Quantization in periodic time
The common quantization that we just described is suitable for cavities – where the assumption
of periodicity in space is sensible. For propagating fields it is convenient to work in the
frequency domain and not the spatial frequencies (k’s) domain - i.e. instead of 𝑎̂𝒌,𝒔 = 𝑎̂(𝒌), to
use 𝑎̂(𝜔). What most do (Loudon, Yariv) is associating 𝜔𝑘 = 𝑐|𝒌| and write :
ℏ𝜔
̂ + (𝒓, 𝑡) = 𝑖 ∑ √ 𝑘 𝑎̂𝜔 ,𝒔 (𝑡)𝑒 𝑖𝒌∙𝒓 𝜺𝒌,𝒔
𝑬
𝑘
2𝜀0 𝑉
𝜔𝑘 ,𝒔
ℏ𝜔𝑘
= 𝑖 ∑√
𝑎̂ (𝜔𝑘 , 𝑡)𝑒 𝑖𝒌∙𝒓 𝜺𝒌,𝒔
2𝜀0 𝑉 𝒔
𝜔𝑘 ,𝑠
→ 𝑎̂𝒔 (𝜔𝑘 , 𝑡) ? ?
Moreover - How to propagate the evolution of the field in some medium? even in CW, when
there is no evolution in time, the field in the entrance (z=0) is different than in the end (z=L). So
what most do (for example Yariv) is write z=ct and then propagate with the Hamiltonian from
t=0 to t=L/c (although this is space and not time). This is far from rigorous and easily wrong – we
now lose one variable out of the two (only t) :
̂ + (𝑐𝑡, 𝑡)
𝑬
This doesn’t work for dispersive materials where each frequency travels at different frequency
so it is
𝑧(𝜔) = 𝑐𝑛(𝜔)𝑡
Let’s see why it SEEMS like it works: as we know the propagator in space is the momentum 𝐺̂ .
When we propagate with the Hamiltonian times zn/c instead of t, it’s like we propagated in z
with an operator that is the Hamiltonian divided by c which is:
̂
𝐻
𝜔𝑚
1
1
†
†
(𝑧)𝑎̂𝒎 (𝑧) + ] = ∑ ℏ𝑘𝑚 [ 𝑎̂𝒎
(𝑧)𝑎̂𝒎 (𝑧) + ] = 𝐺̂
= ∑ℏ
𝑛 [ 𝑎̂𝒎
𝑐
𝑐
2
2
𝒎
𝒎
So accidently, we are doing the right thing. It’s like 64/16=4.
Here’s a way to do it properly, by B. Huttner, S. Serulnik and Y. Ben-Arye PRA 42, 5594 (1990)
HSB’s solution relies on the fact that while the energy density does change in dispersive
̂(𝑧, 𝑡) does not. This agrees with classical
medium, the energy FLUX – the pointing vector 𝑺
treatments that show the flux is the important quantity, for example Yariv’s classical analysis of
SHG and PDC.
SO our H(z), G(z) integrate over flux=Energy/time, i.e. integrate over time T
This means that we need to assume periodicity not in space, but in time – which makes much
more sense in many cases (pulses, and even in CW – the memory time of your system). T is
taken to be larger than all the other timescales in the system we analyze.
Frequency doesn’t change in different media, so instead of composing the field from eignstates
of momentum with temporally-dependent coefficients:
𝑎̂𝒌,𝒔 (𝑧, 𝑡) = 𝑎̂𝒌,𝒔 (𝑡)𝑒 𝑖𝒌∙𝒛
†
(𝑡) creates a photon with momentum k at time t
𝑎̂𝒌,𝒔
we compose the field from eignstates of frequency 𝑒 −𝑖𝜔𝑚𝑡 , with spatially-dependent
coefficients:
𝑎̂𝝎,𝒔 (𝑧, 𝑡) = 𝑎̂𝝎,𝒔 (𝑧)𝑒 −𝑖𝝎𝒕
†
(𝑧) creates a photon with frequency  at position z
𝑎̂𝝎,𝒔
So 𝑎̂𝒌† (𝑧)|0⟩ is now a state |1𝑘 (𝑧)⟩, i.e. one quanta of energy in a plane wave EM mode (in the
direction z, and area A) in which a photon will pass through z during the time T. So the photon is
STILL delocalized.
