5 ms·

> The catch is that appearance of classical objects is still unsolved problem in QM. What? Decoherence explains classical objects.

_0w8t 5y ago
While quantum decoherence explains appearance of classical objects in simple models, predicting the results of real experiments still not possible. And given this has not changed much as far as I know for the last 20 years, I started to suspect that decoherence alone may not be enough.
- willis936 5y ago
  Objective collapse theories are being worked on. https://youtu.be/FP6iyVJ70OU https://youtu.be/FP6iyVJ70OU
tsimionescu 5y ago
No, it doesn't. Decoherence explains why self-interference of a single wave-function doesn't happen after interaction with the environment. But it doesn't explain two important things. 1, which is very practical, is that it doesn't explain why the wave function always decoheres in the same way. In experiments, you can always chose the basis of measurement, and get some definite results in that particular basis, while the properties in another basis remain indefinite. But, in classical reality, all objects have definite properties in the same basis (say, positions in the same Cartesian coordinate system). This is known as the preferred basis problem, and decoherence can't explain it. 2, which is somewhat more philosophical, decoherence doesn't explain the quantitative relationship between the wave function and the probability of observing a particular outcome. Indeed, in MWI it's very hard to even define probability in a coherent way, since all outcomes actually happen and there is an uncountable (in the mathematical sense) number of outcomes.
- tome 5y ago
  > In experiments, you can always chose the basis of measurement, and get some definite results in that particular basis, while the properties in another basis remain indefinite. Can you give some examples of such bases? Position is one, for sure. In "Quantum Mechanics 101" we are told that momentum is another, but I'm sceptical that there is a real experiment that can measure momentum without also measuring position.
  - tsimionescu 5y ago
    Even for position, you can measure less than a point-like position. For example, you can check if a particle is inside this box or outside it. After this measurement, the particle will either be inside the box or outside it, but it will not be at some fixed position inside the box (or some fixed position outside it). For example, if it's determined to be outside the box, it will definitely not interact with any particle that was inside the box at the same time, but it may interact with any particle anywhere else in the entire universe - and according to MWI, it WILL interact with every other particle everywhere else in the universe, in some part of state-space.
- CuriouslyC 5y ago
  The many worlds interpretation works better with quantized probability. Imagine a huge computer running a monte carlo simulation with a probabilistic current state, and a set of transition states (the wave function). With quantized probability and a universe with finite "memory" the amplitude of the wave function basically maps to the amount of system memory being utilized for a portion of the simulated state space.
  - tsimionescu 5y ago
    Well, quantized probability doesn't really make sense, there are extraordinarily improbable events happening every second. It's easy to create an event that has less probability than many things that have never been observed (flip 10,000 coins - whatever result you get has probability 1/2^10k). You could perhaps create a version of QM with a quantized amplitude for the wave function in some basis. It's also important to remember that the Schrodinger equation of a system has an infinity of solutions - for any solution, any linear function of that solution is also a solution. You need to choose a particular basis of measurement - choose a decomposition of the solutions - to be able to apply the Born rule and get from the wave function to a probability distribution.
    - cottager2 5y ago
      Why is quantized probability nonsensical? Isn’t it equivalent to asserting the universe is not a continuum?
      - tsimionescu 5y ago
        By quantized probability, I assume you mean something like "there is a minimal probability p for any possible event; if the probability we compute for some event is q < p, then that event is impossible". If this is the right definition, then it's easy to disprove: for any p you chose, I can construct a series of events that each have probability q < p, but one of which is guaranteed to happen. This was the meaning of the examples I gave with the 10 thousand coins - the probability of any particular result after flipping the coins is extraordinarily small, and yet one of the results is guaranteed to happen. Edit to add: to be clear, even if the universe itself were quantized, all measurable physical quantities, probability still wouldn't need to be quantized, as it's not a physical quantity, it's just a mathematical abstraction. Still, even in QM, not all physical quantities are quantized. For example, space (position) is not quantized in QM, and neither is time. They are both continuous quantities in all of the equations normally used. Planck time and Planck distance are only the shortest possible distances to measure precisely, given the Heisenberg uncertainty principle, but that doesn't require them to actually be quantized. In contrast, the quantization of mass, energy, spin etc are actually necessary for the theory to work, they are not just measurement artifacts.
Koshkin 5y ago
Classical objects are not explained in Quantum Mechanics, they are a fundamental part of the theory itself. (Measurement is an interaction with an apparatus which is a classical object.)
- blix 5y ago
  The notion of a 'Classical Object' in this particular framework of QM is an assumption, not a prediction. Since all objects are composed of quantum particles, all objects are quantum objects. So a complete theory of QM needs to be able to explain, not assume, why some objects behave classically.

