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Question: Let's say we added a single particle X (or, well, multiple) to our otherwise quantum world whose position and momentum we could pin down with arbitrar

Question: Let's say we added a single particle X (or, well, multiple) to our otherwise quantum world whose position and momentum we could pin down with arbitrary accuracy. What would happen if we used that particle to measure another particle Y? Normally Y has some fundamental uncertainty around its position and momentum, but would having access to X let us bypass that limit? If not, what would happen to the total uncertainty in the system after the measurement? (If that would allow us to break the limit, then it would seem like this really is a measurement limitation in some sense, so I'm assuming the answer is that that's not the case, but then I'm wondering what would happen afterward.)

marcosdumay 4y ago
> Let's say we added a single particle X (or, well, multiple) to our otherwise quantum world whose position and momentum we could pin down with arbitrary accuracy. That particle X does not exist. We can't add it to the world.
- dataflow 4y ago
  It's not like I was claiming it exists. I'm asking, if you did observe one tomorrow, what would be the theory's best prediction of what would happen next? And yes, I get the whole Fourier transform uncertainty making the math inconsistent, but that's not an answer for me here. Like if you asked Newton "if gravity didn't travel instantaneously, what would be your theory's best prediction?", he would probably be able to give you a better answer of what he expects the consequences would be than "that's impossible, the math would be inconsistent".
  - marcosdumay 4y ago
    If you observe one tomorrow, then the theory is all wrong and people will all yell "Hey, new physics", celebrate, and try to discover why nobody has seen anything like it before. There is no "best prediction", the entire theory is invalid. It would be like asking Newton "hey, if gravity didn't exist at all, and things traveled at path-dependent trajectories, what would your theory predict?" The answer is that it doesn't predict anything.
  - vgatherps 4y ago
    Your question itself assumes away quantum mechanics. There is fundamentally no fully certain particle in QM, the theory cannot meaningfully make predictions about something that's fully certain. I've been out of school a while so might be wrong but maybe, just maybe, you could torture some mathematics into giving you some infinities if you really wanted to get a "prediction"? But it's sort of like asking "how would linear algebra work if all nonzero matrices were invertible?" Well, all matrices aren't invertible, some definition of matrix that allows for nonzero matrices to be inverted is just different.
    - coderenegade 4y ago
      I think the more general question they're asking is whether it's possible to hack the math to break the uncertainty principle. As an example, if you take a Monte Carlo estimate of some quantity, you have uncertainty associated with that estimate. But there are ways to reduce that uncertainty using additional information (e.g. control variates, sampling tricks, partially solving the problem analytically, etc.). The example they're giving for a hypothetical particle with zero uncertainty might not be a great one, but I think the idea itself shouldn't be dismissed outright.
      - whatshisface 4y ago
        You could write down a theory where position and momentum commuted using the typical operator formalism, you just wouldn't get out reality.
        coderenegade 4y ago
        I think that misses the point of the exercise. What the OP is asking is more analogous to using negative energy in conjunction with existing models to see what would happen. It's not reflective of reality, but it tells you something about the limits of the model.
green_on_black 4y ago
Bypass the limit? No. Partially because you need to propagate that measurement somehow to your brain. But assuming that we have the "pin down perfectly" part (idk how that would even be possible): Mostly because the position/momentum of the measured particle would immediately become uncertain right after the measurement (it becomes a cloud of probabilities again). The total uncertainty in the system would likely decrease, since you can think of it like wavefunction collapse. Of course, everything depends on how exactly the situation works mathematically, which we haven't actually defined yet. To be clear, it is very well known that it isn't a measurement limitation. It's similar to asking "what the length of an oval" is, since there are many ways to measure an oval. It's not a "well-defined" question, and in fact, quantum mechanics requires it to be not-well-defined. For more info, check out the fourier transform and how that necessitates the heisenberg uncertainty principle. I (and many other physics people) find that people have a very hard time even trying to accept the nature of reality when it comes with uncertainty, and I think that's very normal. But according to our best known models, uncertainty is the nature of reality, and not because we can't measure enough.
- dataflow 4y ago
  Thanks. Yes, I get the Fourier transform thing (and Bell's theorem etc.), but imagine asking Newton the consequences of finite-speed gravity. He could either imagine how to reconcile that with his theory and hypothesize what the outcome might be (e.g., he might say if the sun exploded then you'd find out 8 minutes later), or he could tell you it's intrinsically infinite and that your question is like asking "what the length of an oval" is. I find one of these answers more satisfactory than the other. Now, in this particular case: I understand superdeterminism would mean the entire world could be deterministic and yet consistent with QM, correct? And I understand a reason why the world might be superdeterministic is that everything is already correlated/entangled together (say, from the big bang), thus making this in fact inherently a "measurement problem" in that we don't have any unentangled measurement particles available... right? If you buy that so far, then here's where I'm trying to go with this: if QM's response to this is "well, if you did obtain such an unentangled particle, you could use it to reduce the uncertainty in your next measurement beyond your current limits", then it seems to me QM is in favor of the world being superdeterministic, and the uncertainty we face is more coincidental than fundamental. Whereas if QM's response was "well, even if you had such a particle, you couldn't use it to reduce the uncertainty in your next measurement", then it seems to me QM believes the measurement limit is more fundamental than coincidental. Given the situation seems like the former to me, is there any reason to bet against the world being superdeterministic? If anything, it seems to me that superdeterminism has the big bang going for it, no?

