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> Of course you can model quantum measurements! No, you can't. You can statistically model the results of multiple quantum measurements in the aggregate but y
by lisper 2y ago
> Of course you can model quantum measurements!
No, you can't. You can statistically model the results of multiple quantum measurements in the aggregate but you cannot model the physical process of a single measurement because there is a fundamental disconnect between the physics of quantum measurements and TMs, namely, TMs are deterministic and quantum measurements are not. There is a reason that the Measurement Problem is a thing.
- 11101010001100 2y agoSure, the 'problem' is that while the schrodinger equation is deterministic, we can only 'measure' the amplitudes of the solution. Is the wavefunction epistemic or ontological?
- lisper 2y agoNo, 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 agoRight I did hijack the thread a bit, but for me, the distribution is more than enough. The rest is just interpretation.
- lisper 2y agoWell, 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 agoIsn't this just circling back to the original ontic vs epistemic though -> map vs territory?
- lisper 2y agoNo, 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 agoTechnically measurement devices are described by wavefunctions too.
- lisper 2y agoWell, yeah, maybe. There's a reason that the Measurement Problem is called what it is.
- 11101010001100 2y agoI'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.
- hackandthink 2y agoThis 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 agoPower 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 agoThe 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 agoI 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...
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- gregw2 2y ago"God does not play dice with the universe" said Einstein. But he hasn't met my Dungeon Master...
- Y_Y 2y agoA statistical model is still a model. Lots of physics works that way. Newtonian mechanics might be deterministic, but because you often don't have perfect information on initial state that's not a useful model. For example in statistical mechanics you work with ensembles of microstates. That doesn't mean thermodynamics is fake and only F=ma is real. Models are tools for understanding the behaviour of systems, not a glimpse into god's simulation source code where the hidden variables are.
- lisper 2y agoFor QM it can be shown that there are no hidden variables, at least no local ones [1]. Quantum randomness really cannot be modeled by a TM. You need a random oracle. [1] https://en.wikipedia.org/wiki/Bell%27s_theorem https://en.wikipedia.org/wiki/Bell%27s_theorem
- Y_Y 2y agoThat is indeed what I was referring to. To clarify, plenty of classical physical models work only with distributions too. You don't need a random oracle because your model doesn't predict a single microstate. It wouldn't be possible or useful to do so. You can model the flow of heat without an oracle to tell you which atoms are vibrating.
- lisper 2y agoYes, all this is true, but I think you're still missing the point I'm trying to make. Classical mechanics succumbs to statistics without any compromises in terms of being able to make reliable predictions using a TM. But quantum mechanics is fundamentally different in that it produces macroscopic phenomena -- the results of quantum measurements -- that a TM cannot reproduce. At the most fundamental level, you can always make a copy of the state of a TM, and so you can always predict what a given TM is going to do by making such a copy and running that instead of the original TM. You can't make a copy of a quantum state, and so it is fundamentally impossible to predict the outcome of a quantum measurement. So a TM cannot generate a random outcome, but a quantum measurement can.
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