆˆ ˆ

1ˆ ˆˆ ˆ .2

1ˆ ˆ .

o

T new T

ik r rT j j o

jo

T T o oo

E r TE r T

d kE r k k q r r e

E r r r q r r

ˆ ˆˆ ˆ

ˆ

T new T

T

H r TH r T

H r

(Canonical momentum operator is not gauge invariant)

(E-field operator is not gauge invariant)

(Magnetic field operator is gauge invariant)

or

r



ˆ ˆˆ ˆˆˆ

ˆˆ ˆˆ ˆˆ ˆ ˆ ˆ

new

T o

new T o

p TpT

p qA r

mv TmvT T p qA r T p

The relation:

In the new gauge, the kinetic momentum and the canonical momentum are the same!!!!

Therefore, in the new gauge the particle kinetic energy equals: 2

2 ˆ1 ˆ2 2new

pmvm

Implies:

In the old gauge:

2

2ˆˆ

1 ˆ2 2

T op qA rmv

m



23

3

32

ˆ.ˆ2

12

)(ˆ).(ˆ21)(ˆ).(ˆ

21)(ˆ.ˆ)ˆ(

2ˆ

ˆˆˆˆ

ojoj

oTTooTo

new

rrkkd

rHrHrErErdrErrqrqVm

p

THTH

ˆ ˆ( ).ˆ T o oi qA r r r

T e

We have:

Dipole self-energy (infinite!)

We will use:

)(ˆ).(ˆ

21)(ˆ).(ˆ

21)(ˆ.ˆ)ˆ(

2ˆˆ 3

2rHrHrErErdrErrqrqV

mpH oTTooTo

PHIH FH

or

r

Gauge Transformation: The Dipole Hamiltonian


Cavity Quantum Optics

21

Cavity

ˆ ˆ ˆ ˆP F IH H H H The Hamiltonian is:

2211

2221112

ˆˆ

ˆ2ˆˆ

NN

eeeerVqm

pHP

Consider a two-level system in an optical cavity

21ˆˆˆ aaH oF

)(ˆˆˆ ooI rErrqH

Recall that:

)(ˆˆ2

)(ˆ rUaairEo

o

)(ˆˆ2

)(ˆo

o

oo rUaairE



21

Cavity 22112211

particleparticleˆ

1ˆ1ˆ

eeeeHeeee

HH

I

II

We can write the interaction Hamiltonian in a better form:

aakk

aaeekeekHI

ˆˆˆˆ

ˆˆˆ*

21*

12

2 1ˆ( ) .

2o

o oo

k q i U r e r r e

21

12

ˆˆ

eeee

Define:

We have:

Now make the rotating wave approximation:

ˆˆˆˆˆˆˆˆˆ *2211 akakaaNNH o



21

Cavity


The above time-independent Hamiltonian must have eigenstates and eigenenergies

Assume a solution for the eigenstate of the form:

2 1 1e n e n Then:

*2 2 1

1 1 2

*2 2 1 1

ˆ 1 1

( 1) 1 1

1 1 ( 1) 1

o

o

o o

H n e n k n e n

n e n k n e n

n k n e n k n n e n

Take the bra first with: and then with: ne 2 11 ne

EH ˆ

2*

1

11 ( 1)o

o

n k nE

k n n

EH ˆ


Cavity Quantum Optics: Eigenenergies and Eigenstates

21

Cavity Detuning: oo )( 12

We will assume zero detuning from now onwards

2

2

( 1) 1

( 1) 1o

o

E n n k n

E n n k n

The two eigenvalues and eigenstates are:

*2 1

*2 1

1( 1) 12

1( 1) 12

kn e n e nk

kn e n e nk

...............3,2,1,0n

Each value of “n” will give two eigenstates and eigenenergies

2*

1

11 ( 1)o

o

n k nE

k n n


Spectrum of Eigenenergies

21

Cavity

Suppose there was no electron-photon interactionSuppose zero-detuning

Then we would have the following eigenstates and eigenenergies:

1 1

2 1

2 2

2 1

2 2

2 1

2 2

Energies States

0

1 0

2 1

1 1 1

o

o

o

o

ε e

ε eε e

ε eε e

ε n e nε n e n


Spectrum of Eigenenergies: Rabi Splitting

21

Cavity

1 1

2

2

2

2

2

2

Energies States

0

11

2 22 2

1 1

o

o

o

o

ε e

ε kε k

ε kε k

ε n k n nε n k n n

When the electron-photon interaction is turned on, we have the following eigenstates and eigenenergies:

The interaction lifts the degeneracy between the previously degenerate pair of states and splits their energies

