chapter 5: quantization of radiation in cavities and …chapter 5: quantization of radiation in...

Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)

1

Chapter 5: Quantization of Radiation in Cavities and Free Space

5.1 Classical Electrodynamics

5.1.1 Classical Cavity Electrodynamics

Maxwell’s equations for electromagnetism are,

4 ),(

)(),(),(

3 ),(

),(

2 ),(),()(

1 0),(

t

trErtrJtrH

t

trHtrE

trtrEr

trH

o

o

o

Here, r is the relative dielectric constant. Consider fields inside a cavity. We need to find the

eigenmodes of the radiation inside a cavity. The cavity is assumed to be absolutely lossless. We also

assume that 0),( trJ

and 0),( tr

.

From Maxwell’s equations one can drive the wave equation,

2

2

2

2

2

2

),(1),(

)(

1

),()(),(

),(),()3(

t

trE

ctrE

r

t

trE

c

rtrE

t

trHtrE o

The above equation can be solved to obtain the confined radiation modes, and their frequencies, inside

the cavity. It is easier and better to work with the vector and scalar potentials, ),( trA

and ),( tr

,

instead of ),( trE

and ),( trH

fields. These potentials are introduced below.

Cavity


2

5.1.2 Vector and Scalar Potentials

We introduce vector and scalar potentials ),( trA

and ),( tr

to describe ),( trE

and ),( trH

fields,

),(),(

),(),(

),(

trAtrH

trt

trAtrE

o

Since 0),()( trEro

(no free charges),

0),()(),()(

trrtrArt oo

5.1.3 Gauge Transformations and Gauge Fixing

The fields ),( trE

and ),( trH

don’t change if we make the following transformations,

),(),('),(

),(),('),(

trFt

trtr

trFtrAtrA

where ),( trF

is any scalar function. So there is some degree of freedom in the choice of ),( trA

and

),( tr

. In other words, the choice of ),( trA

and ),( tr

is not unique. One can impose an additional

condition on the potentials to make them unique. In non‐relativistic electrodynamics, one imposes the

additional condition 0),()( trAro

. Introduction of a condition like this is called “gauge fixing”,

and this particular choice of gauge is called the coulomb gauge. In the coulomb gauge,

),(),()(

),(),()(),()(),()( 0

trtrr

trtrrtrArot

trEr

o

oo

The electric field can be divided into two parts,

),(),(),( trEtrEtrE TL

In the coulomb gauge,,

),(),( trtrEL

),(),( trAt

trET

It follows that,


3

),(),()( trtrEr Lo

0),()( trEr To

),( trEL

is called the longitudinal electric field and its source is the charge density ),( tr

(whether

static or time dependent). ),( trET

is called the transverse electric field and its source is the time‐

dependent current density ),( trJ

. The vector potential can also be divided into two parts,

),(),(),( trAtrAtrA TL

In the coulomb gauge 0),( trAL

and the vector potential is entirely transverse. In non‐relativistic

quantum electrodynamics, ),( trET

is associated with “photons”. In this Chapter, since ),( tr

will

almost always be zero, ),( trEL

will also be zero, and ),( trE

should be understood to be ),( trET

. In

the coulomb gauge (with 0),( tr

and 0),( trJ

), we have,

),(),(

),(),(

trAtrH

t

trAtrE

o

Maxwell’s equations give,

2

2

2

),(1),(

)(

1

t

trA

ctrA

r

Note that ),( trE

and ),( trA

satisfy the same wave equations.

5.1.4 Cavity Eigenmodes and Eigenvalues

To proceed further, we need to find the eigenmodes of the operator that appears in the wave equation,

)(

1

r

Let these eigenmodes be )(rUn and let the corresponding eigenvalues be 22 cn . Therefore,

)()()(

12

2rU

crU

r nn

n

These eigenfunctions must also satisfy the requirement that 0)()( rUr no

, which follows from

the coulomb gauge condition, 0),()( trAro

. We will assume that the eignemodes can be chosen

to be completely real functions.


4

Orthogonality Relation for the Eigenmodes: Consider two different eigenmodes, )(rUn and )(rUm

.

These satisfy,

)()()(

12

2rU

crU

r nn

n

)()()(

12

2rU

crU

r mm

m

Take the dot product of both sides of the first equation with )(rUm to get,

)()()()()(2

2rUrUr

crUrU nm

nnm

Using the vector identity,

)()()()( CACACA

and integrating over all space we get,

)()()()()( 32

23 rUrUrrd

crUrUrd nm

nnm

Integrate by parts twice on the left hand side to obtain,

0)()()(

)()()()())((

32

2

2

2

2

233

rUrUrrdcc

rUrUrc

rdrUrUrd

nmnm

nmm

nm

Therefore, if nm then,

0)()()(3 rUrUrrd nm

(5)

Therefore, eigenvectors corresponding to different eigenvalues are orthogonal in the sense of (5) above.

