4 ms·

Note that there is a very important property of entangled particles that is hardly ever mentioned in this kind of exposition, which IMHO casts a lot of light on

by lisper 5y ago

Note that there is a very important property of entangled particles that is hardly ever mentioned in this kind of exposition, which IMHO casts a lot of light on what is really going on, and that is that entangled particles do not self-interfere the way non-entangled particles do. For more details see: https://flownet.com/ron/QM.pdf https://flownet.com/ron/QM.pdf

jfengel 5y ago
I don't think the paper justifies the statement as you put it, though perhaps you can point out what I'm missing. I don't think you can tell just from looking at the particle itself whether it has an entangled partner somewhere in the universe. It is, however, possible to use the entangled partners to create systems with decidedly counter-intuitive properties that change the way the un-involved partner interacts. That's also the essence of Bell's Theorem. It only works when you're controlling the experiment as a whole and thus not transmitting information faster than light... though you can set up the experiment in a way that makes the conventional transmission of information incredibly obscure. Bell's Theorem requires you to jump through a lot of hoops to exactly mimic that, which is why it took a long time to definitively rule out other interpretations of the experiments.
- 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.

by lisper 5y ago

jfengel 5y ago

I don't think the paper justifies the statement as you put it, though perhaps you can point out what I'm missing. I don't think you can tell just from looking at the particle itself whether it has an entangled partner somewhere in the universe. It is, however, possible to use the entangled partners to create systems with decidedly counter-intuitive properties that change the way the un-involved partner interacts. That's also the essence of Bell's Theorem. It only works when you're controlling the experiment as a whole and thus not transmitting information faster than light... though you can set up the experiment in a way that makes the conventional transmission of information incredibly obscure. Bell's Theorem requires you to jump through a lot of hoops to exactly mimic that, which is why it took a long time to definitively rule out other interpretations of the experiments.

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.