# FTL-gedanken experiment

#### UChr

FTL-gedanken experiment.

I never really got feedback on my last proposal in 'Communication systems and entanglement'- so I try here with a slightly modified version:

A source produces entangled photons against Alice: p beam - and against Bob: s beam.
These are used to transmit from Alice to Bob.

Transmitter:
Setting T (0): Polarized beam splitter PBS (v) followed by two detectors. PBS (v) forcing the photons to choose between being polarized in the direction v degrees or perpendicular = direction v+90. T (0) maintaining an agreed time and should create an interference pattern at Bob's receiver.

Setting T (1): Polarized beam splitter PBS (v +45), followed by two detectors. PBS (v +45) forces the photons to choose between being polarized in the direction v +45 degrees or in the direction v-45. T (1) maintains the same scheduled time and should not create an interference pattern with Bob.

Starts with a PBS (v+90). The transmitted beam encounters a device to read any interference pattern - a double slit or (if the double slit is problematic), an interferometer (for example Mach Zehnder with BS = a half silvered mirror). The reflected beam is stopped by a detector.
The distance between the source and Alice's detectors are less than the distance from the source and into the beginning of the receiver so that the photons will be measured at Alice place before they reach the receivers PBS (v+90).

T (0): p-photon v degrees so is the corresponding s-photon perpendicular = v+90. All of these are transmitted by PBS (v+90) and would like to form an interference pattern.
p-photon v+90  s-photon v, ie. reflected by the PBS (v+90) and detected.
Together, the system works here as a 'half Coincidence counter': Of the 'entangled' only the desired reach the double slit / interferometer and can form an interference pattern.
Noise will not be stopped. But since this is a gedanken experiment imagined the noise to be minimal.

T (1): p-photon v+45 degrees, so is the corresponding s-photon perpendicular = v-45. Half of those are transmitted by PBS (v+90) and would like to form an interference pattern.
p-photon v-45 degrees then the corresponding s-photon perpendicular = v+45. Half of those are transmitted by PBS (v+90) and would like to form an interference pattern.
Because of reflection should be a half-wave difference between p: v+45 and p: v-45, so the two patterns are shifted half-wave - as - for example, a fringe pattern and an anti-fringe pattern. Together equalize each other.

Example: Walborn et al: Double-slit quantum eraser.
With Quarter Wave Plates and polarizer set to theta =v.
FIG 4 shows a fringe pattern.

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#### DrChinese

Gold Member
FTL-gedanken experiment.

I never really got feedback on my last proposal in 'Communication systems and entanglement'- so I try here with a slightly modified version:

A source produces entangled photons against Alice: p beam - and against Bob: s beam.
These are used to transmit from Alice to Bob.

Transmitter:
Setting T (0): Polarized beam splitter PBS (v) followed by two detectors. PBS (v) forcing the photons to choose between being polarized in the direction v degrees or perpendicular = direction v+90. T (0) maintaining an agreed time and should create an interference pattern at Bob's receiver.
...
And I keep telling you that:

a) Entangled photons do not produce interference patterns, and I provided a reference from Zeilinger on that point. It was pointed out that entangled photons behave as incoherent light, and so the double slit pattern will not appear.

b) The pattern never changes, even when you detect a photon as a wave; you only see an interference pattern when you perform coincidence counting.

c) The ordering of measurements on entangled photons is always immaterial to the results.

#### UChr

And I keep telling you that:

a) Entangled photons do not produce interference patterns, and I provided a reference from Zeilinger on that point. It was pointed out that entangled photons behave as incoherent light, and so the double slit pattern will not appear.

b) The pattern never changes, even when you detect a photon as a wave; you only see an interference pattern when you perform coincidence counting.

c) The ordering of measurements on entangled photons is always immaterial to the results.
>>a) But how / why get Walborn then interference?

And this is only a problem for a double slit – not an interferometer.

>>b) Of course there is a problem with noise - and that only a fraction – for example 1 % - of s-photons reaching the double slit - but what else?

>>c) Maybe - but I prefer a traditional sequence – and maybe it's because of the coincidence counter.

#### Cthugha

>>a) But how / why get Walborn then interference?

And this is only a problem for a double slit – not an interferometer.
Two-photon interference is something different than single-photon interference. The commonly known interference patterns from the double slit or Michelson interferometer are single photon interferences. They require a certain coherence time/length. Two photon interferences are what is seen in experiments on entanglement and show up due to properties of the entangled two-photon state. Therefore you always need coincidence counting to see them and they show up only in the coincidence counts. Using coincidence counting is not just a matter of filtering out noise in these experiments. Note that single- and two-photon interference are complementary. You cannot have both at the same time.

