ece 407 –spring 2009 –farhan rana –cornell universityece 407 –spring 2009 –farhan rana...

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ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Chapter 5: Quantization of Radiation in Cavity and Free Space

In this lecture you will learn:

• Classical Electromagnetism• Quantization of Radiation in Cavity• Quantization of Radiation in Free Space• Field Commutation Relations• Photon Number States• Photon Momentum • Photon Spin


Classical Electrodynamics: Maxwell’s Equations

4 ),(

)(),(),(

3 ),(

),(

2 ),(),()(

1 0),(

ttrE

rtrJtrH

t

trHtrE

trtrEr

trH

o

o

o

o

Maxwell’s Equations:

Cavity0),( trJ

Assume: (no interaction with matter)0),( tr

2

2

2

2

2

2

),(1),(

)(1

),()(),(

),(),()3(

t

trE

ctrE

r

t

trE

c

rtrE

ttrH

trE o

2


Vector and Scalar Potentials

),(),(

),( trt

trAtrE

Since:

Since: 0),()( trEro

0),()(),()( trrtrArt oo

0),( trHo

),(),( trAtrHo

trt

trAtrE

ttrA

trE

t

trHtrE o

,.

),(

0.

),(

),(),(

Since:


Gauge Transformations

),(),('),(

),(),('),(

trFt

trtr

trFtrAtrA

The fields and remain unchanged if one makes the following gauge transformations:

),( trE

),( trH

The choice of and is not unique),( trA

),( tr

One can impose an additional condition to make the potentials uniqueThis is called gauge fixing

),(),()(

),(),()(),()(),()(

trtrr

trtrrtrArt

trEr

o

ooo

Coulomb Gauge: 0),()( trAro

In coulomb gauge (assuming ):

Scalar potential is related to the charge density

0),( tr

3


Transverse and Longitudinal Fields

Electric field can be divided into two parts:

),(),(),( trEtrEtrE TL

where: 0),(. trEr To

trtrEr Lo ,),(.

Since: ),(),()( trtrEro

We must have:

Vector potential can also be divided into two parts:

),(),(),( trAtrAtrA TL

where: 0),(. trAr To

0),(. trAr Lo

Since in the coulomb gauge: 0),()( trAro

),(),( trAtrA T

(i.e. vector potential is entirely transverse in the coulomb gauge)


Transverse and Longitudinal Fields

),(),( trtrEL

),(),( trAt

trET

),(),(

),( trt

trAtrE

Recall that:

Assume coulomb gaugeSince: ),(),()( trtrEro

0),()( trAro

And:

),(),()(),()( trtrrtrEr oo

Therefore:

This means:

In the coulomb gauge: i) the gradient of the scalar potential represents the longitudinal component of the

electric field ii) the vector potential represents the transverse component of the electric field

Coulomb gauge condition

4


Radiation Modes in a Cavity

Cavity

In the coulomb gauge with:

),(),(

),(),(

trAtrH

ttrA

trE

o

0),( trJ

0),( tr

Maxwell’s Equations give:

2

2

2

),(1),(

)(1

t

trA

ctrA

r

We need to find the eignemodes of the operator: )(

1r

)()()(

12

2rU

crU

r nn

n

In other words, we need to solve the eigenvalue equation (subject to appropriate boundary conditions):

0),()( trAro

0)()( rUr no

eigenvalueeigenvector

Additional constrain on the solutions:


Radiation Modes in a Cavity

Cavity

Orthogonality of the eigenmodes:

mnnm rUrUrrd )()()(3

Normalization of the eigenmodes:

1)()( 3rdrUrU nn

mnnnm rUrUrrd )()()(3

Average permittivity seen by the mode

Field expansion in the eigenmodes:

n

nno

n rUtq

trA )()(

),(

Completeness of the eigenmodes:

The eigenmodes form a complete set in the space of all functions that are transverse and also satisfy the boundary conditions appropriate to the cavity

5


Cavity

Radiation Modes in a Cavity: Time Development

n

nno

n rUtq

trA )()(

),(

Plug the expansion in the wave equation:

m mo

mm

nn

n

no

n

mm

m

nn

no

n

rU

c

r

t

tqrUr

c

tq

rUt

tq

c

rrU

tq

t

trA

c

rtrA

)()()()()(

)(

)()()(

)()(

),()(),(

22

2

2

2

2

2

2

2

2

2

Multiplying both sides by , and integrate over all space:)(rU j

2

22 )(

)(t

tqtq j

jj

..)( cceqtqti

jjj


Radiation Modes in a Cavity: Hamiltonian

Cavity

The classical expression for the field energy is:

nmmn

m

m

no

n

o

mnmnmono

o

oo

rUrUtqtq

rUrUtqtqr

rd

trHtrHtrEtrErrdH

)()()()(1

)()()()()(

21

),(),(21

),(),()(21

0

3

3

2

2

2

23

3

)()()()()(

)()(

c

rUc

rrUrdrUrU

rUrUrd

mmnm

mm

nmn

mn

Note that:

mm

mm tqtq

H )(22

)( 222

Finally, one obtains:

