Classical states and decoherence

[lucas_]

Its simple. It's got to do with the difference between a proper and an improper mixed state. The mixed state is Σ pu |u><u| + pd |d><d|. If you observe that with the up-down observable you will get |u><u| with probability pu and |d><d| with probability pd. Now is that because it was in state |u><u| with probability pd and similarly for |d><d|? If so the observation did nothing - no change. In interpretations like BM or GRW that actually is the case and is how they resolve the measurement problem. But they may not be true - the observation may have changed the mixed state to a pure one - there is no way of telling. Remember this is a mixed state - not a superposition - interference terms have been suppressed.

Thanks
Bill

is the "state |u><u| with probability pd and similarly for |d><d|" the improper mixed state in the example (noting that entangled electron with spin up and spin down are entangled so you can't use proper mixed state).. but why did you use proper mixed state in Σ pu |u><u| + pd |d><d|?

 

[bhobba]

is the "state |u><u| with probability pd and similarly for |d><d|" the improper mixed state in the example (noting that entangled electron with spin up and spin down are entangled so you can't use proper mixed state).. but why did you use proper mixed state in Σ pu |u><u| + pd |d><d|?

Both proper and improper mixed states are mathematically the same, pu |u><u| + pd |d><d| is proper or improper - you can't tell from observing it - only by knowing how it was prepared. |u><u| and |d><d| are pure - not mixed.

Thanks
Bill