The equal-space commutation relation is now:
[ 𝑎̂𝒋 (𝑧)𝑎̂𝒌† (𝑧)] = 𝛿𝑗,𝑘
The result in free space or linear medium looks TRIVIALLY similar – c T*A (with A being the mode
area) replaces the volume V, and EE/sqrt(n) as expected classically. So as before:
̂ (𝑧, 𝑡) = 𝑬
̂ + (𝑧, 𝑡) + 𝑬
̂ − (𝑧, 𝑡)
𝑬
∗
̂ − (𝑧, 𝑡) = (𝑬
̂ + (𝑧, 𝑡))
𝑬
ℏ𝜔𝑚
̂ + (𝑧, 𝑡) = 𝑖 ∑ √
𝑬
𝑎̂ (𝑧, 𝑡)
2𝜀0 𝑐 𝑛(𝜔𝑚 )𝑇𝐴 𝑚
𝒎
𝜔𝑚 =
2𝜋𝑚
𝑇
IS THE SMALLEST FREQUENCY “SLICE” – OUR QUANTIZATION “UNIT”,
Now the Pointing vector
̂(𝑧, 𝑡) =
𝑺
1 −
̂ + (𝑧, 𝑡) + 𝐻. 𝐶. ]𝑧̂
[𝑬 (𝑧, 𝑡)𝑩
𝜇0
The intensity (see the definition earlier) of this vector is the photon flux through the plane z :
̂(𝑧, 𝑡)| = ∑ ℏ
|𝑺
𝒎,𝒎′
√𝜔𝑚 𝜔𝑚′ †
[ 𝑎̂𝒎 (𝑧)𝑎̂𝒎′ (𝑧)𝑒 𝑖(𝜔𝑚−𝜔𝑚′ )𝑡 + 𝐻. 𝐶. ]
2𝑇𝐴
Now we integrate over time to get H(z) (Instead of integrating over z to get H(t)), to get the
total number of photons that passed through the point z during the entire time T (instead of
how many photons are in the entire volume at the time t):
1 𝑇
1
1
†
̂(𝑧, 𝑡)|𝑑𝑡 = ∑ ℏ𝜔𝑚 [ 𝑎̂𝒎
̂ (𝑧) = ∫ |𝑺
(𝑧)𝑎̂𝒎 (𝑧) + ]
𝐻
2 0
2
2
𝒎
So now the operator
†
̂ (𝑧) = ∑ 𝑁
̂𝑚 (𝑧) = ∑ 𝑎̂𝒎
(𝑧)𝑎̂𝒎 (𝑧)
𝑁
𝑚
𝑚
Measures how many photons pass through the plane z during the entire time T, (instead of
̂𝑚 (𝑧) measures
measuring how many photons are in the entire volume at the time t), and each 𝑁
how many photons at frequency 𝜔𝑚 pass through the surface z.
It is easy to show that this Hamiltonian indeed satisfies:
̂ (𝑧, 𝑡))
𝜕 (𝑬
𝜕𝑡
𝑖
̂ (𝑧, 𝑡), 𝐻
̂ (𝑧)]
= − [𝑬
ℏ
So (trivial):
̂ + (𝒛, 𝑡) = 𝑖 ∑ √
𝑬
𝒎
ℏ𝜔𝑚
𝑎̂ (𝑧)𝑒 −𝑖𝜔𝑚 𝑡 𝜺𝒌,𝒔
2𝜀0 𝑐𝑇𝑛(𝜔𝑚 )𝐴 𝒎
Now to the momentum:
According to I Abram PRA 35 4661 (1987) the momentum operator is the Minkovsky form:
̂ + (𝑧, 𝑡) + 𝐻. 𝐶. ]
̂ (𝑧, 𝑡) = [𝑫− (𝑧, 𝑡)𝑬
𝒈
Leading, (in linear medium) to:
𝐺̂ (𝑧) = ∑ ℏ
𝒎
𝑛(𝜔𝑚 )𝜔𝑚 †
1
[ 𝑎̂𝒎 (𝑧)𝑎̂𝒎 (𝑧) + ]
𝑐
2
So using
̂ (𝑧, 𝑡))
𝜕 (𝑬
𝜕𝑧
=
𝑖
̂ (𝑧, 𝑡), 𝐺̂ (𝑧)]
[𝑬
ℏ
We get:
ℏ𝜔𝑚
̂ + (𝒛, 𝑡) = 𝑖 ∑ √
𝑬
𝑎̂ (0)𝑒 −𝑖𝜔𝑚(𝑡−𝑛(𝜔𝑚)𝑧/𝑐) 𝜺𝒌,𝒔
2𝜀0 𝑐𝑇𝑛(𝜔𝑚 )𝐴 𝒎
𝒎
̂ 𝒎 (𝟎) is again the slowly varying amplitude, that will change if we add
And now 𝒂
potentials/momenta
THIS was true with the regular quantization as well (that we slowly propagate in z,t with H and
G) but now we can define them (H,G) properly even in dispersive and even nonlinear medium.
