On Monday, August 12, 2019 at 7:43:51 PM UTC-5, smitra wrote: > > > Thing is that the interference we can observe at some position x on the > screen is Re[<A|x><x|B>], which for general x is nonzero despite the > fact that <A|B> = 0. > > Saibal >
To a probability (or measure) theorist, one making the conversion from classical summing rules to quantum summing rules, this should be easier: https://arxiv.org/pdf/gr-qc/9401003.pdf Where quantum theory differs from classical mechanics (in this view) is in its dynamics, which of course is stochastic rather than deterministic. As such, the theory functions by furnishing probabilities for sets of histories. More formally, it associates to a set A of histories a non-negative real number |A|, which I will call its quantum measure |A|; and it is this measure that enters into the sum-rules we will be concerned with. ... *Notions such as state-vectors and observables never appear, except for the sake of computational convenience.* ... In the two-slit experiment, for example, the probability that a particular detector will register the arrival of the electron is (proportional to) the measure |C| of the set C of all electron world lines which in fact pass close enough to that detector to trigger it. When we contemplate also blocking off one or the other slit, there are (for a fixed detector) three sets of histories to consider: the set A of histories which arrive at the detector after traversing the “first” slit, the corresponding set B for the “second” slit, and the original set C = A ∐ B, the disjoint † union of A and B. It is of course characteristic of quantum probability that the interference term I(A, B) := |A ∐ B| − |A| − |B| between the slits is not zero. The surprising thing (once one has gotten used to the fact of interference itself) is that this violation of the classical probability sum-rules is in a certain sense so mild, since the corresponding sum-rule for three alternatives remains valid. In any case, the important thing from the standpoint of interpretation is that the electron follows one and only one path, not somehow two at once. If probabilities are involved, it is only because the path is not determined in advance, just as it is initially undetermined in a classical stochastic process. Given the failure of the sum rule I(A, B) = 0, it is clear that quantum probabilities cannot be interpreted in the same manner that classical ones are wont to be interpreted, in terms of (actual or fictitious) ensemble frequencies. ... @philipthrift -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/9698bb02-41aa-4cd3-88af-e09e572c5016%40googlegroups.com.