This is called Rabi splitting and equals: 2R k n


Quantum Rabi Oscillations

21

Cavity

Suppose: 20t e n

n

210 1 12

t e n n n

Need to find: t

Proceed as follows:

2

2

ˆ

1 1

2 1

0

12

cos 1 sin 1 1

o

o

Hi t

k ki n t i n t i n t

i n t

t e t

e e n e n

k kke n t e n i n t e nk

tnk

tne

tnk

tne

1sin1

1cos

221

222

This implies:

12 nk

R


Vacuum Rabi Oscillations

21

Cavity n=0

Suppose:

210 0 1 12

t e

1sin0cos 12

2et

kkkiet

ket

tεi

Then:

These are called vacuum Rabi oscillations!

Vacuum Rabi frequency is:

2VRk


Single Two-Level System in a Cavity

21


* 1 2ˆ ˆ ˆ ˆ ˆˆ ˆ2 2z oH a a k a k a

2 1ˆ ˆz N N

Write this as:

Where:

*ˆ ˆ ˆ ˆ ˆˆ ˆ2 z oH a a k a k a

Just use:

The commutator algebra of is similar to that of Pauli matrices for spin-1/2 systems

ˆ ˆ, ,z

ˆ ˆˆ

2x yi

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, 2 , 2 , 2

ˆ ˆ ˆ, 2x y z y z x z x y

z

i i i


21

21

*

2 2 2

1 1 1

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ2

ˆ ˆ ˆˆ ˆ

z o

z zj j jj j j

H J k J a k a J a a

J J J

2 Two-Level Systems in a Cavity and Spin Systems

Any state of the atoms (or the two-level systems) can be written using the following basis set:

,

,

,

,

But it is more insightful to use a different basis set …………


2 Two-Level Systems in a Cavity and Spin Systems

21

21

*

2 2 2

1 1 1


ˆ ˆ ˆˆ ˆ

z o

z zj j jj j j

H J k J a k a J a a

J J J

Think “addition of angular momenta” in undergraduate quantum mechanics textbooks

Consider the three triplet states and the one singlet state:

ˆ , 2 ,

1 1ˆ , , 0 , ,2 2

ˆ , 2 ,

z

z

z

J

J

J

1 1ˆ , , 0 , ,2 2zJ

2 2 2 2

2

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ 2z x y

z

J J J J

J J J J J

Multiplet with eigenvalue 8 for the operator2J


Triplets:

Singlet:


2 Two-Level Systems in a Cavity: Radiation Cascades

,

1 , ,2

,

1 , ,2

n

21

21

Triplet Multiplet Singlet

J

J



*

2 2 2

1 1 1


ˆ ˆ ˆˆ ˆ

z o

z zj j jj j j

H J k J a k a J a a

J J J

Dark entangled state !!


2 Two-Level Systems in a Cavity

21

21

1 , ,2

ˆ ˆ 0

S

J S J S

The singlet state is a “dark eigenstate” – it does not interact with photons inside the cavity

Within the multiplet of the three triplet states, we can write the Hamiltonian as:

*

*

2 1 0ˆ 2 1 0 1 2 2

0 2 2 2

o

o

o

n k n

H k n n k n

k n n

,

1 , , 12, 2

n

n

n

1 , ,2

n

System eigenstates

The three eigenenergies, assuming zero detuning, are:

2 2 * 3

2 2 * 3

o

o

o

n k nn

n k n


2 Two-Level Systems in a Cavity: Matter Entanglement Generation

21

21

Suppose at time t=0, the quantum state was prepared such that both two-level systems were in the upper state:

Photon

detector

0 , 0t

Suppose that the Rabi period and the decoherence times are both much longer than then photon lifetime in the cavity

Suppose at time t = T a photon is detected at the cavity output waveguide

Then what is post-detection?

1 , , 02

t T

t T

(with a very high likelihood)

, 1t T (less likely, under the stated assumptions)

*ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ2 z oH J k J a k a J a a


N Two-Level Systems in a Cavity

21

21

21

21

21

*

1 1 1


ˆ ˆ ˆˆ ˆ

z o

N N Nz zj j j

j j j

H J k J a k a J a a

J J J

Several multiplets are possible corresponding to all possible eigenvalues of the operator:

2 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2z x y zJ J J J J J J J J

One multiplet is particularly interesting, whose first state is:

2ˆ , 2 ,ˆ , ,

2 2, 2,

2

z

J M N N N M N

J M N M M N

N N M MJ M N M N

, , , ,.......N N

2,N N

4,N N

,N N

Within this multiplet:



21

21

21

21

21

Suppose at time t=0, the quantum state was prepared such that all two-level systems were in the lower state and there was one photon inside the cavity:

Find the system eigenstates, and then the state at time = t ?