And
̂ + (𝒛, 𝑡) =
𝑩
𝑛(𝜔𝑚 ) +
̂
𝑬 (𝒛, 𝑡)
𝒄
Now to continuous: when things are not period, and if you want (like I did) to work with integral
and not summation, we take T (instead of L^3) to infinity.
B. Dayan PRA 76, 043813 (2007)
I followed the similar procedure in momentum and space as in:
B. Huttner and S. M. Barnett, Phys. Rev. A 46, 4306 (1992), E. Schmidt et al., J. Mod. Opt. 45, 377
(1998), and K. J. Blow, R. Loudon, S. J. D. Phoenix, and T. J. Shepherd, Phys. Rev. A 42, 4102
(1990).
So we use:
∑
→
1
∫ 𝑑𝜔
∆𝜔
And
𝛿𝑗,𝑘 → ∆𝜔 𝛿(𝜔 − 𝜔′)
We get:
𝑎𝑚 (𝑧) → √∆𝜔 𝑎(𝜔)
And so
[𝑎(𝜔), 𝑎† (𝜔′)] = 𝛿(𝜔 − 𝜔′)
Leading to:
∞
𝑛(𝜔)
ℏ𝜔
−𝑖𝑤(𝑡 −
𝑧)
𝑐
𝐸 + (𝑧, 𝑡) = 𝑖 ∫ 𝑑𝜔 √
𝑎(𝑧, 𝑤)𝑒
4𝜋𝜀0 𝑐 𝑛(𝜔) 𝐴
0
∞
𝑛(𝜔)
ℏ
−𝑖𝑤(𝑡 −
𝑧)
𝑐
𝐴+ (𝑧, 𝑡) = ∫ 𝑑𝜔 √
𝑎(𝑧, 𝑤)𝑒
4𝜋𝜀0 𝑐 𝜔 𝑛(𝜔) 𝐴
0
∞
𝑛(𝜔)
ℏ𝜔 𝑛(𝜔)
−𝑖𝑤(𝑡 −
𝑧)
𝑐
𝐵+ (𝑧, 𝑡) = 𝑖 ∫ 𝑑𝜔 √
𝑎(𝑧,
𝑤)𝑒
3
4𝜋𝜀0 𝑐 𝐴
0
∗
̂ (𝑧, 𝑡) = 𝑬
̂ + (𝑧, 𝑡) + 𝑬
̂ − (𝑧, 𝑡) , 𝑬
̂ − (𝑧, 𝑡) = (𝑬
̂ ∓ (𝑧, 𝑡))
𝑬
Final word on wave packets:
So when you analyze a problem you need to break it down to the right operators – cavity,
atomic transitions etc. – they all have a span of frequencies and directions, but we do not need
to solve the multimode problem – we need to find the natural modes of the system and work
with them.
If we neglect the bandwidth of our interaction compared to the optical wavelength – i.e assume,
as is typically the case, that the span of frequencies that are involved in the effect we deal with
is much smaller than the optical frequency itself, then in most of the practical situations we can
use creation and annihilation operators that are wave packets (what is neglected is the change
in the pre-factor √𝜔𝑘 ):
𝑎̂𝑳𝒂𝒔𝒆𝒓 (𝒓, 𝑡) = ∑ 𝑔(𝒌, 𝒔)𝑎̂𝒌,𝒔 (0)𝑒 𝑖(𝒌∙𝒓−𝜔𝑘 𝑡)
𝒌,𝒔
Where 𝑔(𝒌, 𝒔) is the direction and polarization distribution of the laser/cavity etc..
For example – a pulse would be a coherent summation over the bandwidth (if transform limited)
+ phase noise (if not). A CW laser should be considered as a series of transform limited pulses at
the laser’s coherence length (1/BW) (i.e. in the spectrum it is infinitely narrow phase noise over
the BW).
So we are free to solve Gaussian Optics and define a combination of Ks to our mode,
and even a combination of s – but it needs to be the correct one that suits our
experiment. In any case: A MODE IS A MINIMAL PHASE-SPACE UNCERTAINTY AREA.
Which means time/frequency and space/momentum are Fourier transform pairs.