#### DrChinese

Gold Member
>>a) But how / why get Walborn then interference?

And this is only a problem for a double slit – not an interferometer.

>>b) Of course there is a problem with noise - and that only a fraction – for example 1 % - of s-photons reaching the double slit - but what else?

>>c) Maybe - but I prefer a traditional sequence – and maybe it's because of the coincidence counter.
a) Because they use coincidence counting. And there is no double slit pattern.

b) Noise has nothing to do with it. In a perfect situation where all photons are perfectly entangled, nothing ever changes.

c) I appreciate that you prefer a certain ordering. And yet, ordering makes no difference. Ever.

#### UChr

Two-photon interference is something different than single-photon interference. The commonly known interference patterns from the double slit or Michelson interferometer are single photon interferences. They require a certain coherence time/length. Two photon interferences are what is seen in experiments on entanglement and show up due to properties of the entangled two-photon state. Therefore you always need coincidence counting to see them and they show up only in the coincidence counts. Using coincidence counting is not just a matter of filtering out noise in these experiments. Note that single- and two-photon interference are complementary. You cannot have both at the same time.
additional related DrChinese >>c) The ordering of measurements on entangled photons is always immaterial to the results.<<

In two photon interference experiments - for example Zeilinger p 290; Walborn; Kim - meetings, s-photons a double slit before it or its p-partner has been measured. Interference / non-interference are therefore part of the photon-twins shared history before they are measured. (Further: if the meeting with a double slit temporarily suspend entanglement(?) - as a PBS does - there is no temporal surprise at all, that the results are independent of the measured order.)

Because I measure the p-photons before the s-photons reach an interference-device - and interference is not a part of the common history - there are probably talking about single-photon interference in my gedanken experiment.

#### Cthugha

In two photon interference experiments - for example Zeilinger p 290; Walborn; Kim - meetings, s-photons a double slit before it or its p-partner has been measured. Interference / non-interference are therefore part of the photon-twins shared history before they are measured.
This is not true. It does not matter whether photons in one arm meet a double slit or MZ-interferometer or something similar in one arm before or after the photons in the other arm are detected.

As I said before, single- and two-photon interference are complementary (see Abouraddy et al., Phys. Rev. A 63, 063803 (2001)). So that means if you are indeed talking about single photon interference, you have already broken entanglement.

#### UChr

This is not true. It does not matter whether photons in one arm meet a double slit or MZ-interferometer or something similar in one arm before or after the photons in the other arm are detected.
Do you have an example? - In the 3 cases I have mentioned are both photons measured after double slit - but alternately relative to each other.

#### DrChinese

Gold Member
Do you have an example? - In the 3 cases I have mentioned are both photons measured after double slit - but alternately relative to each other.
Your Walborn reference shows this clearly, so what is the question? That is what delayed choice experiments demonstrate: ordering is irrelevant, exactly as predicted.

Gold Member

#### UChr

Your Walborn reference shows this clearly, so what is the question? That is what delayed choice experiments demonstrate: ordering is irrelevant, exactly as predicted.

re Walborn: ‘Experimental setup and procedure: … The double-slit and the quarter-wave plates are placed in path s, 42 cm from the BBO crystal. Detectors Ds and Dp are located 125 cm and 98 cm from the BBO crystal …
The delayed erasure setup is similar, with two changes: (i) detector Dp amd POL1 were placed at a new distance of 2 meters from BBO crystal …’

So there is thus no question of a measurement where Dp is closer than the double slit – as in my experiment.

(One way to understand this Walborn experiment is that the meeting with the double slit locks relationship between s and p with regard to interference and polarization - and therefore it does not matter when you measure them, as it only further happens that through Dp selects a particular subset to consider.)

#### SpectraCat

re Walborn: ‘Experimental setup and procedure: … The double-slit and the quarter-wave plates are placed in path s, 42 cm from the BBO crystal. Detectors Ds and Dp are located 125 cm and 98 cm from the BBO crystal …
The delayed erasure setup is similar, with two changes: (i) detector Dp amd POL1 were placed at a new distance of 2 meters from BBO crystal …’

So there is thus no question of a measurement where Dp is closer than the double slit – as in my experiment.

