Talk:Superluminal communication
Using qubits for superluminal communication
I'm probably missing something here, but it seems to me that you really could use qubits for superluminal communication. The article says:
- "If Alice wishes to transmit a "0", she measures (...), collapsing Bob's state (...). To transmit "1", Alice does nothing to her qubit."
But I was thinking, instead of just measuring or not measuring, couldn't Alice measure one qubit to transmit a 0, and measure a different qubit to transmit a 1? It could go something like this: Alice and Bob create a state which is an entanglement of three qubits. Then, after it's created, Alice keeps the first and third qubit. Bob takes the second qubit and goes to some far-off distance. So now:
- Alice has and
- Bob has
Then, to transmit a zero, Alice performs a measurement on only , and leaves the other two qubits alone. To transmit a one, Alice performs a measurement on only . Then, after Alice performs her measurement, Bob measures . The three-qubit state could be initally set up so that:
- has a high probability (greater than .5) of being measured as
- If is measured as measured as , then, subsequently, has a high probability of being measured as
- has a high probability of being measured as
- If is measured as measured as , then, subsequently, has a high probability of being measured as
There would be a probability of error in the transmission, but this error would be less than .5. One could compensate for that error by using many sets of three qubits to transmit each bit. Or maybe one could compensate with classical error-correcting codes.
Would all of that work? I'm guessing that I've missed something, since I'm not an expert on quantum mechanics. Most of what I know is from taking a one-semester class on quantum computing. --Navigatr85 03:10, 22 August 2007 (UTC)
The example
If Bob creates copies and measures them, wouldn't they all have the same spins on each axis? Conversely, how is Bob measuring one of his copies different from Alice measuring her copy?
Contradiction with No-cloning theorem
The page for no-cloning theorem states "The no cloning theorem does not prevent superluminal communication via quantum entanglement, as cloning is a sufficient condition for such communication, but not a necessary one", yet this page says that it does. Who is correct? J0lt C0la (talk) 22:42, 4 October 2009 (UTC)
A thought experiment – using entanglement
A suggestion for a communication system using quantum entanglement:
A(lice) and B(ob) are measuring corresponding particles from pairs of quantum mechanically entangled photons - so-called Bell-couples.
The distance between the transmitter and the source are suitably much shorter than between the source and the receiver, so with synchronized watches the transmitter will detect its twin particle before the other twin reaches the receiver. The transmitter can change its measuring setup by inserting a mirror or not – choice situation, T (1) or T (0). It keeps its choice for an agreed period – for instance 1 / 300.000 sec. The receiver has a fixed preset setup. It should by measuring its part of the pairs with at least 99% probability, guess the Transmitters choice.
As the exchange between the entangled particles takes place instantaneous it will for a growing distance between A and B create superluminal communication.
A suggestion for transmitter and receiver:
The building block is a polarizing beam splitter = PBS (v) that transmits light polarized in the direction v and reflects light polarized perpendicular to v = v +/- 90 degrees. The beam splitter is not a measurement on the photons. The entangled pair should therefore remain in a superposition although one of the twins encounters one or more PBS’s.
Transmitter:
T (0) = no mirror: a polarizing beam splitter = PBS (0) followed by two photo detectors D1 and D2. PBS (0) transmits light polarized in direction 0 = horizontal polarized light and reflects vertical polarized. T (1) = mirror inserted: it starts with a polarizing beam splitter = PBS (0). The horizontal beam (0 degree) meets again a PBS (0) followed by two photo detectors D1 and D2. The vertical beam meets a PBS (+45) followed by two photo detectors D3 and D4. (The second PBS (0) could be omitted, but the symmetry makes it simpler to place the detectors correctly.)
Receiver.
R = T (0) = a polarizing beam splitter = PBS (0) followed by two photo detectors D1 and D2.
How the system should work.
The entangled pair remains in a superposition until a measurement is made. T (0): The detectors ensure that every photon chose one of the two options and not just remain in a superposition of both. Because of superposition, we can expect that PBS (0)'s two routes are equally attractive. T (1): The PBS (0) splits, as for T (0) before, the beam into two beams with the distribution 50% to 50%. The horizontal beam (0 degree) meets next a PBS (0). This will not split the beam further, because this whole part of the beam is already horizontally polarized, and the distribution of the detectors D1 and D2 are expected to be 50% and 0%. The vertical beam meets next a PBS (+45). The vertically polarized beam must choose between +45 degrees = diagonally to the right or perpendicular to it = diagonally to the left. It will be a fifty - fifty situation. We should therefore expect that the detectors D3 and D4 end with the distribution: 25 % and 25 %.
When the transmitter has measured its photon in the entangled pair as polarized in a certain direction stops the common superposition. The second photon will result in theory; behave as polarized perpendicular to the measured.
In case T (0) would beam towards the receiver has the following distribution: Horizontal / vertical = 50 % / 50 %. The receiver will allow horizontally polarized photons pass along the horizontal path toward detector D1 and let vertically polarized photons pass through the vertical path to D2. We should therefore get the distribution: D1 / D2 = 50% / 50%
In case T (1) would beam towards the receiver have the following distribution: Horizontal / vertical / diagonally to the right / diagonally to the left = 0 % / 50 % / 25 % / 25 %. The vertically polarized photons will pass through the vertical path to detector D2. So from this part of the beam: D1 / D2 = 0 % / 50 %.
The beam polarized diagonally to the right and left must choose between vertical and horizontal. This is a fifty - fifty situation. So from this part: D1 / D2 = 25 % / 25 %. Totally we should get: D1 / D2 = 25 % / 75 % - ie in case T (0) is the probability of D2 = 0.5. In case T (1) = 0.75.
The experiment ought to be fairly easy to test. UChr (talk) 13:49, 17 May 2011 (UTC)