8 ms·

>Collapse is only mysterious to people who never learned thermofield theory, or non-equilibrium quantum dynamics. Oh yeah? Then what is the physical explanatio

>Collapse is only mysterious to people who never learned thermofield theory, or non-equilibrium quantum dynamics. Oh yeah? Then what is the physical explanation for collapse?

btilly 6y ago
What is the physical evidence that collapse actually happens? Flash news. Nobody has ever produced any. To the contrast we have lots of lines of evidence that an observer described by quantum mechanics should, upon observing a quantum experiment, be thrown into a superposition of observers. Each of which appears to have observed collapse. The notion is utterly repugnant to our biases so many reject the idea out of hand. But as we create ever more complex but controlled systems, we can perform ever more elaborate experiments verifying that quantum mechanics works exactly as predicted. At some point if we take seriously the idea that the most successful scientific theory of all time is an accurate description of ourselves, then we have to accept that perhaps there is no collapse after all.
- tsimionescu 6y ago
  But this doesn't answer the question. If you claim that all of these possible observers 'exist', how does this have a physical meaning? This is what I never understood about MWI, in what physical sense can the many worlds be said to exist? Where are they in our universe? What direction would we have to travel to find them? Do they exert gravity on us? If not, then how can we claim that they exist in a physical sense?
  - akvadrako 6y ago
    Existence isn't based on something affecting our world, obviously - that's just absurdly self-centered. But anyway, the other worlds do effect our world - that's why we get interference patterns in double slit experiments.
    - ClumsyPilot 6y ago
      > "Existence isn't based on something affecting our world, obviously - that's just absurdly self-centered." This is very unfair. This is a niche field with contested interpretations, don't make people feel stupid for asking fair questions. It's obvious what the other person meant: what does 'our world' and 'other worlds' mean, and how do you know it's not just a figment of your imagination, as a scientific theory must be falsifiable -> i.e. measurable and provable / disprovable somehow. You should at least point people to reading material before making fun of them.
      - akvadrako 6y ago
        I wasn’t making fun of anyone but your “obvious“ reading seems wrong. I’m pretty sure he suggested that something only exists physically if it has some measurable effect on us.
    - tsimionescu 6y ago
      In physical terms, we do generally define existence that way - for example, we say that time and space didn't exist 'before' the big bang, because there was nothing that could have a position or change. I was thinking of the same notion of existence and how it can be applied to MWI - essentially existence in the physical sense must mean that something is measurable, that it has some effect on the world (perhaps in the past or in the future).
      - akvadrako 6y ago
        I don't think we do define existence that way. Say you and your friend both go to opposite ends of the visible universe; due to inflation you'll never be able to communicate again. I suspect most people would say their friend continues to exist. This is very analogous to the many worlds situation.
        tsimionescu 6y ago
        Thinking about the extreme distances and time spans that entails makes it difficult, and of course relativity has its own "unreasonable" results. Still, they do exist in your past, and they also can assign coordinates in space-time to your current position, even though they are outside your light-cone. On the other hand, you can't meaningfully speak of them existing "now" in relativity, as there is no consistent definition of what "now" means for observers that are space-like separated. I guess the best answers about MWI is that the other versions of these particles continue to exist at different coordinates in Hilbert space, and that they do interact with each other in observable ways, such as the interference patterns in double-slit experiments.
  - nabla9 6y ago
    Different worlds don't exist in extra space or dimension. They are orthogonal quantum states of the whole universe.
  - ben_w 6y ago
    Speaking as a barely informed enthusiast, we can say they exist in the Occam’s Razor sense that the maths is much less complicated when we assume they do. I think there’s also an experimental setup, whose name I forget, but which is essentially nested Schrödinger's cat setups: Alice is in a box, Bob is in a box which contains Alice’s box, Carol is outside; Alice goes into superposition of |Alice+> and |Alice->, Bob opens the box and Carol can now demonstrate that Bob is in a superposition of |observing Alice+> and |observing Alice-> instead of the combination of 100%|observing> and a superposition of |Alice+> and |Alice->.
    - tsimionescu 6y ago
      The maths is the same whether we interpret it as many worlds, wave function collapse, and others.
      - ben_w 6y ago
        Well of course. It they were different — or at least if they gave different conclusions — we could rule some of them out.
  - Enginerrrd 6y ago
    No you're thinking of MWI all wrong. Your conception of the universe you exist in as being non-quantum is fundamentally flawed. The universe with the superposition of all the possible observers exists more purely in hilbert space. Sean Carroll has even started to put together a model for how spacetime could emerge from that hilbert space. So the universes all exist in the same place, since they are the same universe. Your idea of what an observation is, is just an eigenvalue of that corresponding operator.
  - AgentME 6y ago
    An object that moves far enough away from us is said to leave the observable universe, because with the continual expansion of space, it or anything it interacts with would have to travel faster than light back toward us in order to have any effect on us. Should we say that objects that leave the observable universe continue to exist? Should we amend our theories to include a new fall-off effect separate from gravity that says things stop existing when they exit our observable universe?
- gmkiv 6y ago
  I happen to agree with you, the observer is a quantum system must get entangled with the quantum system, but that still doesn't explain probabilities. If you prepare a system - say sqrt(1/3) spin-down + sqrt(2/3) spin-up, and then observe it, repeatedly, your subjective experience is that you saw spin-down 1/3 of the time, and spin-up 2/3 of the time. I don't understand how purely unitary evolution can explain this. Does it?
  - LittleTester 6y ago
    No it can't. There have been many attempts and they don't work. The Born rule is independent of unitary evolution. The closest one can get is to declare that the quantum state is fundamentally a statistical object (i.e. the only information in it is observation probabilities) and then with certain assumptions about the size of the state space you can show that the Born rule is the only possible rule for connecting the state to statistics consistent with the unitary dynamics. So under the assumption that the state encodes probabilities, state space assumptions and consistency with unitary evolution you get the Born rule. However this is not the same as the Born rule arising dynamically from unitary evolution alone.
  - Enginerrrd 6y ago
    Isn't your subjective experience just one probabilistic eigenvalue of a particular combination of operators corresponding to your observation? How does unitary evolution break down here?
    - LittleTester 6y ago
      It's not unitary evolution breaking down, just that the Born rule isn't a consequence of unitary evolution. They're separate independent hypotheses. In most derivations of QM from an axiomatic basis they're consequences of separate combinations of axioms.
      - Enginerrrd 6y ago
        Thanks. Do you by chance have a good source for a gentle introduction into axiomatic QM? Like undergrad level is fine, I've taken basic QM and worked through Griffith's intro book on my own, and I've had a lot of math. I'd love to read more but my google results aren't turning up a good definitive introduction.
- LittleTester 6y ago
  Well in the most common family of interpretations "collapse" isn't an actual physical process, just Bayesian updating. So you wouldn't expect to find physical evidence of it in that sense. It's true that from the perspective of an external superobserver the quantum state evolves to contain terms for each observer observation state. However since all interference observables turn out to be non-physical for macroscopic systems we get a superselection rule and so the probabilities for different macrostates are classical probabilities and thus reflect simple ignorance of the observer's post measurement state. There's very little motivation for reading the quantum state "ontically" in the way you are doing.
nabla9 6y ago
Wave function collapse is not experimentally verified or observed physical phenomenon (so far). It's postulated by some interpretations of QM. Apparent wave function collapse can be explained using quantum decoherence.

