4 ms·

What evidence do you believe shows that?

bowsamic 4y ago
Basically that non-local hidden variables and superdeterminism do not seem like likely solutions to the violation of Bell's inequalities compared to just accepting that nature is random. You have to do a fair amount of reaching in order to hang onto the notion that the universe is deterministic
- zmgsabst 4y ago
  Except that we know entanglement is non-local: this suggests that non-locality rather than non-determinism is the correct resolution to Bell’s inequalities. Non-determinism is an extraneous assumption if we’re forced to accept non-locality for other reasons. Edit: I’ve hit my posting cap, so replying in edit — If we know particles are entangled and subject to non-local effects, we’re already concluding that there’s a hidden non-local variable: the sum of all entanglements we don’t know about. You can model that as a randomness distribution for calculations, but that doesn’t make the universe non-deterministic. Edit 2: I think where we diverge is in the belief you’re measuring “nothing” and so it’s strange you’re getting slightly varying results. I’ll have to think about that aspect some more. I’ll leave off here with a big thanks for giving such detailed replies!
  - bowsamic 4y ago
    Except non-local deterministic theories are much more complicated and almost impossible to use. That's why a lot of physicists regard the result as showing that reality is both non-local and non-deterministic. Very few suspect a non-local hidden variables theory. I think that non-determinism would be an extraneous assumption given the fundamental randomness apparent in quantum measurements. In my opinion, hidden variables are something extra to add to take account of this, and indeed it's very difficult to use Bohmian mechanics with relativity etc.
  - bowsamic 4y ago
    (replying to the edit) It's still just very unlikely. My work is based around reducing the fundamental quantum noise in gravitational wave detectors. This is the noise that arises from the following process: if you set up many copies of an experiment where you measure the electric field E in the vacuum state |0> (no photons) you will get a spread of results around E = 0. This is actually the largest noise affecting advanced gravitational wave detectors such as LIGO at their most sensitive frequencies (the classical noise sources have been reduced so much). If I have a situation where I take the quantum vacuum and get a different, random result every time I measure it, it seems like quite a stretch to conclude that it is _really_ deterministic. Further, if quantum states are described by wave functions then they must be intrinsically random due to the mathematics of the Fourier transform.
    - effie 4y ago
      > If I have a situation where I take the quantum vacuum and get a different, random result every time I measure it, it seems like quite a stretch to conclude that it is _really_ deterministic. If you take number generator and get a different random result every time you call it, you have no idea whether it is deterministic (algorithmic) or not. One can try to do statistical tests, but these can reveal only weak algorithms, not the sufficiently good ones. > Further, if quantum states are described by wave functions then they must be intrinsically random due to the mathematics of the Fourier transform. That is a very strange statement. Which part of mathematics of the Fourier transform suggest wave functions are intrinsically random?
      - bowsamic 4y ago
        If I was given a random number generator, _and_ there was no way to check if it were deterministic by experiment, I would lean on the side of not constructing a mechanism to justify that it is deterministic. Unless there is an experiment to prove otherwise, we should just accept that it plainly presents itself to us as random. The mathematics of the Fourier transform enforces that the frequency spread times the time spread is equal to a constant. Therefore if the quantum state of a system is fully described by its wavefunction, then the system absolutely cannot have predetermined values of conjugate variables. It takes many more wavenumbers to describe a wavefunction well localised in space.

Basically that non-local hidden variables and superdeterminism do not seem like likely solutions to the violation of Bell's inequalities compared to just accepting that nature is random. You have to do a fair amount of reaching in order to hang onto the notion that the universe is deterministic

zmgsabst 4y ago
Except that we know entanglement is non-local: this suggests that non-locality rather than non-determinism is the correct resolution to Bell’s inequalities. Non-determinism is an extraneous assumption if we’re forced to accept non-locality for other reasons. Edit: I’ve hit my posting cap, so replying in edit — If we know particles are entangled and subject to non-local effects, we’re already concluding that there’s a hidden non-local variable: the sum of all entanglements we don’t know about. You can model that as a randomness distribution for calculations, but that doesn’t make the universe non-deterministic. Edit 2: I think where we diverge is in the belief you’re measuring “nothing” and so it’s strange you’re getting slightly varying results. I’ll have to think about that aspect some more. I’ll leave off here with a big thanks for giving such detailed replies!
- bowsamic 4y ago
  Except non-local deterministic theories are much more complicated and almost impossible to use. That's why a lot of physicists regard the result as showing that reality is both non-local and non-deterministic. Very few suspect a non-local hidden variables theory. I think that non-determinism would be an extraneous assumption given the fundamental randomness apparent in quantum measurements. In my opinion, hidden variables are something extra to add to take account of this, and indeed it's very difficult to use Bohmian mechanics with relativity etc.
- bowsamic 4y ago
  (replying to the edit) It's still just very unlikely. My work is based around reducing the fundamental quantum noise in gravitational wave detectors. This is the noise that arises from the following process: if you set up many copies of an experiment where you measure the electric field E in the vacuum state |0> (no photons) you will get a spread of results around E = 0. This is actually the largest noise affecting advanced gravitational wave detectors such as LIGO at their most sensitive frequencies (the classical noise sources have been reduced so much). If I have a situation where I take the quantum vacuum and get a different, random result every time I measure it, it seems like quite a stretch to conclude that it is _really_ deterministic. Further, if quantum states are described by wave functions then they must be intrinsically random due to the mathematics of the Fourier transform.
  - effie 4y ago
    > If I have a situation where I take the quantum vacuum and get a different, random result every time I measure it, it seems like quite a stretch to conclude that it is _really_ deterministic. If you take number generator and get a different random result every time you call it, you have no idea whether it is deterministic (algorithmic) or not. One can try to do statistical tests, but these can reveal only weak algorithms, not the sufficiently good ones. > Further, if quantum states are described by wave functions then they must be intrinsically random due to the mathematics of the Fourier transform. That is a very strange statement. Which part of mathematics of the Fourier transform suggest wave functions are intrinsically random?
    - bowsamic 4y ago
      If I was given a random number generator, _and_ there was no way to check if it were deterministic by experiment, I would lean on the side of not constructing a mechanism to justify that it is deterministic. Unless there is an experiment to prove otherwise, we should just accept that it plainly presents itself to us as random. The mathematics of the Fourier transform enforces that the frequency spread times the time spread is equal to a constant. Therefore if the quantum state of a system is fully described by its wavefunction, then the system absolutely cannot have predetermined values of conjugate variables. It takes many more wavenumbers to describe a wavefunction well localised in space.