Quadratures
The quadrature operators Q,P (the BIG ones, Mandel 1034):
𝑄̂ =
𝑃̂ =
𝑎̂† + 𝑎̂
√2
𝑖(𝑎̂† − 𝑎̂)
√2
I apologize for using Q,P – once again, nobody asked me before re-using the same letters – it is a
convention. They are a scaled version of the canonical variables q,p, and they should not be
confused with the P representation and the Q representation. Some cal them X1 X2. The
normalization varies between 1 (Mandel&Wolf) to ½ (the green book). I like the symmetric
1/√2, but sometimes I neglect the normalization for simplicity.
[𝑄̂ , 𝑃̂] = 𝑖
And their uncertainty:
2
2
√⟨(Δ𝑄̂ ) ⟩ √⟨(Δ𝑃̂) ⟩ ≥ 1/2
The electric field can be presented as:
ℏ𝜔𝑘
†
𝐸̂ (𝒓, 𝑡) = 𝑖 ∑ √
𝜺 {𝑎̂ 𝑒 𝑖(𝒌∙𝒓−𝜔𝑘𝑡) − 𝑎̂𝒌,𝒔
𝑒 −𝑖(𝒌∙𝒓−𝜔𝑘 𝑡) } =
2𝜀0 𝑉 𝒌,𝒔 𝒌,𝒔
𝒌,𝒔
ℏ𝜔𝑘
= 𝑖 ∑√
𝜺 {𝑎̂ (cos(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡) + 𝑖 𝑠𝑖𝑛(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡))
2𝜀0 𝑉 𝒌,𝒔 𝒌,𝒔
𝒌,𝒔
†
− 𝑎̂𝒌,𝒔
(cos(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡) − 𝑖 𝑠𝑖𝑛(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡))}
ℏ𝜔𝑘
= −∑√
𝜺 {𝑃̂ cos(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡) +𝑄̂𝒌,𝒔 𝑠𝑖𝑛(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡)}
𝜀0 𝑉 𝒌,𝒔 𝒌,𝒔
𝒌,𝒔
i.e. they are the cosine and sine parts of the field – hence the Q-P Wigner presentation
We saw that coherent has minimal uncertainty of 1 in each
The angles=phases are arbitrary – we can define
𝑄̂ (𝜃) = 𝑎̂𝟎† 𝑒 𝑖𝜃 + 𝑎̂𝟎 𝑒 −𝑖𝜃
𝑃̂(𝜃) = 𝑎̂𝟎† 𝑒 𝑖(𝜃+𝜋/2) + 𝑎̂𝟎 𝑒 −𝑖(𝜃+𝜋/2) = 𝑄̂ (𝜃 + 𝜋/2)
That will give the same with
ℏ𝜔𝑘
𝐸̂ (𝒓, 𝑡) = − ∑ √
𝜺 {𝑃̂ cos(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡 + 𝜃) +𝑄̂𝒌,𝒔 𝑠𝑖𝑛(𝒌 ∙ 𝒓 − 𝜔𝑘 𝑡 + 𝜃)}
𝜀0 𝑉 𝒌,𝒔 𝒌,𝒔
𝒌,𝒔
Coherent state (reminder):
⟨𝛼|𝑄̂ |𝛼⟩ = ⟨𝛼|𝑎̂† + 𝑎̂|𝛼⟩ = √2 𝑅𝑒(𝛼)
⟨𝛼|𝑃̂|𝛼⟩ = ⟨𝛼|𝑖(𝑎̂† − 𝑎̂)|𝛼⟩ = √2 𝐼𝑚(𝛼)
Which means
𝛼=
𝑄̅ + 𝑖𝑃̅
√2
THE DRAWING OF THE QUADRATURES, THE MEANING OF THE AREA IS NOISE
Let’s start DOING things to light.
The Beam-Splitter Formalism:
1
2
0
3
Classically, we view it as a 3-port problem: incoming wave v1 gives rise to two waves v2=rv1,
v3=tv1, that conserve energy thanks to the 3rd rule above.