0 , , ,...... 1 , 1t N N

The state will couple only with the state when interaction with photons is turned on

So in this subspace, the Hamiltonian is:

, 1N N 2, 0N N

*

2ˆ

2 o

N NkH

NNk

Eigenenergies, assuming zero detuning, are:

2

2

o

o

N Nk

N Nk

Rabi splitting = 2VR N k

Coupling is now times stronger!!N



1

2cos , 1 sin 2, 0

N

i t N k N kkt e t N N i t N Nk

The state at time t is then:

21

21

21

21

21


N Two-Level Systems and Dicke’s Superradiance

21

21

21

21

21

Consider N two-level systems all prepared in the upper state

The rate of spontaneous emission observed from this collection is expected to be proportional to N

But certain highly correlated and entangled matter states can radiate at a rate proportional to ~N2

Robert H. Dicke, "Coherence in Spontaneous Radiation Processes,” Physical Review, 93, 99–110 (1954). From Chapter 10:

Suppose 0 ,t N N ˆ ˆ0 0t J J t N

Therefore:

Photon Flux ˆ ˆ ˆ ˆ, ,r rE r t E r t J t J tc c

ˆ ˆ, ,r rE r t E r t Nc c



21

21

21

21

21

Consider N two-level systems now all prepared in the highly correlated state (assuming N is even):

Robert H. Dicke, "Coherence in Spontaneous Radiation Processes,” Physical Review, 93, 99–110 (1954).

0 0,t M N

21

What does this state looks like:

ˆ0 0, , , , ,.... , , , , ,....t M N S

N/2 N/2The operator makes a superposition such that the resulting state is completely symmetrical across all the two-level systems and is properly normalized

S

22ˆ ˆ0 04 4

N N Nt J J t

Now:

And:

Why is this the case?

2ˆ ˆ, ,r rE r t E r t Nc c


Optical Transitions and Fermi’s Golden Rule

L

m

Stimulated Absorption:

mmmmmo

mmmLL akakaaNNH ˆˆˆˆˆˆˆˆˆ *

net L 0

mmoLm

mmoLLmmm

mmoLLIm

nk

neaken

neHenR

2

2

2

2

ˆˆ12

ˆ12n

Stimulated and Spontaneous Emission: Hmn

mmmmmo

mmmHH akakaaNNH ˆˆˆˆˆˆˆˆˆ *

net H 0

mmoHm

mmoHHmmm

mmoHHIm

nk

neaken

neHenR

12

ˆˆ12

ˆ12

2

2*

2

Agrees with the Einstein’s thermodynamic argument


Spontaneous Emission in Free Space

1 2

z

y

x 2 1

kεV

etkatkaε

ikdVtrE jrki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ .

3

3

31 1 2 2 3

*

,

ˆ ˆ ˆ ˆ ˆ2

ˆ ˆˆ ˆ +

k j jj

j j j jj k

d kH N N V a k a k

k a k k a kV

The Hamiltonian is:

The coupling parameter is (assuming only the z-component of the field couples):

Initial and final states:

02initial e

jke ,1final 1 Many possible final states

2 1ˆ ˆ ˆ.2

kj j z

ok i ε k e q e z e

ε


Spontaneous Emission in Free Space1 2

z

y

x 2 1

22 2 2 1,

,3 22

2 1 2 13

222 1

2

2 ˆ( ) 1 0

2 1ˆ ˆ .22

2 3 2

sp I kk jj k

kj k

j o

o

R FS e H e

d kV q ε k e z eε V

q e z eD

ε

Use Fermi’s Golden Rule:

Photon Density of States:

032

2

03

2

3

3

242

22

cdkcdkkkd

32

2

cD

Number of modes (of both polarizations) per unit volume per unit frequency interval


Spontaneous Emission in a Cavity: The Purcell Effect

21

Cavity

E. M. Purcell "Spontaneous emission probabilities at radio frequencies" Phys. Rev. 69, 681 (1946)



21

CavityConsider a lossless optical cavity containing a two level system

Can we find the spontaneous emission rate inside the cavity?

The above question does not make sense because in a lossless closed cavity there are only Rabi oscillations (no Markovian optical transitions)

We need to incorporate loss into the cavity!

So now consider a lossy optical cavity containing a two level system

And now we ask the same question: what is the spontaneous emission rate for the two level system inside the cavity?

