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MBar2269
  • Member for 1 year, 9 months
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How can the Copenhagen and Everett interpretations of quantum mechanics make the same predictions?
Is the idea then that there is no way, even in principle, to have S and M be isolated from the wider environment such that there is no record of prior measurements or is it theoretically possible to construct such an experiment and test the two different predictions?
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How can the Copenhagen and Everett interpretations of quantum mechanics make the same predictions?
So $\lvert| \frac{1}{2}, {\frac{1}{2}}^{*}$ and $-\lvert| \frac{1}{2}, {\frac{1}{2}}^{*}$ should be considered separate, orthogonal states because they correspond to different worlds? Don't you then have to add some postulate to specify when a new world is created?
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How can the Copenhagen and Everett interpretations of quantum mechanics make the same predictions?
I am sorry but I cannot edit my comment. I meant to say $\lvert \frac{1}{2}, {\frac{1}{2}}^{*} \rangle$ and $\lvert -\frac{1}{2}, {\frac{1}{2}}^{*} \rangle$ would be the same. The picture I have in my head for the measuring device is of a needle pointing either to $\frac{1}{2}$ or $-\frac{1}{2}$, depending on what it has most recently measured, leaving no trace of any prior measurements.
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How can the Copenhagen and Everett interpretations of quantum mechanics make the same predictions?
An important difference between your calculation and mine seems to be that you assume that the state of the measuring device after the second measurement depends on both measurement outcomes, whereas I assumed it only depends on the most recent measurement so, for instance, $\lvert \frac{1}{2}, {\frac{1}{2}}^{*} \rangle$ and $\lvert \frac{1}{2}, {-\frac{1}{2}}^{*} \rangle$ would be the same state under my assumption. Am I right in thinking that's a key difference? If so, could you justify your assumption?
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How does the covariant derivative satisfy the Leibniz rule?
Shift index slightly further to the right for clarity
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