Communication systems and entanglement

[IllyaKuryakin]

Has anyone else picked up the similar themes in these 'new peoples' posts, and what Varon seems to be asking. I wonder if they're the same person, just with different usernames.

I picked up Varon seems to have an unhealthy obsession with interpretations of QM, and he later posted not knowing which one is correct is insanity, at least to him.

Said in jest Stevie. No offence meant, I hope none taken.

 

[UChr]

I will still present my second suggestion (Despite possible frustration)
The idea - as previously mentioned: interference / non-interference.

I compare with the Walborn's experiment: 'A double-slit quantum eraser' which has been discussed in the forum.
I transmit from ‘p to s’: p measured first - and even before the s-twin reaches the 'double slit'.
Transmitter - position 0:
Instead of Dp I place a polarizing beam splitter PBS(0) that transmits light polarized in the direction 0 degree = horizontal and reflects vertical light (90 degrees). This is followed by two detectors Dp(0) and Dp(90).
At Walborn should a coincidence counter between Dp(0) and Ds gives fringes - and between Dp(90) and Ds gives anti-Fringe (or vice versa).

Receiver:

When photon p is measured (0 or 90) the twin s will be set perpendicular - ie vertically / horizontally.
The receiver starts with a PBS(0).

All the now horizontal = ‘Dp (90)’ are transmitted and meets now a double slit with a detector. That should give anti-Fringes by this detector.

All the now vertical = initial horizontal = Dp(0) are reflected and meets another double slit with a detector. This should give fringes.

All in all: Interference by both double slit.



Transmitter position 1 - follow - if the above seem ok.

 

[DrChinese]

I will still present my second suggestion (Despite possible frustration)
The idea - as previously mentioned: interference / non-interference.

I compare with the Walborn's experiment: 'A double-slit quantum eraser' which has been discussed in the forum.
I transmit from ‘p to s’: p measured first - and even before the s-twin reaches the 'double slit'.
Transmitter - position 0:
Instead of Dp I place a polarizing beam splitter PBS(0) that transmits light polarized in the direction 0 degree = horizontal and reflects vertical light (90 degrees). This is followed by two detectors Dp(0) and Dp(90).
At Walborn should a coincidence counter between Dp(0) and Ds gives fringes - and between Dp(90) and Ds gives anti-Fringe (or vice versa).

Receiver:

When photon p is measured (0 or 90) the twin s will be set perpendicular - ie vertically / horizontally.
The receiver starts with a PBS(0).

All the now horizontal = ‘Dp (90)’ are transmitted and meets now a double slit with a detector. That should give anti-Fringes by this detector.

All the now vertical = initial horizontal = Dp(0) are reflected and meets another double slit with a detector. This should give fringes.

All in all: Interference by both double slit.



Transmitter position 1 - follow - if the above seem ok.

I don't follow your example. Are you trying to say that the transmitter does something (it is not clear what) and the receiver sees something different (also not clear what)? Because the pattern will actually be the same regardless. But can you clarify?

 

[San K]

It is also true, as UChr says, that the beams can be reassembled to restore entanglement. And that is true EVEN IF the entanglement was broken on the other end already. (Shock!)

you mean entanglement is restored non-locally?...you mean entanglement is restored between "unconnected" photons separated by time-space and not connected to/by any common source?

in my opinion the entanglement is not restored (and cannot be restored).

the re-appearance of the fringes (or anti-fringes) can be explained by sub-samples...(via coincidence counter)

 

[DrChinese]

you mean entanglement is restored non-locally?...

As "non-locally" as it was broken, yes. No one actually knows the point in time at which the superposition ceases. We speak as if it ends when the first observation is performed, and in a sense that is true. But really, that is more of a convenience for discussion than anything else.

 

[UChr]

This is only position '0'.

The transmitter polarizes p-photons horizontally or vertically.

Receiver PBS sends the now horizontally polarized s-photons at a double slit and vertically polarized against another.