21

CavityWaveguide

z = 0




21

CavityWaveguide

z = 0

Consider a cavity coupled to a waveguide

Step 1: Assume there is no two-level system

tiL

p

g

po oetzb

vtai

dttad

,0ˆˆ

21ˆ

L

eadvL

LeadLetzb

tiL

g

titi

Lo

o

o

o

o

ˆ

2ˆ

2,0ˆ

2

2

2

2

L

eadLv

taidt

tad tiL

pgpo

o

o

ˆ

21ˆ

21ˆ 2

2

L

eea

iidL

veeata

poo

o

po

ttiti

Lpopg

tti

22

2

2 ˆ22

1ˆˆ

Solution is:

L

eai

idLv

tati

Lpopg

o

o

ˆ22

1ˆ2

2

As t→∞:


tiL

pg

tiR oo etzbta

vetzb

),0(ˆˆ1),0(ˆ 2

1

CavityWaveguide

z = 0

Lpo

poR

tiL

po

po

g

tiR

g

aii

a

Lea

iid

vL

Lead

vL o

o

o

o

ˆ22

ˆ

ˆ22

2ˆ

2

2

2

2

2

As t→∞:

L

eai

idLv

tati

Rpopg

o

o

ˆ22

1ˆ2

2

We had:

L

eai

idLv

tati

Lpopg

o

o

ˆ22

1ˆ2

2

L

ai

idLv

a Rpopg

o

o

1ˆ22

1ˆ2

2

SP

HP



21

CavityWaveguide

z = 0

Now see how the interaction Hamiltonian gets modified:

L

ai

idLv

ik

La

iidL

vkakak

Rpopg

Rpopg

o

o

o

o

1ˆˆ22

1

1ˆˆ22

1ˆˆˆˆ

2

2

*

2

2

*

Step 2: Include the two-level system

Initial and final states:

02initial e

final 1 1 Re

Assume that:1

2p VRpk


A continuum of possible final states



2 22 1222

2

2 22

2Cavity2 1 2

1 2 2

1 2

o

o

psp

o p

p

o p

k dR

k

Spontaneous Emission in a Cavity

Using Fermi’s Golden Rule:

One may use Fermi’s Golden Rule provided:1

2p VRpk

212

2212 ˆ).(

2ˆ.)(

2ezerUqkezerUiqk o

o

oo

o

o

21

CavityWaveguide

z = 0

The coupling parameter was:

Define mode volume Vp as:

1)().(1)().( *3* rUrUrdV

rUrUp

oo


Purcell Enhancement of Spontaneous Emission in a Cavity

21

CavityWaveguide

z = 0

2222 1

2 22

1 262Cavity3 2 1 2

pposp

o p o p

q e z eR

ε V

We have:

At resonance we have:

22

2 1 22

62Cavity3 2

posp VR p

o p

q e z eR

ε V

In free space we had:

22

2 12

23 2 o

q e z eR FS D

ε

The ratio of the spontaneous emission rate in the cavity w.r.t. free space is:

6 1p

pV D

Cavity quality factor:

pQ

1VR p


Purcell Enhancement of Spontaneous Emission in a Cavity

21

CavityWaveguide

z = 0

2 23

43

p

QV

n

Purcell Enhancement Factors:

1) Spontaneous emission enhancement factor in a dielectric cavity of average index ncompared to in vacuum:

2) Spontaneous emission enhancement factor in a dielectric cavity of average index ncompared to in an infinite dielectric media of the same refractive index n :

2 33

43

p

QV

n

6 1p

pV D

pQ


Spontaneous Emission in a Cavity with Decoherence Included

2 2

2 1222

2Cavity2 1 2

o

o

psp

o p

k dR

Using Fermi’s Golden Rule (our previous result):

The effect of decoherence of the material polarization can be included by broadening the energy conserving delta function:

2 22

2 2 2222 2

22

2 222

12 1Cavity2 11 2

1 1 22 1 1 2

o

o

psp

o p

p

o p

k d TRT

Tk

T

2

2

2 2VR p

p

T

T

on resonance

21

CavityWaveguide

z = 0

Assume: 21,

2pVR

Tk


Spontaneous Emission Rate in a Cavity: Different Regimes

21

CavityWaveguide

z = 0

The more general case: 21,

2pVR

Tk

p1

2

1T

o

p1

2

1T

2

2

2 2VR p

p

T

T

on resonance

o

o

2 22

2 2 2222 2

22

2 222

12 1Cavity2 11 2

1 1 22 1 1 2

o

o

psp

o p

p

o p

k d TRT

Tk

T

Purcell RegimeIncreasing the cavity Q increases the spontaneous emission rate


Spontaneous Emission Rates in a Cavity: Different RegimesIn vacuum we had:

22

2 12

23 2sp

o

q e z eR FS D

ε

2222 1

2 22

1 262Cavity3 2 1 2

pposp

o p o p

q e z eR

ε V

In a cavity when:

we had:

21

CavityWaveguide

z = 0

2 21, ,

2p pVR

T Tk

In a cavity when:

we had:2 2

1, ,2 p p

VRT T

k

22

2 2 22

2 222 1 22

2 2 22

1Cavity 21

12 33 2 1

spo

o

o p o

k TRT

q e z e TTε V T

22

2VRT

on resonance

Purcell Regime


Enhancement of Spontaneous Emission in a Cavity

21

CavityWaveguide

z = 0

2 23

2

24 13

p p

QV

nT

Cavity Spontaneous Enhancement Factors:

1) Spontaneous emission enhancement factor in a dielectric cavity of average index ncompared to in vacuum:

2) Spontaneous emission enhancement factor in a dielectric cavity of average index ncompared to in an infinite dielectric media of the same refractive index n :

2 33

2

24 13

p p

QV

nT


Weak, Intermediate, and Strong Coupling in Cavity Quantum Optics

21

CavityWaveguide

z = 0

Weak Coupling:

2

2 1 1,2VR

p

kT

Strong Coupling:

2

21 1,2 VR

p

kT

Vacuum Rabi frequency is much smaller than the decoherence and the photon decay rates

Vacuum Rabi frequency is much larger than the decoherence and the photon decay rates

Intermediate Coupling:

2

21 12VR

p

kT

Vacuum Rabi frequency is much smaller than the decoherence and the photon decay rates


Spontaneous Emission Spectrum in the Strong Coupling Limit

21

CavityWaveguide

z = 0 ˆ , oi tLb z t e

ˆ , oi tRb z t e

2

ˆ 1 ˆˆ ˆˆ dd t ii t k N t a t G t

dt T

ˆ 1 ˆˆ ˆ 0,

2og i t

o Lp p

vda t ii a t k t b t edt

Material polarization:

Cavity field equation:

ˆ

ˆ ˆˆ ˆ2dd N t i k t a t k a t tdt

Population equation:

2

21 1,2 VR

p

kT

Assume a strong coupling regime:

1ˆ ˆˆ0, 0,o oi t i tR L

g pb z t e a t b z t e

v

Equation for the field outside the cavity:


21

CavityWaveguide


ˆ , oi tRb z t e

1ˆ ˆˆ0, 0,o oi t i tR L

g pb z t e a t b z t e

v

Equation for the field outside the cavity:

The outside the cavity field correlation function becomes:

ˆ ˆ ˆ ˆ0, 0, oo i ti tR Rb z t e b z t e a t a t

Initial state at time t:

10 1t e

Possible states of the cavity system as the time evolves:

2 1 10 1 0e e e

Within this reduced Hilbert space:

ˆ ˆ ˆ 0dN t a t a t



21

CavityWaveguide


ˆ , oi tRb z t e

therefore we can use the simpler linear equations:

2

2

ˆ 1 ˆˆ ˆˆ

1 ˆˆˆ

dd t ii t k N t a t G t

dt T

ii t k a t G tT

ˆ 1 ˆˆ ˆ 0,

2g i t

o Lp p






21

CavityWaveguide


ˆ , oi tRb z t e


2

ˆ 1 ˆˆd t ii t k a t

dt T

ˆ 1 ˆ ˆ

2op

d a t ii a t k tdt

Consider the averaged equations:

Solution subject to the conditions:

is: ˆˆ 0 1 0 0t a t

*

ˆ sinoi t t kka t t i e t tk

2

2

21 1,2

1 122

0

VRp

p

o

kT

T


21

CavityWaveguide


ˆ , oi tRb z t e

The correlation functions become:

2

ˆ ˆ 1 ˆ ˆ ˆˆ

ˆˆ

d a t t ii a t t k a t a td T

a t G t

ˆ ˆ 1 ˆ ˆ ˆ ˆ2

ˆˆ 0,

op

g i tL

p

d a t a t ii a t a t k a t td

va t b t e

0

0



21

CavityWaveguide


ˆ , oi tRb z t e


2

ˆ ˆ 1 ˆ ˆ ˆˆd a t t ii a t t k a t a t

d T

ˆ ˆ 1 ˆ ˆ ˆ ˆ2o

p


Given the linearity of the equations, it is not difficult, but tedious, to show that:

*ˆ ˆa t a t t t

*


2

2

21 1,2

1 122

0

VRp

p

o

kT

T

Where:



21

CavityWaveguide


ˆ , oi tRb z t e

*ˆ ˆa t a t t t

*


2

2

21 1,2

1 122

0

VRp

p

o

kT

T

The correlation function is not stationary!!!

