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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

by 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).
      - lisper 8y ago
        Are you distinguishing between a "mixed state" and an "ensemble" and a "mixture of pure states"? Because AFAIK these are all the same thing. If you think they are different, what distinguishes them? If you agree that they are all the same, then what you are saying is exactly the same as what I am saying, except for the "no entanglement" part. Whether or not you have entanglement or a mixed state depends on what point of view you choose to take. If you have an EPR state, then the joint system is in a pure state, but each individual particle is in a mixed state when viewed as an isolated system. That is what a mixed state is. A mixed state is nothing more or less than the state of a proper subset of an entangled system.
        kgwgk 8y ago
        A mixed state can be produced as a subset of a entangled system. It can also be produced as a mixture of pure independent states. They need not correspond to the same physical situation: you can prepare the same mixture state in different ways. But these ensembles can not be distinguished later (unless you “remember” the lost information about the precise state for each element!). For example, you can get unpolarized light as a mixture of left- and right-polarized light, or as a mixture of horizontal- and vertical-polarized light, or getting one photon from entangled pairs (and discarding the other). I’m sure that you can find a way to describe that as an entanglement with something else, but there is no need to do so. In particular, when Alice measures a photon in an entangled pair the entanglement is broken. Bob’s photon is no longer entangled with Alice’s photon. Hopefully you will agree that Alice’s photon is now in a pure state. Again, you may want to make things much more difficult but it’s not required to prevent FTL signaling.
        lisper 8y ago
        > I’m sure that you can find a way to describe that as an entanglement with something else, but there is no need to do so. That depends on your goals. If all you care about is making correct predictions of experimental results, then yes, one point of view is as good as another. But if you care about actually understanding what is going on then explaining measurement in terms of entanglement is progress. > when Alice measures a photon in an entangled pair the entanglement is broken No. The entanglement remains, and is incorporated into a (much) larger system of mutually entangled particles, including Alice herself. As part of this process, decoherence happens, which makes the original entanglement impossible to detect as a practical matter. But it doesn't go away. > Hopefully you will agree that Alice’s photon is now in a pure state. No, I will not agree. When Alice measures the photon, the photon must be absorbed by some detector, and so can longer be said to exist at all in any meaningful sense.

by kgwgk 8y ago

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).
    - lisper 8y ago
      Are you distinguishing between a "mixed state" and an "ensemble" and a "mixture of pure states"? Because AFAIK these are all the same thing. If you think they are different, what distinguishes them? If you agree that they are all the same, then what you are saying is exactly the same as what I am saying, except for the "no entanglement" part. Whether or not you have entanglement or a mixed state depends on what point of view you choose to take. If you have an EPR state, then the joint system is in a pure state, but each individual particle is in a mixed state when viewed as an isolated system. That is what a mixed state is. A mixed state is nothing more or less than the state of a proper subset of an entangled system.
      - kgwgk 8y ago
        A mixed state can be produced as a subset of a entangled system. It can also be produced as a mixture of pure independent states. They need not correspond to the same physical situation: you can prepare the same mixture state in different ways. But these ensembles can not be distinguished later (unless you “remember” the lost information about the precise state for each element!). For example, you can get unpolarized light as a mixture of left- and right-polarized light, or as a mixture of horizontal- and vertical-polarized light, or getting one photon from entangled pairs (and discarding the other). I’m sure that you can find a way to describe that as an entanglement with something else, but there is no need to do so. In particular, when Alice measures a photon in an entangled pair the entanglement is broken. Bob’s photon is no longer entangled with Alice’s photon. Hopefully you will agree that Alice’s photon is now in a pure state. Again, you may want to make things much more difficult but it’s not required to prevent FTL signaling.
        lisper 8y ago
        > I’m sure that you can find a way to describe that as an entanglement with something else, but there is no need to do so. That depends on your goals. If all you care about is making correct predictions of experimental results, then yes, one point of view is as good as another. But if you care about actually understanding what is going on then explaining measurement in terms of entanglement is progress. > when Alice measures a photon in an entangled pair the entanglement is broken No. The entanglement remains, and is incorporated into a (much) larger system of mutually entangled particles, including Alice herself. As part of this process, decoherence happens, which makes the original entanglement impossible to detect as a practical matter. But it doesn't go away. > Hopefully you will agree that Alice’s photon is now in a pure state. No, I will not agree. When Alice measures the photon, the photon must be absorbed by some detector, and so can longer be said to exist at all in any meaningful sense.