As with Walburn should cause interference in both places.

 

[UChr]

Additional explanation:

Instead of letting a coincidens counter split the set into two subsets splits (Receiver)-PBS in the same two subsets. Therefore, coincidens couter spared.

 

[SpectraCat]

This is only position '0'.

The transmitter polarizes p-photons horizontally or vertically.

Receiver PBS sends the now horizontally polarized s-photons at a double slit and vertically polarized against another.

As with Walburn should cause interference in both places.

No ... in the Walborn DCQE experiment, there is no interference pattern shown in the single-photon measurements for the s-photon when the QWP's are in place. The interference pattern can only be recovered by looking at the coincidence counts between the s- and p-photon measurements when there is a properly oriented polarizer in the p-photon beam.

 

[UChr]

I've apparently just confused by referring to Walburn.

In my gedanken experiment, there is no coincidence counter or QWP's.

Position 0 would cause interference:

The transmitter polarizes p-photons horizontally or vertically.

Receiver PBS sends the now horizontally polarized s-photons at a double slit and vertically polarized against another.

Position 1 should then erase this by changing the transmitter - comes later - if 0 seem ok.

 

[SpectraCat]

I've apparently just confused by referring to Walburn.

In my gedanken experiment, there is no coincidence counter or QWP's.

Position 0 would cause interference:

The transmitter polarizes p-photons horizontally or vertically.

Receiver PBS sends the now horizontally polarized s-photons at a double slit and vertically polarized against another.

Position 1 should then erase this by changing the transmitter - comes later - if 0 seem ok.

You still seem to be missing an important point .. the fringes and anti-fringes are features of the coincidence counts ONLY, which arise from setting the p-branch polarizer to the angle of the QWP over EITHER the left-slit (45º, for fringes), or the right slit (135º, for anti-fringes). So you need both the QWP's and the coincidence counters to observe them.

 

[UChr]

None QWP’s:

One DS receives exclusively horizontally polarized photons - this should lead to interference - and the other DS receives only vertically - and again interference (horisontally - vertically = half wave = fringes and anti-fringes).

Walburn Re: This is the type of interference the QWP s later erase.

 

[SpectraCat]

None QWP’s:

One DS receives exclusively horizontally polarized photons - this should lead to interference - and the other DS receives only vertically - and again interference (horisontally - vertically = half wave = fringes and anti-fringes).

Walburn Re: This is the type of interference the QWP s later erase.

No No No No .. read my posts again .. fringes and anti-fringes ONLY arise due to BOTH the QWP's AND the coincidence counting .. your setup has neither. Also, read the recent threads in this forum on DCQE .. that may help you understand why what you are proposing won't work.

 

[UChr]

Yes and no. It is my assumption that when horizontally gives fringes - vertically should give (because of reflection - half-wave) anti-fringes.

Walburn re:

If you look at arXiv – text, I think about interference shown on figure 2.

Grad.physics text, p 6 schedule: ‘Detected polarization for photon p’ : x / y. (my horizontally / vertically).

American Scientist: p 340 Volumen 91. ‘Now imaging that we repeat Yong’s experiment with many horizontally polarized photons. Behind the slits we insert two QWP […] Remarkably, the interference fringes will disappear.’

 

[SpectraCat]

Yes and no. It is my assumption that when horizontally gives fringes - vertically should give (because of reflection - half-wave) anti-fringes.

You need to define what you mean by fringes and anti-fringes, because it is necessarily something different than in the DCQE experiments. In your setup, you have two different double-slits (without QWP's) with two different detecting screens. Each apparatus does a single-photon double-slit experiment .. why are you expecting to see something different at the detection screens, and what specifically is the difference that you are expecting to see?

Also, I guess you realize that you need to send many many photons to observe an interference pattern, right? You haven't mentioned that explicitly, and from some of your posts I get the impression you think you will see "interference patterns" as discrete events.

