Mach-Zender, loss of coherence and the Quantum Ereaser:
Let’s look at a Mach-Zender interferometer, and propagate from the right – present the
annihilation (detection) operators at the output as a function of the input annihilation
operators:
c
f
a
𝜃
b
e
d
𝑐=
𝑖𝑎 + 𝑏
,
𝑑=
𝑖𝑎 + 𝑏 + 𝑖𝑎 − 𝑏
= 𝑖𝑎 ,
2
𝑓=
√2
𝑎 + 𝑖𝑏
√2
,
For BALANCED Mach-Zender:
𝑒=
𝑎 + 𝑖𝑏 − 𝑎 + 𝑖𝑏
= 𝑖𝑏 ,
2
Direct mapping – whatever comes in a (b) gets out in e (f)
Now we add a phase (=delay assuming single mode):
𝑒=
𝑖𝑎 + 𝑏 + 𝑖𝑒 𝑖𝜃 (𝑎 + 𝑖𝑏)
,
2
𝑒=𝑖
𝑓=
𝑓=
𝑒 𝑖𝜃 (𝑎 + 𝑖𝑏) − 𝑎 + 𝑖𝑏
,
2
𝜃
𝜃
𝜃
(1 + 𝑒 𝑖𝜃 )
(1 − 𝑒 𝑖𝜃 )
𝑎+
𝑏 = 𝑖𝑒 𝑖 2 [𝑎 cos − 𝑏 sin ] ,
2
2
2
2
𝜃
𝜃
𝜃
(𝑒 𝑖𝜃 − 1)
(1 + 𝑒 𝑖𝜃 )
𝑎+𝑖
𝑏 = 𝑖𝑒 𝑖 2 [𝑎 sin + 𝑏 cos ]
2
2
2
2
If b is Vac and a is WHATEVER we get probability cos(teta/2)^2 and sin(teta/2)^2
We can see it also be propagating from the other direction:
Start with the creation operators on vacuum at the input, and propagate the creation operators
to see the final state, and then check the detection probabilities. The advantage – you can
ignore the Vacuum input (if there is one), as it doesn’t lead to a creation operator at the end! In
this case:
𝑎† →
𝑒 𝑖𝜃 𝑑† + 𝑖𝑐 †
→
𝑒 𝑖𝜃 𝑓 † + 𝑖𝑒 𝑖𝜃 𝑒 † + 𝑖𝑒 † − 𝑓 †
(𝑒 𝑖𝜃 − 1) †
(𝑒 𝑖𝜃 + 1) †
=
𝑓 +𝑖
𝑒
2
2
2
√2
We clearly see the interference here as well – whatever composes a will come to f,e
alternatingly: sin
𝜃
2
𝑓 † + cos
𝜃
2
𝑒†
Now a which-path detector is introduced by using the transformation, including phase on the darm:
𝑎† →
→
𝑒 𝑖𝜃 𝑑† ℎ† + 𝑖𝑐 † 𝑔†
√2
, 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎𝑛 𝐸𝑁𝑇𝐴𝑁𝐺𝐿𝐸𝐷 𝑆𝑇𝐴𝑇𝐸 𝑜𝑓 𝑐𝑜𝑢𝑟𝑠𝑒
𝑒 𝑖𝜃 (𝑓 † + 𝑖𝑒 † )ℎ† + 𝑖(𝑒 † + 𝑖𝑓 † )𝑔†
= 𝑒 𝑖𝜃 𝑓 † ℎ† + 𝑖𝑒 𝑖𝜃 𝑒 † ℎ† − 𝑓 † 𝑔† + 𝑖𝑒 † 𝑔†
2
g
c
a
f
𝜃
e
d
h
Now the final states always include TWO photons: GE, GF, HE, HF.
Do this properly and get:
⟨ |𝑓 † 𝑓| ⟩ =
1 1
+ = 1/2
4 4
NO INTERFERENCE!
