6 ms·

Sure, the 'problem' is that while the schrodinger equation is deterministic, we can only 'measure' the amplitudes of the solution. Is the wavefunction epistemic

• lisper 2y ago
  No, this has nothing to do with ontology vs epistemology. That's a philosophical problem. The problem here is that a measurement is a sample from a random distribution. A TM cannot emulate that. It can compute the distribution, but it cannot take a random sample from it. For that you need a random oracle (https://en.wikipedia.org/wiki/Random_oracle https://en.wikipedia.org/wiki/Random_oracle).
  • 11101010001100 2y ago
    Right I did hijack the thread a bit, but for me, the distribution is more than enough. The rest is just interpretation.
    • lisper 2y ago
      Well, no. The measurements are the things that actually happen, the events that comprise reality. The distribution may be part of the map, but it is definitely not the territory.
      • 11101010001100 2y ago
        Isn't this just circling back to the original ontic vs epistemic though -> map vs territory?
        • lisper 2y ago
          No, because the original map-vs-territory discussion had to do with the wave function: > Is the wavefunction epistemic or ontological? https://news.ycombinator.com/item?id=42383854 https://news.ycombinator.com/item?id=42383854 Now we're talking about measurements which are indisputably a part of the territory.
          • 11101010001100 2y ago
            Technically measurement devices are described by wavefunctions too.
            • lisper 2y ago
              Well, yeah, maybe. There's a reason that the Measurement Problem is called what it is.
              • 11101010001100 2y ago
                I'm replying here since we appear to have reached the end. Presumably measurement involves interaction with 3 or more degrees of freedom (i.e., an entangled pair of qubits and a measurement device). This is something, for most types of interactions (exclude exactly integrable systems for the moment), classical or quantum, we cannot analytically write down the solution. We can approximately solve these systems with computers. All that to say, is that any solution to any model of an 'individual' measurement will be approximate. (Of course, one of the key uses of quantum computing is improving upon these approximate solutions.) So what type of interaction should you pick to describe your measurement? Well, there is a long list and we can use a quantum computer to check! I guess part of the point I am trying to make, is when you open the box of a measurement device, you enter the world of many body physics, where obtaining solutions to the many-body equations of motion IS the problem.
                • lisper 2y ago
                  > We can approximately solve these systems with computers. Yes, but with quantum measurements you cannot even approximate. Your predictions for e.g. a two-state system with equal amplitudes for the two states will be exactly right exactly half of the time, and exactly wrong the other half.
                  • 11101010001100 2y ago
                    I guess I don't have an issue with being wrong if we treat 'measurement' like a black box.
  • hackandthink 2y ago
    This is not really about quantum computing. A classical probabilistic Turing samples from a random distribution: "probabilistic Turing machines can be defined as deterministic Turing machines having an additional "write" instruction where the value of the write is uniformly distributed" I remember that probabilistic Turing machines are not more powerful than deterministic Turing machines, though Wikipedia is more optimistic: "suggests that randomness may add power." https://en.wikipedia.org/wiki/Probabilistic_Turing_machine https://en.wikipedia.org/wiki/Probabilistic_Turing_machine
    • lisper 2y ago
      Power is not the point. The point is just that probabilistic TM's (i.e. TMs with a random oracle) are different. For example, the usual proof of the uncomputability of the halting problem does not apply to PTMs. The proof can be generalized to PTMs, but the point is that this generalization is necessary. You can't simply reduce a PTM to a DTM.
      • Y_Y 2y ago
        The problem is about physics, not Turing machines. You don't need to make a random choice as part of your physical model, the model only makes predictions about the distribution. You can't represent the continuous dynamical manifolds of classical or quantum mechanics on a TM either, but that's ok, because we have discrete models that work well.
        • hackandthink 2y ago
          I am asking myself: Does a probabilistic Turing machines needs aleatory uncertainty? (would have called this ontological but (1) disagrees) Epistemic uncertainty would mean her: We don't know which deterministic Turing machine we are running. Right now, I see no way to use this in algorithms. (1) https://dictionary.helmholtz-uq.de/content/types_of_uncertainty.html https://dictionary.helmholtz-uq.de/content/types_of_uncertai...
          • 2y ago
  • gregw2 2y ago
    "God does not play dice with the universe" said Einstein. But he hasn't met my Dungeon Master...