Morbert
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Where we disagree is I don't think such a variable is needed "to explain why a single outcome (rather than all outcomes at once) realizes at all". We can show that, when presented with a sample space of experimental outcomes, QM rules out the possibility that all outcomes (or even more than one) will occur at once, since the probability is 0, as shown in my last post. Similarly, we can show that "no outcome occurs" also has a probability ##p(\varnothing) = \mathrm{tr}_{sD}([I_{sD}-\sum_i\Pi_i(t)] \rho_s\otimes\rho_D) = 0##.Demystifier said:You represent outcomes with projectors. For example, in the case of particle position, the projector would be something like ##|x\rangle\langle x|##. That's enough for computing the probability of position ##x##. However, the formalism you outlined talks about probabilities of position, not about position itself. A formalism that talks about position itself should have a real-valued variable ##x##, and the formalism you outlined does not have such a variable. (The operator ##\hat{x}## would not count because it's a hermitian operator, not a real-valued variable.)
Compare this formalism with classical stochastic mechanics. There one has a function ##x(t)##, which is a stochastic (not deterministic) function of ##t##. Such a quantity is missing in the quantum formalism.
I.e. We can interpret QM as returning probabilities for possible outcomes, without a variable corresponding to the "true outcome", and this interpretation won't suffer from problems like implying all outcomes might occur at once.
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