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S Oct 13, 2018 at 16:56 history bounty ended StarBucK
S Oct 13, 2018 at 16:56 history notice removed StarBucK
Oct 10, 2018 at 9:38 vote accept StarBucK
Oct 10, 2018 at 0:37 answer added Luzanne timeline score: 2
Oct 9, 2018 at 18:00 history tweeted twitter.com/StackPhysics/status/1049721310463619077
Oct 9, 2018 at 16:11 answer added Arnold Neumaier timeline score: 1
Oct 9, 2018 at 14:56 history edited StarBucK CC BY-SA 4.0
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S Oct 9, 2018 at 14:41 history bounty started StarBucK
S Oct 9, 2018 at 14:41 history notice added StarBucK Draw attention
Oct 9, 2018 at 14:37 history edited StarBucK CC BY-SA 4.0
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Feb 28, 2018 at 11:49 comment added Bardeen Sum of qi/pji is pj
Feb 28, 2018 at 11:38 comment added Bardeen you are correct qi here is taken to be $$\ p_{i|j}$$
Feb 28, 2018 at 11:32 comment added Bardeen the two psis are going to become pho multiplied by $P_j $ in two sides and the sum of $\frac{1}{p_{j|i}}$ is $p_j$
Feb 28, 2018 at 11:07 comment added StarBucK @001 I don't understand what you mean precisely. For example for me the term should be $\frac{q_i}{p_{j|i}}$, where $q_i$ is the statistical probability (not quantum) to have the state $|\psi_i\rangle$. And $p_{j|i}$ is the quantum probability to be in the state $j$ after measurement when we prepared the state $|\psi_i\rangle$ at the beginning.
Feb 28, 2018 at 10:58 comment added Bardeen they are averaging out the expectation value for each pure state giving each a weight $$\frac{p_{i|j}}{p_{j|i}}$$
Feb 28, 2018 at 10:30 history asked StarBucK CC BY-SA 3.0