5 ms·

>> C was a pure state and now B is described by the same density matrix [after teleportation, i.e. after the whole procedure including the two-bit message recep

by kgwgk 8y ago

>> C was a pure state and now B is described by the same density matrix [after teleportation, i.e. after the whole procedure including the two-bit message reception and corresponding transformation] > No, it isn't. All of B's measurement outcomes are the same, but the density matrix is necessarily different because B is in an entangled state. If this were not the case then it would lead to FTL communication. The density matrix describing C at the beginning and B at the end is the same (otherwise how could all measurement outcomes be the same?) and B is not in an entangled state. I don't see how this leads to FTL communication, the procedure requires Alice to send a (non-FTL) message to Bob. >> B is in a different mixed state afterwards [after Alice's measurement] > That depends on what you mean. The only difference is that afterwards, B is entangled with C as well as A, so now the joint A-B system is in a mixed state whereas before the A-B system was pure. B is no longer entangled with the A/C system after Alice's measurement. > In that respect, yes, B is in a different state. But viewed in isolation, B's state is exactly the same after the A-C entanglement as before. For the sake of the argument, let's say that Alice knows that the original state C is |up> and that after performing her measurement she sends to Bob the message "you don't have to do any transformation, the state of B is now equal to C, i.e. |up>" (this case will happen eventually if we repeat the experience several times). If, according to you, the state of B is exactly the same after Alice's measurement as it was at the beginning of the procedure, how do you explain that the state of B is |up> at the end? Or what would you say is happening to B, if anything at all?

lisper 8y ago
> The density matrix describing C at the beginning and B at the end is the same No, it isn't. C starts in a pure state. B is created and always remains in a mixed state. > (otherwise how could all measurement outcomes be the same?) Because you cannot distinguish a mixed state from a pure state when measuring only a single particle. You can distinguish them if you make measurements on multiple particles in the same state: pure states produce first-order interference. Mixed states don't. (This is just the old QM trope about "measurement destroys interference", which is not quite true. Entanglement "destroys" interference, except that it doesn't really, it just destroys first-order interference.) > B is no longer entangled with the A/C system after Alice's measurement. Yes, it is. The only way to "undo" an entanglement is to bring the originally entangled particles back together at the same location, i.e. it is essentially impossible. If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) by performing an operation on its partner then you could do FTL signaling by observing the presence or absence of first-order interference. > how do you explain that the state of B is |up> at the end It isn't. It's in a mixed state. It is no different than regular EPR with only two particles. If Alice measures A and gets the result "up" then Bob is guaranteed to get "down", but that does NOT mean that B is in the |down> state. It is still in a mixed state. |down> is a pure state. It is important to remember that mixed states and pure states cannot be distinguished by making measurements on a single particle, so if there is only one particle this is a philosophical distinction, not a scientific one. But if the experiment can be repeated, then the fact that B is mixed can be experimentally demonstrated, with all the usual caveats about quantum weirdness. The point is that QT does not introduce any new weirdness that wasn't already there in EPR.
- kgwgk 8y ago
  In case anyone else is reading this, I want to make clear that I'm giving the standard description of quantum teleportation. https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_presentation https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p... provides a clear description of the initial states of C and A/B, the operations performed by Alice and Bob and the final state of B (which is identical to the initial state of C). Clearly you don't find this description correct but I would be surprised if you can write down an alternative description that makes sense. You're free to have your own theory, but I don't understand why do you feel that you need one in the first place. What is the problem with measurement breaking entanglement? > If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) by performing an operation on its partner then you could do FTL signaling by observing the presence or absence of first-order interference. Let's say Alice and Bob have entangled spins in the pure state (leaving out the normalization term): |up>_alice * |down>_bob - |down>_alice * |up>_bob According to QM, if Alice measures the spin along the z-axis one of the following alternatives will happen: a) she gets a positive measurement, her spin is in the pure state |up>, Bob's spin is in the pure state |down>, there is no entanglement between the spins or b) she gets a negative measurement, her spin is in the pure state |down>, Bob's spin is in the pure state |up>, there is no entanglement between the spins How do you think this allows for FTL signaling?
  - lisper 8y ago
    > How do you think this allows for FTL signaling? I already told you: with a single particle, it's not possible. But with multiple particles you can tell the difference between a pure state and a mixed state by observing the presence or absence of first-order interference. So to communicate, you send a stream of EPR pairs to Alice and Bob. Alice transmits by either measuring batches of particles or not measuring them, and Bob receives by observing the presence or absence of first-order interference on his end. Fun fact: I was able to get a patent on this idea :-) https://patents.google.com/patent/US7126691B2/en https://patents.google.com/patent/US7126691B2/en
    - kgwgk 8y ago

