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I've seen the other questions, but almost all of them deal with manipulating quantum states so that the other person reads a message in the data. However, you cannot force a quantum state on something without breaking entanglement.

So here is my idea. I know that it can't work, but I don't know why. If someone could point out the flaw, I would appreciate it. I do not claim to be an expert.

So there is the double slit experiment. If I pass a single photon through a double slit, it will create an interference pattern. This is the evidence that it is in a quantum state because a single particle should not be interfering with itself. In reading article above, this also works my photon happens to be entangled with another photon.

Now in the quantum eraser section of the linked article above, it goes on to mention that if you apply a filter to one of the photons (i.e. measure it), than the interference pattern of both collapses. Thus, by measuring the quantum state of one, both collapse.

So, here is how I propose FTL communication.

Method

Lets say I create streams of quantum entangled photons. I will split up the pairs of quantum entangled photons and send one half to $A$ and the other half to $B$ in order. Thus, the first photon that $A$ receives will be entangled with the first photon that $B$ receives.

At $B$'s end, each photon will be run through the double slit to detect whether an interference pattern exists. If there was an interference pattern, then this is encoded as a binary $0$. If there isn't an interference pattern, then this is a $1$. It doesn't matter which slit the photon is known to travel through - the simple absence of the interference pattern is enough.

At $A$'s end, to communicate, I will encode my message in binary and then encode each bit in photon stream. If the next bit is a $1$, I will measure the next photon, thus collapsing its quantum state and eliminating the interference pattern. If the next bit is a $0$, I will not measure the next photon.

Example

$A$ needs to send 1101. $A$ would measure the first, second, and forth photons. At $B$'s end, when these photons are run through the double slit, you would see that the first, second, and forth have no interference pattern, so you would know they are $1$s. The third retains an interference pattern, so it would be a $0$.

Two-Way Communication

This allows $A$ to communicate to $B$ instantly. For $B$ to communicate back, you could create another set of streams where $A$ runs the photons through the double slit and $B$ measures to collapse the quantum state on some photons. Alternatively, odd numbered photons can be used for the reverse communication, so long as both $A$ and $B$ agree.

Summary

So, what is wrong with this method? It does not try to force a photon to go through a slit, but rather uses the act of observing or measuring to collapse the quantum state.

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This has been answered before. I recommend not to read the question I link to, which is horrible, but the chosen answer, which is extremely well written and in accord with what I have heard our Professor for Quantum Optics say. Double-double-slit with entangled photons

P.S.: In general, on FTL communication: The crux is usually that entanglement is not as miraculous as it is thought to be. When you determine the state of one entangled particle at your end, it just means that you know which state the other one is in - but that's it! You can't SEND information, you just KNOW the state the other one is in. Same for the other end - he KNOWS now that your particle is the opposite of his, but neither of you could determine beforehand which state it would collapse into!

In your experiment the problem is, I think (and I'm not sure because I'm too tired), that interference would either show up in all cases or in none, so you couldn't choose at one station whether the other would measure interference or not.

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  • $\begingroup$ But here is an excerpt from the article I linked (maybe the article is wrong?) Initially, neither detector shows an interference pattern. Since we control the polarization of photons passing through the slits and we know the polarization accepted by each slit, we can deduce which way the photons travelled (counter-clockwise through the left; clockwise through the right). Thus no interference patterns are detected. $\endgroup$ – Trenin Dec 11 '15 at 13:02
  • $\begingroup$ cont. However, if we rotate the polarizing filter in front of detector A so that the polarizations of the photons that hit the detector are the same (that is, we can no longer distinguish between clockwise and counter-clockwise polarizations), then the interference pattern appears at both detectors! $\endgroup$ – Trenin Dec 11 '15 at 13:02
  • $\begingroup$ I guess what I was trying to do in this experiment is use the act of measuring to collapse the quantum state. It assumes that you can detect that a photon is still in a quantum state by seeing its interference pattern. Thus, by measuring one entangled photon at one location, the quantum state of the other collapses. I know you can't force a quantum state, but the act of measuring seems to give rise to the "spooky action at a distance". $\endgroup$ – Trenin Dec 11 '15 at 14:52
  • $\begingroup$ The question is: Is the act of measuring thus detectable FTL? If so, then the act of measuring can be used as the communication. The results of measuring become irrelevant. $\endgroup$ – Trenin Dec 11 '15 at 14:53
  • $\begingroup$ @Trenin In direct response to your last: "Is the act of measuring thus detectable FTL? If so then the act of measuring can be used as the communication. The results of measuring become irrelevant." The answer is yes but no. While the result of your measurement will affect my measurement, the only way I can know IF MY MEASUREMENT WAS AFFECTED OR NOT is to compare notes with you. And I can only do that at the speed of light. Think about that a bit, it's probably the most boiled-down explanation possible. $\endgroup$ – JPattarini Dec 17 '15 at 16:44
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I'm going to use electrons in place of photons because they have two options for their spin, but the problem is equivalent. The problem with the communication method is as follows:

You wanted 1 to be conveyed by A measuring the electron, 0 to be conveyed by A not measuring it.