Since these states are distinguishable, we must calculate the PROBABILITY of each, and so E and
F have separate probabilities and can not interfere. We see that by the fact that we are forced
to detect G,H to get a number and not be left with a state after we detect E or F. This is called
“tracing-over”.
Since each such state has only ONE possible path that leads to it, there is no interference when
we compute the probability of any of them, since:
What interferes in quantum mechanics are, always, the PASTS that lead to each single state as Feynman said.
M. O. Scully, K. Drühl, “Quantum eraser - A proposed photon correlation experiment concerning
observation and ‘delayed choice’ in quantum mechanics“, Phys. Rev. A, 25, 2208-2213 (1982)
Now we put the Quantum eraser, (which is not what I thought), which will allow us to detect the
photons emitted to G,H without having which-path information:
c
k
j
g
a
f
𝜑
h
e
𝜃
d
𝑒 𝑖𝜃+𝜑 𝑓 † (𝑘 † + 𝑖𝑗 † ) + 𝑖𝑒 𝑖𝜃+𝜑 𝑒 † (𝑘 † + 𝑖𝑗 † ) − 𝑓 † (𝑗 † + 𝑖𝑘 † ) + 𝑖𝑒 † (𝑗 † + 𝑖𝑘 † )
(𝑒 𝑖𝜃+𝜑 − 𝑖)𝑓 † 𝑘 † + (𝑖𝑒 𝑖𝜃+𝜑 − 1)𝑒 † 𝑘 † + (𝑖 − 𝑒 𝑖𝜃+𝜑 )𝑒 † 𝑗 † + (𝑖𝑒 𝑖𝜃+𝜑 − 1)𝑓 † 𝑗 †
𝜋
𝜋
𝜋
𝜋
(𝑒 𝑖𝜃+𝜑+ 2 + 1) 𝑓 † 𝑘 † + 𝑖 (𝑒 𝑖𝜃+𝜑+ 2 − 1) 𝑒 † 𝑘 † − (1 + 𝑒 𝑖𝜃+𝜑+ 2 ) 𝑒 † 𝑗 † + 𝑖 (𝑒 𝑖𝜃+𝜑+ 2 − 1) 𝑓 † 𝑗 †
Now the Final states are JE, JF, KE, KF – AND WE HAVE INTERFERENCE AGAIN:
While indeed EACH one of these events is distinguishable (and so no interference exists
between E and F – THERE IS NO WAY TO RECOVER THE COHERENCE=“SECRECY” AND REDUCE
THE HILBERT SPACE BACK), each one of the has TWO paths leading to it and so each one of them
expresses a SECOND-ORDER interference – the probability for COINCIDENCES (easily derived by
DRAWING the phazors, and where they are zero) is NONLOCAL:
𝑃(𝐹𝐾) ∝ sin (
𝜃+𝜑+
𝑃(𝐸𝐾) ∝ cos (
𝑃(𝐸𝐽) ∝ sin (
2
𝜋 2
2)
𝜋
𝜃+𝜑+2
2
𝜋
𝜃+𝜑+2
𝑃(𝐹𝐽) ∝ cos (
2
)
2
)
𝜋
𝜃+𝜑+2
2
2
2
)
𝑆𝑜 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐𝑎𝑙𝑙𝑦 − 𝑖𝑓 𝑤𝑒 𝑘𝑛𝑜𝑤 𝐽 𝑜𝑟 𝐾 𝑜𝑐𝑐𝑢𝑟𝑒𝑑 𝑎𝑛𝑑 𝑤𝑒 𝑝𝑜𝑠𝑡 − 𝑠𝑒𝑙𝑒𝑐𝑡,
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝐸 𝑜𝑟 𝐹 𝑏𝑒ℎ𝑎𝑣𝑒 𝑙𝑖𝑘𝑒 𝑆𝑖𝑛𝑒 2 , 𝐶𝑜𝑠𝑖𝑛𝑒 2 𝑜𝑟 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑤𝑎𝑦 𝑎𝑟𝑜𝑢𝑛𝑑,