 

[UChr]

You need to define what you mean by fringes and anti-fringes, because it is necessarily something different than in the DCQE experiments. In your setup, you have two different double-slits (without QWP's) with two different detecting screens. Each apparatus does a single-photon double-slit experiment .. why are you expecting to see something different at the detection screens, and what specifically is the difference that you are expecting to see?

Also, I guess you realize that you need to send many many photons to observe an interference pattern, right? [...]

OK – fringes both places.

Yes - many - many. (Maybe many is enough. I will believe that it is enough to measure 4 to 5 selected locations at the detection screen to observe the difference between constructive and destructive interference or not - but it's technology - so let it just be many - many.).

 

[UChr]

Summary:

T(0) = Transmitter position 0:
A PBS(0) which polarizes p-photons horizontally or vertically.

Receiver:
A PBS(0) which sends the now horizontally polarized s-photons at a double slit and vertically polarized against another.

After a while, the many many photons forming interference pattern on both detector screens.
---
New: T(1) = Transmitter position 1:

A PBS(45) which polarizes p-photons 45 or -45 degrees to the horizontal = diagonally right or left / diagonally positive or negative.

Effect on the s-photons:
a) The s-photons will be polarized diagonally negative or positive.
b) Because of the difference between transmission and reflection: a half wave difference between diagonally negative and positive.

When s-photons meetings the receiver PBS(0) they will have a fifty - fifty chance of being polarized horizontally or vertically:
ie the beam against both double slits will consist of a mix of photons coming from diagonally negative or positive = with a half wave difference.

One must therefore expect no / very little interference in this case.

(It should be possible to do something similar with a source of circularly polarized entangled photons. - could possibly be discussed later) 

 

[DrChinese]

Summary:

T(0) = Transmitter position 0:
A PBS(0) which polarizes p-photons horizontally or vertically.

Receiver:
A PBS(0) which sends the now horizontally polarized s-photons at a double slit and vertically polarized against another.

After a while, the many many photons forming interference pattern on both detector screens.
---
New: T(1) = Transmitter position 1:

A PBS(45) which polarizes p-photons 45 or -45 degrees to the horizontal = diagonally right or left / diagonally positive or negative.

Effect on the s-photons:
a) The s-photons will be polarized diagonally negative or positive.
b) Because of the difference between transmission and reflection: a half wave difference between diagonally negative and positive.

When s-photons meetings the receiver PBS(0) they will have a fifty - fifty chance of being polarized horizontally or vertically:
ie the beam against both double slits will consist of a mix of photons coming from diagonally negative or positive = with a half wave difference.

One must therefore expect no / very little interference in this case.

(It should be possible to do something similar with a source of circularly polarized entangled photons. - could possibly be discussed later) 

In neither case will there be any interference. Entangled photons do NOT exhibit self interference UNLESS which path information is ERASED. This cannot be made to happen at will.

 

[UChr]

In neither case will there be any interference. Entangled photons do NOT exhibit self interference UNLESS which path information is ERASED. This cannot be made to happen at will.

'which path information' - no QWP so where?

 

[DrChinese]

'which path information' - no QWP so where?

What does the quarter wave plate have to do with anything? Or allow me to rephrase: a QWP does not erase anything. A pair of them "might" but only with coincidence counting and without the beam splitter. But the setup you described won't give any interference.

 

[UChr]

? # 47: ' UNLESS which path information is ERASED.´
There is no which path information to erase.

But the setup you described won't give any interference.

Could you please write why T(0) should not work? - (because T(1) is not intended to give interference).

 

[DrChinese]

? # 47: ' UNLESS which path information is ERASED.´
There is no which path information to erase.

Ah, but there is! You just don't plan to look at it.

See Zeilinger, page 290, figure 2, there is no direct interference pattern for entangled photons:

Experiment and the foundations of quantum physics (1999)

 

[DrChinese]

By the way, I fell victim to this little twist when I was first looking at this setup. I was sure I had found a way to signal FTL... but nature is tricky!