(One way to understand this Walborn experiment is that the meeting with the double slit locks relationship between s and p with regard to interference and polarization - and therefore it does not matter when you measure them, as it only further happens that through Dp selects a particular subset to consider.)
This has been mentioned before, but it bears repeating: for a given experimental setup, the order in which the entangled photons pass the different optical elements, including the double slit (with or without QWP's) and the polarizer in the Dp branch are completely irrelevant to the results of the experiment. You can place them wherever you like along the beampaths, so the photons encounter them in any order ... nothing about the experimental results (i.e. the coincidence statistics and the one-photon measurements) will change one little bit. That is why we say that for such experiments, the results depend on the entire context of the experiment.

So, for your example, it doesn't matter if the s-photon hits the double slit first, or the p-photon hits the polarizer first, the selecting of particular subsets in the coincidence measurements works out to give precisely the same results.

#### UChr

Why I think the order may be important in some types of experiments:

S-photons encounter a PBS (0) - polarize horizontally / vertically (and the vertically are detected).
P-photon encounters a PBS (45) - polarization diagonal positive / negative (and both the diagonal positive and negative are detected).
Both interrupts entanglement and both causes in addition a difference of half a wave between the transmitted and reflected.

If s first meetings PBS (0): It will transmit Beam with roughly the same wavelength shift as before for all.

If p first meetings PBS (45): s-beam will be oriented diagonally negative / positive - with a difference of half a wave - and when this beam subsequent meetings PBS (0): half of each type will be transmitted - so this time the resulting beam consists of a fifty-fifty blend with a half wave difference.

#### UChr

As I said before, single- and two-photon interference are complementary (see Abouraddy et al., Phys. Rev. A 63, 063803 (2001)). So that means if you are indeed talking about single photon interference, you have already broken entanglement.
As far as I understand it compares 'Abouraddy et al' picture all forms (entangled and noise) with the image the entangled form alone - ie noise filtered off through coincidence counter.
The surprising result is that although the interference of the entangled alone increases so can the complete picture show less interference.
This must surely assume that the noise constitutes a considerable part. If virtually all were entangled could this difference probably does not occur? OR??

#### UChr

And I keep telling you that:

a) Entangled photons do not produce interference patterns, and I provided a reference from Zeilinger on that point. It was pointed out that entangled photons behave as incoherent light, and so the double slit pattern will not appear. [...]
I have a problem with Zeilinger p. 290 - Fig. 2 + Fig. 3.

Is the experiment in Fig. 3 only a gedanken experiment - or is it done of B. Dopfer in 1998?
I have not been able to find online a more accurate description of it - not even at Zeilinger's website.
When I want to look at it, it is because the allegation in Fig. 2: ’The beams of particle 1 then pass a double-slit assembly. Because of the perfect correlation of the two particles particle 2 can serve to find out which slit particle 1 passed […]’

I have hitherto understood that for a photon to interfere with itself should it pass both slits - and not just the one.
It seems that this double slit must be very close to the source and with long distance between the two slit?

(PS - I understand Which Path information in the direction of that someone disturbs a path (more or less) and thereby disappears / (decrease) interference.)

#### DrChinese

Gold Member
I have a problem with Zeilinger p. 290 - Fig. 2 + Fig. 3.

Is the experiment in Fig. 3 only a gedanken experiment - or is it done of B. Dopfer in 1998?
I have not been able to find online a more accurate description of it - not even at Zeilinger's website.
When I want to look at it, it is because the allegation in Fig. 2: ’The beams of particle 1 then pass a double-slit assembly. Because of the perfect correlation of the two particles particle 2 can serve to find out which slit particle 1 passed […]’

I have hitherto understood that for a photon to interfere with itself should it pass both slits - and not just the one.
It seems that this double slit must be very close to the source and with long distance between the two slit?

(PS - I understand Which Path information in the direction of that someone disturbs a path (more or less) and thereby disappears / (decrease) interference.)
The distance to the slits is not really a factor, as you can route the beam wherever you like. The issue is that Alice COULD learn which path info - even if you aren't trying to - and that means Bob must act accordingly. And vice versa. And again, these photons are incoherent but I must admit I don't know all the rules on that.

Perhaps someone else can help out on this point?

#### UChr

Zeilinger Fig 2.: "Particle 1 is either emitted into beams a or a' "

so again: 'I have hitherto understood that for a photon to interfere with itself should it pass both slits - and not just the one.' ??

#### SpectraCat

Zeilinger Fig 2.: "Particle 1 is either emitted into beams a or a' "

so again: 'I have hitherto understood that for a photon to interfere with itself should it pass both slits - and not just the one.' ??
Yup .. and that's why there is no interference pattern in the one photon measurements for entangled photons.

#### UChr

Ok - so to get right around 'Zeilinger':
I put a lens in front of the transmitter - PBS + detectors - (so the PBS matches the lens focal plane).
And one problem less.