𝑎𝑛𝑑 𝑜𝑓𝑐𝑜𝑢𝑟𝑠𝑒 𝑑𝑒𝑡𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸 𝑜𝑟 𝐹 𝑤𝑖𝑙𝑙 𝑠ℎ𝑜𝑤 𝑡ℎ𝑒 𝑆𝑖𝑛𝑒 2 , 𝐶𝑜𝑠𝑖𝑛𝑒 2 𝑖𝑛 𝐾, 𝐽
This scheme was used to create a Quantum Memory with atomic ensembles for long quantum
communication channels (using entanglement swapping between the nodes), in the so-called
“DLCZ protocol”: Theoretical: Nature414_413_(2001) Long Q communication with Atomic
ensambles, L.-M. Duan*², M. D. Lukin³, J. I. Cirac* & P. Zoller
Experimental Realization: Nature423_731_(2003) Generation nonclassical photons for Q
communication atomic ensambles, A. Kuzmich, W. P. Bowen, A. D. Boozer, A. Boca, C. W. Chou,
L.-M. Duan
(The issue of COHERENCE will become clearer once we go back to multimode)
Reminders:
The quadrature operators Q,P (the BIG ones, Mandel 1034):
𝑄̂ =
𝑃̂ =
𝑎̂† + 𝑎̂
√2
𝑖(𝑎̂† − 𝑎̂)
√2
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)
The Quadratures of the electromagnetic field are the sine (Q) and cosine (P) of the wave, and
a coherent state is accordingly:
𝛼=
𝑄̅ + 𝑖𝑃̅
√2
COHERENT STATES in a BS:
1
2
0
3
𝑄̂𝟐 + 𝑖 𝑃̂𝟐 = 𝑡(𝑄0 + 𝑖𝑃0) + 𝑖𝑟(𝑄1 + 𝑖𝑃1)
𝐷0 (𝛼)𝐷1 (𝛽)|𝑣𝑎𝑐⟩ = 𝐷2 (𝑡𝛼 + 𝑖𝑟𝛽)𝐷3 (𝑡𝛽 + 𝑖𝑟𝛼)
and their phase-space meaning, what happens in a beam splitter, displacement by mixing with
a strong LO on a t~1 beamsplitter, and tomography by Homodyne detection:
⟨𝐼3 − 𝐼2 ⟩ ∝ |𝐿𝑂|⟨ 𝑎̂𝟎† 𝑒 𝑖𝜃 + 𝑎̂𝟎 𝑒 −𝑖𝜃 ⟩ = |𝐿𝑂| 𝑄̂𝟎 (𝜃)
⟨Δ(𝐼3 − 𝐼2 )2 ⟩ = ⟨(𝐼3 − 𝐼2 )2 ⟩ − ⟨𝐼3 − 𝐼2 ⟩2 ∝ |𝐿𝑂|2 ⟨Δ 𝑄̂𝟎 (𝜃)2 ⟩
The importance of the Vacuum port
1st point: Detector efficiency: The BS model for Loss : 90% efficient detectors are modeled as
100% detectors with BS before |𝒕|𝟐 = 𝟎. 𝟗, |𝒓|𝟐 = 𝟎. 𝟏, and so they bring in 10% vacuum noise
(so we are limited at measuring 10dB squeezing with such detectors).
The best measurements are interferometric, so let’s look at a Mach-Zender interferometer.
c
f
a
b
e
𝜃
d
𝑒=
𝑖𝑎 + 𝑏 + 𝑖𝑒 𝑖𝜃 (𝑎 + 𝑖𝑏)
,
2
𝑒=𝑖
𝑓=
𝑓=
𝑒 𝑖𝜃 (𝑎 + 𝑖𝑏) − 𝑎 + 𝑖𝑏
,
2
𝜃
𝜃
𝜃
(1 + 𝑒 𝑖𝜃 )
(1 − 𝑒 𝑖𝜃 )
𝑎+
𝑏 = 𝑖𝑒 𝑖 2 [𝑎 cos − 𝑏 sin ] ,
2
2
2
2
𝜃
𝜃
𝜃
(𝑒 𝑖𝜃 − 1)
(1 + 𝑒 𝑖𝜃 )
𝑎+𝑖
𝑏 = 𝑖𝑒 𝑖 2 [𝑎 sin + 𝑏 cos ]
2
2
2
2
Indeed for equal paths (𝜃 = 0) we have identity, 𝑒 = 𝑖𝑎, 𝑓 = 𝑖𝑏
In this case the measurement of tiny changes in the path is done by looking at the “Dark Port”.