Let's try to see what happens at B if A measures the electron. There is a 50% chance of A measuring spin 1/2, and a 50% of -1/2. What will B see? Well, there is a 50% chance that he sees of -1/2 (corresponding to A getting 1/2) and a 50% chance that he sees 1/2. A cannot control what the measurement will be, so A cannot control what value B will see.

What if A doesn't measure the electron? B is first to measure it. It has a 50/50 chance of spin 1/2 or -1/2.

The point: Even if A measures the electron, there is no measurable difference at B's location. In either case, B sees a 50/50 chance of +1/2 or -1/2. He doesn't know whether he is the first one to measure the particle or not. A cannot change the experimental outcome of measurement at B's location.

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  • $\begingroup$ In your example, you are concerned with the values of the measurements. Mine does not. It only is concerned with the face that it was measured because when it is measured, the interference pattern of both sides collapses. $\endgroup$ – Trenin Dec 11 '15 at 13:01
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While the other two answers are technically correct, I believe the question hinges on a misunderstanding of what is really required to see an interference pattern in the first place.

The question implies that simply by not measuring the path of A through the slits, an interference pattern for B can be observed. The missing part of this scenario is that interference can only be observed as the sum of many observations. B will appear as a single dot on the detecting plate regardless of whether the path of A is measured or not. No single observation can be determined to fit either with an interference pattern or with the absence of one. After many observations where which-way information is not collected on many A's, many B's will have formed an interference pattern, however this is clearly not helpful if one's goal is communication.

What you'd actually observe on the detecting plate for B if you alternated measurement/non-measurement of A (expected to produce a 010101 in your proposed scenario) is a wide distribution of marks with no discernible interference pattern. Only by working backwards and time-correlating the B marks with the A marks that were not measured would you be able to pick out the interference pattern formed by those specific B's. The remaining marks on the detecting plate would NOT form an interference pattern, and would correspond to the times that which-way information was collected about A.

Of course, in order to time-correlate those measurements the folks running the A detector would have to share their data with the B detector scientists... no faster than the speed of light!

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  • $\begingroup$ OK, so if I understand you, a single photon can not show an interference pattern. You need many. So then why couldn't I use 100, or 1000, or 1 billion photons in each set? Take a look at the "simpler" experiment in my other question. I asked it in a different question because answers like yours seem to find flaws in my experiment, and explaining better here would be too big for a comment. In the linked article of that question, it seems pretty cut and dry; measuring at A affects the interference pattern at B. $\endgroup$ – Trenin Dec 11 '15 at 14:57
  • $\begingroup$ @Trenin part of why you're having trouble with this is the popular media portrayal of these types of experiments - I echo what another commenter on your other question said in that you'd be less confused by just reading the primary literature. Think of it this way: If you send me 100 photons without which-way info, I will see an interference pattern. If you then send me 100 photons WITH which-way info, I will not. You can absolutely send me groups of 100 and switch back and forth between which-way/no which-way knowledge, (continued...) $\endgroup$ – JPattarini Dec 16 '15 at 22:17
  • $\begingroup$ @Trenin but as I have no way of knowing which order you will send them I'll just be collecting what appears to be a random distribution of marks on my detection plate. I would have to have you supply me the times that you fired each set for me to be able to deconstruct an interference pattern, if one exists, from the giant bell-curve of detections. (Continued...) $\endgroup$ – JPattarini Dec 16 '15 at 22:17
  • $\begingroup$ @Trenin Now, if I knew you were going to only send 100 photons at a time and we agreed I was going to throw the detection plate away between groups of 100 (which is what I think you are getting at), then you could send me a series of interference/no interference/interference pictures... but you'd still be bound by the speed of light, since the light would still have to travel from you to me for me to detect it. This is then no different from using different polarizations of light to encode a message, which we do routinely. $\endgroup$ – JPattarini Dec 16 '15 at 22:18
  • $\begingroup$ OK, so lets say that I have for a detection plate something that samples for a milli-second and then refreshes. So you can watch this detection plate and see an interference pattern on it (or not) and every milli-second there is a new plate. $\endgroup$ – Trenin Dec 17 '15 at 19:26

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