3 ms·

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/Quan

by 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
  People get patents for perpetual motion machines as well... Do you really think your idea works? This is what I think: the quantum state of your entangled pair is a pure state and if you look at the quantum state of the photon sent to the "receiver" it is described by the density matrix obtained by tracing out the other photon : 1/2 0 0 1/2 If you measure in the "transmiter" the horizontal polarization, what you get in the "receiver" is a 50/50 mixture of pure states |H> and |V> which is described by the same density matrix. If you measure in the "transmiter" the 45-degrees polarization, what you get in the "receiver" is a 50/50 mixture of pure states |45> and |-45> which is again described by the same density matrix. No matter what yo do or do not measure at the "transmiter" you will get the same measurements at the "receiver" because the density matrix describing the state is always the same (unpolarized light).
  - lisper 8y ago
    > Do you really think your idea works? Of course not. That's the whole point.
    - kgwgk 8y ago
      I'm lost. It doesn't work according to QM. And you agree that it doesn't work. How does it pose a problem for QM again?
      - lisper 8y ago
        I never said it did. I said that if you can transform a mixed state into a pure state then that poses a problem for QM because it would lead to FTL signaling. But you can't, so it doesn't.
        kgwgk 8y ago
        Every time you do your measurement in the “transmiter” you will have a pure state in the receiver and it doesn’t lead to FTL signaling. What you have when you repeat the procedure is an ensemble of states: half of them the pure state |H> and half of them the pure state |V> (or half of them |H> and half of them |V>). There is no entanglement, only (a mixture of) pure states. And it doesn’t lead to FTL signaling. At the end of the “teleporting” procedure you have a pure state B which is identical to the original pure state C. And it doesn’t lead to FTL signaling. Maybe your problem arises from the following misunderstanding: > If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) In the example I gave above, the density matrix for the photon arriving at the “detector” doesn’t change when the measurement is made at the “transmitter” and entanglement is broken. But the same density matrix describes different situations. When there is entanglement the photon is part of a pure state and the density matrix is obtained by tracing out the rest of the system. Later there is no entanglement and the density matrix describes a statistical mixture of pure states (50% |H> and 50% |V>, or whichever is the apropriate basis for the measurement that was made on the other photon destroying entanglement).

by kgwgk 8y ago

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
People get patents for perpetual motion machines as well... Do you really think your idea works? This is what I think: the quantum state of your entangled pair is a pure state and if you look at the quantum state of the photon sent to the "receiver" it is described by the density matrix obtained by tracing out the other photon : 1/2 0 0 1/2 If you measure in the "transmiter" the horizontal polarization, what you get in the "receiver" is a 50/50 mixture of pure states |H> and |V> which is described by the same density matrix. If you measure in the "transmiter" the 45-degrees polarization, what you get in the "receiver" is a 50/50 mixture of pure states |45> and |-45> which is again described by the same density matrix. No matter what yo do or do not measure at the "transmiter" you will get the same measurements at the "receiver" because the density matrix describing the state is always the same (unpolarized light).
- lisper 8y ago
  > Do you really think your idea works? Of course not. That's the whole point.
  - kgwgk 8y ago
    I'm lost. It doesn't work according to QM. And you agree that it doesn't work. How does it pose a problem for QM again?
    - lisper 8y ago
      I never said it did. I said that if you can transform a mixed state into a pure state then that poses a problem for QM because it would lead to FTL signaling. But you can't, so it doesn't.
      - kgwgk 8y ago
        Every time you do your measurement in the “transmiter” you will have a pure state in the receiver and it doesn’t lead to FTL signaling. What you have when you repeat the procedure is an ensemble of states: half of them the pure state |H> and half of them the pure state |V> (or half of them |H> and half of them |V>). There is no entanglement, only (a mixture of) pure states. And it doesn’t lead to FTL signaling. At the end of the “teleporting” procedure you have a pure state B which is identical to the original pure state C. And it doesn’t lead to FTL signaling. Maybe your problem arises from the following misunderstanding: > If you could cause an entangled particle to become unentangled (i.e. to transition from a mixed state to a pure state) In the example I gave above, the density matrix for the photon arriving at the “detector” doesn’t change when the measurement is made at the “transmitter” and entanglement is broken. But the same density matrix describes different situations. When there is entanglement the photon is part of a pure state and the density matrix is obtained by tracing out the rest of the system. Later there is no entanglement and the density matrix describes a statistical mixture of pure states (50% |H> and 50% |V>, or whichever is the apropriate basis for the measurement that was made on the other photon destroying entanglement).