Typically this is not a good point for measurement since the light in the dark port depends
quadratically on the phase: assume just a small phase 𝜀 and that b is |vac> and a is a local
oscillator (LO) coherent state: 𝛼 = |𝐿𝑂|𝑒 𝑖𝛽 , then the expectation value is of the dark port is
(taking Taylor expansion, keeping only up to 2nd order):
2
𝑐𝑜𝑠 2 (𝑥) = 12(1+cos(2𝑥)) ∝ 12(1+1−2𝑥2 )=(1 − 𝑥 2 ), and (𝜃 = 𝜀)
⟨𝐿𝑂, 𝑣𝑎𝑐|𝑓 † 𝑓 |𝐿𝑂, 𝑣𝑎𝑐⟩ ≈
⟨𝐿𝑂, 𝑣𝑎𝑐| 𝑏̂ † 𝑏̂ (1 −
𝜀2
𝜀
𝜀2
|𝐿𝑂, 𝑣𝑎𝑐⟩
) + (𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂ ) + 𝑎̂† 𝑎̂
4
2
4
AND WE DON'T NEED ALL THAT BECAUSE ONLY THE LAST TERM SURVIVES THE VACUUM IN b:
≈ |𝐿𝑂|2
𝜀2
4
𝜋
The better working point for an interferometer is around 𝜃 = 2 , where there are no dark ports.
In this case subtracting the currents will give the change that results from tiny phase shifts:
𝐼𝑓 − 𝐼𝑒 ∝ 𝑓 † 𝑓 − 𝑒 † 𝑒
𝑓 †𝑓
𝜃 2
𝜃
𝜃
𝜃 2
= 𝑏̂ † 𝑏̂ (cos ) + cos sin (𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂ ) + 𝑎̂† 𝑎̂ (sin )
2
2
2
2
𝑒 †𝑒
𝜃 2
𝜃
𝜃
𝜃 2
= 𝑎̂† 𝑎̂ (cos ) − cos sin (𝑏̂ † 𝑎̂ + 𝑎̂† 𝑏̂ ) + 𝑏̂ † 𝑏̂ (sin )
2
2
2
2
Using
𝜃 2
2 (sin ) = 1 − cos 𝜃 ,
2
𝜃
𝜃
2 cos sin = sin 𝜃 ,
2
2
𝜃 2
2 (cos ) = 1 + cos 𝜃
2
We get:
𝐼𝑓 − 𝐼𝑒 ∝ 𝑓 † 𝑓 − 𝑒 † 𝑒
= (𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂ ) sin 𝜃 + (𝑏̂ † 𝑏̂ − 𝑎̂† 𝑎̂ ) cos 𝜃
𝜋
Exactly at 𝜃 = 2 :
𝐼𝑓 − 𝐼𝑒 ∝ 𝑓 † 𝑓 − 𝑒 † 𝑒
= 𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂
Again, if a is some LO : 𝛼 = |𝐿𝑂|𝑒 𝑖𝛽 the expectation value is:
⟨𝐼𝑓 − 𝐼𝑒 ⟩ ∝ ⟨𝑓 † 𝑓 − 𝑒 † 𝑒 ⟩ = ⟨𝛼|𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂ |𝛼⟩ = |𝐿𝑂|⟨𝑏̂ † 𝑒 𝑖𝛽 + 𝑏̂𝑒 −𝑖𝛽 ⟩ = |𝐿𝑂|⟨𝑄̂𝑏 (𝛽)⟩
Exactly like Homodyne ! we get the quadrature of the EMPTY !! input port b.
