6 ms·

"...making a measurement on some entangled property of A will give you a random result and B will have the complementary result." "...once you make a measureme

"...making a measurement on some entangled property of A will give you a random result and B will have the complementary result." "...once you make a measurement you lose entanglement." One entangled pair can transmit exactly one bit of information exactly once. The measurement necessarily destroys the pair. Do I understand correctly?

convery 4y ago
No, as you can't influence the outcome of the measurement. So both parties reads a bit, which is random, they just happen to be opposite of eachother. No useful information is transmitted.
- ratg13 4y ago
  I think what the person is struggling to understand is the phrase "lose entanglement". I also wonder what is meant here. Are the particles no longer entangled after they are measured once, or what is being said here?
  - b112 4y ago
    I'm struggling with the idea that reading a bit at A is random, and B is random, but that A is the inverse/opposite of B. That means B is not-random to me. I have a feeling that terms being used are specifically defined terms of art. EG, don't mean what they mean in common english, but instead, have a different technical meaning.
    - themoonisachees 4y ago
      B basically isn't random (that's why quantum entanglement is intersting) but you have 3 options on when to read its state ( which you can only do once): Option 1 is that you read it before the state of A is read, in which case my undestanding is you are effectively A in this scenario. Option 2 is that the state of A has been read, but it is unkown to you yet, because of speed of light reasons. You read B, and can infer the state of A, but because A is random, no information has been transmitted. The "sender" cannot control which state of A they read, and had you not known about it, the state of B might as well be random. Option 3 is that you read B after recieving the state of A, in which case there is something useful in that only you can make sure that only the sender knew of the state of A, meaning you are effectively authenticating a message. But still no information has been transmitted faster than light, it's just a secure channel.
    - Viliam1234 4y ago
      The inverse/opposite of "50% chance head, 50% chance tails" is "50% chance tails, 50% chance head". Which is the same thing. If you are looking at A alone, A is random. If you are looking at B alone, B is random. If you are looking at A and B together, they are entangled, i.e. their probabilities are "50% chance A head B tails, 50% chance A tails B head". Random means something you can't predict (other than statistically), until it actually happens. If you know the state of the entire universe except for the particle B, you can't predict A. If you know the state of the entire universe except for the particle A, you can't predict B.
  - Viliam1234 4y ago
    Correct. After you measure the particles once, they are no longer entangled. It is actually quite difficult to keep the particles entangled, because the rest of the universe keeps trying to measure them. (I am not a physicist, and I have no idea how they keep the particles entangled.)
8note 4y ago
You can't know that the bit is ready to be checked without separate communication
- yreg 4y ago
  You could have pre-agreed reading time, but the bit you "send" is also random.
  - moffkalast 4y ago
    But hold on, this could be practical in some specific cases regardless. Say you have two fleets of spaceships on the opposite ends of the solar system, both having plans A and B for attack. They want to surprise the enemy by being unpredictable so despite the enemy knowing about the two plans, if you decide randomly which fleet does which you'd still have an advantage. Maybe one fleet is larger so they could focus forces where they need to be if they knew the plan ahead of time. But if you choose randomly by default you could have both fleets do plan A, which wouldn't work. But if one measures the entangled pair they both get a mutually exclusive random result and thus can make an unpredictable plan work without a pre-set decision of who does what. A weird far fetched example to be sure, but I'd imagine cryptography nerds could find a matching case for some kind of encryption or whatever.
    - yreg 4y ago
      I think it would work, but for practical purposes it isn't different from picking which fleet will commit to which plan in advance, sealing it to an envelope and opening it at the agreed time. Your use case fails to take advantage of the fact that the quantum states collapse at the time of reading.
    - toast0 4y ago
      There's something here. I think (if I understand correctly, which is kind of iffy) it's also the case that if either party reads the property early, the entanglement is used up, and the on time read is no longer correlated. That may not be very helpful for a battle plan, but if there was concerns about enemy infiltration, the entanglement could be intentionally used early, resulting in neither ship knowing what the other is doing, although the ship that read on time wouldn't know they didn't know.
denton-scratch 4y ago
Not exactly; nothing is "transmitted".