Now suppose we want to implement this to our quantum mechanical creation and annihilation
operators, which of course obey:
[ 𝑎̂𝒋 𝑎̂𝒌† ] = 𝛿𝑗,𝑘
But if we write a2=ra1, a3=ta1 we get (except for the trivial cases where t or r are 1 and the
other 0, so this is not a BS at all):
[ 𝑎̂𝟐 𝑎̂𝟐† ] = |𝑟|2 [ 𝑎̂𝟏 𝑎̂𝟏† ] = |𝑟|2 ≠ 1
[ 𝑎̂𝟑 𝑎̂𝟑† ] = |𝑡|2 [ 𝑎̂𝟏 𝑎̂𝟏† ] = |𝑡|2 ≠ 1
[ 𝑎̂𝟐 𝑎̂𝟑† ] = 𝑟𝑡 ∗ [ 𝑎̂𝟏 𝑎̂𝟏† ] = 𝑟𝑡 ∗ ≠ 0
First of all – you can’t turn one mode into two or the other way (but yes a linear combination) so
there is no 1X2 device, only 2X2, so it is wrong to ignore the other input.
Specifically, the problem is the Vacuum – H=Sum(hw(n+1/2)). Even for VAC we have not only
Energy, but INFINITE energy !! These are the vacuum fluctuations (VF), and what is called the
zero-point energy (ZPE). One cannot overestimate the fundamental importance of this in
Nature, and it is part of the answer to what is quantum in quantum optics.
So it is like in Gollum’s riddle to Bilbo:
“It cannot be seen (a|0>=0), cannot be felt, cannot be heard, cannot be smelt.
It lies behind stars, and under hills, and empty holes (modes) it fills”
The answer is “Dark”, and to see that vacuum=dark indeed creeps everywhere in a way that is
fundamentally important to the quantum description of Nature, including the most simple nontrivial optical elements – the Beam Splitter (BS):
So the reason is the fact that we ignored the NOTHING – Gollum’s Dark.. – the vacuum that fills
the empty modes and comes in through the empty port:
r,t is from one side, and r’,t’ is from the other – assuming there are no losses, it’s matrix is
unitary, which means that:
|𝑟| = |𝑟 ′ |
|𝑡| = |𝑡 ′ |
|𝑟|2 + |𝑡|2 = 1
𝑟𝑡 ∗ + 𝑟′∗ 𝑡 ′ = 0
It is SO easy, HW 
So the correct transformation is:
𝑎̂𝟐 = 𝑟 𝑎̂𝟏 + 𝑡′ 𝑎̂𝟎
𝑎̂3 = 𝑡 𝑎̂𝟏 + 𝑟′ 𝑎̂𝟎
And now indeed
[ 𝑎̂𝟐 𝑎̂𝟐† ] = |𝑟|2 [ 𝑎̂𝟏 𝑎̂𝟏† ] + |𝑡|2 [ 𝑎̂𝟎 𝑎̂𝟎† ] = 1
[ 𝑎̂𝟑 𝑎̂3† ] = |𝑡|2 [ 𝑎̂𝟏 𝑎̂𝟏† ] + |𝑟|2 [ 𝑎̂𝟎 𝑎̂𝟎† ] = 1
[ 𝑎̂𝟐 𝑎̂𝟑† ] = 𝑟𝑡 ∗ [ 𝑎̂𝟏 𝑎̂𝟏† ] + 𝑟′∗ 𝑡′[ 𝑎̂0 𝑎̂0† ] = 0
So the vacuum field plays a fundamental role and required for internal consistency, and it has
consequences in quantum electrodynamics that have no classical counterpart in classical
electrodynamics and cannot be ignored. It is from here trivial to show that this is the correct
transformation and that, for example, the number of photons is conserved n0+n1=n2+n3.