I.e. – in time it means a TRANSFORM-LIMITED PULSE (CW can be considered as a series
of transform limited pulses, each with a random phase compared to the previous one).
In space/momentum a mode is any shape with MINIMAL BEAM PARAMETER PRODUCT
(BPP) = Lambda/pi
To summarize it ALL :
1) A photon is a unit of energy in an EO mode
2) We are free to define that mode according to the physical apparatus / what we
measure in our experiment (pulse/cavity/click or homodyne.. after which
filters…)
3) E and B fields are a sum of a,a+ with pre-factors (sqrt(hw) or 1/that), and the
energy H and momentum G just count photons with the appropriate pre factor +
Vacuum energy.
We know Fock
̂|𝑛⟩ = 𝑛
⟨𝑛|𝑁
What is their E?
ℏ𝜔𝑘
†
(𝑡)𝑒 −𝑖𝒌∙𝒓 }|𝑛𝑘0 ,𝑠0 ⟩ =
⟨𝑛𝑘0 ,𝑠0 |𝐸̂ |𝑛𝑘0 ,𝑠0 ⟩ = 𝑖 ∑ √
𝜺 ⟨𝑛
|{𝑎̂ (𝑡)𝑒 𝑖𝒌∙𝒓 − 𝑎̂𝒌,𝒔
2𝜀0 𝑉 𝒌,𝒔 𝑘0 ,𝑠0 𝒌,𝒔
𝒌,𝒔
= 𝑡ℎ𝑒𝑦 𝑛𝑒𝑣𝑒𝑟 𝑚𝑎𝑡𝑐ℎ … = 0
?? not the EO wave we expect , and the same will be for all the operators of that form A,B,D…
Let’s see what state IS the EO wave we expect - the eignstates of a (Choose ONE k,s):
∞
|𝛼⟩ = ∑ 𝐶𝑛 |𝑛⟩
𝑛=0
Apply a on both sides:
∞
∞
𝑎|𝛼⟩ = ∑ 𝐶𝑛 √𝑛|𝑛 − 1⟩ = (𝑤𝑒 𝑤𝑎𝑛𝑡) = 𝛼 ∑ 𝐶𝑛 |𝑛⟩
𝑛=0
𝑛=0
𝐶𝑛 √𝑛 = 𝛼𝐶𝑛−1
𝐶𝑛 =
𝐶𝑛 =
𝛼
√𝑛
𝐶𝑛−1
𝛼2
√𝑛(𝑛 − 1)
=
𝛼𝑛
√𝑛!
𝐶𝑛−2 = ⋯
𝐶0
Normalization will give:
∞
|𝐶0
|2
∑
𝑛=0
|𝛼|2𝑛
2
= |𝐶0 |2 𝑒 |𝛼|
𝑛!
The result is the COHERENT state (defined per k):
1
|𝛼⟩ = 𝑒 −2
|𝛼|2
∑
𝛼𝑛
√𝑛!
|𝑛⟩
As coherent states are eignvectors of the annihilation operator, let’s see the E field:
ℏ𝜔𝑘
†
⟨𝛼𝑘0 ,𝑠0 |𝐸̂|𝛼𝑘0 ,𝑠0 ⟩ = 𝑖 ∑ √
𝜺 ⟨𝛼
𝑒 −𝑖(𝒌∙𝒓−𝜔𝑘𝑡) }|𝛼𝑘0 ,𝑠0 ⟩ =
|{𝑎̂ 𝑒 𝑖(𝒌∙𝒓−𝜔𝑘𝑡) − 𝑎̂𝒌,𝒔
2𝜀0 𝑉 𝒌,𝒔 𝑘0 ,𝑠0 𝒌,𝒔
𝒌,𝒔
We’ll operate the creation on the left (getting 𝛼 ∗ ) and the annihilation on the right, and
naturally only the right k,s will contribute so we drop the notation from now:
ℏ𝜔𝑘
⟨𝛼|𝐸̂ |𝛼⟩ = 𝑖 √
𝜺 [𝜶𝑒 𝑖(𝒌∙𝒓−𝜔𝑘𝑡) − 𝛼 ∗ 𝑒 −𝑖(𝒌∙𝒓−𝜔𝑘𝑡) ]
2𝜀0 𝑉 𝒌,𝒔
Taking
𝛼 = |𝛼|𝑒 𝑖𝜃
We get
ℏ𝜔𝑘
⟨𝛼|𝐸̂ |𝛼⟩ = 2|𝛼|√
𝜺 𝑠𝑖𝑛(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡 + 𝜃)
2𝜀0 𝑉 𝒌,𝒔
Which is what we expect classically!
