4 ms·

Decoherence may be enough. The idea is that the starting wavefunction of us and some experiment (say, an electron) is a tensor product | us > ⊗ | electron

by evanb 2y ago

Decoherence may be enough. The idea is that the starting wavefunction of us and some experiment (say, an electron) is a tensor product | us > ⊗ | electron in superposition > and unitary time evolution governed by the Schrödinger equation evolves that state to | us seeing spin up in a detector > ⊗ | spin up > + | us seeing spin down > ⊗ | spin down > (I elided the amplitudes since we're just interested in the idea and not some particular values). In each 'branch' of the wavefunction (meaning: in each addend) we're left wondering "wow, how did the wavefunction collapse to the value that I am seeing", because we're inside the wf. But that's just a question we have that we've been tricked into by our limited sight; if we could see the whole wf of the universe, with our own superposition included, we'd see there is no collapse but just unitary Schrödinger evolution forever.

GoblinSlayer 2y ago
Quantum physics doesn't use normalization much and these formulas work better without normalization. Normalization can be provided by the Bayes formula, so postulating it is redundant. Also I heard QFT has some problem with global normalization, because everything must be local there, so normalization is replaced by Stokes' formula: the flux of state norm through closed spacetime surface is zero, i.e. the norm only conserves locally. And the evolution formula U(t)|0,0> = |1,t> + |2,t> can be seen as local conservation of the norm.
tsimionescu 2y ago
This runs into two separate but maybe related problems. One is called the preferred basis problem: while you're right that the world could be described by |us seeing spin up in a detector> ⊗ |spin up> + |us seeing spin down> ⊗ |spin down> it can equally validly be described by (|us seeing spin up in a detector> + |us seeing spin down>) ⊗ (|spin up> + |spin down>) + (|us seeing spin up in a detector> - |us seeing spin down>) ⊗ (|spin up> - |spin down>) (again leaving off all of the coefficients - each of the states should have a 1/sqrt(2) coefficient to be a valid superpositon). But even though in the math these are perfectly equivalent and valid solutions, we only ever observe solution 1 and never solution 2. We don't even have words to describe what decomposition 2 would mean. The second problem is one of arriving from these values to actual probabilities. In this interpretation, all values of the wave function are phsycially realized, so all possible outcomes have probability 1. Now, in a simple binary with equal probabilities on both sides (if the electron is equally like to be spin up or spin down), this can sort of work. But if the Spin Up state has a higher amplitude than the spin down state (which is easy to do in experiments), then we have a problem. If the electron is both spin up and spin down every time we run this experiment, how come we observe the spin up version more often than the spin down version? In more technical terms, how can we justify the exact shape of the Born rule without postulating any extra processes? In some variants of many worlds, they do add an extra postulate: that the amplitude of the wavefunction of a particular state corresponds to the number of "worlds" in which that state occurs. An observer can then deduce what are the odds that they live in one of the worlds where the outcome is spin up versus the worlds where it is spin down. But more modern versions of MWI like to say that no separate "worlds" actually exist (since that brings about a whole host of other problems), but then justifying these probabilities, without even more complicated postulates, comes back.
- GoblinSlayer 2y ago
  >If the electron is both spin up and spin down every time we run this experiment, how come we observe the spin up version more often than the spin down version? Because after a series of measurements your state asymptotically approaches |we observe the spin up version more often than the spin down version>, which is in fact a mere tautology: if it didn't, you wouldn't conclude it. You don't even need any interpretation for this, you can just calculate it, and asymptotic certainty is trivial to interpret. In this sense MWI is a shut up and calculate interpretation, the only difficulty is to formulate your question in a computable form. If your question has no computable form, it probably doesn't match reality.
  - tsimionescu 2y ago
    No, why would it? If we never collapse the wave function, we'll still have states where we only saw spin up, states where we only saw spin down, states where we saw more spin up, and states where we saw more spin down. And all of these are understood to be real in the MWI, so some observers will be guaranteed to exist that calculate that some state has a high amplitude, but they will never experience it, no matter how many billions of times they run the same experiment.
    - GoblinSlayer 2y ago
      Marginal outcomes are possible in any interpretation. It's not absolute certainty, only exponential asymptotic certainty.
      - tsimionescu 2y ago
        These outcomes are marginal in many interpretations, but they are certain to happen in MWI (since all possible results of the wave function actually happen). And the point I was trying to make is that, if you don't postulate the Born rule in addition to the Schrodinger equation, then you don't even have a way to say that this "happens for fewer observers". But then, if you do have to postulate the Born rule, then the MWI has no more explanatory power than any other interpretation, it's not just "following the math", and it still needs a definition for what consitutes a measurement for purposes of applying the Born rule.