If )(rUnis normalized such that,

1)()( 3rdrUrU nn

then,

nmnnm rUrUrrd )()()(3

where n is the average relative dielectric constant seen by the mode )(rUn.


5

5.1.5 Field Expansion using Eigenmodes

Since the eigenmodes form a complete basis set in the space of all transverse vector fields, one can

always expand any transverse vector field, and in particular ),( trA

, in terms of these eigenmodes,

n

nno

n rUtq

trA )()(

),(

Substituting the above expansion in the wave equation we get,

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 )(rU j

, and integrate over all space, we get,

2

22 )(

)(t

tqtq

jjj

The above equation has the solution,

ttqttqtq j

j

jjjj

sin

0cos0)(

Or, equivalently, in complex time‐harmonic notation,

..)( cceqtqti

jjj

5.1.6 Field Hamiltonian

The classical expression for the field energy is,

nmmn

m

m

no

n

omnmn

mono

o

oo

rUrUtqtq

rUrUtqtqr

rd

trHtrHtrEtrErrdH

)()()()(1

)()()()()(

2

1

),(),(2

1),(),()(

2

1

0

3

3

Note that,


6

2

2

2

23

3

)()()()()(

)()(

c

rUc

rrUrdrUrU

rUrUrd

mmnm

mm

nmn

mn

Therefore,

mm

mm tqtq

H )(22

)( 222

If one defines )()( tqtp mm then,

mm

mm tqtp

H )(22

)( 222

From Maxwell’s equation we had found,

)()( 2

2

2tq

t

tqmm

m

So the variables )(tpm and )(tqm satisfy the equations,

)()(

)()( 2

tptqdt

d

tqtpdt

d

mm

mmm

Comparing the above equations with those for a simple harmonic oscillator, we see that they are

identical!

5.2 Cavity Quantum Electrodynamics

5.2.1 Quantization of Cavity Fields

For classical fields we derived the following expression for the energy,

m

mmm tqtpH

22ˆ

222

And the dynamics are given by,


7

tptqdt

d

tqtpdt

d

mm

mmm2

We need a fundamental relation (not derivable from any other theory) to quantize the field. Note that

each field mode behaves like a simple harmonic oscillator (SHO). Not just that the expression for total

energy matches that of a SHO, but the dynamics are also similar. So we can try quantizing by the

following steps:

(a) Let the dynamic variables )(tqm and )(ˆ tpm become operators )(ˆ tqm and )(ˆ tpm

(b) Let the equal‐time commutation relation between )(ˆ tqm and )(ˆ tpm be itptq mm )(ˆ),(ˆ

(c) For different modes let the commutation relations be 0)(ˆ),(ˆ tptq nm

The Hamiltonian operator of the field becomes,

m

mmm tqtpH

2

ˆ

2

ˆˆ222

As in the case of SHO, define creation and destruction operators tamˆ and tamˆ for each field mode

as,

tpitqta

tpitqta

mmmm

m

mmmm

m

ˆˆ2

1

ˆˆ2

1ˆ

It follows that,

mnnm tata ˆ,ˆ

and,

mmmm tataH

2

1ˆˆˆ

The operator time dependence in the above expressions should be interpreted in the Heisenberg sense.

So, for example, the Hamiltonian in the Schrodinger picture would be,

mmm

mmmm naaH

2

1ˆ2

1ˆˆˆ

where mn is the number operator for the mode m .


8

5.2.2 Energy Eigenstates and Eigenenergies

Since all radiation eigenmodes behave independently, we consider one mode only. The Hamiltonian for

a single mode is,

2

1ˆˆmmm aaH

Let the eigenstates of this Hamiltonian be mn| . Following the discussion in Chapter 4, we have,

mmmm nnnaa ˆˆ

mmm nnna 11ˆ

mmm nnna 1ˆ

and,

mmm

nnnH

2

1ˆ

(i) m

n is an energy eigenstate with energy

2

1nm

(ii) The energies of different ergenstates are separated by multiples of m . The eigenenergies are,

,.........2

13,

2

12,

2

1,

2

1mmmmmmm

(iii) The ground state m

0 has energy m2

1

5.2.3 The Photon

On quantizing the electromagnetic radiation we see that the energy of a single mode of the field with

frequency m can only take on values separated from each other by multiples of m (i.e. the mode

energy can only be increased or decreased in multiples of m ). This is in contrast with classical

electrodynamics where a radiation mode can take any energy value by increasing or decreasing the field

amplitude in a continuous fashion. This minimum amount of energy (i.e. m ) is associated with the

term “photon”. A state with a single photon is the smallest possible excitation above the ground state

m0 of a field mode. A state with a single photon is

m1 and has energy

mm

2

1. A state


9

with two photons is m

2 and has energy

mm

2

12 . A state with n photons is

mn . The

ground state m0| of a field mode has no photons but still has energy equal to m2

1.