Now – how many photons are in a coherent state ?
2
̂|𝛼𝑘 ,𝑠 ⟩ = ⟨𝛼𝑘 ,𝑠 | ∑ 𝑎̂𝒌† 𝑎̂𝒌 |𝛼𝑘 ,𝑠 ⟩ = |𝛼𝑘 ,𝑠 |
⟨𝛼𝑘0 ,𝑠0 |𝑁
0 0
0 0
0 0
0 0
So |𝛼|2 is the number of photons, and the phase of alpha is the OPTICAL phase – NOT TO BE
CONFUSED with the quantum phase, please!
2
𝑃(𝑛) = |⟨𝑛|𝛼⟩|2 = |𝐶𝑛 |2 = 𝑒 −|𝛼|
= 𝑒 −𝑛̅
NOTE: We see that the distribution is Poissonian –
No knowledge from one event to the other.
THIS IS IMPORTANT, WE’LL REMEMBER THIS
𝑛̅𝑛
𝑛!
|𝛼|2𝑛
𝑛!
We’ll get more intuition about them, and the relation to harmonic oscillators, if we see how
they look in position and momentum: Remember that from:
qˆ, pˆ   i
we get: eiq0 p / q  q  q0
ˆ
, but more importantly now:
q | pˆ  i  q  q  ,
and the fact that there are Fourier Transform of each other.
So let’s obtain the wavefunction of the ground state:
y (q) = q 0
From
𝑎̂(𝑡) =
1
√2ℏ𝜔
(𝜔 𝑞̂(𝑡) + 𝑖 𝑝̂ (𝑡))
we get:
SO
is a (normalized) Gaussian around zero, so it’s first moment
<q>=0
and uncertainty
∆𝑞 = √⟨𝑞 2 ⟩ − ⟨𝑞⟩2 = 𝜎 = √ℏ⁄2𝜔
is just its second moment = sigma^2 =
<q2>=h/2w
PROOF (basic):
q̂ =
òqq
q
y (q) = q 0
ò q1 q1
ò q1 ò q2
q̂2 = 0
= y (q1)*
q1 ò q2 q2 q2 0
q1 q2 y (q2)
= ò q2y (q)*y (q)
We know that the p representation is the Fourier transform of the q representation.
HW : Show that
Similarly, here we get
is a (normalized) Gaussian around zero, so its first moment is
<p>=0
and uncertainty
∆𝑝 = √⟨𝑝2 ⟩ − ⟨𝑝⟩2 = 𝜎 = √𝜔ℏ⁄2
Is the second moment = sigma^2 =
<p2>=wh/2
and SO;
which is the minimum uncertainty allowed by the Heisenberg uncertainty relation.
Let’s now find the q,p representation of the coherent states â    
(similarly to what we did for vacuum):
We see that  is a ground state shifted by:
q0  
2

Re  
in the position coordinate and by
p0   2  Im  
in the momentum representation.
So coherent state is simply a displaced vacuum!
We can now define the coherent state using displacement operators. Using:
𝑎̂(𝑡) =
1
√2ℏ𝜔
(𝜔 𝑞̂(𝑡) + 𝑖 𝑝̂ (𝑡))
𝑞̂(𝑡) = √ℏ/2𝜔 ( 𝑎̂(𝑡) + 𝑎̂† (𝑡))
𝑝̂ (𝑡) = −𝑖√ℏ𝜔/2 ( 𝑎̂(𝑡) − 𝑎̂† (𝑡))
So, in position space we displace using:
e
iq0 pˆ /
e

Re  aˆ †  aˆ

.
Similarly, a displacement in momentum space is:
e
 ip0 qˆ /
e

i Im  aˆ †  aˆ

.
1é
ù
ˆ , Bˆ   const , then up to a constant phase e 2ë Â,B̂û
You showed: If  A


ˆ
ˆ
ˆ ˆ
e A e B  e A B
Hence we have (up to a constant phase factor):
Where:

 / 2   q0  ip0 /  
Summarizing:
†
a = D̂a 0 = eaâ -a â 0
*
ˆ Dˆ  Dˆ .
Also, using the Baker-Hausdorff formula we get D
 
 
Notice that indeed we get the previous result:
D̂a 0 = e
=e
=e
1 2
- a
a â † -a *â
2
1 2
- a
a â †
2
e
1 2
- a
2
å
e e
0
0
an
n!
n
Let’s see how these displaced vacuum states vary in time
e iĤt a
=e
iw tn̂
=å
å
an
n!
