Chapter 5: Quantization of Radiation in Cavities and …Chapter 5: Quantization of Radiation in...
Transcript of 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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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.
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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.
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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 .
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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.
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
In the Schrodinger picture,
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
25
tkatkatkatkakkd
VS LLRR ,ˆ,ˆ,ˆ,ˆˆ2
ˆ3
3
In the Schrodinger picture,
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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,
Quantum Optics for Photonics and Optoelectronics (Farhan Rana, Cornell University)
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.