6


Radiation Modes in a Cavity: Hamiltonian

Cavity

mm

mm tqtq

H )(22

)( 222

The Hamiltonian is:

)()( tqtp mm Let:

then:

mm

mm tqtp

H )(22

)( 222

)()(

)()( 2

tptqdtd

tqtpdtd

mm

mmm

The time-development is according to the equations:Compare to SHO


Quantization of Radiation in a Cavity

Cavityi)

ii) Postulate the equal-time commutation relation:

iii) For different modes postulate:

)(ˆ)( tqtq mm

)(ˆ)( tptp mm

itptq mm )(ˆ),(ˆ

0)(ˆ),(ˆ tptq nm

Quantization of the radiation is done in the following steps:

m

mmm tqtpH

2

ˆ

2

ˆˆ222

The Hamiltonian operator is:

7


Creation and Destruction Operators

Cavity m

mmm tqtpH

2

ˆ

2

ˆˆ222

The Hamiltonian operator is:

tpitqta

tpitqta

mmmm

m

mmmm

m

ˆˆ2

1

ˆˆ2

1ˆ

Define creation and destruction operators for each radiation mode as:

It follows that:

mnnm tata ˆ,ˆ

mmmm tataH

21

ˆˆˆ

Or in the Schrodinger picture:

mmm

mmmm naaH

21

ˆ21

ˆˆˆ


Energy Eigenstates and Eigenvalues

Cavity

Consider the Hamiltonian for a single radiation mode:

21

ˆˆmmm aaH

The eigenstates are:mn

mmmm nnnaa ˆˆ

mmm nnna 11ˆ mmm nnna 1ˆ

mmm nnnH

21ˆ

The eigenenergies are:

,.........21

3,21

2,21

,21

mmmmmmm

The ground state has energym0 m2

1

m

nm

m n

an 0

!

)ˆ(

8


Energy Eigenstates and Eigenvalues

Cavity

The Photon:

A photon is not a wave, not a particle, it is the smallest value of energy that can be taken away or added to a radiation field

The state represents a state with ‘n’ photons

The ground state has no photons (but still has energy!?!)

mn

m0

Photon number operator: mmm aan ˆˆˆ

mmm nnnn ˆ

Photon creation and destruction operators:

mmm nnna 11ˆ mmm nnna 1ˆ


Multimode StatesA multimode quantum state with 3 photons in mode m , 4 photons in mode n, 5 photons in mode p, is written as:

pnm 5|4|3||

|5|ˆ

|4|ˆ

|3|ˆ

p

n

m

n

n

n

Or:

5,4,3|| pnm nnn

...........0|0|0|0| 321

The ground state is the ground state of all the modes:

..............0|0|3|0|| 4321

A state with 3 photons in mode 2 can be written in two different ways:

23||

9


Multimode States

A state with one photon in mode m and one in mode n is:

1,1|1|1|0|| nmnmnm nnaa

1|1|2

11,0|0,1|

2

1

1|0|0|1|2

10|0|

2

1|

nmnmnm

nmnmnm

nnnnnn

aa

A state which is a superposition of one photon in mode m and one in mode n is:

Completeness of the photon number states:

0 0321321

1 2

1,,,,,,n n

nnnnnn

Or when working in a reduced Hilbert space:

0 01or1

n nmmmm

m

nnnn

Orthogonality of the photon number states:

ppnmnm pp


Time Development

mti

mti

m

mmmm

ae

taeta

taHtadt

tadi

m

m

ˆ

0ˆˆ

ˆˆ,ˆˆ

Time development of creation and destruction operators follow from the Heisenberg equation:

mti

mti

m

mmmm

ae

taeta

taHtadt

tadi

m

m

ˆ

0ˆˆ

ˆˆ,ˆˆ

And:

