5 ms·

Are you suggesting that either wave collapse does not occur, or that instead of wave collapse, that there is a wave "narrowing" which is imperceptible to the ob

by j1vms 8y ago

Are you suggesting that either wave collapse does not occur, or that instead of wave collapse, that there is a wave "narrowing" which is imperceptible to the observer (who observes it as a particle)?

nabla9 8y ago
It's more correct to say "apparent wave function collapse" because wave function collapse as a real physical phenomenon or event is just a hypothesis in some interpretations. Mathematically "appeared collapse" is explained by decoherence. It provides an explanation for the observation of wave function collapse when the system and the environment mix in irreversible way.
- phkahler 8y ago
  >> It's more correct to say "apparent wave function collapse" because wave function collapse as a real physical phenomenon or event is just a hypothesis in some interpretations. Wave function collapse is fiction IMO. There is no observable difference between a particle whose wave function has collapsed and one that isn't. If there was an observable difference we could place an emitter of entangled particle pairs mid-way between two observers (Alice and Bob) arbitrarily far apart. Alice could either turn on her measurements of the particles that arrive on here end or leave it off. This would cause Bob to receive either a stream of particles with collapsed wave functions, or a stream of uncollapsed ones. Alice can now transmit information to Bob faster than the speed of light using Morse code, which is not possible. The reality is that Bob can't tell anything about the superposition of states (or lack of) by observing his particles.
  - kgilpin 8y ago
    > receive either a stream of particles with collapsed wave functions, or a stream of uncollapsed ones I don’t think there’s any way for Bob to determine whether the particle stream has been measured by Alice or not.
    - phkahler 8y ago
      >> I don’t think there’s any way for Bob to determine whether the particle stream has been measured by Alice or not. That's why I wrote: "There is no observable difference between a particle whose wave function has collapsed and one that isn't." If there was a way to tell it would violate relativity by allowing FTL communication.
  - kgwgk 8y ago
    > There is no observable difference between a particle whose wave function has collapsed and one that isn't. Don't you think that there is an observable difference between a particle that has been already observed (i.e the wave function has collapsed) and a particle that has not been observed yet (i.e. the wave function has not collapsed)?
    - scotty79 8y ago
      The observed one has just narrower wave function. That narrow that we can be pretty sure that next measurement of the same thing will fall close to the first one.
      - kgwgk 8y ago
        If we measure, say, the spin of an electron along the z-axis the only possible outcomes are UP and DOWN. If we measure UP, the next measurement of the same thing can only be UP. The system has a narrow wave function indeed. (On the other hand, if the system was originally prepared in a pure state along a perpendicular axis and has not been observed yet the probability of measuring UP or DOWN is 50/50. The difference between these scenarios is quite observable.)
        scotty79 8y ago
        Yes. Spin UP and DOWN are so "far" from each other that detector that can tell them apart reshapes the probability so much that from that point on this detector will always register UP. Think about location. If you send electron through half-mirror and put a screen behind it and see electron hitting the screen then you cut down your wave function to at most half of the size. You'll never see this photon in front of the mirror because that small interaction with the screen narrowed down the possibilities. You'll say that after it hit the screen it must go on in straight line and will hit any subsequent screen as if it was small ball just flying through space and screens. But imagine instead of the screen just a detector made from solenoid that will notice passing electron but not exactly where it is. This way location is still uncertain. Probability wave is not that narrow. The only thing your detection did is reshaping the wave so you no longer have any possibility of finding electron in front of a mirror. My point is that there's no difference in math between moving ball hitting things on the way and probability wave evolving in space time and exchanging momentum and energy with other waves and getting reshaped during these exchanges. We know equations that govern this evolution, we know what rules the exchanges obey and how the waves change shape when they interact. So why even have little balls cluttering our imagination if they are not needed to explain anything?