Indeed if we put nothing (vacuum) in b then we get average of zero.
𝜋
2
To see our sensitivity let’s look at a small phase shift 𝜃 = + 𝜀:
𝐼𝑓 − 𝐼𝑒 ∝ 𝑓 † 𝑓 − 𝑒 † 𝑒
= (𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂ ) cos 𝜀 − (𝑏̂ † 𝑏̂ − 𝑎̂† 𝑎̂ ) sin 𝜀
≈ (𝑎̂† 𝑏̂ + 𝑏̂ † 𝑎̂ ) (1 −
𝜀2
†
†
) − 𝜀(𝑏̂ 𝑏̂ − 𝑎̂ 𝑎̂ )
2
⟨𝐼𝑓 − 𝐼𝑒 ⟩ ∝ ⟨𝑓 † 𝑓 − 𝑒 † 𝑒 ⟩ ≈
𝜀2
(1 − ) |𝐿𝑂|⟨𝑄̂𝑏 (𝛽)⟩ + |𝐿𝑂|2 𝜀
2
= |𝐿𝑂|2 𝜀
Most of the signal comes from the coherent input port, with Linear dependence indeed.
And what about the NOISE ?
It is obvious to see that
Most of the noise will come from the VACUUM input port
Let’s prove:
2
⟨(𝐼𝑓 − 𝐼𝑒 ) ⟩ ∝ ⟨𝛼|(𝑏 † 𝑎 + 𝑎† 𝑏)(𝑏† 𝑎 + 𝑎† 𝑏)|𝛼⟩
= ⟨𝛼|𝑏 † 𝑎 𝑏 † 𝑎 + 𝑏 † 𝑎𝑎† 𝑏 + 𝑎† 𝑏𝑏 † 𝑎 + 𝑎† 𝑏𝑎† 𝑏|𝛼⟩
= ⟨𝛼|𝑏 † 𝑎 𝑏 † 𝑎 + 𝑏 † (1 + 𝑎† 𝑎)𝑏 + 𝑎† 𝑏𝑏 † 𝑎 + 𝑎† 𝑏𝑎† 𝑏|𝛼⟩
Exactly like the calculation in Homodyne, if we assume the LO is large we can neglect the 1 (in
the expectation value) and get:
2
= |𝐿𝑂|2 ⟨𝑏† 𝑏† 𝑒 2𝑖𝛽 + 𝑏 † 𝑏 + 𝑏𝑏 † + 𝑏𝑏𝑒 −2𝑖𝛽 ⟩ = |𝐿𝑂|2 ⟨(𝑄̂ (𝛽)) ⟩
Since:
2
⟨(𝑄̂ ) ⟩ = ⟨(𝑎̂† 𝑒 𝑖𝜃 + 𝑎̂𝑒 −𝑖𝜃 )(𝑎̂† 𝑒 𝑖𝜃 + 𝑎̂𝑒 −𝑖𝜃 )⟩
= ⟨𝑎̂† 𝑎̂† 𝑒 2𝑖𝜃 + 𝑎̂† 𝑎̂ + 𝑎̂ 𝑎̂† + 𝑎̂ 𝑎̂ 𝑒 −2𝑖𝜃 ⟩
So we proved that the NOISE in the subtraction is the quadrature NOISE in the EMPTY port !
So our detection limit will be given by this noise –
2
minimal |𝐿𝑂|2 𝜀 will be proportional to the RMS of |𝐿𝑂|2 ⟨(𝑄̂ (𝛽)) ⟩, i.e. to |𝐿𝑂|/√2
Replacing |𝐿𝑂|2 by the number of photons n we get:
𝜀 ∝ 1/√𝑛
which is called the standard limit
Note that the theoretical limit to phase measurement is (handwavingly) given using Heisenberg
uncertainty by
 ~ 1/ n . We don’t achieve this so called Heisenberg Limit by putting vacuum
in the empty port.
SEEMS LIKE THERE IS NOTHING WE CAN DO ?!
Squeezing of course