10


Field Operators

mmmm

mom

nn

no

n

rUtata

rUtq

trA

)(ˆˆ2

)()(ˆ

),(ˆ

mmmm

mo

m rUtatai

t,trA

,trE

)(ˆˆ2

)(ˆ)(ˆ

mmmm

nomo

o

rUtata

,trA,trH

)()(ˆ)(ˆ2

1

)(ˆ1)(ˆ

The fields are represented by Hermitian operators:

tpitqa

tpitqta

mmmm

m

mmmm

m

ˆˆ2

1

ˆˆ2

1ˆ


Field Operators

Consider a state with a large number of photons in mode m:

mmn 99 1010

Lets find the expectation value of the electric field:

0

ˆˆ2

)(ˆ)(ˆ2

)(ˆ

rUeaeai

rUtatai,trE

nn

tin

tin

no

n

nn

nnno

n

nn

Clearly all photon number states will result in a zero expectation value for all the fields

11


Vacuum Energy

The ground state of a mode m has energy m0 2m

mm

mH 02

0ˆ

This means:

mmom

mmm

trHtrHtrEtrErrd

H

21

0),(),(21

),(),()(21

0

21

0ˆ0

03

The total vacuum energy is then:

m

m

2

And is likely very large (perhaps infinite!)……….is that a problem?


Classic Electrodynamics in Free Space

t

trEtrH

ttrH

trE

trE

trH

o

o

o

o

,,

,,

0,.

0,.

Maxwell’s equation in free space are:

The above equations give:

2

2

22

2

2

22

2

2

2

,1,

,1,,.

,1,

t

trE

ctrE

t

trE

ctrEtrE

t

trE

ctrE

12


Electrodynamics in Free Space in the Coulomb Gauge

The fields are:

trAtrH

ttrA

trE

o ,,

,,

0, trA

Coulomb gauge

The wave equation is: 2

2

22 ,1

,t

trA

ctrA

We need to find the eignemodes of the operator:2

)()(2

22 rU

crU k

kk

In other words, we need to solve the eigenvalue equation (subject to appropriate boundary conditions):

0),( trA 0)(. rUk

eigenvalueeigenvector

Additional constrain on the solutions:


Free-Space Eigenmodes

V

ekrU

rki

k

)( V

Universe

The eigenmodes are:

Unit vector indicating field polarization

The eigenmodes are normalized in a very large box (universe) of volume V

All physical results should be independent of the volume V

To find the eigenvalue, plug the eigenmode into the wave equation:

)()(2

22 rU

crU k

kk

kc

kcc

kk

k

k

k

2

2

.

Change of notation: frequency only depends on the magnitude of k

Eigenvalues:

13


Free-Space Eigenmodes: Polarization

V

Universe

V

ekrU

rki

k

)(

0)(. rUk

The condition implies:

0)(ˆ. kk

k

)(ˆ1 k

)(ˆ2 k

There are two independent orthogonal directions for each : k

V

ekrU

rki

jkj

)(ˆ,

Add one more label to the eigenmodes for each polarization direction:


Free-Space Eigenmodes

V

Universe

V

ekrU

rki

jkj

)(ˆ,

Orthogonality and normalization of the eigenmodes:

kkj

rkirki

jkkj V

ek

V

ekrdrUrUrd

,'

'3

,*

',3 )(ˆ.)'(ˆ.

Expansion of fields in the eigenmodes:

k j

j

rki

o

j

k jkj

o

j kεV

e

ε

tkqrU

ε

tkqtrA

2

1

.2

1, ˆ

,,,

trAtrA ,, * ktkqktkq jjjj

ˆ,ˆ,

A convenient choice is:

kk

tkqtkq

jj

jj

ˆˆ

,,

14


Periodic Boundary Conditions

V

Periodic boundary conditions:

L

L

L

Lxkixki xx ee L

nkx

2 ,3,2,1,0 n

Lykiyki yy ee

L

mky

2 ,3,2,1,0 m

Lzkizki zz ee L

pkz

2 ,3,2,1,0 p

kd

V

k

332

There is one allowed wavevector in a k-space volume of VL

3

3

3 22

There are allowed wavevectors per unit volume of k-space 32

V

One can convert summations over allowed wavevectors into integrals:

2

1

.