convery 4y ago

No, as you can't influence the outcome of the measurement. So both parties reads a bit, which is random, they just happen to be opposite of eachother. No useful information is transmitted.

ratg13 4y ago
I think what the person is struggling to understand is the phrase "lose entanglement". I also wonder what is meant here. Are the particles no longer entangled after they are measured once, or what is being said here?
- b112 4y ago
  I'm struggling with the idea that reading a bit at A is random, and B is random, but that A is the inverse/opposite of B. That means B is not-random to me. I have a feeling that terms being used are specifically defined terms of art. EG, don't mean what they mean in common english, but instead, have a different technical meaning.
  - themoonisachees 4y ago
    B basically isn't random (that's why quantum entanglement is intersting) but you have 3 options on when to read its state ( which you can only do once): Option 1 is that you read it before the state of A is read, in which case my undestanding is you are effectively A in this scenario. Option 2 is that the state of A has been read, but it is unkown to you yet, because of speed of light reasons. You read B, and can infer the state of A, but because A is random, no information has been transmitted. The "sender" cannot control which state of A they read, and had you not known about it, the state of B might as well be random. Option 3 is that you read B after recieving the state of A, in which case there is something useful in that only you can make sure that only the sender knew of the state of A, meaning you are effectively authenticating a message. But still no information has been transmitted faster than light, it's just a secure channel.
  - Viliam1234 4y ago
    The inverse/opposite of "50% chance head, 50% chance tails" is "50% chance tails, 50% chance head". Which is the same thing. If you are looking at A alone, A is random. If you are looking at B alone, B is random. If you are looking at A and B together, they are entangled, i.e. their probabilities are "50% chance A head B tails, 50% chance A tails B head". Random means something you can't predict (other than statistically), until it actually happens. If you know the state of the entire universe except for the particle B, you can't predict A. If you know the state of the entire universe except for the particle A, you can't predict B.
- Viliam1234 4y ago
  Correct. After you measure the particles once, they are no longer entangled. It is actually quite difficult to keep the particles entangled, because the rest of the universe keeps trying to measure them. (I am not a physicist, and I have no idea how they keep the particles entangled.)

8note 4y ago

You can't know that the bit is ready to be checked without separate communication

yreg 4y ago
You could have pre-agreed reading time, but the bit you "send" is also random.
- moffkalast 4y ago
  But hold on, this could be practical in some specific cases regardless. Say you have two fleets of spaceships on the opposite ends of the solar system, both having plans A and B for attack. They want to surprise the enemy by being unpredictable so despite the enemy knowing about the two plans, if you decide randomly which fleet does which you'd still have an advantage. Maybe one fleet is larger so they could focus forces where they need to be if they knew the plan ahead of time. But if you choose randomly by default you could have both fleets do plan A, which wouldn't work. But if one measures the entangled pair they both get a mutually exclusive random result and thus can make an unpredictable plan work without a pre-set decision of who does what. A weird far fetched example to be sure, but I'd imagine cryptography nerds could find a matching case for some kind of encryption or whatever.
  - yreg 4y ago
    I think it would work, but for practical purposes it isn't different from picking which fleet will commit to which plan in advance, sealing it to an envelope and opening it at the agreed time. Your use case fails to take advantage of the fact that the quantum states collapse at the time of reading.
  - toast0 4y ago
    There's something here. I think (if I understand correctly, which is kind of iffy) it's also the case that if either party reads the property early, the entanglement is used up, and the on time read is no longer correlated. That may not be very helpful for a battle plan, but if there was concerns about enemy infiltration, the entanglement could be intentionally used early, resulting in neither ship knowing what the other is doing, although the ship that read on time wouldn't know they didn't know.

denton-scratch 4y ago

Not exactly; nothing is "transmitted".