- #1

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But suppose the state prepared is

[tex]|\psi\rangle = \sqrt{\frac{1}{3}}|A\rangle + \sqrt{\frac{2}{3}}|B\rangle[/tex]

If I perform the experiment 1000 times, then there will be a superposition of 2^1000 me's, one for each possible set of results of the observations. But this can't be right. In such a case a typical "me" in the ensemble will have observed approximately 50% A's and 50% Bs. But the answer should be 1/3 and 2/3.

The only way for the statistical nature of quantum mechanics to be due to such an all-possible-outcomes-happen picture is for the result of one an observation of the state above to result in 1 me that observed A and

__2__me's that observed B. Then everything counts up correctly.

Do I understand this right? So if we have instead

[tex]|\psi\rangle = \sqrt{\frac{79}{162}}|A\rangle + \sqrt{\frac{83}{162}}|B\rangle[/tex]

an experiment results in 79 me's that get A and 83's me that get B?

If this is really the many-worlds answer to the statistical aspect of observations, then it is even kookier than I thought.