3

3ˆ

,

2,

jj

rki

o

j kεV

e

ε

tkqkdVtrA

Therefore our field expansion in the eignemodes becomes:


Some Useful Relations

'

2

'2

3'.3

3

33.'3

rrekd

kkerd

rrki

rkki

''33 rfrfrrrd

Delta functions:

''33 kgkgkkkd

Plane wave integrations:

More delta functions:

kd

V

k

332

'2

'2

'

33

',

',3

3

',

kkV

kgkgkdV

kgkg

kk

kk

kkk

15


Free-Space Eigenmodes: Completeness

V

Universe

V

ekrU

rki

jkj

)(ˆ,

Completeness of the eigenmodes:

'ˆˆ1

2

)(ˆ)(ˆ2

)(ˆ)(ˆ'

'

3

3

2

1

'

3

3

2

1

'2

1

*,,

rrV

ekk

kd

Ve

kkkd

V

ek

V

ekrUrU

rrki

j

rrki

jj

k j

rki

j

rki

jk j

kjkj

Take the dot product from left and right sides with Cartesian unit vectors and :ae be

'

2ˆ.'.ˆˆ.'.ˆ

'

23

32

1

*,,

rr

Ve

k

kkkderreerUrUe

ab

rrkiba

abbak j

bkjkja

Transverse delta function


Expansion of Fields in the Eigenmodes

2

1

.

3

3ˆ

,

2

,,

jj

rki

o

j kεV

e

ε

tkqkdV

ttrA

trE

2

1

.

3

3ˆ

,

2,

jj

rki

o

j kεV

e

ε

tkqkdVtrA

2

1

.

3

3ˆ

,

2

,,

jj

rki

oo

j

okεki

V

e

ε

tkqkdV

trAtrH

The fields can be written as:

All fields are real:

trHtrH

trEtrE

trAtrA

,,

,,

,,

*

*

*

16


Free-Space Radiation Modes: Time Development

2

1

.

3

3ˆ

,

2,

jj

rki

o

j kεV

e

ε

tkqkdVtrA

Plug the above into the wave equation:

2

2

22 ,1

,t

trA

ctrA

to get:

2

1

.

3

32

1

.2

3

3ˆ

,

2ˆ

,

2 ss

rki

o

s

ss

rki

o

s kεV

e

ε

tkqkdVkε

V

e

ε

tkqk

kdV

Multiply on both sides with and integrate on both sides to get: kεV

ej

rki

ˆ.

tkq

tkqckdt

tkqd

jk

jj

,

,,

2

222

2


Field Energy

Ve

tkqtkqkεkikεkic

Ve

kεkεtkqtkqkd

Vkd

Vrd

trHtrHtrEtrEε

rdH

rkki

srsr

r s

rkki

srsr

oo

.'2

.'

3

3

3

33

3

,','ˆ.ˆ2

'ˆ.ˆ,',21

2

'

2

,.,2

,.,2

1ˆ.ˆ s

sr kk

2ˆˆ kkεkikεki ss

r

Integrate over all space and note the following identities:

Field energy is:

CBDADBCA

DCBA

....

.

to get:

j

jjk

jj tkqtkqtkqtkqkdVH

2

,,

2

,,

2

23

3

17


Field Energy

j

jjk

jj tkqtkqtkqtkqkdVH

2

,,

2

,,

2

23

3

Define:

tkqtkp jj ,,

j

jjk

jj tkqtkqtkptkpkdVH

2

,,

2

,,

2

23

3

Time development equations:

,tkpt

,tkq

,tkqωt

,tkp

jj

jkj

2 tkptkp

tkqtkq

jj

jj

,,

,,

Not exactly like a traditional SHO – the variables are complex


Field Momentum

The classical expression for the momentum of the electromagnetic field is:

trHtrErdP oo ,,3

jjj tkqtkqki

kdVP ,,

2 3

3

Using:

2

1

.

3

3ˆ

,

2

,,

jj

rki

o

j kεV

e

ε

tkqkdV

ttrA

trE

2

1

.

3

3ˆ

,

2

,,

jj

rki

oo

j

okεki

V

e

ε

tkqkdV

trAtrH

One gets:

s

sr kikεkikε

ˆˆ

18


Field Angular Momentum

The classical expression for the angular momentum of the electromagnetic field around the reference point is:

trHtrErrrdJ ooo ,,3

trHtrErrdJ oo ,,3

or

If the reference point is chosen to be the origin then:


Quantization of Radiation in Free Space

),(ˆ),( tkqtkq jj

),(ˆ),( tkptkp jj

First step in field quantization is the promotion of observables to operators:

tkptkp

tkqtkq

jj

jj

,,

,,

But these classical variables obey:

tkptkp

tkqtkq

jj

jj

,ˆ,ˆ

,ˆ,ˆ

2

1

.

3

3ˆ

,

2,

jj

rki

o

j kεV

e

ε

tkqkdVtrA

2

1

.