by kgwgk 8y ago

lisper 8y ago

> The density matrix describing C at the beginning and B at the end is the same No, it isn't. C starts in a pure state. B is created and always remains in a mixed state. > (otherwise how could all measurement outcomes be the same?) Because you cannot distinguish a mixed state from a pure state when measuring only a single particle. You can distinguish them if you make measurements on multiple particles in the same state: pure states produce first-order interference. Mixed states don't. (This is just the old QM trope about "measurement destroys interference", which is not quite true. Entanglement "destroys" interference, except that it doesn't really, it just destroys first-order interference.) > B is no longer entangled with the A/C system after Alice's measurement. Yes, it is. The only way to "undo" an entanglement is to bring the originally entangled particles back together at the same location, i.e. it is essentially impossible. If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) by performing an operation on its partner then you could do FTL signaling by observing the presence or absence of first-order interference. > how do you explain that the state of B is |up> at the end It isn't. It's in a mixed state. It is no different than regular EPR with only two particles. If Alice measures A and gets the result "up" then Bob is guaranteed to get "down", but that does NOT mean that B is in the |down> state. It is still in a mixed state. |down> is a pure state. It is important to remember that mixed states and pure states cannot be distinguished by making measurements on a single particle, so if there is only one particle this is a philosophical distinction, not a scientific one. But if the experiment can be repeated, then the fact that B is mixed can be experimentally demonstrated, with all the usual caveats about quantum weirdness. The point is that QT does not introduce any new weirdness that wasn't already there in EPR.

kgwgk 8y ago
In case anyone else is reading this, I want to make clear that I'm giving the standard description of quantum teleportation. https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_presentation https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p... provides a clear description of the initial states of C and A/B, the operations performed by Alice and Bob and the final state of B (which is identical to the initial state of C). Clearly you don't find this description correct but I would be surprised if you can write down an alternative description that makes sense. You're free to have your own theory, but I don't understand why do you feel that you need one in the first place. What is the problem with measurement breaking entanglement? > If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) by performing an operation on its partner then you could do FTL signaling by observing the presence or absence of first-order interference. Let's say Alice and Bob have entangled spins in the pure state (leaving out the normalization term): |up>_alice * |down>_bob - |down>_alice * |up>_bob According to QM, if Alice measures the spin along the z-axis one of the following alternatives will happen: a) she gets a positive measurement, her spin is in the pure state |up>, Bob's spin is in the pure state |down>, there is no entanglement between the spins or b) she gets a negative measurement, her spin is in the pure state |down>, Bob's spin is in the pure state |up>, there is no entanglement between the spins How do you think this allows for FTL signaling?
- lisper 8y ago
  > How do you think this allows for FTL signaling? I already told you: with a single particle, it's not possible. But with multiple particles you can tell the difference between a pure state and a mixed state by observing the presence or absence of first-order interference. So to communicate, you send a stream of EPR pairs to Alice and Bob. Alice transmits by either measuring batches of particles or not measuring them, and Bob receives by observing the presence or absence of first-order interference on his end. Fun fact: I was able to get a patent on this idea :-) https://patents.google.com/patent/US7126691B2/en https://patents.google.com/patent/US7126691B2/en
  - kgwgk 8y ago