The operator mmm aan ˆˆˆ is called the number operator for the mode m because it gives the number

of photons in a state,

mmm nnnn ˆ

ma and ma are called photon creation and destruction operators because they increase or decrease

the number of photons in a state,

mmm nnna 11ˆ

mmm nnna 1ˆ

The state m

n can be written as,

m

nm

m n

an 0

!

)ˆ(

The states m

n are called photon number states since they are eigenstates of the number operator

mmm aan ˆˆˆ .

A photon is not a particle and a photon is not a wave. A photon is just the smallest possible energy

excitation of a field mode.

5.2.4 Multimode States

A 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||

So,

|5|ˆ

|4|ˆ

|3|ˆ

p

n

m

n

n

n

Or sometimes | may be written as,


10

5,4,3|| pnm nnn

The vacuum state 0| means the ground state of all modes,

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

If we have a state with 3 photons in mode 2 then,

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

The above notation is too cumbersome. Instead one typically writes,

23||

One only writes down the states that have non‐zero photon numbers or are of interest in some other

way. Few examples are given below.

(1) A state with p photons in mode m ,

m

pm pp

a

|0|

!

ˆ|

(2) A state with one photon in mode m and one in mode n ,

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

(3) A state which is a linear superposition of one photon in mode m and me in mode n ,

1|1|2

11,0|0,1|

2

1

1|0|0|1|2

10|0|

2

1|

nmnmnm

nmnmnm

nnnnnn

aa

Completeness Relation for the Photon Number States: A full blown completeness relation for photon

states look like

0 0321321

1 2

1,,,,,,n n

nnnnnn

But in any practical problem one is usually dealing with photons in one or a few modes. So for one mode

(say the mode m) the completeness relation becomes,

0 01or1

n nmmmm

m

nnnn


11

Orthogonality Relation for the Photon Number States: We need to find the value of the inner product

nm pp . Since the corresponding mode spatial functions )(rUm

and )(rU n

are not orthogonal in

the sense nmnm rdrUrU

3)(),( , it is not immediately obvious what nm pp should be. But

since m

p and n

p are eigenstates of a Hermitian operator (i.e. H ), we have,

ppnmnm pp

5.2.5 Time Development of Creation and Destruction Operators

The time development of the creation and destruction operators follows from the Heisenberg equation,

mti

mti

m

mmmm

aetaeta

taHtadt

tadi

mm ˆ0ˆˆ

ˆˆ,ˆˆ

and,

mti

mti

m

mmmm

aetaeta

taHtadt

tadi

mm ˆ0ˆˆ

ˆˆ,ˆˆ

5.2.6 Field Operators

The fields are physical observables and are therefore represented by operators. We can write ),(ˆ trA

and ),(ˆ trE

in Heisenberg picture using,

n

nno

n rUtq

trA )()(ˆ

),(ˆ

mmmm

mo

m

mmmm

mom

rUtataεε

ωi

t

,trA,trE

rUtataεεω

)(ˆˆ2

)(ˆ)(ˆ

)(ˆˆ2


12

mmmm

nmo

o

rUtataεεω

,trA,trH

)()(ˆ)(ˆ2

1

)(ˆ1)(ˆ

0

Note that all field operators are Hermitian.

Now consider a quantum state with a billion photons in mode m , i.e m

mn99 1010 .

Let’s find the average electric field for this state,

0

)(ˆ)(ˆ2

)(ˆ

0

rUtataεε

ωi,trE m

mmm

m

m

There are a large number of photons in the state m

910 but it has a zero average value for the

electric field. Therefore, the photon number state is likely not what emerges from, say, your cell phones.

5.2.7 Vacuum Fluctuations

The ground state of a mode m

0 has energy 2m even though there are no photons in this state.

The energy is due to what are called vacuum fluctuations; in the ground state, the electric and magnetic

fields have zero average values but averages of the squares of the fields are not zero. In fact, restricting

ourselves to just one mode, a direct computation shows,

mmom trHtrHtrEtrErrd

2

10),(),(

2

1),(),()(

2

10 0

3

The total vacuum energy, including contributions from all modes, is then m

m 2 , which is infinite.

This infinity is not a problem since in experiments only the differences in energies are measured and not

absolute energies.

We don’t know whether the sum m

m 2 is valid for values of m all the way up to . When m

becomes very large, the modes )(rUm vary very fast in space and over length scales comparable to or

shorter than the Planck scale of 10‐33 cm. Nobody knows if quantum electrodynamics is valid at such

short spatial scales.


13

5.3 Electrodynamics in Free Space

5.3.1 Classical Electrodynamics in Free Space

Maxwell’s equations in free space are,

t

trEtrH

t

trHtrE

trE

trH

o

o

o

,,

,,

0,.

0,.

The above equations result in the wave equation,

2

2

2

,1,

t

trE

ctrE

Assuming, as before, coulomb gauge and all fields to be transverse (divergence free), let,

trAtrH

t

trAtrE

o ,,

,,

where,

0,, trAtrE

It follows that,

2

2

2

,1,

t

trA

ctrA

trAtrAtrAtrA

,

,,,2

2

So the wave equation in free space becomes,

2

2

22 ,1

,t

trA

ctrA

5.3.2 Free-Space Eigenmodes and Eigenvalues


14

We need to find eigenmodes of the operator 2 . Let these be rUk

with eigenvalues 22 ck , i.e.,

rUc

,trU kk

k

2

22 )(

The eigenvectors must also satisfy the coulomb gauge condition,

00, rUtrA k

The eigenvectors can be written in terms of plane waves,

V

ekrU

rki

k

)(

The eigenmodes are normalized in very large box of volume V . All physical results will turn out to be

independent of V . The right hand side above represents a plane wave propagating in the direction of

the vector k and with the vector potential (and the electric field) polarized in the direction of the unit

vector )(ˆ k

. Note that,

rUkV

ekrU k

rki

k

222 )(ˆ

Therefore, the eigenvalue of the plane wave solution is,

22

2

. kkkck

Since k does not depend on the direction of k

but only on its magnitude, we will write it as k . The

constrain 0 rUk

implies 0)(ˆ. kk

. The direction of field polarization is perpendicular to the

direction of propagation of the plane wave given by the vector k. For each direction k

, there are two

independent and mutually orthogonal directions perpendicular to k that )(ˆ k

can have. One can

choose any two such directions for the polarization unit vectors and label them )(ˆ1 k

and )(ˆ2 k

. For

example, if xkk ˆ

(i.e. pointing in the x‐direction) then one can choose ykε ˆˆ1

and zkε ˆˆ2

. Or,

one may also choose,

zykε ˆˆ2

1ˆ1

zykε ˆˆ2

1ˆ2


15

The plane wave eigenmodes form a complete set in the space of all transverse vector fields. Therefore,

the field trA ,

can be expanded in this complete set,

k j

j

rki

o

jkε

V

e

ε

tkqtrA

2

1

.ˆ

,, (1)

Since trA ,

is real (i.e., trAtrA ,, * ), one must have,

ktkqktkq jjjj

ˆ,ˆ,

It is convenient to choose,

kk

tkqtkq

jj

jj

ˆˆ

,,

Keep in mind that since k can be pointing in any direction, the unit vectors )(ˆ1 k

and )(ˆ2 k

have no

simple relationship with the Cartesian unit vectors x , y and z .

5.3.3 Periodic Boundary Conditions

Free space is infinite. For the purpose of calculations, it is useful to assume that free space is a large box

of side L on each side, and volume V equal to 3L . We also assume that each facet of that box is

connected with the opposite facet. Since the eigenmodes must be single‐valued everywhere, the

following periodic boundary conditions must hold,

xLxkixxki ee ˆ.ˆ.

L

nkx

2 ,3,2,1,0 n

yLykiyyki ee ˆ.ˆ.

L

mky

2 ,3,2,1,0 m

zLzkizzki ee ˆ.ˆ.

L

pkz

2 ,3,2,1,0 p


16

Thus, the vector k can only have certain discrete values. There is only one such allowed value in a cube

of volume 32 L in k‐space. Therefore, noting that there are 32V different allowed k values

per unit volume in k‐space, the summation over k of the form

k , can be replaced by the integral,

kdV

k

332

The field trA ,

can now be written as,

2

1

.

3

3ˆ

,

2,

jj

rki

o

jkε

V

e

ε

tkqkdVtrA

Orthogonality Relation for the Eigenmodes: The orthogonality relation for the eigenmodes is,

kkrss

rki

r

rkiδδkε

V

e.kε

V

erd

,'

.'.3 ˆ'ˆ

Sometimes one is interested in the Cartesian components of trA ,

(i.e. trA,trA,trA zyx ,,,

).

These can be found as follows,

2

1

.

03

3ˆ.ˆ

,

2,.ˆ,

jj

rkij

x kεxV

e

ε

tkqkdVtrAxtrA

z.kεkε

y.kεkε

x.kεkε

kεkεkε

kε.kε

jjz

jjy

jjx

jzjyjx

jj

ˆˆ

ˆˆ

ˆˆ

1

1ˆˆ222

Finally, the electric and magnetic fields are,

2

1

.

3

3ˆ

,

2

,,

jj

rki

o

jkε

V

e

ε

tkqkdV

t

trAtrE

2

1

.

3

3ˆ

,

2

,,

jj

rki

oo

j

okεki

V

e

ε

tkqkdV

trAtrH

5.3.4 Time Development

We can plug the expansion for trA ,

in the equation,


17

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

Multiplying by kεe jrki

ˆ. on both sides, integrating over all space, and using the orthogonality relation

for the modes, we get,

tkq

tkqckdt

tkqd

jk

jj

,

,,

2

22

2

2

5.3.5 Field Energy

The energy of the field can be found as follows,

V

etkqtkqkεkikεki

c

V

ekεkεtkqtkq

kdV

kdVrd

trHtrHtrEtrEε

rdH

rkki

srsr

r s

rkki

srsr

oo

.'2

.'

3

3

3

33

3

,','ˆ.ˆ2

'ˆ.ˆ,',2

1

2

'

2

,.,2

,.,2

First we carry out the integration over all space. From the exponentials one obtains a factor of ',kkV

that can be used to get rid of the integration/summation over 'k, and one can replace 'k

everywhere

by k

. Also, note the following relations,

kk

tkqtkq

jj

jj

ˆˆ

,,

1ˆ.ˆ s

sr kk

2ˆˆ kkεkikεki ss

r

The final result is,


18

.

2

,,

2

,,

2

23

3

j

jjk

jj tkqtkqω

tkqtkqkdVH

The expression for energy has complex quantities. So it is a little different from what you have seen in

the past. If one defines,

tkqtkp jj ,,

then,

,tkpt

,tkq

,tkqωt

,tkp

jj

jkj

2

and the expression for the energy becomes,

.

2

,

2

,

2

2

2

2

3

3

j

j

k

j tkqω

tkpkdVH

5.3.6 Field Momentum

In classical electromagnetism the momentum of the electromagnetic field is,

trHtrErdP oo ,,3

Using the relation,

s

sr kikεkikε

ˆˆ

one obtains,

jjj tkqtkqki

kdVP ,,

2 3

3

5.3.7 Field Angular Momentum

Classical electromagnetic fields have a well‐defined angular momentum. Recall that the angular

momentum of a particle with momentum tp and position vector tr with respect to a point or is

given as,

tprtrtL o


19

The classical expression for the momentum of the electromagnetic field is (from previous Section),

trHtrErdP oo ,,3

The classical expression for the angular momentum of the field is,

trHtrErrrdJ ooo ,,3

Usually the point of reference is the origin (given the homogeneity of free space) and the above

expression is written as,

trHtrErrdJ oo ,,3

5.4 Quantum Electrodynamics in Free Space

5.4.1 Quantlzation of Radiation in Free Space

The first step in the quantization of any system is the promotion of observables to operators and the

imposition of commutation relations. In quantum electrodynamics the observables are the field

operators, trE ,ˆ and trH ,ˆ

, and these operators must therefore be Hermitian. In classical

electromagnetism,

tkptkp

tkqtkq

jj

jj

,,

,,

The above conditions were necessary for the fields to be real. In quantum electrodynamics, the field

operators will be Hermitian provided,

tkptkp

tkqtkq

jj

jj

,ˆ,ˆ

,ˆ,ˆ

To quantize, we impose the following equal‐time commutation relations,

',,',,ˆkkrssr itkptkq

It follows that,

',,',,ˆkkrssr itkptkq

We define creation and destruction operators as before,


20

,tkpi,tkqωω

,tka

,tkpi,tkqωω

,tka

jjkk

j

jjkk

j

ˆˆ2

1ˆ

ˆˆ2

1ˆ

This gives,

0'ˆˆ

'ˆˆ',

,tka,,tka

,tka,,tka

sr

kkrssr

5.4.2 Field Hamiltonian

The field energy can be written as,

j

jjk

jj tkqtkqω

tkptkpkdVH

2

,,ˆ

2

,ˆ,ˆ

2ˆ 2

3

3

and in terms of the creation and destruction operator it becomes,

2

1,ˆ,ˆ

2ˆ

,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 vacuum energy is,

j

kk kd

Vkd

V

3

3

3

3

222

The vacuum energy density is therefore,

kkd

3

3

2


21

5.4.3 Time Development of Creation and Destruction Operators

The time development of the creation and destruction operators follows from the Heisenberg equation,

kaetkaetka

tkaHtkadt

tkadi

rti

rti

r

rkrr

kk

ˆ0,ˆ,ˆ

,ˆˆ,,ˆ,ˆ

and,

kaetkaetka

tkaHtkadt

tkadi

rti

rti

r

rkrr

kk

ˆ0,ˆ,ˆ

,ˆˆ,,ˆ,ˆ

5.4.4 Field Operators

The field operator trA ,ˆ is,

kεV

etkatka

εω

kdVtrA j

rki

jjj

ok

ˆ,ˆ,ˆ22

,ˆ .

3

3

and, of course, the field operator is Hermitian, trAtrA ,ˆ,ˆ . The operators for the electric and

magnetic fields are,

kε

V

etkatka

εi

kdV

t

trAtrE j

rki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ,ˆ .

3

3

kεkiV

etkatka

ε

kdV

trAtrH j

rki

jjj

okoo

ˆ,ˆ,ˆ2

1

2

,ˆ,ˆ .

3

3

5.4.5 Energy Eigenstates and Photons

The energy eigenstate with m photons in mode jk,

is jk

m, , so that,

jkkjk

mmmH,, 2

1ˆ

5.4.6 Momentum of a Photon

The field momentum operator is,


22

rdtrHtrEεP oo

3,ˆ,ˆˆ

Using the eigenmode expansions for the field operators one gets,

jjjjj tkatkatkatka

kkdVP ,ˆ,ˆ,ˆ,ˆ

22

ˆ3

3

The only non‐zero terms are,

jjj

jjj

jjjjj

jjjjj

tkatkakkd

V

tkatkakkd

V

tkatkatkatkakkd

V

tkatkatkatkakkd

VP

,ˆ,ˆ2

2

1,ˆ,ˆ

2

,ˆ,ˆ,ˆ,ˆ22

,ˆ,ˆ,ˆ,ˆ22

ˆ

3

3

3

3

3

3

3

3

In the Schrodinger picture,

jjj kakak

kdVP

ˆˆ

2

ˆ3

3

Now we can check if the energy eigenstates (i.e. the photon number states) are also momentum

eigenstates. We need to evaluate,

jk

mP,

ˆ

The result is,

jkjk

mkmmP,,

ˆ

Therefore, the momentum associated with a single photon in the radiation mode with wavevector k is

k .

5.4.7 Angular Momentum and Spin of a Photon

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

trHtrErrdJ oo ,,3

The above expression consists of two contributions,


23

a) The orbital angular momentum

b) The intrinsic angular momentum or the spin angular momentum

These two parts are neither easily separable nor separately conserved. The reason is that spin angular

momentum for a massive particle, say an electron, is clearly defined by looking at the angular

momentum in the rest frame of the particle. Massless particles, like photons, have no rest frame and

this procedure does not work. Nevertheless, if one looks at radiation states, like plane waves (with ideal

infinite phase fronts), that are unlikely going to have any orbital component, then the angular

momentum found using the expression above can tell us something about the intrinsic angular

momentum of radiation. To see this more clearly we proceed as follows. We assume that the fields are

transverse (i.e. divergence free). We start from the classical expression for the angular momentum

above and perform a few manipulations,

sssoo

oo

oo

o

oo

trArtrErdtrAtrErd

trAtrErrdtrArtrErd

trAtrErrdtrArtrErd

trAtrErrd

trHtrErrdJ

,,,,

,,.,.,

,,.,.,

,,

,,

33

33

33

3

3

The first part is usually identified with the intrinsic or the spin angular momentum and the second part

with the orbital angular momentum. It is tempting to write,

LSJ

However, as mentioned above, this decomposition runs into problems. For plane waves the second part

can be shown to be zero. So for plane waves we will take the expression for the intrinsic angular

momentum to be,

trAtrErdS o ,,3

Upon quantization, the fields become operators given by,

kεV

etkatka

εω

kdVtrA j

rki

jjj

ok

ˆ,ˆ,ˆ22

,ˆ .

3

3

kε

V

etkatka

εi

kdV

t

trAtrE j

rki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ,ˆ .

3

3

Using the above expressions, the operator S becomes,

tkatkatkatkakkikd

VSj

jjj ,ˆ,ˆ,ˆ,ˆˆˆ22

ˆ

,3

3


24

Using a convenient convention for the unit polarization vectors,

kkk

kk jj

ˆˆˆ

ˆˆ

21

we obtain,

tkatkatkatkakikd

VS ,ˆ,ˆ,ˆ,ˆˆ2

ˆ21123

3


kakakakakikd

VS

21123

3ˆˆˆˆˆ

2

ˆ

The operator S is diagonal in the wavevector indices but not diagonal in the polarization indices. Recall

that the operator ka

1ˆ creates a single photon with wavevector k

and with linear polarization in the

direction of the unit vector k1 . Similarly, ka

2ˆ creates a single photon with wavevector k

and with

linear polarization in the direction of the unit vector k2 . This means that a photon in a linearly

polarized plane wave state is not in an eigenstates of S. We define two new destruction (and

corresponding adjoint creation) operators as follows,

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

Note the following commutation relations,

0,'ˆ,,ˆ,'ˆ,,ˆ

,'ˆ,,ˆ,'ˆ,,ˆ',',

tkatkatkatka

tkatkatkatka

RLLR

kkLLkkRR

The operator kaRˆ on the creates a single photon with wavevector k

in a linear superposition of

polarization vectors k1 and k2 with a 90‐degrees phase difference between the two

polarizations. In fact, kaRˆ creates a single photon that is right‐hand circularly polarized. Similarly,

kaLˆ creates a single photon that is left‐hand circularly polarized. In terms of these new operators, the

expression for S becomes,


25

tkatkatkatkakkd

VS LLRR ,ˆ,ˆ,ˆ,ˆˆ2

ˆ3

3


kakakakakkd

VS LLRR

ˆˆˆˆˆ

2

ˆ3

3

The operator S is now diagonal in all indices. Consider a single photon state 0ˆ kaR

. It is

eigenstate of the operator S with eigenvalue k ,

kS ˆˆ

We say that the angular momentum of a right‐circularly polarized single photon plane wave state is

k . Similarly, the state 0ˆ kaL is an eigenstate of S

with eigenvalue k and therefore the

angular momentum of a left‐circularly polarized single photon plane wave state is k . The direction of

the angular momentum is always along the axis of propagation. The spin of a particle is related to its

intrinsic angular momentum and is measured in units of . Therefore, the spin of a photon can have two values, +1 or ‐1, and these correspond to photons with right‐circular or left‐circular polarization

states.

5.4.8 Position of a Photon

For photons position is not an observable, and there is no position operator for photons, and there are

no position eigenstates. Any attempt to define position eigenstates for photons ends up violating

Lorentz invariance. In fact, the same difficulty arises in the case of other massive particles, such as

electrons, but for a massive particle one can always find a rest frame in which the particle is at rest (or

moving very slowly compared to the speed of light) and then one can define approximate position

eigenstates and the position operator in a non‐relativistic setting. But photons, being massless, are

always travelling at the speed of light and therefore no position operator can be rigorously defined.

However, one may define the position of a photon in a measurement sense and ask the following

question, “If there is a photon in an eigenmode rUm of the radiation, what is the probability of

detecting the photon at location r at time t in a time interval t when a photo detector is placed at

r.” We will discuss these questions in more detail when we discuss photon detection in later Chapters.

Here we present a discussion of what goes wrong when trying to localize a photon in the detection

sense.

Consider the single photon state,

0ˆ1 11,ka

k


26

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? With this as the motivation,

consider the following state,

The operator 'rb creates a single photon in a maximal superposition of plane wave states and with

a polarization in the direction of the Cartesian unit vector be . The question arises how localized is the

photon at the location 'r. If it is really localized at 'r

then we could perhaps write a position eigenstate

for the photon as,

0'' rr bb

One way to answer the above question is to try destroying the photon at a different location r and in

polarization along ae . Physically, this action corresponds to detecting a photon at location r and with

polarization along ae , given the state 0'rb . So we evaluate the following matrix element,

baba rrrr '0'0

If the photon created by 'rb is localized, and position eigenstates exist for photons, then we should

expect the answer to look like,

''ˆ.ˆ'0'0 33 rrrreerrrr abbababa

We evaluate this matrix element below,

0'ˆˆ.'ˆ2

'0'

''.2

13

3

kaV

eek

kdVr j

rki

bjj

b


27

'2

ˆ).(ˆ)(ˆ.ˆ

0ˆ'ˆ0ˆ).'(ˆ)(ˆ.ˆ

0ˆ'ˆ'ˆ,ˆ0ˆ).'(ˆ)(ˆ.ˆ

0'ˆˆ0ˆ).'(ˆ)(ˆ.ˆ

0'0

'

23

3

2

1

'

','

2

1

''2

1

'

2

1

''2

1

'