(
a e iwt
n!
)
n
n
n
= a e iwt
A “new” alpha - Just a phase – as we expect!
Again, do not confuse – this is an OPTICAL phase of alpha – not state phase !
The reason coherent states are important – they are the result of classical sources:
classically the interaction energy from some current density 𝒋(𝒓, 𝑡) is the integration over the
volume:
𝑉(𝑟, 𝑡) = ∫ 𝒋(𝒓, 𝑡) ∙ 𝑨(𝒓, 𝑡) 𝑑3 𝑟
ℏ
†
̂ (𝒓, 𝑡) = ∑ √
(𝑡)𝑒 −𝑖𝒌∙𝒓 } 𝜺𝒌,𝒔
𝑨
{𝑎̂ (𝑡)𝑒 𝑖𝒌∙𝒓 + 𝑎̂𝒌,𝒔
2𝜔𝑘 𝜀0 𝑉 𝒌,𝒔
𝒌,𝒔
Substituting and using the Fourier components of the current:
𝑱(𝒌, 𝑡) = ∫ 𝒋(𝒓, 𝑡) 𝑒 𝑖𝒌𝑟 𝑑3 𝑟
We get for every k :
ℏ
†
(𝑡) 𝜺𝒌,𝒔 ∙ 𝑱(𝒌, 𝑡)]
𝑉(𝒌, 𝑡) = √
[𝑎̂ (𝑡) 𝜺𝒌,𝒔 ∙ 𝑱(𝒌, 𝑡) + 𝑎̂𝒌,
2𝜔𝑘 𝜀0 𝑉 𝒌,
So if we define:
𝛽𝑘 (𝑡) =
−𝑖
√2𝜔𝑘 𝜀0 𝑉ℏ
𝜺𝒌,𝒔 ∙ 𝑱(𝒌, 𝑡)
We get:
𝑉(𝒌, 𝑡) = 𝑖ℏ [𝛽𝑘 (𝑡)𝑎̂𝒌† (𝑡) − 𝛽𝑘∗ (𝑡)𝑎̂𝒌, (𝑡)]
And so the infinitesimal time-evolution operator at each point in time is:
𝑈(𝑡 + Δ𝑡, 𝑡) = 𝑒
𝑖
− 𝑉(𝑡)Δ𝑡
ℏ
=𝑒
[𝛽𝑘0 (𝑡)𝑎̂𝒌† (𝑡)−𝛽𝑘∗ 0 (𝑡)𝑎̂𝒌 (𝑡)]Δ𝑡
= 𝐷(𝛽𝑘 (𝑡)Δ𝑡)
Serge has shown that the displacement operator creates a coherent state (that was his
definition), and consecutive operations of displacements just add, so if we begin at vacuum, we
end up after time T with the coherent state:
𝑇
{𝛼𝑘 (𝑇)} = ∫ 𝛽𝑘 (𝑡) 𝑑𝑡
0
Similarly, this can (and probably will) be shown for coherent ensemble of atoms (Laser is a
mixture of those).
So Coherent state is really generated by classical sources (or approximation of)
SUMMARY:






Coherent states are the classical fields
(we’ll see how they are used as an over-complete basis – the P representation. Yes,
the letter P is WAY over-used. If P can be considered to be probability-density-like the
state of the light will be classical)
alpha^2 is the number of photons, the phase of alpha is the OPTICAL phase = the
phase of the cosin/sine
The probability of exactly n photon P(n) is Poissonian = memory-less, NO
CORRELATIONS BETWEEN PHOTONS (we’ll se more about it)
Most importantly, and by definition: They are the eign vectors of the annihilation
operator
As such – they are Pointer states – resilient to measurement/interaction by the
environment
HOW COME ?
Apples
Poissonian is REAL no knowledge from one event to the other.
It is also a direct result of the requirement that a coherent state is an eignvector of the
annihilation operator.
But how can that be if we have one less photon, that the state still has the same number of
photons ?
Because the number is uncertain to begin with, and the detection of a photon gives exactly the
right amount of knowledge. Like the beggar and the rich person. Now we even see that a state
can have larger number – the tree example:
1/sqrt(2) [vac+|100>]
After one operation of a we will collapse to the state |99> - which has MORE photons… (once
we normalize to include the knowledge that we got a photon.
Decay is caused by NOT getting apples on you.
So sum will look like decay with up jumps, some like lower jumps, and coherent will have no
jumps.
Nature Photonics