  - TheOtherHobbes 8y ago
    The observable difference is that a particle's properties have been observed. An uncollapsed state is a placeholder - constrained by statistics, but with no definite value. A collapsed state has a definite value along some measurement basis. Entanglement is a different phenomenon. Uncollapsed entangled pairs remain unmeasured by anyone. Either Alice or Bob can collapse the state of an entangled pair by measuring one item in the pair. This constrains the possible results the other person can find when they measure the other object (if that's a word that even applies in QM...) in the pair. The correlation isn't obvious unless the results are compared, so neither experimenter can see it on their own. But there has to be collapse - or decoherence, or a measurement, or some kind of interaction - for the correlations to become observable at all.
  - scotty79 8y ago
    Copenhagen interpretation is not that easy to debunk. It is not backed by any additional math beyond all of the quantum physics math that is solid but backs multiple interpretations. In physics if something doesn't have its own math then it is just an opinion. You can't prove it, you can't debunk it, you can have different one. Ultimately it doesn't matter. For me Copenhagen interpretation becomes absurd from the moment it states that observer matters. Each observer is just lots of particles and the fact that they arranged to have eyes and brain can't possibly matter for the basic laws of physics.
    - consz 8y ago
      You don't need eyes and a brain to observe, a photon can be an observer.
      - scotty79 8y ago
        That's not obvious to everybody. Even in this post there is mention that Shrodinger's cat is somehow different from Shrodinger's scientist.
    - nabla9 8y ago
      Copenhagen interpretation does not imply human or conscious observer. That's called quantum mystism that usually uses Copenhagen interpretation as a starting point. What Heisenberg and others assumed was that that there exists so called 'Heisenberg cut' between somewhere between quantum and macroscopic phenomenon.
      - scotty79 8y ago
        That's not obvious to everybody. Even in this post there is mention that Shrodinger's cat is somehow different from Shrodinger's scientist. And I've seen serious physicist lecturers claim that even recorded measurement didn't cause collapse because recording was not seen by anyone. This is completely bonkers and misticism is the right word for it. People who invented the thing are to blame. Why call something 'observation' if you mean interaction with macroscopic object. And why make the fuss about Shrodinger's cat?
scotty79 8y ago
Yes. Also you never really observe a particle. You observe interactions. You see that this electron bounced off of this ion. You see energy was exchanged because that energy triggers chain of events that in the end influences macroscopic object to the point you notice it with your own eyes and can write down as measurement.
- AstralStorm 8y ago
  What do you mean? Observing particle A is another way of saying measuring state of particle B after interaction with A, specifically comparing state B1 to state B0 expected when no interaction has occurred. Using words "not really" is like saying I should've used exact particle names and state matrices instead of placeholders.
  - scotty79 8y ago
    Math is solid. My objection is to calling it "observing a particle". It's like watching animated movie and saying "I observe moving Donald Duck". There's no Donald Duck. What you see is 24 completely separate slightly differnt pictures per second of something that give you an illustion that Donald Duck exists and he is moving. It's perfectly fine to say Donald Duck is moving when you are considering Earth orbiting the Sun, but when you are talking about particles and quantum it misleads you into thinking Schrodinger's cat should be dead and alive before you open the box. Conversly, thinking about "paricles" not as specks of dust but as whole fuzzy clouds may give you a lot of new intuitive understanding of what's actually happening. Like when you drop planetary model of atom and you consider orbitals. Then you can understand chemical bonds better. Things that didn't make sense earlier, like benzene make now perfect sense, orbital hybridisation brings new explanations.