by evanb 2y ago

GoblinSlayer 2y ago

Quantum physics doesn't use normalization much and these formulas work better without normalization. Normalization can be provided by the Bayes formula, so postulating it is redundant. Also I heard QFT has some problem with global normalization, because everything must be local there, so normalization is replaced by Stokes' formula: the flux of state norm through closed spacetime surface is zero, i.e. the norm only conserves locally. And the evolution formula U(t)|0,0> = |1,t> + |2,t> can be seen as local conservation of the norm.

tsimionescu 2y ago

This runs into two separate but maybe related problems. One is called the preferred basis problem: while you're right that the world could be described by |us seeing spin up in a detector> ⊗ |spin up> + |us seeing spin down> ⊗ |spin down> it can equally validly be described by (|us seeing spin up in a detector> + |us seeing spin down>) ⊗ (|spin up> + |spin down>) + (|us seeing spin up in a detector> - |us seeing spin down>) ⊗ (|spin up> - |spin down>) (again leaving off all of the coefficients - each of the states should have a 1/sqrt(2) coefficient to be a valid superpositon). But even though in the math these are perfectly equivalent and valid solutions, we only ever observe solution 1 and never solution 2. We don't even have words to describe what decomposition 2 would mean. The second problem is one of arriving from these values to actual probabilities. In this interpretation, all values of the wave function are phsycially realized, so all possible outcomes have probability 1. Now, in a simple binary with equal probabilities on both sides (if the electron is equally like to be spin up or spin down), this can sort of work. But if the Spin Up state has a higher amplitude than the spin down state (which is easy to do in experiments), then we have a problem. If the electron is both spin up and spin down every time we run this experiment, how come we observe the spin up version more often than the spin down version? In more technical terms, how can we justify the exact shape of the Born rule without postulating any extra processes? In some variants of many worlds, they do add an extra postulate: that the amplitude of the wavefunction of a particular state corresponds to the number of "worlds" in which that state occurs. An observer can then deduce what are the odds that they live in one of the worlds where the outcome is spin up versus the worlds where it is spin down. But more modern versions of MWI like to say that no separate "worlds" actually exist (since that brings about a whole host of other problems), but then justifying these probabilities, without even more complicated postulates, comes back.

GoblinSlayer 2y ago
>If the electron is both spin up and spin down every time we run this experiment, how come we observe the spin up version more often than the spin down version? Because after a series of measurements your state asymptotically approaches |we observe the spin up version more often than the spin down version>, which is in fact a mere tautology: if it didn't, you wouldn't conclude it. You don't even need any interpretation for this, you can just calculate it, and asymptotic certainty is trivial to interpret. In this sense MWI is a shut up and calculate interpretation, the only difficulty is to formulate your question in a computable form. If your question has no computable form, it probably doesn't match reality.
- tsimionescu 2y ago
  No, why would it? If we never collapse the wave function, we'll still have states where we only saw spin up, states where we only saw spin down, states where we saw more spin up, and states where we saw more spin down. And all of these are understood to be real in the MWI, so some observers will be guaranteed to exist that calculate that some state has a high amplitude, but they will never experience it, no matter how many billions of times they run the same experiment.
  - GoblinSlayer 2y ago
    Marginal outcomes are possible in any interpretation. It's not absolute certainty, only exponential asymptotic certainty.
    - tsimionescu 2y ago
      These outcomes are marginal in many interpretations, but they are certain to happen in MWI (since all possible results of the wave function actually happen). And the point I was trying to make is that, if you don't postulate the Born rule in addition to the Schrodinger equation, then you don't even have a way to say that this "happens for fewer observers". But then, if you do have to postulate the Born rule, then the MWI has no more explanatory power than any other interpretation, it's not just "following the math", and it still needs a definition for what consitutes a measurement for purposes of applying the Born rule.