3

3ˆ

,ˆ

2,ˆ

jj

rki

o

j kεV

e

ε

tkqkdVtrA

These ensure that the field operators are Hermitian

Second step in field quantization is the imposition of commutation relations:

Possible choices are:

',

',

,',,ˆ

,',,ˆ

kkrssr

kkrssr

itkptkq

itkptkq

Only this one works!

19


Creation and Destruction Operators

',,',,ˆ kkrssr itkptkq

We impose:

But then:

',,',,ˆ,',,ˆ kkrssrsr itkptkqtkptkq

We choose creation and destruction operators as follows:

,tkpi,tkq,tka

,tkpi,tkq,tka

jjkk

j

jjkk

j

ˆˆ2

1ˆ

ˆˆ2

1ˆ

It follows that:

0'ˆˆ

'ˆˆ ',

,tka,,tka

,tka,,tka

sr

kkrssr


Field Hamiltonian

21

,ˆ,ˆ2

ˆ

,ˆ,ˆ,ˆ,ˆ22

,ˆ,ˆ,ˆ,ˆ

,ˆ,ˆ,ˆ,ˆ42

2

,,ˆ

2

,ˆ,ˆ

2ˆ

3

3

3

3

3

3

23

3

tkatkakd

VH

tkatkatkatkakd

V

tkatkatkatka

tkatkatkatkakd

V

tkqtkqtkptkpkdVH

jjj

k

jjjjj

k

jjjj

jjjjj

k

j

jjk

jj

The Hamiltonian becomes:

The vacuum energy is:

jk

k kdV

kdV

3

3

3

3

222Infinite!

20


Time Development of Creation and Destruction Operators

kaetkaetka

tkaHtkadt

tkadi

rti

rti

r

rkrr

kk

ˆ0,ˆ,ˆ

,ˆˆ,,ˆ,ˆ

Time development follows from the Heisenberg equation:

kaetkaetka

tkaHtkadt

tkadi

rti

rti

r

rkrr

kk

ˆ0,ˆ,ˆ

,ˆˆ,,ˆ,ˆ

And:


Field Operators

kεV

etkatka

εkd

VtrA j

rki

jjj

ok

ˆ,ˆ,ˆ22

,ˆ.

3

3

kε

V

etkatka

εi

kdV

ttrA

trE j

rki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ,ˆ

.

3

3

kεkiV

etkatka

εkd

VtrA

trH j

rki

jjj

okoo

ˆ,ˆ,ˆ2

1

2

,ˆ,ˆ

.

3

3

The field operators are:

Energy eigenstates:

jkkjk nnnH ,, 21ˆ

The photon number states are also energy eigenstates:

21


Energy Eigenstates

In the Schrodinger picture:

21

ˆˆ2

ˆ3

3kaka

kdVH jj

jk

The photon number states, defined as:

0

!

ˆ, n

kan

nj

jk

are also energy eigenstates:

jkkjk nnnH ,, 21ˆ


Field Momentum and Photon Momentum

The field momentum operator is:

jjjoo tkqtkpki

kdVtrHtrErdP ,ˆ,ˆ

2,ˆ,ˆˆ

3

33

jjjjj tkatkatkatka

kkdVP ,ˆ,ˆ,ˆ,ˆ

22ˆ

3

3

This becomes:

jjj

jjj

jjjjj

jjjjj

tkatkakkd

V

tkatkakkd

V

tkatkatkatkakkd

V

tkatkatkatkakkd

VP

,ˆ,ˆ2

21

,ˆ,ˆ2

,ˆ,ˆ,ˆ,ˆ22

,ˆ,ˆ,ˆ,ˆ22

ˆ

3

3

3

3

3

3

3

3

Keeping only the non-zero terms:

22


Field Momentum and Photon Momentum

jjj tkatkak

kdVP ,ˆ,ˆ

2ˆ

3

3

The field momentum operator is:

The field energy eigenstates (the photon number states) are also field momentum eigenstates:

,,3

3

, 11ˆˆ2

1ˆqq

jjjq qkakak

kdVP

,,3

3

,ˆˆ

2ˆ

qqj

jjq nqnnkakakkd

VnP

A photon of wavevector carries a momentum equal to q q

In the Schrodinger picture:

jjj kakak

kdVP

ˆˆ

2ˆ

3

3


Spin and Helicity of Particles: Classical Relativistic Physics

The total angular momentum of a particle consists of an orbital part and an intrinsic part which is called the spin:

The orbital part,

has no component along the direction of momentum, i.e. .

