4 ms·

It's still the consensus that we can't (even theoretically) send a bit of information this way, right? Like, no quantum telegraph? So we measure the particles

It's still the consensus that we can't (even theoretically) send a bit of information this way, right? Like, no quantum telegraph? So we measure the particles and find that they are in inverse states. I assume there's a reason for this, but why are we sure that we're not just reading off a random number generator that's set from the same seed?

johncolanduoni 10y ago
Quantum teleportation is actually kind of the opposite. You transmit a quantum state (which if your particle is as simple as a photon, can be used to create a copy of the particle) by using an already-separated entangled pair and sending two classical bits any boring old way (fiber, wire, smoke signals). If you don't get the classical information on the receiving end, any attempts to read the quantum state will garble the quantum transmission such that it'll be indistinguishable from random noise. Since those classical bits are limited by the speed of light, so is the overall transmission of the state.
- trentlott 10y ago
  So why does the reciver need the two bits? I'm not prone to study too deeply because I've learned (apparently) very bad analogies or schemes for entanglement.
  - johncolanduoni 10y ago
    The teleportation process involves using a measurement to entangle the receiver's side of the pair with the original state in a useful way. Since measurements are probabilistic (and the measurement is along two qubits) there are four outcomes of this measurement with four different kinds of "garbling" of the transmission (one of which is no garbling). The receiver needs the two bits to choose which of four kinds of garbling occurred, so it can correct them with the appropriate post-processing (if necessary).
    - mratzloff 10y ago
      So it's kind of like a traditional error-correcting code, just sent along a separate channel?
      - johncolanduoni 10y ago
        Sort of. The only issue with that interpretation is that either piece of data in isolation conveys none of the original information, even probabilistically. A better analogy would be sending two classical bits by different channels, with the desired bit being the XOR of the two, and one of the bits being chosen by a random number generator.
tgb 10y ago
Third time I'm writing this in this thread, but it causes so much confusion: in order to quantum teleport one qubit, you transmit two classical bits of information, say by ethernet. So there's no way to take this to work as a sneaky means of sending a classical bit. It's my sacred duty to keep repeating this fact until the whole internet knows. (It's even in the first sentence of the wikipedia page on quantum teleportation!)
etatoby 10y ago
> why are we sure that we're not just reading off a random number generator that's set from the same seed? We aren't. In fact, according to the decoherence / many-world interpretation (and possibly others) that's exactly what happens. When you "measure" (aka. entangle yourself with) one of the two photons, you end up (find yourself) in one of the infinitely many possible "worlds", in which both photons have a single, definite state (always the opposite of each other, in this setup.) It's as simple as that. I keep recommending this sequence of blog posts as the best introduction to the topic: http://lesswrong.com/lw/r5/the_quantum_physics_sequence/ http://lesswrong.com/lw/r5/the_quantum_physics_sequence/

johncolanduoni 10y ago

Quantum teleportation is actually kind of the opposite. You transmit a quantum state (which if your particle is as simple as a photon, can be used to create a copy of the particle) by using an already-separated entangled pair and sending two classical bits any boring old way (fiber, wire, smoke signals). If you don't get the classical information on the receiving end, any attempts to read the quantum state will garble the quantum transmission such that it'll be indistinguishable from random noise. Since those classical bits are limited by the speed of light, so is the overall transmission of the state.

trentlott 10y ago
So why does the reciver need the two bits? I'm not prone to study too deeply because I've learned (apparently) very bad analogies or schemes for entanglement.
- johncolanduoni 10y ago
  The teleportation process involves using a measurement to entangle the receiver's side of the pair with the original state in a useful way. Since measurements are probabilistic (and the measurement is along two qubits) there are four outcomes of this measurement with four different kinds of "garbling" of the transmission (one of which is no garbling). The receiver needs the two bits to choose which of four kinds of garbling occurred, so it can correct them with the appropriate post-processing (if necessary).
  - mratzloff 10y ago
    So it's kind of like a traditional error-correcting code, just sent along a separate channel?
    - johncolanduoni 10y ago
      Sort of. The only issue with that interpretation is that either piece of data in isolation conveys none of the original information, even probabilistically. A better analogy would be sending two classical bits by different channels, with the desired bit being the XOR of the two, and one of the bits being chosen by a random number generator.

tgb 10y ago

Third time I'm writing this in this thread, but it causes so much confusion: in order to quantum teleport one qubit, you transmit two classical bits of information, say by ethernet. So there's no way to take this to work as a sneaky means of sending a classical bit. It's my sacred duty to keep repeating this fact until the whole internet knows. (It's even in the first sentence of the wikipedia page on quantum teleportation!)

etatoby 10y ago

> why are we sure that we're not just reading off a random number generator that's set from the same seed? We aren't. In fact, according to the decoherence / many-world interpretation (and possibly others) that's exactly what happens. When you "measure" (aka. entangle yourself with) one of the two photons, you end up (find yourself) in one of the infinitely many possible "worlds", in which both photons have a single, definite state (always the opposite of each other, in this setup.) It's as simple as that. I keep recommending this sequence of blog posts as the best introduction to the topic: http://lesswrong.com/lw/r5/the_quantum_physics_sequence/ http://lesswrong.com/lw/r5/the_quantum_physics_sequence/