How to find the spectrum??


21

CavityWaveguide


ˆ , oi tRb z t e


*

2

*

2

2

2

ˆ ˆ .

ˆ ˆ2

.

2

.

2

.

2

i

i

i ti t

i t

dt d a t a t e c cS

dt a t a t

dt d t t e c c

dt t

dt t e d t e c c

dt t

dt t e c c

dt t

2 2

2 22 2

4

o o

k

k k

2

2

21 1,2

1 122

0

VRp

p

o

kT

T

For non-stationary correlation functions, the spectral density is:


21

CavityWaveguide


ˆ , oi tRb z t e


2VRk

Spontaneous emission spectrum

Linewidth (FWHM) = 2

1 12 pT

2 2

2 22 2

4

o o

kS

k k



21

CavityWaveguide


ˆ , oi tRb z t e

1n

1 0e

1n VR

o

2 0e

1 1e

1 1

2

2

2

2

2

2

Energies States

0

11

2 22 2

1 1

o

o

o

o

ε e

ε kε k

ε kε k


Dressed states


Transmission through a Strongly Coupled Cavity

21

o

Consider an empty-cavity transmission experiment:

A resonance in transmission will be observed at the frequency o

Consider a cavity transmission experiment when the contains a two-level system:

What will be the frequencies of transmission resonances now??

o

21



2

2 1 1,VRp

kT

Assume strong coupling:

Input state:

ˆ , i t i tR inb z t e e CW coherent state of frequency

Output state:

ˆ , i t i tR outd z t e e

Statement of the problem:

Find as a function of the frequency out

in

21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e


2

ˆ 1 ˆˆ ˆˆ dd t ii t k N t a t G t

dt T

ˆ 1 1 ˆ ˆˆ ˆ 0, 0,

2 2g gi t i t

o R Lp p p p

v vda t ii a t k t b t e d t edt

Material polarization:

Cavity field equation:

1

ˆ ˆ 1 ˆˆ ˆˆ ˆ2 2d dN

d N t N t i k t a t k a t t F tdt T

Population equation:

1ˆ ˆˆ0, 0,

1ˆ ˆˆ0, 0,

i t i tR L

g p

i t i tL R

g p

d z t e a t d z t ev

b z t e a t b z t ev

Output field equations:

21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e



Transmission through a Strongly Coupled Cavity: Small Input Power

Possible states of the cavity system when the input power is very small:

2 1 10 1 0e e e



2

2

ˆ 1 ˆˆ ˆˆ

1 ˆˆˆ

dd t ii t k N t a t G t

dt T

ii t k a t G tT

ˆ 1 ˆˆ ˆ 0,

2g i t

o Lp p


So we can write:

21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e


ˆ i tt e

ˆ i ta t e

2

ˆ 1 ˆˆd t ii t k a t

dt T

ˆ 1 1 ˆ ˆ2 2

ˆ ˆ 0, 0,

op p

g gi t i tR L

p p

d a t ii a t k tdt

v vb t e d t e

Assume in steady state:

ˆ , i t i tR inb z t e e

21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e



22

22

gin

po

p

ii vT

ki iT

2

22

gin

po

p

ki v

ki iT

21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e



2

ˆ ˆ 1 ˆ ˆ ˆˆd a t t ii a t t k a t a t

d T

ˆ ˆ 1 1 ˆ ˆ ˆ ˆ2 2

ˆ ˆˆ ˆ 0, 0,

op p

g gi t i tR L

p p


v va t b t e a t d t e

* iine

0

Need:

1ˆ ˆ ˆ ˆ0, 0, i ti tR R

g pd z t e d z t e a t a t

v

21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e



21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e

Since a steady state does exist, and correlation function is going to be stationary, one can solve via Fourier transformation and obtain:

2

222

22

ˆ ˆ 2 giin

po

p

ivT

d a t a t eki i

T

Spectral density of the input:

2ˆ ˆ0, 0, 2i ti t iR R ind b z t e b z t e e

Spectral density of the output:

2ˆ ˆ0, 0, 2

1 ˆ ˆ

i ti t iR R out

i

g p

d d z t e d z t e e

d a t a t ev



21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e

Spectral density of the inputSpectral density of the output

=

2

222

22

1

po

p

iT

ki iT

2

2

21 1,2

1 122

0

VRp

p

o

kT

T

222

22 22 2

1 1o

po o

T

k k

The transmission spectrum is given by:



21

o ˆ , i tLb z t e

ˆ , i tRb z t e

ˆ , i tLd z t e

ˆ , i tRd z t e

2VRk

The transmission at goes to zero as T2 goes to ∞

The zero transmission is a result of interference in transmission via the two Rabi-split cavity resonances



Transmission through a Strongly Coupled Cavity: Photon Blockade


Spectrum of Eigenenergies: Rabi Splitting

21

Cavity

When the electron-photon interaction is turned on, we have the following eigenstates and eigenenergies:

The interaction lifts the degeneracy between the previously degenerate pair of states and splits their energies

This is called Rabi splitting and equals: 2R k n

1 1

2

2

2

2

2

2

Energies States

0

11

2 22 2

1 1

o

o

o

o

ε e

ε kε k

ε kε k



Transmission through a Strongly Coupled Cavity: First Photon Entry

21

21

Initial state Final state

0E oE k

o

k

Maximum transmission of the first incident photon if its energy is: oE k

So you choose your laser photon energy to be:

oE k

Abs

orpt

ion

spec

tra

o k Energy reqd:


21

Initial state

oE k

21

Final state

2 2oE k

o

Maximum transmission of the second incident photon if its energy is:

Transmission through a Strongly Coupled Cavity: Second Photon Entry

2 1o k

Energy reqd:

But you have already chosen your laser photon energy to be: oE k 2 1o k

k

2 1k

• Second photon is less likely to enter the cavity when the first photon is already present because the second photon is now detuned!

• Photons are more likely to enter the cavity one at a time

Abs

orpt

ion

spec

tra


Transmission through a Strongly Coupled Cavity: Anti-Bunching and Bunching

o k

o Bunched light


Decoherence in a Two-Level System: Heisenberg PictureConsider a two-level system with the Hamiltonian:

2211 ˆˆˆ NNH

1 2

2 1

2 1

ˆ ˆ0 0

ˆ ˆ

ˆ ˆ

dN t dN tdt dt

d t i tdt

d t i tdt

Suppose the system is not interacting with anything:

Now lets try to introduce pure decoherence that destroys the superpositon between the upper and lower levels:

2

2ˆˆˆ

ˆˆˆ

Ttti

dttd

Ttti

dttd

12

The decay terms destroy the quantum mechanical consistency of the equations!


12

t

Tittt

2

11ˆˆ

t

Tittt

2

11ˆˆ

tNtNttNttN

tT

tNtNttNttN

tT

tttttt

1212

21212

2

ˆˆˆˆ

21ˆˆˆˆ

21ˆ,ˆˆ,ˆ

Solve the equations to order t in time:

And find the equal-time commutator to order t in time:

Wrong!

Decoherence in a Two-Level System: Heisenberg Picture


Modeling Decoherence: Langevin Sources12

ˆ ˆG t G t

2

2

ˆ 1 ˆˆ ˆ

ˆ 1 ˆˆ ˆ

d t i t t G tdt T

d t i t t G tdt T

Add Langevin noise sources:

We make the assumptions:

a) Noise operators act in their own Hilbert space

b) System operators at time t commute with the noise operators at time t’ if t’ > t

c)

d)

e)

f)

ˆ ˆ 0G t G t

1 2 1 1 2ˆ ˆG t G t A t t t 1 2 1 1 2

ˆ ˆG t G t B t t t

1 2 1 1 1 2ˆ ˆ,G t G t A t B t t t

1 2 1 2ˆ ˆ ˆ ˆ 0G t G t G t G t

Assume now that: 1 2ˆ ˆ0dN t dN t

dt dt


Modeling Decoherence: Langevin Sources 12 The solution to order t in time is:

2

2

1 ˆˆ ˆ 1

1 ˆˆ ˆ 1

t t

t

t t

t

t t t i t G t dtT

t t t i t G t dtT

STEP 1: Multiply the second with the first and take the average: tttN ˆˆˆ 2

2 22

2 ˆ ˆˆ ˆ ˆ ˆ1

ˆ ˆ " "

t t t t

t tt t t t

t t

N t t N t t t G t dt G t dt tT

dt dt G t G t

22

2

22

2

22 2

2 2

ˆ 2 ˆ ˆˆ " "

ˆ 2 ˆ

ˆ2 2ˆ ˆ

t t t t

t t

dN tt N t t dt dt G t G t

dt T

dN tN t A t

dt T

dN tA t N t N t

T dt T


Modeling Decoherence: Langevin Sources 12

STEP 2: Multiply the second with the first and take the average:

tttN ˆˆˆ1

tNTdt

tNdtNT

tB 12

11

2ˆ2ˆˆ2

STEP 3: Take the commutator of the first and the second:

2

2 1 2 12

2ˆ ˆ ˆ ˆ, , 1

ˆ ˆ,

2ˆ ˆ ˆ ˆ 1

ˆ ˆ,

t t t tt t

t t t tt t

t t t t t t tT

dt dt G t G t

N t t N t t N t N t tT

dt dt G t G t

2 12 1

2 2

ˆ ˆ' ' 2 2ˆ ˆ ˆ ˆ' , " ' ' ' "dN t dN tG t G t N t N t t t

dt dt T T


Modeling Non-Radiative Transitions: Heisenberg Picture

Consider the Hamiltonian:

2211 ˆˆˆ NNH

Suppose we include non-radiative transitions:

1

21

122

ˆˆ

ˆˆ

TtN

dttNd

TtN

dttNd

12

1T

Check to see if the equations remain consistent:

122

122

1ˆˆ

ˆˆ

TttNttN

TtN

dttNd

122

22

21

22

22

21ˆˆ

ˆˆ21ˆˆ

TttNttN

tNtNT

ttNttNNow square both sides:

Not consistent!


12

1TTry fixing by adding Langevin noise sources:

tFT

tNdt

tNdNˆ

ˆˆ

122 tFtF NN ˆˆ

Since:

0ˆˆ 21 tNtNdtd

We must have:

tFT

tNdt

tNdNˆ

ˆˆ

121

The equations give:

111

22 ˆ1ˆˆ tFdtT

ttNttN Ntt

t

Now square both sides:

ttt NN

ttt

ttt N

tFtFdtdt

dttFT

ttNT

ttNttN

2121

111

21

22

ˆˆ

ˆ1ˆ221ˆˆ

1 2 1 1 2ˆ ˆN NF t F t C t t t



12

1T

ttt NN

ttt

ttt N

tFtFdtdt

dttFT

ttNT

ttNttN

2121

111

21

22

ˆˆ

ˆ1ˆ221ˆˆ

Take the average:

2 21

2ˆ ˆ 1 tN t t N t C t tT

2

1

ˆaverage relaxation rate

N tC t

T

The noise source therefore represents the shot noise associated with the non-radiative transitions

tFT

tNdt

tNdNˆ

ˆˆ

122

tFT

tNdt

tNdNˆ

ˆˆ

121



Cavity Quantum Dynamics: Complete Set of Equations


The Hamiltonian is:

The Heisenberg equations are:

2 2

1

1 2

1

ˆ ˆ ˆˆ ˆˆ ˆ

ˆ ˆ ˆˆ ˆˆ ˆ

N

N

d N t N t i k t a t k a t t F tdt T

dN t N t i k t a t k a t t F tdt T

2 12

2 12

ˆ 1 ˆˆ ˆˆˆ

ˆ 1 ˆˆ ˆ ˆˆ

d t ii t k a t N t N t G tdt T

d t ii t k N t N t a t G tdt T

21

Cavity

Waveguide

z = 0

ˆ , oi tLb z t e

ˆ , oi tRb z t e

1T


Field Equations:

ˆ 1 1 ˆˆ ˆ2

ˆ 1 1 ˆˆ ˆ2

o inp p

o inp p

da t ii a t k t S tdt

da t ii a t k t S tdt

1ˆ ˆ ˆˆ0, 0,

1ˆ ˆ ˆˆ0, 0,

o o

o o

i t i tout g R g L

p

i t i tout g R g L

p

S t v b z t e a t v b z t e

S t v b z t e a t v b z t e


21

Cavity

Waveguide

z = 0

ˆ , oi tLb z t e

ˆ , oi tRb z t e

1T



2

1

Cavity

Waveguide

z = 0

ˆ , oi tLb z t e

ˆ , oi tRb z t e

1TOptical pumping can be included in a similar fashion with a corresponding Langevin noise source: PT

The Heisenberg equations become:

2 1 2

1

1 1 2

1

ˆ ˆ ˆ ˆ ˆˆ ˆˆ ˆ

ˆ ˆ ˆ ˆ ˆˆ ˆˆ ˆ

N PP

N PP

d N t N t N t i k t a t k a t t F t F tdt T T

dN t N t N t i k t a t k a t t F t F tdt T T

ˆ ˆP PF t F t

1 2 1 1 2

1 11 average pumping rate

ˆ ˆ

ˆP P

P

F t F t D t t t

N tD t

T

ece 407 –spring 2009 –farhan rana –cornell university...5 ece 407 –spring 2009 –farhan...

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