J L S

L r p

. 0L p

So we define helicity as the projection of the total angular momentum along the direction of momentum as,

.J pH

p

Helicity therefore must necessarily be related to the intrinsic angular momentum of particles

23


Spin, Helicity, Chirality and all thatMassive Particles Massless Particles

Spin angular momentum is conserved in timeSpin angular momentum is not Lorentz invariant

Helicity: Projection of the angular momentum along the momentum direction,

Helicity is conserved in timeHelicity is not Lorentz invariant

Chirality: Depends on whether the particle state transforms as the left or right handed representation of the Poincare group

Chirality is not conserved in timeChirality is Lorentz invariant

.J pH

p

Helicity: Projection of the angular momentum along the momentum direction,

Helicity is conserved in timeHelicity is Lorentz invariant

Chirality: Depends on whether the particle state transforms as the left or right handed representation of the Poincare group

Chirality is conserved in timeChirality is Lorentz invariant

Chirality and helicity and spin angular momentum are all the same thing

.J pH

p

J L S

J L S


Spin, Helicity, Chirality and all thatMassive Particles Massless Particles

Helicity eigenstates for spin S particle can have the following helicity values:

-S, S

(S integer or half-integer)

The spin eigenstates follow from the 1-dimensional representations of the rotation group SO2 with an additional topological constrain

Helicity eigenstates for spin S particle can have the following helicity values:

-S, -S+1, -S+2, ……., 0, …….. S-2, S-1, S

(S integer or half-integer)

The spin eigenstates follow from the (2S+1)-dimensional representations of the rotation group SO3

24


Field Angular Momentum

The classical expression for the angular momentum of the electromagnetic field is:

trHtrErrdJ oo ,,3

The above expression contains:i) Orbital angular momentumii) Intrinsic or spin angular momentum

These are not separately conserved

If one looks at plane waves, which will not carry any orbital angular momentum, then their angular momentum will tell us something about the intrinsic angular momentum of the field

sssoo

oo

oo

o

oo

trArtrErdtrAtrErd

trAtrErrdtrArtrErd

trAtrErrdtrArtrErd

trAtrErrd

trHtrErrdJ

,,,,

,,.,.,

,,.,.,

,,

,,ˆ

33

33

33

3

3

This part is zero for plane waves


Photon Spin

So for a plane wave the intrinsic angular momentum is given by:

trAtrErdS o ,,ˆ 3

Upon substituting the eigenmode expansions for the field operators:

tkatkatkatkakk

ikdVS

jjjj ,ˆ,ˆ,ˆ,ˆˆˆ

22

ˆ

,3

3

Assume we have chosen the polarization unit vectors such that:

kkk

kkk

kk jj

ˆˆˆ

ˆˆˆ

ˆˆ

21

21

tkatkatkatkaki

kdVS ,ˆ,ˆ,ˆ,ˆˆ

2

ˆ21123

3

We obtain:

k

)(ˆ1 k

)(ˆ2 k

25


Photon Spin

kakakakaki

kdVS

21123

3ˆˆˆˆˆ

2

ˆ

k

)(ˆ1 k

)(ˆ2 k

tkatka ,ˆ,ˆ 12

A product like does not seem to be a number operator

What are the eigenstates of ?(Clearly eigenstates of a definite linear polarization are not eigenstates of )

ka

1ˆi) The operator creates a photon with wavevector and polarization )(ˆ1 k

k

ii) The operator creates a photon with wavevector and polarization

ka

2ˆ)(ˆ2 k

k

Recall that:

2

,ˆ,ˆ,ˆ

2

,ˆ,ˆ,ˆ

21

21

tkaitkatka

tkaitkatka

L

R

2

,ˆ,ˆ,ˆ

2

,ˆ,ˆ,ˆ

21

21

tkaitkatka

tkaitkatka

L

R

Define two new creation and destruction operators:

S

S


Photon Spin

2

,ˆ,ˆ,ˆ

2

,ˆ,ˆ,ˆ

21

21

tkaitkatka

tkaitkatka

L

R

2

,ˆ,ˆ,ˆ

2

,ˆ,ˆ,ˆ

21

21

tkaitkatka

tkaitkatka

L

R

2,1,21 11

21

02

ˆˆ0ˆ

kkR ikaika

ka

The operators and create photons in superposition of polarizations with ±90-degrees phase shift

kaR

ˆ kaL

ˆ

A photon with a right-hand circular polarization state

0,'ˆ,,ˆ,'ˆ,,ˆ

,'ˆ,,ˆ,'ˆ,,ˆ ',',

tkatkatkatka

tkatkatkatka

RLLR

kkLLkkRR

Equal-time commutation relations:

k

)(ˆ1 k

)(ˆ2 k

26


Photon Spin

kakakakaki

kdVS

21123

3ˆˆˆˆˆ

2

ˆ

kakakakak

kdVS LLRR

ˆˆˆˆˆ

2

ˆ3

3

Consider the photon in a right-hand circular polarization state:

2,1,21

, 1121

02

ˆˆ0ˆ1 kkRRk i

kaikaka

Lets see if this state is an eigenstate of : S

RkRk kS ,, 1ˆ1ˆ

Consider the photon in a left-hand circular polarization state:

2,1,21

, 1121

02

ˆˆ0ˆ1 kkLLk i

kaikaka

LkLk kS ,, 1ˆ1ˆ


Photon Spin

RkRk kS ,, 1ˆ1ˆ

LkLk kS ,, 1ˆ1ˆ

The operator represents the intrinsic, or spin angular, momentum of a photonS

i) The direction of spin is always the direction of wave propagation

ii) The magnitude of spin for a single photon state is 1 in units of

iii) The sign of spin for a single photon state can be +1 or -1 for right or left circularly polarized photons

27


Particle Position and Relativistic Quantum MechanicsQuestion: How does one localize a particle?Answer: Try superposing all possible momentum eigenstates

Question: Would the above recipe work without violating Lorentz invariance

.33

3 22

p rid p

r d p p p r e p

Suppose:

ˆ ˆ

3

.3

3 2

0

0

2

o

o

H Hi t i t

o

p r r E pi i t

t r

t e t e d p p p r

d pr t e e

Then:

2 4 2 2E p m c p c

r t

Turns out that is non-zero even when , i.e outside the light cone

or r ct

Particles cannot be localized. There is no position eigenket. Position is not an observable!


Photon Position

Can one define the position of a photon?Can a photon be even localized?

Consider the single photon state:

0ˆ1 11, kak

The photon is created in a “plane wave state” and is therefore spread out in space

What if we create a photon in a superposition of plane wave states to make it more localized at location ?

kaV

eek

kdVr j

rki

ajj

a

ˆˆ.ˆ2

.2

13

3

r

raDoes create a photon polarized in the direction at location ?r

ae

0ra

28


Photon Position

0'' rr ba

Try to destroy that photon at another location and in another polarization and see what happens

0'rr ba

If we have managed to destroy the photon we should get the vacuum back again:

3ˆ ˆ0 ' 0 . 'b b a br r e e r r

Lets see what we get:

' '2 2

1 1'

' '2 2

1 1'

ˆ ˆ ˆ ˆˆ ˆ0 ' 0 . ( ) ( '). 0 ' 0

ˆ ˆˆ ˆ . ( ) ( ').

ik r ik r

a b a j b jjk k

ik r ik r

a j bjk k

e er r e k k e a k a k

V V

e ee k k e

V V

, '

'2

1

'3

3 2

ˆ ˆ0 ' 0

ˆ ˆˆ ˆ . ( ) ( ).

2

j jk k

ik r ik r

a j j bjk

ik r ra b

ab

a k a k

e ee k k e

V V

k kd k e

Vk

'ab r r

Not very localized

3ˆ ˆ' . 'a a bbr r e e r r

?

j

baabbsaj

k

kkekekε

2ˆ.ˆ.ˆ


The Transverse Delta Function

'3

3 2

'3

3 2

ˆ' 12

ˆ ˆ' . ' .

2

'

ik r r

ab a b

ik r ra b

ab

ab

d k k k er r

Vk

r r e r r e

k kd k e

Vk

r r

33 2

ˆ ˆ. .1' ' 3

4 ' '

a bab ab ab

r r e r r er r r r

r r r r

3', ' . ',TF r t d r r r F r t

29


kεV

etkatka

εkd

VtrA j

rki

jjj

ok

ˆ,ˆ,ˆ22

,ˆ.

3

3

kε

V

etkatka

εi

kdV

ttrA

trE j

rki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ,ˆ

.

3

3

The field operators are:

Equal-Time Field Commutation Relations

ˆ ˆˆ ˆˆ ˆ, , . , , .a a a bE r t E r t e A r t A r t e Let:

3 2. '3

1

3'

3 2

ˆ ˆ ˆ ˆ ˆ ˆ, , ', . .2

2

'

i k r ra o b a j j b

j

ik r ra bab

ab

d kA r t E r t i e e ε k ε k e

k kd ki e

k

i r r

The equal-time commutation relation is:

Fields don’t really present independent degrees of freedom at different spatial distances closer than the smallest size of a photon!