 

[UChr]

By the way, I fell victim to this little twist when I was first looking at this setup. I was sure I had found a way to signal FTL... but nature is tricky!

Nice to hear. Yes very tricky.

Re 290, Figure 2: I have some questions - that might help me further:

To remove the suspicion of Which Path information I could try to modify my experimental set-up:
I could use optical fibers to transport p - photons from the source until the transmitter's PBS and also from source until the receiver’s PBS, and I could replace both the Double Slits with DS - substitutes:
1: a half-silvered mirror followed by optical fibers - adjusted in length equal to the difference between transmitted and reflected - and compiled to emulate a DS

2: an interferometer.
Arises then similar problems - or should this solve the WP – problem?

Re Walborn:
Figure 2 is without QWP and without polarizer - but shows interference. How does this work?

With the two QWP’s in place: Should then just a laserbeam with diagonal (positive) polarized photons cause interference - or is the presence of entanglement decisive?

 

[DrChinese]

Nice to hear. Yes very tricky.

Re 290, Figure 2: I have some questions - that might help me further:

To remove the suspicion of Which Path information I could try to modify my experimental set-up:
I could use optical fibers to transport p - photons from the source until the transmitter's PBS and also from source until the receiver’s PBS, and I could replace both the Double Slits with DS - substitutes: ...

To get interference for Alice, Bob must be detected as a wave and not a localized particle. Sending Bob through a fiber will turn Bob particle-like and that means no interference for Alice. And vice versa! Again, you must look at the entire context. It is not easy to detect both members of a pair as wave-like and even in principle this cannot be done on demand.

I didn't try to work through your examples all the way as these Quantum Eraser discussions get sidetracked almost immediately with the details of the setup and it gets impossible to move past that. The point is always the same as I mention: you don't get interference from photons in which it is possible, in principle, to know the which path information - regardless of whether you actually know it or not.

 

[UChr]

To get interference for Alice, Bob must be detected as a wave and not a localized particle. Sending Bob through a fiber will turn Bob particle-like and that means no interference for Alice. And vice versa!

then no fiber - I thought maybe it was more smart - but it is a gedanken experiment, so no need to unnecessary practical difficulties.

So transmitter PBS(0) or PBS(45).
Receiver: a PBS(0) followed by two Mach-Zehnder interferometers - with for example a BS = half silvered mirror.

I didn't try to work through your examples all the way as these Quantum Eraser discussions get sidetracked almost immediately with the details of the setup and it gets impossible to move past that.

Maybe one of those who has studied the Walborn experiment, answer my two questions?

(Walborn figure 2 is without QWP and without polarizer - but shows interference. How does this work?

With the two QWP’s in place: Should then just a laserbeam with diagonal (positive) polarized photons cause interference - or is the presence of entanglement decisive?)

 

[SpectraCat]

Which Walborn experiment? Please link to the paper you are referencing.

 

[DrChinese]

then no fiber - I thought maybe it was more smart - but it is a gedanken experiment, so no need to unnecessary practical difficulties.

So transmitter PBS(0) or PBS(45).
Receiver: a PBS(0) followed by two Mach-Zehnder interferometers - with for example a BS = half silvered mirror.



Maybe one of those who has studied the Walborn experiment, answer my two questions?

(Walborn figure 2 is without QWP and without polarizer - but shows interference. How does this work?

With the two QWP’s in place: Should then just a laserbeam with diagonal (positive) polarized photons cause interference - or is the presence of entanglement decisive?)

Have you looked at this?

http://grad.physics.sunysb.edu/~amarch/

http://grad.physics.sunysb.edu/~amarch/Walborn.pdf

 

[UChr]

Which Walborn experiment? Please link to the paper you are referencing.

arXiv:quant-ph/0106078v1 13 juni 2001