## Method and Apparatus For Transmitting Signals Faster than Light?? (Based on EPR/QM)

Method and Apparatus For Transmitting Signals Faster than Light?? (Based on EPR/QM)

Does the following apparatus make sense? Please discuss and let me know where the logic or science is falling short, if it is falling short.

45Revolver: Method and Apparatus for Transmitting Signals Faster than
the Speed of Light

Feel free to email me for the paper and the figures:
REVOLVER45@gmail.com

Consider an apparatus with a source for two interacting spin ½ photons
at the center, as shown in Figure 1. On either side of the source is a
double slit at distance d1, and beyond each double slit is a screen at
distance d2.

The two interacting spin ½ photons separate and travel in opposite
directions. Each passes on through a double slit-double-slit1 and
double-slit2, and an interference pattern is observed on each of the
two screens, screen1 and screen2.

Suppose then that a measurement device is then placed in front of the
upper slit in the double-slit to the left of the photon source, as
shown in Figure 2. Now photons must either go through the upper slit
or lower slit in double-slit1, and the interference pattern disappears
on screen1 to the left of the photon source.

Because momentum is conserved, whenever one photon passes through the
upper slit in double-slit1 (situated to the left of the source), the
other photon must pass through the lower slit in double-slit2 (situated
to the right of the source), as shown in Figure 2. And whenever one
photon passes through the lower slit in double-slit1, the other photon
will pass through the upper slit in double-slit2. Thus the
interference pattern will also disappear on screen2 to the right of the
source, whenever a measurement device is placed in front of the upper
slit in double slit1 to the left of the source.

Thus the placement of the detector in front of double slit1 to the
left of the photon source affects the interference pattern on screen2
to the right of the source.

Because d1 and d2 can be changed without affecting the basic physics,
a situation in which d1 is very, very large with d2 just a little bigger (d1 ~ d2) is conceivable. In such a scenario, by
selectively placing a measurement device in front of double slit1 and
removing it at a controlled interval, the interference pattern at
screen2 will be created and destroyed at the same controlled interval,
thereby sending a signal.

As d1 becomes greater and greater as
compared to d2, the signal can be sent faster and faster, transcending
the velocity of light and approaching instantaneous.

There could be many practical physical embodiments of such a device or apparatus for sending signals faster than the speed of light.

One preferred embodiment would be a central photon source, and fiber
optics run to two distant locations on either side of the photon
source. A double slit or other interference apparatus would be
employed at each of the two distant locations. A removable measurement
device would be located at one or both locations, and its use or
none-use in the setup could instantaneously affect the interference
pattern at the other location.

45Revolver: Method and Apparatus for Transmitting Signals Faster than
the Speed of Light

Feel free to email me for the paper and the figures:
REVOLVER45@gmail.com

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 Recognitions: Science Advisor I'm not sure, but my guess would be that no interference pattern is ever seen in the total pattern of photons on the screen in such a setup, even if no measurement is made of which slit each one through; instead, the interference pattern would only be seen in the some sort of coincidence-counting, like if you only looked at the subset of photons hitting the screen on the right side that are paired to photons that hit the bottom portion of the left screen, or something along those lines. My guess here is based on an analogy with the delayed choice quantum eraser experiment which is discussed on this thread.

 Quote by JesseM I'm not sure, but my guess would be that no interference pattern is ever seen in the total pattern of photons on the screen in such a setup, even if no measurement is made of which slit each one through; instead, the interference pattern would only be seen in the some sort of coincidence-counting, like if you only looked at the subset of photons hitting the screen on the right side that are paired to photons that hit the bottom portion of the left screen, or something along those lines. My guess here is based on an analogy with the delayed choice quantum eraser experiment which is discussed on this thread.
Thanks Jesse,

It seems strange to me that no interference pattern would ever be seen, as if a photon passes through both slits, it must interfere with itself.

If we pictured millions or billions of photons an interference pattern would emerge.

Or then how could a photon pass through both slits and not interfere with itself?

