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Jun 14, 2020 at 8:22 vote accept Vyassa Baratham
Sep 15, 2016 at 2:24 comment added Zhengyan Shi @VyassaBaratham That's right. Also note that one density operator can represent different mixed states... So the mapping from mixed states to density operator is not one-to-one.
Sep 13, 2016 at 18:46 vote accept Vyassa Baratham
Jun 14, 2020 at 8:22
Sep 13, 2016 at 18:45 comment added Vyassa Baratham Thanks, I think I'm starting to understand this. The density operator measures more than just the probability of each measurement outcome in a particular basis - it's really a statistical distribution of pure states (but perhaps not eigenstates), and you'd need some knowledge about how the system was prepared in order to write the density operator out (or some measurement procedure more complicated than "observe all the outcomes and sum their self outer products weighted by probability")
Sep 13, 2016 at 7:46 history answered Zhengyan Shi CC BY-SA 3.0