Equal-Time Field Commutation Relations

kεV

etkatka

εi

kdV

ttrA

trE j

rki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ,ˆ

.

3

3

kεkiV

etkatka

εkd

VtrA

trH j

rki

jjj

okoo

ˆ,ˆ,ˆ2

1

2

,ˆ,ˆ

.

3

3

ba

aa

etrHtrH

etrEtrE

ˆ.,ˆ,ˆ

ˆ.,ˆ,ˆ

Let:

Then:

'

2,'ˆ,,ˆ

32

'.3

32

rrci

ekd

citrHtrE

cabcc

rrkicabc

cba

30


Heisenberg Equations for Field Operators

One can now find the Heisenberg equations for the electric and magnetic field operators:

ˆ , ˆ ˆˆ, , ,

ˆ , ˆ ˆˆ, , ,

o

o

E r t ii E r t H H r t

t

H r t ii H r t H E r t

t

Have you seen these equations before??


Unequal-Time Field Commutation Relations

','ˆ,,ˆ trEtrE ba

What does this commutator tell us?

Such commutation relation show whether accurate simultaneous measurements on fields at different locations and times are possible

Field operators must commute for space-like intervals, i.e. when:

222 'ttcrr

''''

'4

','ˆ,,ˆ

22

2ttcrrttcrr

rrci

ct

trEtrE

obaab

ba

Commutator is non-zero only on the light cone, i.e. when:What does that mean?

'ttcrr

31


Radiation in a Cavity

Cavity

mmmm

momo

mmmm

mo

n

mmmm

mom

rUtatatrH

rUtataitrE

rUtatatrA

)()(ˆ)(ˆ2

1),(ˆ

)()(ˆ)(ˆ2

),(ˆ

)()(ˆ)(ˆ2

),(ˆ

Field operators are:

We consider only one mode for simplicity:

)()(ˆ)(ˆ2

1),(ˆ

)()(ˆ)(ˆ2

),(ˆ

)()(ˆ)(ˆ2

),(ˆ

rUtatatrH

rUtataitrE

rUtatatrA

ooo

o

o

oo

21

ˆˆˆ aaH o


Photon Number States

nnnaann

n

an

n

ˆˆˆ

0!

)(

Photon number states are defined as:

These are eigenstates of the photon number operator

nnnH o

21ˆ

11ˆ

1ˆ

nnna

nnna

0|ˆ||ˆ| nannan

Since:

0),(ˆ),(ˆ ntrHnntrEn

Average values of fields in number states are zero!

32


What Quantum States are Generated by Oscillating Currents?

Cavity

Maxwell’s Equations:

t

trHtrE

ttrE

rtrJtrH

o

o

,,

,,,

Hamiltonian Description:

rAtrJrdaatH o

ˆ.,ˆˆˆ 3

trStrJ ocos,

Schrodinger Picture

)(ˆˆ2

)(ˆ rUaarAoo

taaJaatH ooo cosˆˆˆˆˆ )(.2

3 rUrSrdJoo

o

Initial State:

00 t



Cavity

taaJaatH ooo cosˆˆˆˆˆ

Start from the Schrodinger equation:

ttHt

ti

ˆ

aaH oo ˆˆˆ

Let:

t

tietH

t

ti

tet

tHi

o

tHi

o

o

ˆ

ˆ

ˆ

33



Cavity

0

ˆˆ2

cosˆˆ

cosˆˆ

cosˆˆˆˆ

ˆ

ˆˆ

ˆˆ

ˆ

* atat

iio

otiti

o

tHi

oo

tHi

ooo

tHi

o

et

teaeaJ

t

ti

tteaeaJt

ti

tetaaJet

ti

ttaaJHt

tietH

t

ti

ttHt

ti

oo

oo

o

We get:

teJ

it io 2

Ignore non-resonant terms


0

0

0

0

0

ˆˆ

ˆˆ

ˆˆ

ˆˆˆˆ

ˆ

ˆˆˆ

ˆ

*

*

*

*

*

atat

eateat

tattat

tHi

tHi

atattHi

atattHi

tHi

e

e

e

eeee

ee

tet

toitoi

ooo

o

o


Cavity

The state of the radiation is then:

tio oteJ

it2

What is this state??

34


ece 407 –spring 2009 –farhan rana –cornell universityece 407 –spring 2009 –farhan rana...

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