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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 th

by 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.
- kgwgk 8y ago
  Fine, but this approach leads nowhere. Before you said that “Before teleportation C is in a pure state”. But that’s surely impossible, because wherever C is coming from it’s entangled with a gazillion things already... If the original state C can be “pure”, then the final state B can be “pure” as well (and identical to C). I think you have a problem with the probabilistic nature of QM, if you only can accept a statistical mixture as being part of an entangled pure system. FTL signaling is just a excuse (and you don’t explain how standard QM would allow for it).
  - lisper 8y ago
    > this approach leads nowhere. It has led to a great deal of mental clarity for me. YMMV. > Before you said that “Before teleportation C is in a pure state”. But thats surely impossible, because wherever C is coming from it’s entangled with a gazillion things already... That's right, and indeed figuring out how a laser actually works is quite challenging. http://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle.html http://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzl... https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle-part-deux.html https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzz... https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle-part-drei.html https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzz... > If the original state C can be “pure”, then the final state B can be “pure” as well (and identical to C). Only if you are willing to explain how B was transformed from its initial state into its final state, and more to the point, when this transition happened. > I think you have a problem with the probabilistic nature of QM Huh? What have I said that leads you to believe that? It's not true. > if you only can accept a statistical mixture as being part of an entangled pure system That has more to do with philosophy than physics. Does an apple really fall because gravity is pulling down on it? No, but it's a good enough approximation to the truth that it's usually not worth quibbling over.
    - kgwgk 8y ago
      > how B was transformed from its initial state into its final state, and more to the point, when this transition happened. There is a full description here: https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_presentation https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p... It's difficult to copy formulas here but essentially A and B are in one Bell state at the beginning. When Alice makes a measurement in the basis of the Bell states for C and A, the result is that B will now be in one of the states alpha|0> + beta|1> alpha|0> - beta|1> beta|0> + alpha|1> beta|0> - alpha|1> Alice knows which one of the four is the pure state of B after the measurement and tells Bob how to rotate the state (if needed) to recover alpha|0> + beta|1> So, again, when Alice does the measurement the state becomes one of the four listed states and when Bob rotates the state (if needed) B is the same state that C was originally. Maybe you agree that this is making correct predictions of experimental results, maybe not. I still don't know when and how does B change along the procedure according to your view. > No, but it's a good enough approximation to the truth that it's usually not worth quibbling over. Talking about "pure" states after a measurement is a good enough approximation to the truth, but you find it worth quibbling over :-)

by kgwgk 8y ago

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.

kgwgk 8y ago
Fine, but this approach leads nowhere. Before you said that “Before teleportation C is in a pure state”. But that’s surely impossible, because wherever C is coming from it’s entangled with a gazillion things already... If the original state C can be “pure”, then the final state B can be “pure” as well (and identical to C). I think you have a problem with the probabilistic nature of QM, if you only can accept a statistical mixture as being part of an entangled pure system. FTL signaling is just a excuse (and you don’t explain how standard QM would allow for it).
- lisper 8y ago
  > this approach leads nowhere. It has led to a great deal of mental clarity for me. YMMV. > Before you said that “Before teleportation C is in a pure state”. But thats surely impossible, because wherever C is coming from it’s entangled with a gazillion things already... That's right, and indeed figuring out how a laser actually works is quite challenging. http://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle.html http://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzl... https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle-part-deux.html https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzz... https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzzle-part-drei.html https://blog.rongarret.info/2018/05/a-quantum-mechanics-puzz... > If the original state C can be “pure”, then the final state B can be “pure” as well (and identical to C). Only if you are willing to explain how B was transformed from its initial state into its final state, and more to the point, when this transition happened. > I think you have a problem with the probabilistic nature of QM Huh? What have I said that leads you to believe that? It's not true. > if you only can accept a statistical mixture as being part of an entangled pure system That has more to do with philosophy than physics. Does an apple really fall because gravity is pulling down on it? No, but it's a good enough approximation to the truth that it's usually not worth quibbling over.
  - kgwgk 8y ago
    > how B was transformed from its initial state into its final state, and more to the point, when this transition happened. There is a full description here: https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_presentation https://en.wikipedia.org/wiki/Quantum_teleportation#Formal_p... It's difficult to copy formulas here but essentially A and B are in one Bell state at the beginning. When Alice makes a measurement in the basis of the Bell states for C and A, the result is that B will now be in one of the states alpha|0> + beta|1> alpha|0> - beta|1> beta|0> + alpha|1> beta|0> - alpha|1> Alice knows which one of the four is the pure state of B after the measurement and tells Bob how to rotate the state (if needed) to recover alpha|0> + beta|1> So, again, when Alice does the measurement the state becomes one of the four listed states and when Bob rotates the state (if needed) B is the same state that C was originally. Maybe you agree that this is making correct predictions of experimental results, maybe not. I still don't know when and how does B change along the procedure according to your view. > No, but it's a good enough approximation to the truth that it's usually not worth quibbling over. Talking about "pure" states after a measurement is a good enough approximation to the truth, but you find it worth quibbling over :-)