Recognitions:

## Method and Apparatus For Transmitting Signals Faster than Light?? (Based on EPR/QM)

 Quote by Dr. RMC Thanks Jesse, It seems strange to me that no interference pattern would ever be seen, as if a photon passes through both slits, it must interfere with itself.
Well, look over the delayed choice quantum eraser experiment, which is also discussed on this thread. In this experiment, you have an entangled pair of photons, one labeled the "signal photon" which ends up at detector D0 (which can also be replaced with a screen as in the double-slit experiment), another labeled the "idler photon" which ends up at either D1, D2, D3, or D4. If it ends up at D3 or D4, then you know which location (analogous to the slits) the pair of photons came from, like in the double-slit experiment where you measure which slit the photons go through; if it ends up at D1 or D2, then the which-path information is lost, like in the double-slit experiment where you don't measure which slit it went through. If you look only at the subset of signal photons whose idlers end up at D3, you don't see an interference pattern, and likewise with D4. On the other hand, if you look at the subset of signal photons whose idlers end up at D1, you do see an interference pattern, and likewise with D2, so in a sense you could say the photon has "interfered with itself", as you say. But what's interesting is that the peaks of the D1 interference pattern line up with the valleys of the D2 interference pattern, so if you look at the sum of the D1 and D2 interference patterns (the subset of signal photons whose idlers ended up at either D1 or D2), you get a pattern that shows no interference. Again, look at that earlier thread I linked to above and the paper I linked to in my post there, which includes a diagram of the setup and a graph of the various signal photon patterns, showing how the peaks of the D0/D1 interference pattern line up with the valleys of the D0/D2 interference pattern.

So like I said, I would guess that something similar happens in the setup you describe--if you look at the proper subsets of photons going to the left screen which are paired with photons going to the right screen, you would get an interference pattern, but when you summed all the possible interference patterns based on such subsets, the peaks of some would line up with the valleys of others in such a way that the total pattern shows no interference. That's just a guess since I don't know how to explicitly calculate the correct QM prediction in this setup, but I can't see any other way that FTL signalling (or even backwards-in-time signalling!) could be avoided here.

 Quote by JesseM Well, look over the delayed choice quantum eraser experiment, which is also discussed on this thread. In this experiment, you have an entangled pair of photons, one labeled the "signal photon" which ends up at detector D0 (which can also be replaced with a screen as in the double-slit experiment), another labeled the "idler photon" which ends up at either D1, D2, D3, or D4. If it ends up at D3 or D4, then you know which location (analogous to the slits) the pair of photons came from, like in the double-slit experiment where you measure which slit the photons go through; if it ends up at D1 or D2, then the which-path information is lost, like in the double-slit experiment where you don't measure which slit it went through. What's interesting is that if you look only at the subset of signal photons whose idlers end up at D3, you don't see an interference pattern, and likewise with D4. On the other hand, if you look at the subset of signal phtons whose idlers end up at D1, you do see an interference pattern, and likewise with D2, so in a sense you could say the photon has "interfered with itself", as you say. But the peaks of the D1 interference pattern line up with the valleys of the D2 interference pattern, so if you look at the sum of the D1 and D2 interference patterns (the subset of signal photons whose idlers ended up at either D1 or D2), you get a pattern that shows no interference. Again, look at that earlier thread I linked to above and the paper I linked to in my post there, which includes a diagram of the setup and a graph of the various signal photon patterns, showing how the peaks of the D0/D1 interference pattern line up with the valleys of the D0/D2 interference pattern. So like I said, I would guess that something similar happens in the setup you describe--if you look at the proper subsets of photons going to the left screen which are paired with photons going to the right screen, you would get an interference pattern, but when you summed all the possible interference patterns based on such subsets, the peaks of some would line up with the valleys of others in such a way that the total pattern shows no interference. That's just a guess since I don't know how to explicitly calculate the correct QM prediction in this setup, but I can't see any other way that FTL signalling (or even backwards-in-time signalling!) could be avoided here.
Thanks Jesse,

I looked over the link and the paper.

If there is never any interference pattern upon the screens in the setup I describe, then the respective photons to the right and left of the source must not be passing through both slits in the double slits.

Hence, even without being measured, the photons are behaving like particles, with definitive positions and momenta.

But if the photons are behaving as waves, then an interference pattern must be produced.

Thanks.

