1. Spin and Orbital Angular
Momentum of a Photon
Michael London and Angela Guzman
Quantum Optics Group FAU, Sept 25 2008

2. Maxwell’s Equations
Source Free Field
 E0
 B0
  E   t B
1
  B  2 t E
c
Vectors to Quantized Field Operators,
ˆ
ˆ ˆ ˆ
F (r , t )  E (r , t ), B(r , t ), A(r , t ) 

3. Plane Wave Mode Expansion
Electric Field Operator
ˆ 1 
E (r , t )   2 o
(iak , s k , s ei ( k r t )  iak , s k*, s ei ( k r t ) )
†
L3 k ,s
Magnetic Field Operator
ˆ 1
B(r , t )   2 o
(iak , s (k   k , s )ei ( k r t )  iak , s (k   k ,s )* e i ( k r t ) )
†
L3 k ,s
2 2 2
where k  ( n1 , n2 , n3 ) and (n1 , n2 , n3 )  0, 1, 2...
L L L

4. Polarization Vectors
Orthonormal Transverse pairs (circular
or linear)
k   0
 k*, s   k , s   ss
k
 k , s   k , s  
k
Commutation Relation for the creation
and annihilation operators: [a , a ]  
ˆ ˆ k ,s
†
k , s k ,k 

5. Total Angular Momentum
Depends on a point ro and is an integral
of the angular momentum density
ˆ o 3 ˆ ˆ ˆ ˆ
J (r , t )   d x(r  ro )  ( E (ro , t )  B(r , t )  B(r , t )  E (ro , t ))
2V
Separate into two parts and determine
ˆ
the Linear Momentum P(ro , t )
ˆ ˆ o 3 ˆ ˆ ˆ ˆ
J (r , t )  J (0, t )  ro   d x( E (ro , t )  B(r , t )  B(r , t )  E (ro , t ))
2V
ˆ o 3 ˆ ˆ ˆ ˆ
P(ro , t )   d x( E (ro , t )  B(r , t )  B(r , t )  E (ro , t ))
2V

6. 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
J (ro , t )  J (0, t )  ro  P(ro , t )
ˆ
P(ro , t )   knk , s
ˆ
k ,s
By using the mode expansion for the Electric and
Magnetic fields the final expression for linear
momentum shows that it depends on the photon
ˆ
number operator: nk , s

7. Photon Number
The photon number operator:
ˆk , s  ak ,s ak ,s
n ˆ† ˆ
The Fock space defines a Orthonormal complete set:
nk ,s nk ,s  ak ,s ak ,s nk ,s  nk ,s nk ,s
ˆ ˆ† ˆ
The total field is written as a product of the states of the
individual modes:
nk1 ,1 , nk1 ,2 , nk2 ,1 , nk2 ,2 ,...  nk1 ,1 nk2 ,2 nk2 ,1 nk2 ,2 ...  {nk ,s }

8. Constant of the Motion
ˆ
The Linear Momentum, P(ro , t ) is a
constant of the motion since the
photon number, nk ,s is a constant.
ˆ
ˆ
The total Angular Momentum, J (ro , t )
ˆ
will on change in time if J (0o , t ) changes
in time.

9. Time Rate of Change of Total
Angular Momentum
Using Maxwell’s equations we get
ˆ ˆ ˆ
t J (ro , t )  t J (0, t )  t (r  P(r , t ))
ˆ ˆ  ˆ ˆ ˆ ˆ
 t J (ro , t )   t J (0, t )  o  d 3 xr  ( t E (r , t )  B(r , t )  E (r , t )   t B(r , t ))
2V
ˆ ˆ ˆ ˆ 1 ˆ ˆ
 t J (ro , t )   t J (0, t )    d 3 xr  ( o E (r , t )  E (r , t )  B(r , t )  B(r , t ))
V
o
ˆ with
Use equal time commutators of E
ˆ with ˆ
and B B

10. Triple Cross Product
ˆ ˆ ˆ ˆ ˆ ˆ 1 ˆ ˆ ˆ
E  E  EiEi  E  E  E 2  (  E ) E
2
Condition from Maxwell’s equation yields:
ˆ
 E  0
The Electric field
ˆ ˆ 1 ˆ ˆ ˆ 1 ˆ ˆ ˆ
r  ( E  ( E )  (r ) E 2  r  (  E ) E    (rE 2 )    ( Er  E )
2 2
The Magnetic field:
ˆ ˆ 1 ˆ ˆ ˆ 1 ˆ ˆ ˆ
r  ( B  ( B)  (r ) B 2  r  (  B) B    (rB 2 )    ( Br  B)
2 2

11. Gauss’s Theorem over Volume
and Surfaces
ˆ 1 ˆ2 1 ˆ2
 t J (r , t )   dS  r ( o E (r , t )  B (r , t )) 
2 s o
1 ˆ ˆ 1 ˆ ˆ
2
dS  ( o E (r , t )r  E (r , t )  B(r , t )r  B(r , t ))
s
o
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:
ˆ 1 ˆ ˆ 1 ˆ ˆ
 t J (r , t )   dS  ( o E (r , t )r  E (r , t )  B(r , t )r  B(r , t ))
2 s o

12. Rate of Change of Total Angular
Momentum
Component form:
ˆ  1
t J l lmp  dS p rm ( o Em E p  Bm Bp )
V
o
Summing over repeated indices, the term
with m ≠ p vanishes in pairs at the
boundary and only m = p remains.

13. Positive and Negative Frequency
Parts
Decompose the Hermitian operators:
ˆ
E (r , t )  E  (r , t )  E  ( r , t )
E  (r , t ) contains annihilation operators and
E  (r , t ) creation operators.
Normal ordering
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
Em E p  Em E p  Em E p  Em E p  En E p

14. Normal Ordering for the fields
Commutation Relation and the normal
ordering procedure
ˆ ˆ
[E  , E  ]  0
m p
Invert the normal ordering for the last term
ˆ ˆ ˆ ˆ
Em E p  Em E p
Correct normal ordering after inverting
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
Em E p  Em E p  Em E p  Em E p  E p Em

15. Normal Ordered Time Rate of
Change of the Total Angular
Momentum
ˆ  ˆ E  1 B B )
ˆ ˆ ˆ
t J l lmp  dS p rm ( o Em p
V
o
m p
Insert the normal ordering terms in the above equation
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
 t J l   lmp  dS p rm ( o ( Em E p  Em E p  Em E p  E p Em ) 
V
1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
( Bm B p  Bm B p  Bm B p  B p Bm ))
o

16. Expectation Value to the total
Angular Momentum
Surface Integral over the boundary
ˆ
E  (r , t )  i  0
ˆ
B  (r , t )  i  0
Expectation Value:
ˆ
 2 t J l (ro , t )  1  0
Constant of the motion.

17. 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.
 o  d 3 x(r  ro )  ( E (r , t )  B(r , t ))   o  d 3 xEi (r , t )((r  ro ) ) Ai ( r , t ) 
V V
 o  dS  E (r , t )(r  ro )  A(r , t )   o  d 3 xE (r , t )  A(r , t )
S V
where A(r , t ) is the magnetic vector potential

18. Orbital Angular Momentum
OAM
ˆ o 3 ˆ ˆ ˆ ˆ
L (r , t )   d x( Ei (r , t )((r  ro )  ) Ai (r , t )  (( r  ro )  Ai ( r , t )) Ei ( r , t ))
2V
o ˆ ˆ ˆ ˆ

2 
S
(dS  E (r , t )(r  ro )  A(r , t )  (r  ro ) A(r , t ) E (r , t ).dS )

19. Spin
• Spin
ˆ o
S   d 3 x( E (r , t )  A(r , t )  A(r , t )  E (r , t ))
2V

20. Decomposed into Two Terms
Total Angular Momentum is now
decomposed into the Intrinsic Spin
and Orbital Angular Momentum
ˆ ˆ ˆ
J  L S
The integral is written over the surface boundary and
can be written as normal order

21. Spin
After using the mode expansion for the
Electric and Magnetic fields which is
integrated over a volume we obtain this
form: ˆ 1
S i
†
 (a
k
ˆ *
s , s
ˆ
a
k , s k , s
  s , s )( k , s   k , s )
2
We choice  k ,1 and  k ,2 to represent orthonormal states or
right and left circular polarization
( k ,s   k*,s )  is s ,s
k
where s , s  1 and  
k

22. Spin
The choice of the polarization is in a
simple form such that the spin
becomes:
ˆ
S    (nk ,1  nk ,2 )
ˆ ˆ
k
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.

23. OAM
The orbital angular momentum is a
constant of motion.
ˆ ˆ ˆ
L  J S
ˆ ˆ
L  J    (nk ,1  nk ,2 )
ˆ ˆ
k
ˆ 1 ˆ ˆ ˆ
L   (ak , s ak , s  ak , s ak , s ) F (r , t ) L F (r , t )
† †
2 s , s

24. 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.