marcosdumay 4y ago

> Let's say we added a single particle X (or, well, multiple) to our otherwise quantum world whose position and momentum we could pin down with arbitrary accuracy. That particle X does not exist. We can't add it to the world.

dataflow 4y ago
It's not like I was claiming it exists. I'm asking, if you did observe one tomorrow, what would be the theory's best prediction of what would happen next? And yes, I get the whole Fourier transform uncertainty making the math inconsistent, but that's not an answer for me here. Like if you asked Newton "if gravity didn't travel instantaneously, what would be your theory's best prediction?", he would probably be able to give you a better answer of what he expects the consequences would be than "that's impossible, the math would be inconsistent".
- marcosdumay 4y ago
  If you observe one tomorrow, then the theory is all wrong and people will all yell "Hey, new physics", celebrate, and try to discover why nobody has seen anything like it before. There is no "best prediction", the entire theory is invalid. It would be like asking Newton "hey, if gravity didn't exist at all, and things traveled at path-dependent trajectories, what would your theory predict?" The answer is that it doesn't predict anything.
- vgatherps 4y ago
  Your question itself assumes away quantum mechanics. There is fundamentally no fully certain particle in QM, the theory cannot meaningfully make predictions about something that's fully certain. I've been out of school a while so might be wrong but maybe, just maybe, you could torture some mathematics into giving you some infinities if you really wanted to get a "prediction"? But it's sort of like asking "how would linear algebra work if all nonzero matrices were invertible?" Well, all matrices aren't invertible, some definition of matrix that allows for nonzero matrices to be inverted is just different.
  - coderenegade 4y ago
    I think the more general question they're asking is whether it's possible to hack the math to break the uncertainty principle. As an example, if you take a Monte Carlo estimate of some quantity, you have uncertainty associated with that estimate. But there are ways to reduce that uncertainty using additional information (e.g. control variates, sampling tricks, partially solving the problem analytically, etc.). The example they're giving for a hypothetical particle with zero uncertainty might not be a great one, but I think the idea itself shouldn't be dismissed outright.
    - whatshisface 4y ago
      You could write down a theory where position and momentum commuted using the typical operator formalism, you just wouldn't get out reality.
      - coderenegade 4y ago
        I think that misses the point of the exercise. What the OP is asking is more analogous to using negative energy in conjunction with existing models to see what would happen. It's not reflective of reality, but it tells you something about the limits of the model.

green_on_black 4y ago

Bypass the limit? No. Partially because you need to propagate that measurement somehow to your brain. But assuming that we have the "pin down perfectly" part (idk how that would even be possible): Mostly because the position/momentum of the measured particle would immediately become uncertain right after the measurement (it becomes a cloud of probabilities again). The total uncertainty in the system would likely decrease, since you can think of it like wavefunction collapse. Of course, everything depends on how exactly the situation works mathematically, which we haven't actually defined yet. To be clear, it is very well known that it isn't a measurement limitation. It's similar to asking "what the length of an oval" is, since there are many ways to measure an oval. It's not a "well-defined" question, and in fact, quantum mechanics requires it to be not-well-defined. For more info, check out the fourier transform and how that necessitates the heisenberg uncertainty principle. I (and many other physics people) find that people have a very hard time even trying to accept the nature of reality when it comes with uncertainty, and I think that's very normal. But according to our best known models, uncertainty is the nature of reality, and not because we can't measure enough.

dataflow 4y ago
Thanks. Yes, I get the Fourier transform thing (and Bell's theorem etc.), but imagine asking Newton the consequences of finite-speed gravity. He could either imagine how to reconcile that with his theory and hypothesize what the outcome might be (e.g., he might say if the sun exploded then you'd find out 8 minutes later), or he could tell you it's intrinsically infinite and that your question is like asking "what the length of an oval" is. I find one of these answers more satisfactory than the other. Now, in this particular case: I understand superdeterminism would mean the entire world could be deterministic and yet consistent with QM, correct? And I understand a reason why the world might be superdeterministic is that everything is already correlated/entangled together (say, from the big bang), thus making this in fact inherently a "measurement problem" in that we don't have any unentangled measurement particles available... right? If you buy that so far, then here's where I'm trying to go with this: if QM's response to this is "well, if you did obtain such an unentangled particle, you could use it to reduce the uncertainty in your next measurement beyond your current limits", then it seems to me QM is in favor of the world being superdeterministic, and the uncertainty we face is more coincidental than fundamental. Whereas if QM's response was "well, even if you had such a particle, you couldn't use it to reduce the uncertainty in your next measurement", then it seems to me QM believes the measurement limit is more fundamental than coincidental. Given the situation seems like the former to me, is there any reason to bet against the world being superdeterministic? If anything, it seems to me that superdeterminism has the big bang going for it, no?