A set of r,r’,t,t’ that satisfies the above conditions and is suitable for DIELECTRIC BS is:
𝑟′ = 𝑟
= 𝑖|𝑟|
𝑡′ = 𝑡
𝑤ℎ𝑒𝑟𝑒 𝑟, 𝑡 𝑖𝑠 𝑟𝑒𝑎𝑙
Alternatively you can have: 𝑟 ′ = −𝑟
= |𝑟|
In practice, it usually doesn’t matter (I never saw a case where it mattered), as long as you are
consistent throughout your calculation
What happens to quadratures in a BS:
2
1
𝑎̂𝟐 = 𝑖𝑟 𝑎̂𝟏 + 𝑡 𝑎̂𝟎
0
3
𝑎̂3 = 𝑡 𝑎̂𝟏 + 𝑖𝑟 𝑎̂𝟎
It is easy to see that in “reverse” we get:
𝑎̂𝟏 = 𝑡 𝑎̂𝟑 − 𝑖𝑟 𝑎̂𝟐
𝑎̂0 = 𝑡 𝑎̂𝟐 − 𝑖𝑟 𝑎̂𝟑
So
𝑄̂𝟐 = 𝑎̂𝟐† + 𝑎̂𝟐 = −𝑖𝑟 𝑎̂𝟏† + 𝑡𝑎̂𝟎† + 𝑖𝑟 𝑎̂𝟏 + 𝑡 𝑎̂𝟎 = 𝑡 𝑄̂𝟎 − 𝑟 𝑃̂𝟏
𝑃̂𝟐 = 𝑖𝑎̂𝟐† − 𝑖 𝑎̂𝟐 = 𝑟 𝑎̂𝟏† + 𝑖𝑡𝑎̂𝟎† + 𝑟 𝑎̂𝟏 − 𝑖𝑡 𝑎̂𝟎 = 𝑡 𝑃̂𝟎 + 𝑟 𝑄̂𝟏
𝑄̂𝟑 = 𝑎̂𝟑† + 𝑎̂𝟑 = −𝑖𝑟 𝑎̂𝟎† + 𝑡𝑎̂𝟏† + 𝑖𝑟 𝑎̂𝟎 + 𝑡 𝑎̂𝟏 = 𝑡 𝑄̂𝟏 − 𝑟 𝑃̂𝟎
𝑃̂3 = 𝑖𝑎̂𝟑† − 𝑖 𝑎̂𝟑 = 𝑟 𝑎̂𝟎† + 𝑖𝑡𝑎̂𝟏† + 𝑟 𝑎̂𝟎 − 𝑖𝑡 𝑎̂𝟏 = 𝑡 𝑃̂𝟏 + 𝑟 𝑄̂0
SO What happens to COHERENT STATES in a BS:
𝑄̂𝟐 + 𝑖 𝑃̂𝟐 = 𝑡(𝑄0 + 𝑖𝑃0) + 𝑖𝑟(𝑄1 + 𝑖𝑃1)
we can also see that by:
𝐷0 (𝛼)𝐷1 (𝛽)|𝑣𝑎𝑐⟩ = 𝑒𝑥𝑝(𝛼𝑎̂𝟎† − 𝛼 ∗ 𝑎̂𝟎 + 𝛽𝑎̂𝟏† − 𝛽 ∗ 𝑎̂𝟏 )|𝑣𝑎𝑐⟩
= 𝑒𝑥𝑝(𝛼𝑡 𝑎̂𝟐† + 𝛼 𝑖𝑟 𝑎̂𝟑† − 𝛼 ∗ 𝑡 𝑎̂𝟐 + 𝑖𝑟𝛼 ∗ 𝑎̂𝟑 + 𝛽𝑡 𝑎̂𝟑† + 𝑖𝛽𝑟 𝑎̂𝟐† − 𝛽 ∗ 𝑡 𝑎̂𝟑 + 𝑖𝛽 ∗ 𝑟 𝑎̂𝟐 )|𝑣𝑎𝑐⟩
= 𝑒𝑥𝑝 ((𝑡𝛼) 𝑎̂𝟐† − (𝑡𝛼)∗ 𝑎̂𝟐 + (𝑖𝑟𝛼) 𝑎̂3† − (𝑖𝑟𝛼)∗ 𝑎̂𝟑 + (𝑡𝛽) 𝑎̂𝟑† − (𝑡𝛽)∗ 𝑎̂𝟑 + (𝑖𝑟𝛽) 𝑎̂2†
− (𝑖𝑟𝛽)∗ 𝑎̂𝟐 ) |𝑣𝑎𝑐⟩
= 𝐷2 (𝑡𝛼 + 𝑖𝑟𝛽)𝐷3 (𝑡𝛽 + 𝑖𝑟𝛼)
QED
So how do you do displacement in the lab?
Displacement by a beam-splitter and a strong Local Oscillator (LO)
The previous derivation shows that by having a strong LO , using abeam-splitter with very small
𝑟 and 𝑟 ≈ 1 you get that the Q,P of your state will be practically the Q0,P0 of the original state,
with - rP1 added to the Q quadrature, and rQ1 added to the P quadrature, but the noise, which
is the same for large LO as for vacuum, will be decreased by r – i.e. a displacement without
adding noise.
Homodyne
To measure the quadrature of a state (which is Hermitian and so an observable) what you do is
Homodyne, which is mixing your state with a strong LO using 50:50 beamsplitter, and SUBTRACT
the currents of the two output detectors. This “isolates” the interference term between the two
beams (since the average intensities are identical in both detectors, but the interference term is
opposite in sign), and in classical optical detection it is used for detection of weak signals (since
it is proportional to the LO amplitude). It is typically better to have a difference between the
frequency of the LO and the signal, in which case the interference term oscillates at the
difference frequency (beating) and so is detected by lock-in amplifier (avoiding the noises
around DC, the 1/f in particular). In this case the measurement is called Heterodyne.
1
2
0
3
1
1
⟨𝐼3 − 𝐼2 ⟩ ∝ ⟨ 𝑎̂𝟑† 𝑎̂𝟑 − 𝑎̂2† 𝑎̂2 ⟩ = ( 𝑎̂𝟏† − 𝑖 𝑎̂𝟎† )( 𝑎̂𝟏 + 𝑖 𝑎̂𝟎 ) − ( 𝑎̂𝟎† − 𝑖 𝑎̂𝟏† )( 𝑎̂𝟎 + 𝑖 𝑎̂𝟏 )
2
2
1
= ⟨ 𝑎̂𝟏† 𝑎̂𝟏 − 𝑖 𝑎̂𝟎† 𝑎̂𝟏 + 𝑖 𝑎̂𝟏† 𝑎̂𝟎 + 𝑎̂𝟎† 𝑎̂𝟎 − 𝑎̂𝟎† 𝑎̂𝟎 + 𝑖 𝑎̂𝟏† 𝑎̂𝟎 − 𝑖 𝑎̂𝟎† 𝑎̂𝟏 − 𝑎̂𝟏† 𝑎̂𝟏 ⟩
2
= 𝑖⟨ 𝑎̂𝟏† 𝑎̂𝟎 − 𝑎̂𝟎† 𝑎̂𝟏 ⟩
Assume a1 is a strong (=classical, stronger than 𝑎̂𝟎 ) local oscillator with amplitude =
|𝐿𝑂|𝑒 𝑖(𝜃+𝜋/2) :
⟨𝐼3 − 𝐼2 ⟩ ∝ |𝐿𝑂|⟨ 𝑎̂𝟎† 𝑒 𝑖𝜃 + 𝑎̂𝟎 𝑒 −𝑖𝜃 ⟩ = |𝐿𝑂| 𝑄̂𝟎 (𝜃)
And then the noise comes only from the measured state and not from the LO
HW! (hint - since we can approximate the expectation value:
⟨ 𝑎̂𝟏 𝑎̂𝟏† ⟩ = ⟨1 + 𝑎̂𝟏† 𝑎̂𝟏 ⟩ ≈ ⟨ 𝑎̂𝟏† 𝑎̂𝟏 ⟩
Leading to:
2
⟨(𝐼3 − 𝐼2 )2 ⟩ ∝ |𝐿𝑂|2 ⟨( 𝑄̂𝟎 (𝜃)) ⟩
⟨Δ(𝐼3 − 𝐼2 )2 ⟩ = ⟨(𝐼3 − 𝐼2 )2 ⟩ − ⟨𝐼3 − 𝐼2 ⟩2
2
∝ |𝐿𝑂|2 ⟨( 𝑄̂𝟎 (𝜃)) ⟩ − |𝐿𝑂|2 ⟨ 𝑄̂𝟎 (𝜃)⟩2 = |𝐿𝑂|2 ⟨Δ 𝑄̂𝟎 (𝜃)2 ⟩
DRAW THE DRAWING – while varying the phase of the LO the state rotates around the origin
and we measure the projection on one axis so we see the shape – like tomography.