Recognitions:
 Quote by Dr. RMC Thanks Jesse, I looked over the link and the paper. If there is never any interference pattern upon the screens in the setup I describe, then the respective photons to the right and left of the source must not be passing through both slits in the double slits.
But what if there is interference when considering various subsets of photons in your setup--like considering the subset of all photons hitting the right screen that are paired with photons hitting the bottom half of the left screen--but when all these patterns are added together, the result is a noninterference pattern?

Perhaps it would be better to talk about this issue in terms of the delayed choice quantum eraser experiment, where it's clearer what subsets we're considering. If we look at the total pattern of signal photons on the screen and see no interference, would you say this means the signal photon/idler photon pairs are not coming from both radioactive atoms A and B (which are analogous to the slits)? If that's what you would say, how do you explain the fact that when we consider only the subset of signal photons whose idlers end up at D1, we do see an interference pattern? Does this mean that the photons in this subset are coming from both atoms?

To make this problem sharper, suppose we replace the beam-splitters BSA and BSB in the fig. 1 of that paper with mirrors, so all the idlers end up at D1 and D2. Again, if you look at the subset of signal photons whose idlers ended up at D1, you see an interference pattern; and if you look at the subset of photons whose idlers ended up D2, you see an interference patter; and every photon falls into one subset or the other. So does this mean every photon did come from both radioactive atoms? But then if you look at the total pattern of signal photons (the sum of photons from the first subset with photons from the second), you see a non-interference pattern. Does this mean every photon did not come from both radioactive atoms?

It seems to me the problem is that your terminology is fuzzy--I don't really see what it means to say that a given photon did or didn't go through both slits/come from both atoms, when a single photon can be viewed both as a member of a subset that does show interference, and simultaneously as a member of a larger subset which does not.

 I guess the basic question is this: Suppose we have a beam of two interacting spin 1/2 photons. The beam of photons separates, with each of the two interacting photons traveling in opposite directions. If in each direction there exists a double slit setup, will an interfence pattern be created on a screen beyond the double slit?

Recognitions:
 Quote by Dr. RMC I guess the basic question is this: Suppose we have a beam of two interacting spin 1/2 photons. The beam of photons separates, with each of the two interacting photons traveling in opposite directions. If in each direction there exists a double slit setup, will an interfence pattern be created on a screen beyond the double slit?
Will the total pattern of all photons hitting the screen create an interference pattern? I think the answer is no, but I also think that certain well-defined subsets of the photons hitting the screen will create an interference pattern.

Did you follow the details of the delayed choice quantum eraser setup well enough to be able to understand the questions I was asking in my previous post? If not, I can try to explain it further; if so, it would be helpful if you would address them.

 Yes--I followed your argument and will address it in a bit. But first I would like to remain focused on my apparatus. Consider a beam of photons on either side of the apparatus I describe. If the beam is incident upon a double slit, how could it not create an interference pattern upon a screen just beyond the double slit? What could possibly cause a beam of photons to not create an interference pattern? Thanks!

Recognitions:
 Quote by Dr. RMC Yes--I followed your argument and will address it in a bit. But first I would like to remain focused on my apparatus. Consider a beam of photons on either side of the apparatus I describe. If the beam is incident upon a double slit, how could it not create an interference pattern upon a screen just beyond the double slit? What could possibly cause a beam of photons to not create an interference pattern? Thanks!
If I'm right that there would be no interference in the total pattern of photons, presumably the "cause" would be that photons which are members of entangled pairs do not behave in quite the same way as photons which are not, in the context of the double-slit experiment. After all, there is some possibility that by performing the right measurement on the other member of the pair, you can determine which slit the first photon went through, even without performing measurements at either of the two slits the first photon could have passed through. If this is correct, then you should be able to show that this is the case using the rules of quantum mechanics, so the ultimate "cause" is just that that's how the laws of quantum mechanics say the photon should behave, trying to translate these mathematical laws into some kind of intuitive verbal explanation may not always be possible.

 Quote by JesseM If I'm right that there would be no interference in the total pattern of photons, presumably the "cause" would be that photons which are members of entangled pairs do not behave in quite the same way as photons which are not, in the context of the double-slit experiment. After all, there is some possibility that by performing the right measurement on the other member of the pair, you can determine which slit the first photon went through, even without performing measurements at either of the two slits the first photon could have passed through. If this is correct, then you should be able to show that this is the case using the rules of quantum mechanics, so the ultimate "cause" is just that that's how the laws of quantum mechanics say the photon should behave, trying to translate these mathematical laws into some kind of intuitive verbal explanation may not always be possible.
Would this then mean that photons which are members of entangled pairs show no interference pattern? And then would this mean that if a photon was once a member of an entagled pair, it could never again behave live a wave and interfere with itself?