2

1

''2

1

rr

V

e

k

kkkd

V

eek

V

eke

kakaV

eek

V

eke

kakakakaV

eek

V

eke

kakaV

eek

V

eke

rr

ab

rrkiba

ab

j

rki

bj

rki

jak

jkkjk j

rki

b

rki

jak

jjk j

rki

b

rki

jak

jk j

rki

b

rki

jak

ba

The answer 'rrab

is called the transverse delta function. It is not localized like a delta function but

is spread out. One can write,

abba

ababrr

errerr

rrrrrr

233

'

ˆ.ˆ.3

'4

1''

The reason we did not just get '3 rrab

, which would correspond to our expectation for position

eigenkets, is that the eigenmodes are not scalars but vectors. One can superpose plane waves and

produce a delta function but vector plane waves cannot be superposed to produce a delta function. This

argument shows that photons cannot be localized.

5.4.9 Field Commutation Relations

The equal‐time commutation relations among the fields express the possibility of simultaneous

measurements. Consider first the electric and magnetic fields,

kε

V

etkatka

εi

kdV

t

trAtrE j

rki

jjj

o

k

ˆ,ˆ,ˆ22

,ˆ,ˆ .

3

3

kεkiV

etkatka

ε

kdV

trAtrH j

rki

jjj

okoo

ˆ,ˆ,ˆ2

1

2

,ˆ,ˆ .

3

3

Let,


28

ba

aa

etrHtrH

etrEtrE

ˆ.,ˆ,ˆ

ˆ.,ˆ,ˆ

The equal‐time commutation relations between these two components are,

2,1

'.3

32 ˆ.ˆˆ.ˆ

2,'ˆ,,ˆ

jbjaj

rrkiba ekεkiekεe

kdcitrHtrE

Noting that for any two vectors, the cross‐product can be written as,

cbabccb

a BABA ,

where abc is the Levi‐Civita symbol with the property that 1123 and abc picks a negative for any

permutation of the indices (e.g. 1321 and 1231 ), one can write,

cabccj

bjaj kiekεkiekε 2,1

ˆ.ˆˆ.ˆ

And,

'

2,'ˆ,,ˆ

32

'.3

32

rrci

ekd

citrHtrE

cabcc

rrkicabc

cba

The result is a derivative of a delta function. The symbol c stands for derivative with respect to the “c”

Cartesian component (“c” could be x, y, or z). The result shows that same components of the electric

and magnetic fields commute at all points but different components of the electric and magnetic fields

do not commute.

Other interesting quantities are the field commutation relations at different locations and at different

times,

','ˆ,,ˆ trEtrE ba

Such commutation relation show whether accurate simultaneous measurements on fields at different

locations and times are possible. Of course, one would expect in light of relativity that if field

measurements are made at locations far enough such that no signal could travel in the time interval

between the measurements then such measurements should not affect each other. In other words, all

field operators must commute for space‐like intervals (i.e. when 222

'ttcrr

). Using the

expression for the electric field operator we get,


29

cceekd

ct

eeeekkckd

k

kkeeee

kd

k

kkeee

kd

ekεekεeeekd

trEtrE

ttirrki

kobaab

ttirrkittirrkibaabk

ko

baab

ttirrkittirrki

o

k

baab

ttittirrki

o

k

jbjaj

ttittirrki

ba

k

kk

kk

kk

kk

.22

22

22

22

ˆ.ˆˆ.ˆ2

','ˆ,,ˆ

''.3

32

2

2

''.''.223

3

2''.''.

3

3

2'''.

3

3

2,1

'''.3

3

To evaluate the above expression we need to evaluate the imaginary part of the following expression,

3. ' '

3

' cos '2

0 0

' ' '2

0

' '2

0

22

1sin

2 4

1

2 ' 4

lim1

02 ' 4

k

k

k

k

i k r r i t t

o k

i k r r i t t

o

i k r r i k r r i t t

o

i k r r i k r r i t tk

o

d ke e

kdk d e ec

idk e e e

c r r

idke e e e

c r r

'

2

lim1 1 1

02 ' ' ' ' '4oc r r r r c t t i r r c t t i

The imaginary part of the above expression is,

''''4

1

'2ttcrrttcrr

rrco

Finally, the field commutator becomes,

''''

'4

','ˆ,,ˆ

22

2ttcrrttcrr

rrc

ic

t

trEtrE

obaab

ba

The above expression shows that the commutator is non‐zero only on the light cone (i.e. when

'ttcrr

) and therefore the fields commute for space‐like intervals.

chapter 5: quantization of radiation in cavities and …chapter 5: quantization of radiation in...

Documents