#### UChr

The distance to the slits is not really a factor, as you can route the beam wherever you like. The issue is that Alice COULD learn which path info - even if you aren't trying to - and that means Bob must act accordingly. And vice versa. And again, these photons are incoherent but I must admit I don't know all the rules on that.

Perhaps someone else can help out on this point?
I found Dopfer (Zeilinger) on the German-speaking part of the network:
http://www.univie.ac.at/qfp/publications/thesis/bddiss.pdf

A quick look at the 'figures' gives the following:

p 36 - Fig. 4.5: interference by 'virtually infinitely distant source' = max interference.

p.37-Fig. 4.6: interference by 'virtually point source' very close to the double slit = no interference.

P 86 - Table 4.7: shows that even with a large deviation from the 'ideal infinitely distant source' and quite close to the 'punctate' more than 90% of the interference contrast are preserved.

This last may explain why, for example, Walborn achieves interference in his attempts.

- And should it be necessary, one can correct it with a lens.

#### Cthugha

I found Dopfer (Zeilinger) on the German-speaking part of the network:

This last may explain why, for example, Walborn achieves interference in his attempts.

- And should it be necessary, one can correct it with a lens.
I absolutely do not see which point you are trying to make. The figures you quote give the distinguishability of the possible photon paths and the corresponding loss of interference visibility for probing the two-photon state in coincidence counting. That does not change anything about the single photon states being incoherent. Single- and two-photon interference are still complementary as discussed on page 44-47. There it is also explained that the distance between the PDC crystal and the double slit is one of the important quantities in this experiment (or better: the effective distance - what matters is the angular size of the PDC crystal as seen by the double slit, not the real distance) as it determines "near-field" vs. "far-field" conditions which correspond to the difference between seeing single-photon interference and two-photon interference.

#### UChr

I absolutely do not see which point you are trying to make.
The problem with Which Path seemed more difficult to circumvent in Zeilinger's brief article than it appears from Dopfers experiment. Further, the distance: 'Doppelspalt - Kristall 40mm' very short - as I had expected - #.

Ad one / two photon(s): I think it was a little tricky you asked me - when the experiment was presented. But ok - my analysis of my experiment:

It starts as a two - photon - experiment.
When p - is measured entanglement is broken and s - photon continues alone - ie as single - photon against his PBS and is detected or continues at double-slit / interferometer - to then finally be detected.

#### DrChinese

Gold Member
The problem with Which Path seemed more difficult to circumvent in Zeilinger's brief article than it appears from Dopfers experiment. Further, the distance: 'Doppelspalt - Kristall 40mm' very short - as I had expected - #.

Ad one / two photon(s): I think it was a little tricky you asked me - when the experiment was presented. But ok - my analysis of my experiment:

It starts as a two - photon - experiment.
When p - is measured entanglement is broken and s - photon continues alone - ie as single - photon against his PBS and is detected or continues at double-slit / interferometer - to then finally be detected.
When p is measured, it's entanglement acts as if it is broken. Conceptually you could, upon suitable preparation, diffract it through a double slit and get interference. But s does not act any different at that time in the sense that you could do the same with it. It will not act as if entanglement has ended until it is measured.

In other words, they act as if they are entangled until both are measured. That creates a context.

#### SpectraCat

In other words, they act as if they are entangled until both are measured. That creates a context.
I don't think that could be true, because it implies FTL communication. In other words, if Alice measures her particle is able to observe 1-photon interference, then she knows that Bob MUST have measured his particle as well. That is basically UChr's argument .. that a device can be built on this principle.

I think that the first measurement breaks the entanglement ... that way, it is impossible for Alice to know if Bob has made a measurement when she makes her own.

#### DrChinese

Gold Member
I don't think that could be true, because it implies FTL communication. In other words, if Alice measures her particle is able to observe 1-photon interference, then she knows that Bob MUST have measured his particle as well. That is basically UChr's argument .. that a device can be built on this principle.

I think that the first measurement breaks the entanglement ... that way, it is impossible for Alice to know if Bob has made a measurement when she makes her own.
Not what I meant.

Clearly, there aren't suitable words to present this in our normal language. We know when both become entangled (approximately anyway). And we know when neither is no longer entangled. But in between, you cannot really make a firm statement as to WHEN (or exactly by what mechanism) things change.

If the Alice's first measurement broke entanglement from every observers' perspective, you COULD send an FTL message to Bob. Because you could choose to make the measurement or not, and Bob could sense that - since entangled particles act a little differently (as we have been discussing).

The fact is, you cannot strictly say either Alice or Bob ended the entanglement.

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