4 ms·

I agree that the cartoon that you criticize is confusing. Starting with the first panel, because it doesn't make clear that Alice doesn't know the color/state o

by kgwgk 8y ago

I agree that the cartoon that you criticize is confusing. Starting with the first panel, because it doesn't make clear that Alice doesn't know the color/state of her photon. If she knows it's yellow she only has to tell Bob to prepare a yellow photon and there is no need for an entangled pair to make it possible. The whole point of the setup is to be able to "teleport" an arbitrary unknown quantum state. But the following doesn't seem correct: > when Bob receives Alice’s bits, he uses that information to apply one of four transformations to his photon. It is Bob’s action that changes his photon’s state into the state of Alice’s original photon. And that is the only time that Bob’s photon changes state. To recover the state being "teleported" by Alice, Bob has to apply one of the four pre-specified "rotations" (unitary transformations) according to the two-bit message he receives. In fact one of the operations is to do nothing, sometimes the state of his photon is already the state of Alice's original photon. Of course this is true only after the measurement performed by Alice. Previously there was no relationship between the state of Bob's photon and the state of Alice's "source" photon: the probability of the state being the same (or related by one of the three other pre-specified rotations) is essentially zero.

lisper 8y ago
> sometimes the state of his photon is already the state of Alice's original photon [but] this is true only after the measurement performed by Alice No, that's wrong. Alice's measurement does not change the state of Bob's particle. There is no measurement Bob can perform on his particle which will tell him whether or not Alice has performed her measurement or not. Quantum teleportation is no different from regular old EPR in this regard. > the probability of the state being the same (or related by one of the other three pre-specified rotations) is essentially zero. Actually, it is precisely zero. The entire claim that the state of Alice's particle is teleported intact to Bob is actually a lie. What is true is that no measurement can distinguish the state of Bob's particle (B) after teleportation from Alice's original (C). But this does not mean that the particles are in the same state. Before teleportation C is in a pure state, and B is in a mixed state (as is A). The Bell measurement that Alice performs on A and C entangles them, and the result is a three-particle mutually entangled system, plus two classical bits that tell you how to tweak B so that its measurement outcomes are the same as C's. But B is in the exact same (mixed) state after Alice's measurement as it was before. BTW, there is a measurement Bob can perform, not on a single B, but on a collection of B's that will reveal that all the B's are in fact in a mixed state. If you want to go down that rabbit hole, read this: http://www.flownet.com/ron/QM.pdf http://www.flownet.com/ron/QM.pdf
- deleted 8y ago
  [deleted]
- kgwgk 8y ago
  > Alice's measurement does not change the state of Bob's particle. > There is no measurement Bob can perform on his particle which will tell him whether or not Alice has performed her measurement or not. These statements are not equivalent. > Actually, it is precisely zero. I meant that it's zero as in "the probability of a random real number in [0 1] to be 0.5 is zero", but not as in "the probability of a random real number in [0 1] to be 42 is zero". > What is true is that no measurement can distinguish the state of Bob's particle (B) after teleportation from Alice's original (C). But this does not mean that the particles are in the same state. C was a pure state and now B is described by the same density matrix so it is necessarily the same pure state. > But B is in the exact same (mixed) state after Alice's measurement as it was before. B is in a different mixed state afterwards and the two-bit message enables Bob to distinguish between the four potential pure states and recover Alice's original pure state C.
  - lisper 8y ago
    > These statements are not equivalent. Why not? > C was a pure state and now B is described by the same density matrix 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. > B is in a different mixed state afterwards 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. 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.
    - 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?

by kgwgk 8y ago

lisper 8y ago

> sometimes the state of his photon is already the state of Alice's original photon [but] this is true only after the measurement performed by Alice No, that's wrong. Alice's measurement does not change the state of Bob's particle. There is no measurement Bob can perform on his particle which will tell him whether or not Alice has performed her measurement or not. Quantum teleportation is no different from regular old EPR in this regard. > the probability of the state being the same (or related by one of the other three pre-specified rotations) is essentially zero. Actually, it is precisely zero. The entire claim that the state of Alice's particle is teleported intact to Bob is actually a lie. What is true is that no measurement can distinguish the state of Bob's particle (B) after teleportation from Alice's original (C). But this does not mean that the particles are in the same state. Before teleportation C is in a pure state, and B is in a mixed state (as is A). The Bell measurement that Alice performs on A and C entangles them, and the result is a three-particle mutually entangled system, plus two classical bits that tell you how to tweak B so that its measurement outcomes are the same as C's. But B is in the exact same (mixed) state after Alice's measurement as it was before. BTW, there is a measurement Bob can perform, not on a single B, but on a collection of B's that will reveal that all the B's are in fact in a mixed state. If you want to go down that rabbit hole, read this: http://www.flownet.com/ron/QM.pdf http://www.flownet.com/ron/QM.pdf

deleted 8y ago
[deleted]
kgwgk 8y ago
> Alice's measurement does not change the state of Bob's particle. > There is no measurement Bob can perform on his particle which will tell him whether or not Alice has performed her measurement or not. These statements are not equivalent. > Actually, it is precisely zero. I meant that it's zero as in "the probability of a random real number in [0 1] to be 0.5 is zero", but not as in "the probability of a random real number in [0 1] to be 42 is zero". > What is true is that no measurement can distinguish the state of Bob's particle (B) after teleportation from Alice's original (C). But this does not mean that the particles are in the same state. C was a pure state and now B is described by the same density matrix so it is necessarily the same pure state. > But B is in the exact same (mixed) state after Alice's measurement as it was before. B is in a different mixed state afterwards and the two-bit message enables Bob to distinguish between the four potential pure states and recover Alice's original pure state C.
- lisper 8y ago
  > These statements are not equivalent. Why not? > C was a pure state and now B is described by the same density matrix 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. > B is in a different mixed state afterwards 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. 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.
  - 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?