In the context of justifying the failure of modeling quantum observables in the 'more natural' way as real functions on the phase space (i.e. similar to the mathematical image modeling classical observables) and somehow justifying the use of the framework of Hilbert space and operators . In the following talk

https://www.youtube.com/watch?v=ODAngTW8deg (see 1:00 to 4:00)

Alain Connes argued that phase space and real functions on it cannot support the co-existence of continuous real observables (i.e. those whose image has the continuum cardinality) and discrete ones (i.e. whose image is finite or countable) by the following argument : Denote by $X$ a phase space and suppose the existence of a discrete observable $D$ and a continuous $C$ one on $X$. We then have necessarily that $X$ has the continuum cardinality. But then we will get at least a real value $\lambda$ in $\mathbb{R}$ which has an (even 'continuous') infinite number of pre-images by $D$. (i.e. a continuous infinite number of states which corresponds to this value..) so the values $\lambda$ will have an 'infinite multiplicity' and he said that this is impossible ! Can anyone help what is the impossible in this situation ? I mean do we have something in the classical paradigm which says that an observable value $\lambda$ (i.e. A measured value of an observable) cannot be hitten an infinite number of times ? (i.e. there cannot be an infinite number of states in the phase space having the value $\lambda$ as a measure of a fixed observable $A$). Is it a physical problem or a mathematical one ?

  • $\begingroup$ It's mathematical, more specifically topological as it breaks continuity. The infinitude of values in the preimage basically causing tearing. $\endgroup$ Dec 3, 2022 at 20:34
  • $\begingroup$ Sir @CyclotomicField but function representation of observable are not supposed to be continuous, so which continuity is broken ? $\endgroup$
    – user536450
    Dec 4, 2022 at 9:38
  • $\begingroup$ He's explaining why a continuous representation doesn't exist. That's why he says what he says. If you want to know why they point it out then you'll have to discuss that with the person who said it. $\endgroup$ Dec 4, 2022 at 11:58
  • $\begingroup$ @CyclotomicField it is then not my question. Since we can consider discrete observables too as discrete real functions..the question is why this generates a mathematical problem as you said. thank you for your help. $\endgroup$
    – user536450
    Dec 4, 2022 at 15:47


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