Suppose we consistently measured one of the entangled photons, thereby liberating the other photon from the entangled state. Woul the liberated photons be capable of behaving like photons and creating an interference pattern after passing hrough a double slit? Because this is another manner in which signals may be sent faster than the speed of light.

Mentor
Blog Entries: 27
 Quote by Dr. RMC Suppose we have a beam of two interacting spin 1/2 photons.
Er... suppose we DON'T have a beam of interacting spin 1/2 photons!

Zz.

Mentor
Blog Entries: 27
 Quote by Dr. RMC Would this then mean that photons which are members of entangled pairs show no interference pattern? And then would this mean that if a photon was once a member of an entagled pair, it could never again behave live a wave and interfere with itself? Suppose we consistently measured one of the entangled photons, thereby liberating the other photon from the entangled state. Woul the liberated photons be capable of behaving like photons and creating an interference pattern after passing hrough a double slit? Because this is another manner in which signals may be sent faster than the speed of light.
Entangled photons do interfere, but not within the entangled unit. It has been shown that entangled photons beat the diffraction limit[1,2], as it should if the entangled glob cannot be considered as separable (mathematically) and thus have a higher energy (freq) as an entity.

Zz.

[1] P. Walther et al., Nature v.429, p.158 (2004).
[2] M.W. Mitchell et al., Nature v.429, p.161 (2004).

 Suppose we consistently measured one of the entangled photons, thereby liberating the other photon from the entangled state. Woudl the liberated photons be capable of behaving like photons and creating an interference pattern after passing hrough a double slit? If so, then signals may be sent faster than the speed of light. If one can intentionally alter an interference pattern from afar at velocities faster than light, then an apparatus may be designed for faster-than-light communication.

Recognitions:
 Quote by Dr. RMC Suppose we consistently measured one of the entangled photons, thereby liberating the other photon from the entangled state.
That's not how it works. The other photon will continue to behave in an entangled way until it is measured; if you compare the first measurement of photon A with the first measurement of photon B you should find the correlations that are characteristic of entanglement, even if photon A was measured long before photon B. Subsequent measurements of particles that were previously entangled would no longer show entanglement though, it's only the first measurement of each that counts.

 Quote by Dr. RMC Method and Apparatus For Consider an apparatus with a source for two interacting spin ½ photons at the center, as shown in Figure 1. On either side of the source is a double slit at distance d1, and beyond each double slit is a screen at distance d2. The two interacting spin ½ photons separate and travel in opposite directions. Each passes on through a double slit-double-slit1 and double-slit2, and an interference pattern is observed on each of the two screens, screen1 and screen2. Suppose then that a measurement device is then placed in front of the upper slit in the double-slit to the left of the photon source, as shown in Figure 2. Now photons must either go through the upper slit or lower slit in double-slit1, and the interference pattern disappears on screen1 to the left of the photon source.
The interference pattern disappears on screen1 because you've, in effect, closed one of the openings in double-slit1.
 Quote by Dr. RMC Because momentum is conserved, whenever one photon passes through the upper slit in double-slit1 (situated to the left of the source), the other photon must pass through the lower slit in double-slit2 (situated to the right of the source), as shown in Figure 2. And whenever one photon passes through the lower slit in double-slit1, the other photon will pass through the upper slit in double-slit2. Thus the interference pattern will also disappear on screen2 to the right of the source, whenever a measurement device is placed in front of the upper slit in double slit1 to the left of the source.
The side that has both slits open will show an interference pattern. The side that has one slit closed will not show an interference pattern.
 Quote by Dr. RMC Thus the placement of the detector in front of double slit1 to the left of the photon source affects the interference pattern on screen2 to the right of the source.
No. Closing one of double slit1's openings will not affect the raw data interference pattern produced by double slit2. Double slit2 produces interference banding (after about 60 to 70k detections) because both its slits are unobstructed. Double slit1, with an obstructed opening, will not.

Recognitions: