3 ms·

See section 4.2, and in particular the paragraph that starts "Here's the kicker..." It is true that you can't tell if a single particle is entangled or not. B

by lisper 5y ago

See section 4.2, and in particular the paragraph that starts "Here's the kicker..." It is true that you can't tell if a single particle is entangled or not. But if you have an ensemble of particles all prepared in the same state then you can tell if that state is entangled or not. Non-entangled (a.k.a. pure) states have a preferred basis that produce self-interference. Entangled (a.k.a.) mixed states do not. (The pure-mixed dichotomy is a little misleading because it depends on your point of view. A single member of an EPR pair is in a mixed state, but the pair as a whole is in a pure state.)

gus_massa 5y ago
I agree with the GP. If you only have a single member of the pair, then you will see the same interference pattern in a double slit experiment than with a not-entangled particle. It doesn't matter if the other particle has collided with a brick, went thru a double slit experiment, went thru a bad double slit experiment, or is flying to Andromeda. (In spite that the calculation to get the correlation with any result in the other experiment may be much harder with the entangled pair than with two non entangled particles.)
- lisper 5y ago
  What can I say? You're wrong. The math shows that you're wrong (as does the elementary argument presented in the paper). Find a physicist and ask them if you don't believe me.
  - gus_massa 5y ago
    I have at least 8 Physicist that I email/zoom regularly (at least twice per month). I also have half a degree in Physics, with at least 2 courses of Quantum Mechanics (all the advanced courses also use QM, but there are 2 courses only about QM). [I also have a degree and a PhD in Math, but it's not too relevant here.] Anyway, I'll read the article thoughtfully and write a long comment tomorrow. Can you take a look tomorrow?
    - lisper 5y ago
      Sure. Just to be clear, the part you are wrong about is this: > If you only have a single member of the pair, then you will see the same interference pattern in a double slit experiment than with a not-entangled particle. This bit: > It doesn't matter if the other particle has collided with a brick, went thru a double slit experiment, went thru a bad double slit experiment, or is flying to Andromeda. is correct. Also, there is an interference pattern in the results of a double-slit run on entangled particles, but it is not "the same" as you get with non-entangled particles, and the procedure you have to go through to observe this interference pattern is radically different.
      - gus_massa 5y ago
        From the last part of 4.2 > Here's how it works. We send a pair of EPR photons through a pair of two-slit apparati each of which has a polarization rotator on one of the slits. On one side of the apparatus (side A) we install a polarization filter which filters out interference on that side and makes it visible. We can filter out interference on the other side (side B) of the apparatus as follows: on side A we keep a record of which photons passed through the filter and which were reflected. On side B we keep a record of where each photon landed on the screen. We then take these two records and combine them: for each photon that was passed through the filter on side A, we take the corresponding photon on side B and note where it landed on the screen. The end result is a (visible) interference pattern. It was there all along, but the only way we can filter it out so we can see it is to combine information from both sides of the experiment. And that is the last nail in the coffin of superluminal communication via entangled photons. Let's suppose you are in lab B and measure where the photons hit the screen. There are no visible interference patters. But just before you call to lab A, it's is nuked from orbit. Now, you are unsure if the people that was generating the photons were sending pairs of entangled photons to A and B, or they were just sending pairs of normal photons. Can you look at the data you collected in B and discover what the people in the generator were doing? Now imagine the same experiment with a rebuild lab A', but you remove the polarization rotator. Now you see the interference pattern. And A' gets nuked again. Can you look at the data you collected in B and discover what the people in the generator were doing? --- I understand that you can take a plane and collect all the pieces of A and A' are reconstruct them, after all Classic Mechanics and Quantum Mechanics without the measurement rule are reversible, so it's theoretically possible, but very impractical. I think this discussed in section 5. For the measurement problem, I prefer the something-something-decoherence solution. I call it something-something-decoherence because there are still a lot of work to be done before it's clear if it's the correct solution. --- About the experiment in 4.2: I'm 99% sure after adding the polarizer at 45° there will not be interference. You can split the polarizer into two smaller polarizers with the same angle, one for each slit. The exchange the order of the rotator+polarizer in one slit to polarizer+rotator. Note that after the exchange, the new polarizer must be rotated 90°, so it's polarizer at "-45°". Now the first thing in one slit is a polarizer at "+45°" and the other is at "-45°", so they will select orthogonal states and not get interference even after rotating one of them. I think this can be fixed using a quarter-wave plate, but the calculation is slightly more complicated. [Sorry for the delay.]

by lisper 5y ago

gus_massa 5y ago

I agree with the GP. If you only have a single member of the pair, then you will see the same interference pattern in a double slit experiment than with a not-entangled particle. It doesn't matter if the other particle has collided with a brick, went thru a double slit experiment, went thru a bad double slit experiment, or is flying to Andromeda. (In spite that the calculation to get the correlation with any result in the other experiment may be much harder with the entangled pair than with two non entangled particles.)

lisper 5y ago
What can I say? You're wrong. The math shows that you're wrong (as does the elementary argument presented in the paper). Find a physicist and ask them if you don't believe me.
- gus_massa 5y ago
  I have at least 8 Physicist that I email/zoom regularly (at least twice per month). I also have half a degree in Physics, with at least 2 courses of Quantum Mechanics (all the advanced courses also use QM, but there are 2 courses only about QM). [I also have a degree and a PhD in Math, but it's not too relevant here.] Anyway, I'll read the article thoughtfully and write a long comment tomorrow. Can you take a look tomorrow?
  - lisper 5y ago
    Sure. Just to be clear, the part you are wrong about is this: > If you only have a single member of the pair, then you will see the same interference pattern in a double slit experiment than with a not-entangled particle. This bit: > It doesn't matter if the other particle has collided with a brick, went thru a double slit experiment, went thru a bad double slit experiment, or is flying to Andromeda. is correct. Also, there is an interference pattern in the results of a double-slit run on entangled particles, but it is not "the same" as you get with non-entangled particles, and the procedure you have to go through to observe this interference pattern is radically different.
    - gus_massa 5y ago
      From the last part of 4.2 > Here's how it works. We send a pair of EPR photons through a pair of two-slit apparati each of which has a polarization rotator on one of the slits. On one side of the apparatus (side A) we install a polarization filter which filters out interference on that side and makes it visible. We can filter out interference on the other side (side B) of the apparatus as follows: on side A we keep a record of which photons passed through the filter and which were reflected. On side B we keep a record of where each photon landed on the screen. We then take these two records and combine them: for each photon that was passed through the filter on side A, we take the corresponding photon on side B and note where it landed on the screen. The end result is a (visible) interference pattern. It was there all along, but the only way we can filter it out so we can see it is to combine information from both sides of the experiment. And that is the last nail in the coffin of superluminal communication via entangled photons. Let's suppose you are in lab B and measure where the photons hit the screen. There are no visible interference patters. But just before you call to lab A, it's is nuked from orbit. Now, you are unsure if the people that was generating the photons were sending pairs of entangled photons to A and B, or they were just sending pairs of normal photons. Can you look at the data you collected in B and discover what the people in the generator were doing? Now imagine the same experiment with a rebuild lab A', but you remove the polarization rotator. Now you see the interference pattern. And A' gets nuked again. Can you look at the data you collected in B and discover what the people in the generator were doing? --- I understand that you can take a plane and collect all the pieces of A and A' are reconstruct them, after all Classic Mechanics and Quantum Mechanics without the measurement rule are reversible, so it's theoretically possible, but very impractical. I think this discussed in section 5. For the measurement problem, I prefer the something-something-decoherence solution. I call it something-something-decoherence because there are still a lot of work to be done before it's clear if it's the correct solution. --- About the experiment in 4.2: I'm 99% sure after adding the polarizer at 45° there will not be interference. You can split the polarizer into two smaller polarizers with the same angle, one for each slit. The exchange the order of the rotator+polarizer in one slit to polarizer+rotator. Note that after the exchange, the new polarizer must be rotated 90°, so it's polarizer at "-45°". Now the first thing in one slit is a polarizer at "+45°" and the other is at "-45°", so they will select orthogonal states and not get interference even after rotating one of them. I think this can be fixed using a quarter-wave plate, but the calculation is slightly more complicated. [Sorry for the delay.]