_0w8t 5y ago

While quantum decoherence explains appearance of classical objects in simple models, predicting the results of real experiments still not possible. And given this has not changed much as far as I know for the last 20 years, I started to suspect that decoherence alone may not be enough.

willis936 5y ago
Objective collapse theories are being worked on. https://youtu.be/FP6iyVJ70OU https://youtu.be/FP6iyVJ70OU

tsimionescu 5y ago

No, it doesn't. Decoherence explains why self-interference of a single wave-function doesn't happen after interaction with the environment. But it doesn't explain two important things. 1, which is very practical, is that it doesn't explain why the wave function always decoheres in the same way. In experiments, you can always chose the basis of measurement, and get some definite results in that particular basis, while the properties in another basis remain indefinite. But, in classical reality, all objects have definite properties in the same basis (say, positions in the same Cartesian coordinate system). This is known as the preferred basis problem, and decoherence can't explain it. 2, which is somewhat more philosophical, decoherence doesn't explain the quantitative relationship between the wave function and the probability of observing a particular outcome. Indeed, in MWI it's very hard to even define probability in a coherent way, since all outcomes actually happen and there is an uncountable (in the mathematical sense) number of outcomes.

tome 5y ago
> In experiments, you can always chose the basis of measurement, and get some definite results in that particular basis, while the properties in another basis remain indefinite. Can you give some examples of such bases? Position is one, for sure. In "Quantum Mechanics 101" we are told that momentum is another, but I'm sceptical that there is a real experiment that can measure momentum without also measuring position.
- tsimionescu 5y ago
  Even for position, you can measure less than a point-like position. For example, you can check if a particle is inside this box or outside it. After this measurement, the particle will either be inside the box or outside it, but it will not be at some fixed position inside the box (or some fixed position outside it). For example, if it's determined to be outside the box, it will definitely not interact with any particle that was inside the box at the same time, but it may interact with any particle anywhere else in the entire universe - and according to MWI, it WILL interact with every other particle everywhere else in the universe, in some part of state-space.
CuriouslyC 5y ago
The many worlds interpretation works better with quantized probability. Imagine a huge computer running a monte carlo simulation with a probabilistic current state, and a set of transition states (the wave function). With quantized probability and a universe with finite "memory" the amplitude of the wave function basically maps to the amount of system memory being utilized for a portion of the simulated state space.
- tsimionescu 5y ago
  Well, quantized probability doesn't really make sense, there are extraordinarily improbable events happening every second. It's easy to create an event that has less probability than many things that have never been observed (flip 10,000 coins - whatever result you get has probability 1/2^10k). You could perhaps create a version of QM with a quantized amplitude for the wave function in some basis. It's also important to remember that the Schrodinger equation of a system has an infinity of solutions - for any solution, any linear function of that solution is also a solution. You need to choose a particular basis of measurement - choose a decomposition of the solutions - to be able to apply the Born rule and get from the wave function to a probability distribution.
  - cottager2 5y ago
    Why is quantized probability nonsensical? Isn’t it equivalent to asserting the universe is not a continuum?
    - tsimionescu 5y ago
      By quantized probability, I assume you mean something like "there is a minimal probability p for any possible event; if the probability we compute for some event is q < p, then that event is impossible". If this is the right definition, then it's easy to disprove: for any p you chose, I can construct a series of events that each have probability q < p, but one of which is guaranteed to happen. This was the meaning of the examples I gave with the 10 thousand coins - the probability of any particular result after flipping the coins is extraordinarily small, and yet one of the results is guaranteed to happen. Edit to add: to be clear, even if the universe itself were quantized, all measurable physical quantities, probability still wouldn't need to be quantized, as it's not a physical quantity, it's just a mathematical abstraction. Still, even in QM, not all physical quantities are quantized. For example, space (position) is not quantized in QM, and neither is time. They are both continuous quantities in all of the equations normally used. Planck time and Planck distance are only the shortest possible distances to measure precisely, given the Heisenberg uncertainty principle, but that doesn't require them to actually be quantized. In contrast, the quantization of mass, energy, spin etc are actually necessary for the theory to work, they are not just measurement artifacts.

Koshkin 5y ago

Classical objects are not explained in Quantum Mechanics, they are a fundamental part of the theory itself. (Measurement is an interaction with an apparatus which is a classical object.)

blix 5y ago
The notion of a 'Classical Object' in this particular framework of QM is an assumption, not a prediction. Since all objects are composed of quantum particles, all objects are quantum objects. So a complete theory of QM needs to be able to explain, not assume, why some objects behave classically.