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Bounty Ended with 100 reputation awarded by JP McCarthy
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Nik Weaver
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The way I interpret your question is: deterministic = pure state on an abelian C${}^*$-algebra, random = arbitrary state on an abelian C${}^*$-algebra, quantum = pure state on an arbitrary C${}^*$-algebra. There's one further level of generality, arbitrary state on an arbitrary C${}^*$-algebra, which gives you statistical ensembles of quantum states.

(Note that under this interpretation you do not have "random $\subset$ quantum", unless your "quantum" includes ensembles.)

The way I interpret your question is: deterministic = pure state on an abelian C${}^*$-algebra, random = arbitrary state on an abelian C${}^*$-algebra, quantum = pure state on an arbitrary C${}^*$-algebra. There's one further level of generality, arbitrary state on an arbitrary C${}^*$-algebra, which gives you statistical ensembles of quantum states.

The way I interpret your question is: deterministic = pure state on an abelian C${}^*$-algebra, random = arbitrary state on an abelian C${}^*$-algebra, quantum = pure state on an arbitrary C${}^*$-algebra. There's one further level of generality, arbitrary state on an arbitrary C${}^*$-algebra, which gives you statistical ensembles of quantum states.

(Note that under this interpretation you do not have "random $\subset$ quantum", unless your "quantum" includes ensembles.)

Source Link
Nik Weaver
  • 42.8k
  • 3
  • 112
  • 213

The way I interpret your question is: deterministic = pure state on an abelian C${}^*$-algebra, random = arbitrary state on an abelian C${}^*$-algebra, quantum = pure state on an arbitrary C${}^*$-algebra. There's one further level of generality, arbitrary state on an arbitrary C${}^*$-algebra, which gives you statistical ensembles of quantum states.