>Collapse is only mysterious to people who never learned thermofield theory, or non-equilibrium quantum dynamics. Oh yeah? Then what is the physical explanation for collapse?

btilly 6y ago

What is the physical evidence that collapse actually happens? Flash news. Nobody has ever produced any. To the contrast we have lots of lines of evidence that an observer described by quantum mechanics should, upon observing a quantum experiment, be thrown into a superposition of observers. Each of which appears to have observed collapse. The notion is utterly repugnant to our biases so many reject the idea out of hand. But as we create ever more complex but controlled systems, we can perform ever more elaborate experiments verifying that quantum mechanics works exactly as predicted. At some point if we take seriously the idea that the most successful scientific theory of all time is an accurate description of ourselves, then we have to accept that perhaps there is no collapse after all.

tsimionescu 6y ago
But this doesn't answer the question. If you claim that all of these possible observers 'exist', how does this have a physical meaning? This is what I never understood about MWI, in what physical sense can the many worlds be said to exist? Where are they in our universe? What direction would we have to travel to find them? Do they exert gravity on us? If not, then how can we claim that they exist in a physical sense?
- akvadrako 6y ago
  Existence isn't based on something affecting our world, obviously - that's just absurdly self-centered. But anyway, the other worlds do effect our world - that's why we get interference patterns in double slit experiments.
  - ClumsyPilot 6y ago
    > "Existence isn't based on something affecting our world, obviously - that's just absurdly self-centered." This is very unfair. This is a niche field with contested interpretations, don't make people feel stupid for asking fair questions. It's obvious what the other person meant: what does 'our world' and 'other worlds' mean, and how do you know it's not just a figment of your imagination, as a scientific theory must be falsifiable -> i.e. measurable and provable / disprovable somehow. You should at least point people to reading material before making fun of them.
    - akvadrako 6y ago
      I wasn’t making fun of anyone but your “obvious“ reading seems wrong. I’m pretty sure he suggested that something only exists physically if it has some measurable effect on us.
  - tsimionescu 6y ago
    In physical terms, we do generally define existence that way - for example, we say that time and space didn't exist 'before' the big bang, because there was nothing that could have a position or change. I was thinking of the same notion of existence and how it can be applied to MWI - essentially existence in the physical sense must mean that something is measurable, that it has some effect on the world (perhaps in the past or in the future).
    - akvadrako 6y ago
      I don't think we do define existence that way. Say you and your friend both go to opposite ends of the visible universe; due to inflation you'll never be able to communicate again. I suspect most people would say their friend continues to exist. This is very analogous to the many worlds situation.
      - tsimionescu 6y ago
        Thinking about the extreme distances and time spans that entails makes it difficult, and of course relativity has its own "unreasonable" results. Still, they do exist in your past, and they also can assign coordinates in space-time to your current position, even though they are outside your light-cone. On the other hand, you can't meaningfully speak of them existing "now" in relativity, as there is no consistent definition of what "now" means for observers that are space-like separated. I guess the best answers about MWI is that the other versions of these particles continue to exist at different coordinates in Hilbert space, and that they do interact with each other in observable ways, such as the interference patterns in double-slit experiments.
- nabla9 6y ago
  Different worlds don't exist in extra space or dimension. They are orthogonal quantum states of the whole universe.
- ben_w 6y ago
  Speaking as a barely informed enthusiast, we can say they exist in the Occam’s Razor sense that the maths is much less complicated when we assume they do. I think there’s also an experimental setup, whose name I forget, but which is essentially nested Schrödinger's cat setups: Alice is in a box, Bob is in a box which contains Alice’s box, Carol is outside; Alice goes into superposition of |Alice+> and |Alice->, Bob opens the box and Carol can now demonstrate that Bob is in a superposition of |observing Alice+> and |observing Alice-> instead of the combination of 100%|observing> and a superposition of |Alice+> and |Alice->.
  - tsimionescu 6y ago
    The maths is the same whether we interpret it as many worlds, wave function collapse, and others.
    - ben_w 6y ago
      Well of course. It they were different — or at least if they gave different conclusions — we could rule some of them out.
- Enginerrrd 6y ago
  No you're thinking of MWI all wrong. Your conception of the universe you exist in as being non-quantum is fundamentally flawed. The universe with the superposition of all the possible observers exists more purely in hilbert space. Sean Carroll has even started to put together a model for how spacetime could emerge from that hilbert space. So the universes all exist in the same place, since they are the same universe. Your idea of what an observation is, is just an eigenvalue of that corresponding operator.
- AgentME 6y ago
  An object that moves far enough away from us is said to leave the observable universe, because with the continual expansion of space, it or anything it interacts with would have to travel faster than light back toward us in order to have any effect on us. Should we say that objects that leave the observable universe continue to exist? Should we amend our theories to include a new fall-off effect separate from gravity that says things stop existing when they exit our observable universe?
gmkiv 6y ago
I happen to agree with you, the observer is a quantum system must get entangled with the quantum system, but that still doesn't explain probabilities. If you prepare a system - say sqrt(1/3) spin-down + sqrt(2/3) spin-up, and then observe it, repeatedly, your subjective experience is that you saw spin-down 1/3 of the time, and spin-up 2/3 of the time. I don't understand how purely unitary evolution can explain this. Does it?
- LittleTester 6y ago
  No it can't. There have been many attempts and they don't work. The Born rule is independent of unitary evolution. The closest one can get is to declare that the quantum state is fundamentally a statistical object (i.e. the only information in it is observation probabilities) and then with certain assumptions about the size of the state space you can show that the Born rule is the only possible rule for connecting the state to statistics consistent with the unitary dynamics. So under the assumption that the state encodes probabilities, state space assumptions and consistency with unitary evolution you get the Born rule. However this is not the same as the Born rule arising dynamically from unitary evolution alone.
- Enginerrrd 6y ago
  Isn't your subjective experience just one probabilistic eigenvalue of a particular combination of operators corresponding to your observation? How does unitary evolution break down here?
  - LittleTester 6y ago
    It's not unitary evolution breaking down, just that the Born rule isn't a consequence of unitary evolution. They're separate independent hypotheses. In most derivations of QM from an axiomatic basis they're consequences of separate combinations of axioms.
    - Enginerrrd 6y ago
      Thanks. Do you by chance have a good source for a gentle introduction into axiomatic QM? Like undergrad level is fine, I've taken basic QM and worked through Griffith's intro book on my own, and I've had a lot of math. I'd love to read more but my google results aren't turning up a good definitive introduction.
LittleTester 6y ago
Well in the most common family of interpretations "collapse" isn't an actual physical process, just Bayesian updating. So you wouldn't expect to find physical evidence of it in that sense. It's true that from the perspective of an external superobserver the quantum state evolves to contain terms for each observer observation state. However since all interference observables turn out to be non-physical for macroscopic systems we get a superselection rule and so the probabilities for different macrostates are classical probabilities and thus reflect simple ignorance of the observer's post measurement state. There's very little motivation for reading the quantum state "ontically" in the way you are doing.

nabla9 6y ago

Wave function collapse is not experimentally verified or observed physical phenomenon (so far). It's postulated by some interpretations of QM. Apparent wave function collapse can be explained using quantum decoherence.