by j1vms 8y ago

Are you suggesting that either wave collapse does not occur, or that instead of wave collapse, that there is a wave "narrowing" which is imperceptible to the observer (who observes it as a particle)?

nabla9 8y ago

It's more correct to say "apparent wave function collapse" because wave function collapse as a real physical phenomenon or event is just a hypothesis in some interpretations. Mathematically "appeared collapse" is explained by decoherence. It provides an explanation for the observation of wave function collapse when the system and the environment mix in irreversible way.

phkahler 8y ago
>> It's more correct to say "apparent wave function collapse" because wave function collapse as a real physical phenomenon or event is just a hypothesis in some interpretations. Wave function collapse is fiction IMO. There is no observable difference between a particle whose wave function has collapsed and one that isn't. If there was an observable difference we could place an emitter of entangled particle pairs mid-way between two observers (Alice and Bob) arbitrarily far apart. Alice could either turn on her measurements of the particles that arrive on here end or leave it off. This would cause Bob to receive either a stream of particles with collapsed wave functions, or a stream of uncollapsed ones. Alice can now transmit information to Bob faster than the speed of light using Morse code, which is not possible. The reality is that Bob can't tell anything about the superposition of states (or lack of) by observing his particles.
- kgilpin 8y ago
  > receive either a stream of particles with collapsed wave functions, or a stream of uncollapsed ones I don’t think there’s any way for Bob to determine whether the particle stream has been measured by Alice or not.
  - phkahler 8y ago
    >> I don’t think there’s any way for Bob to determine whether the particle stream has been measured by Alice or not. That's why I wrote: "There is no observable difference between a particle whose wave function has collapsed and one that isn't." If there was a way to tell it would violate relativity by allowing FTL communication.
- kgwgk 8y ago
  > There is no observable difference between a particle whose wave function has collapsed and one that isn't. Don't you think that there is an observable difference between a particle that has been already observed (i.e the wave function has collapsed) and a particle that has not been observed yet (i.e. the wave function has not collapsed)?
  - scotty79 8y ago
    The observed one has just narrower wave function. That narrow that we can be pretty sure that next measurement of the same thing will fall close to the first one.
    - kgwgk 8y ago
      If we measure, say, the spin of an electron along the z-axis the only possible outcomes are UP and DOWN. If we measure UP, the next measurement of the same thing can only be UP. The system has a narrow wave function indeed. (On the other hand, if the system was originally prepared in a pure state along a perpendicular axis and has not been observed yet the probability of measuring UP or DOWN is 50/50. The difference between these scenarios is quite observable.)
      - scotty79 8y ago
        Yes. Spin UP and DOWN are so "far" from each other that detector that can tell them apart reshapes the probability so much that from that point on this detector will always register UP. Think about location. If you send electron through half-mirror and put a screen behind it and see electron hitting the screen then you cut down your wave function to at most half of the size. You'll never see this photon in front of the mirror because that small interaction with the screen narrowed down the possibilities. You'll say that after it hit the screen it must go on in straight line and will hit any subsequent screen as if it was small ball just flying through space and screens. But imagine instead of the screen just a detector made from solenoid that will notice passing electron but not exactly where it is. This way location is still uncertain. Probability wave is not that narrow. The only thing your detection did is reshaping the wave so you no longer have any possibility of finding electron in front of a mirror. My point is that there's no difference in math between moving ball hitting things on the way and probability wave evolving in space time and exchanging momentum and energy with other waves and getting reshaped during these exchanges. We know equations that govern this evolution, we know what rules the exchanges obey and how the waves change shape when they interact. So why even have little balls cluttering our imagination if they are not needed to explain anything?
- TheOtherHobbes 8y ago
  The observable difference is that a particle's properties have been observed. An uncollapsed state is a placeholder - constrained by statistics, but with no definite value. A collapsed state has a definite value along some measurement basis. Entanglement is a different phenomenon. Uncollapsed entangled pairs remain unmeasured by anyone. Either Alice or Bob can collapse the state of an entangled pair by measuring one item in the pair. This constrains the possible results the other person can find when they measure the other object (if that's a word that even applies in QM...) in the pair. The correlation isn't obvious unless the results are compared, so neither experimenter can see it on their own. But there has to be collapse - or decoherence, or a measurement, or some kind of interaction - for the correlations to become observable at all.
- scotty79 8y ago
  Copenhagen interpretation is not that easy to debunk. It is not backed by any additional math beyond all of the quantum physics math that is solid but backs multiple interpretations. In physics if something doesn't have its own math then it is just an opinion. You can't prove it, you can't debunk it, you can have different one. Ultimately it doesn't matter. For me Copenhagen interpretation becomes absurd from the moment it states that observer matters. Each observer is just lots of particles and the fact that they arranged to have eyes and brain can't possibly matter for the basic laws of physics.
  - consz 8y ago
    You don't need eyes and a brain to observe, a photon can be an observer.
    - scotty79 8y ago
      That's not obvious to everybody. Even in this post there is mention that Shrodinger's cat is somehow different from Shrodinger's scientist.
  - nabla9 8y ago
    Copenhagen interpretation does not imply human or conscious observer. That's called quantum mystism that usually uses Copenhagen interpretation as a starting point. What Heisenberg and others assumed was that that there exists so called 'Heisenberg cut' between somewhere between quantum and macroscopic phenomenon.
    - scotty79 8y ago
      That's not obvious to everybody. Even in this post there is mention that Shrodinger's cat is somehow different from Shrodinger's scientist. And I've seen serious physicist lecturers claim that even recorded measurement didn't cause collapse because recording was not seen by anyone. This is completely bonkers and misticism is the right word for it. People who invented the thing are to blame. Why call something 'observation' if you mean interaction with macroscopic object. And why make the fuss about Shrodinger's cat?

scotty79 8y ago

Yes. Also you never really observe a particle. You observe interactions. You see that this electron bounced off of this ion. You see energy was exchanged because that energy triggers chain of events that in the end influences macroscopic object to the point you notice it with your own eyes and can write down as measurement.

AstralStorm 8y ago
What do you mean? Observing particle A is another way of saying measuring state of particle B after interaction with A, specifically comparing state B1 to state B0 expected when no interaction has occurred. Using words "not really" is like saying I should've used exact particle names and state matrices instead of placeholders.
- scotty79 8y ago
  Math is solid. My objection is to calling it "observing a particle". It's like watching animated movie and saying "I observe moving Donald Duck". There's no Donald Duck. What you see is 24 completely separate slightly differnt pictures per second of something that give you an illustion that Donald Duck exists and he is moving. It's perfectly fine to say Donald Duck is moving when you are considering Earth orbiting the Sun, but when you are talking about particles and quantum it misleads you into thinking Schrodinger's cat should be dead and alive before you open the box. Conversly, thinking about "paricles" not as specks of dust but as whole fuzzy clouds may give you a lot of new intuitive understanding of what's actually happening. Like when you drop planetary model of atom and you consider orbitals. Then you can understand chemical bonds better. Things that didn't make sense earlier, like benzene make now perfect sense, orbital hybridisation brings new explanations.