talk spinoam photon

Spin and Orbital Angular

Momentum of a Photon

Michael London and Angela Guzman

Quantum Optics Group FAU, Sept 25 2008

Maxwell’s Equations

Source Free Field

2

0

0

1

t

t

E

B

E B

B Ec

Vectors to Quantized Field Operators,

ˆˆ ˆ ˆ( , ) ( , ), ( , ), ( , )F r t E r t B r t A r t

Plane Wave Mode Expansion

Electric Field Operator

( ) † * ( )

, , , ,3,

1ˆ( , ) ( )

2

i k r t i k r t

k s k s k s k sk s o

E r t ia e ia eL

Magnetic Field Operator

( ) † * ( )

, , , ,3,

1ˆ( , ) ( ( ) ( ) )

2

i k r t i k r t

k s k s k s k sk s o

B r t ia k e ia k eL

where and 1 2 3

2 2 2( , , )k n n n

L L L

1 2 3( , , ) 0, 1, 2...n n n

Polarization Vectors

Orthonormal Transverse pairs (circular

or linear)

*

, ,

, ,

0

ssk s k s

k s k s

k

k

k

Commutation Relation for the creation

and annihilation operators: †

, , ,ˆ ˆ[ , ]

k s k s k ka a

Total Angular Momentum

Depends on a point and is an integral

of the angular momentum density or

3ˆ ˆ ˆ ˆ ˆ( , ) ( ) ( ( , ) ( , ) ( , ) ( , ))

2

oo o o

V

J r t d x r r E r t B r t B r t E r t

Separate into two parts and determine

the Linear Momentum ˆ( , )oP r t

3ˆ ˆ ˆ ˆ ˆ ˆ( , ) (0, ) ( ( , ) ( , ) ( , ) ( , ))

2

oo o o

V

J r t J t r d x E r t B r t B r t E r t

3ˆ ˆ ˆ ˆ ˆ( , ) ( ( , ) ( , ) ( , ) ( , ))

2

oo o o

V

P r t d x E r t B r t B r t E r t

Defining Linear Momentum

The difference between the classical case and the

field theory case is that the fields are symmetric

Hermitian operators. ˆ ˆ ˆ( , ) (0, ) ( , )o o oJ r t J t r P r t

,,

ˆˆ( , )o k s

k s

P r t kn

By using the mode expansion for the Electric and Magnetic fields the final expression for linear momentum shows that it depends on the photon number operator:

,ˆ

k sn

Photon Number

The photon number operator:

†

, , ,ˆ ˆ ˆ

k s k s k sn a a

The Fock space defines a Orthonormal complete set:

†

, , , , , , ,ˆ ˆ ˆ

k s k s k s k s k s k s k sn n a a n n n

1 1 2 2 1 2 2 2,1 ,2 ,1 ,2 ,1 ,2 ,1 ,2 ,, , , ,... ... { }k k k k k k k k k s

n n n n n n n n n

The total field is written as a product of the states of the

individual modes:

Constant of the Motion

The Linear Momentum, is a

constant of the motion since the

photon number, is a constant.

ˆ( , )oP r t

,ˆ

k sn

The total Angular Momentum,

will on change in time if changes

in time.

ˆ( , )oJ r t

ˆ(0 , )oJ t

Time Rate of Change of Total

Angular Momentum

Using Maxwell’s equations we get

ˆ ˆ ˆ( , ) (0, ) ( ( , ))t o t tJ r t J t r P r t

3ˆ ˆ ˆ ˆ ˆ ˆ( , ) (0, ) ( ( , ) ( , ) ( , ) ( , ))

2

ot o t t t

V

J r t J t d xr E r t B r t E r t B r t

3 1ˆ ˆ ˆ ˆ ˆ ˆ( , ) (0, ) ( ( , ) ( , ) ( , ) ( , ))t o t o

oV

J r t J t d xr E r t E r t B r t B r t

Use equal time commutators of with

and with

ˆE

ˆB

ˆB

Triple Cross Product

21ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ( )2

i iE E E E E E E E E

ˆ0E

Condition from Maxwell’s equation yields:

2 21 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ( ) ( ) ( ) ( ) ( )

2 2r E E r E r E E rE Er E

2 21 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ( ) ( ) ( ) ( ) ( )

2 2r B B r B r B B rB Br B

The Electric field

The Magnetic field:

Gauss’s Theorem over Volume

and Surfaces

2 21 1ˆ ˆ ˆ( , ) ( ( , ) ( , ))

2

1 1ˆ ˆ ˆ ˆ( ( , ) ( , ) ( , ) ( , ))

2

t o

os

o

os

J r t dS r E r t B r t

dS E r t r E r t B r t r B r t

The first term vanishes if we apply surface elements at

(-L/2,y,z) and (+L/2,y,z). The surface term of the cross

product points in opposite directions. So the second

terms remains: 1 1ˆ ˆ ˆ ˆ ˆ

( , ) ( ( , ) ( , ) ( , ) ( , ))2

t o

os

J r t dS E r t r E r t B r t r B r t

Rate of Change of Total Angular

Momentum

Component form:

1ˆ ( )t l lmp p m o m p m p

oV

J dS r E E B B

Summing over repeated indices, the term

with m ≠ p vanishes in pairs at the

boundary and only m = p remains.

Positive and Negative Frequency

Parts

Decompose the Hermitian operators:

ˆ( , ) ( , ) ( , )E r t E r t E r t

( , )E r t

( , )E r t

contains annihilation operators and

creation operators.

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆm p m p m p m p n pE E E E E E E E E E

Normal ordering

Normal Ordering for the fields

Commutation Relation and the normal

ordering procedure

ˆ ˆ[ , ] 0m pE E

Invert the normal ordering for the last term

ˆ ˆ ˆ ˆm p m pE E E E

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆm p m p m p m p p mE E E E E E E E E E

Correct normal ordering after inverting

Normal Ordered Time Rate of

Change of the Total Angular

Momentum 1ˆ ˆ ˆ ˆ ˆ( )t l lmp p m o m p m p

oV

J dS r E E B B

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ( )

1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ))

t l lmp p m o m p m p m p p m

V

m p m p m p p m

o

J dS r E E E E E E E E

B B B B B B B B

Insert the normal ordering terms in the above equation

Expectation Value to the total

Angular Momentum

Surface Integral over the boundary

ˆ( , ) 0iE r t

ˆ( , ) 0iB r t

Expectation Value:

2 1ˆ ( , ) 0t l oJ r t

Constant of the motion.

Decomposition of Total Angular

Momentum into spin and

orbital parts.

In Classical EM theory we can decompose into two part that depend on position while the last term does not.

3 3

3

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

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

o o o i o i

V V

o o o

S V

d x r r E r t B r t d xE r t r r A r t

dS E r t r r A r t d xE r t A r t

where is the magnetic vector potential ( , )A r t

Orbital Angular Momentum

OAM

3ˆ ˆ ˆˆ ˆ( , ) ( ( , )(( ) ) ( , ) (( ) ( , )) ( , ))2

ˆ ˆˆ ˆ( ( , )( ) ( , ) ( ) ( , ) ( , ). )

2

oi o i o i i

V

oo o

S

L r t d x E r t r r A r t r r A r t E r t

dS E r t r r A r t r r A r t E r t dS

Spin

• Spin

3ˆ( ( , ) ( , ) ( , ) ( , ))

2

o

V

S d x E r t A r t A r t E r t

Decomposed into Two Terms

Total Angular Momentum is now decomposed into the Intrinsic Spin and Orbital Angular Momentum

ˆˆ ˆJ L S

The integral is written over the surface boundary and

can be written as normal order

Spin

After using the mode expansion for the

Electric and Magnetic fields which is

integrated over a volume we obtain this

form: † *

,, , , ,,

1ˆˆ ˆ( )( )

2s sk s k s k s k s

s sk

S i a a

*

,, ,( ) s sk s k s

is

, 1s s

,1k

,2kWe choice and to represent orthonormal states or

right and left circular polarization

where k

k and

Spin

The choice of the polarization is in a simple form such that the spin becomes:

,1 ,2

ˆˆ ˆ( )

k kk

S n n

The spin is diagonal in the photon number state. It is

written as the difference of the right and left polarization.

The spin is a constant of the motion since the photon

number is a constant.

OAM

The orbital angular momentum is a

constant of motion.

ˆˆ ˆL J S

,1 ,2

ˆ ˆˆ ˆ( )

k kk

L J n n

† †

, , , ,,

1ˆ ˆ ˆ ˆ( ) ( , ) ( , )

2 k s k s k s k ss s

L a a a a F r t L F r t

Conclusion

Spin and Orbital Angular Momentum depend on

the photon number and are therefore constants

of the motion.

The commutation relations shows that neither

spin nor orbital angular momentum generate

rotations.

To further investigate the physical significance on

should consider the interaction of matter with

the radiation field.