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https://arxiv.org/abs/1405.7907

In particular I would like to understand this paragraph, right before chapter 6:

This route to the Born Rule has a simple physical interpretation. Take the wave function and write it as a sum over orthonormal basis vectors with equal amplitudes for each term in the sum (so that many terms may contribute to a single branch). Then the Born Rule is simply a matter of counting – every term in that sum contributes an equal probability.

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Aug 7, 2022 at 18:44
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    $\begingroup$ you need to add more context: people aren’t gonna read a paper to answer your question. $\endgroup$ Commented Aug 7, 2022 at 19:13

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It helps to see what is going on if we simplify the notation by leaving out the environment and observer.

If you have a state representing the superposition of two outcomes with unequal amplitudes, like:

$\sqrt{a}\left|x\right>+\sqrt{b}\left|y\right>$

then we want to rewrite it as a sum of orthogonal equal-amplitude terms:

$\left(\left|x\right>_1+\left|x\right>_2+\ldots+\left|x\right>_a\right)+\left(\left|y\right>_1+\left|y\right>_2+\ldots+\left|y\right>_b\right)$

where there are $a$ terms with an $\left|x\right>$ as part of the state and $b$ terms with a $\left|y\right>$. By Pythagoras theorem in $n$ dimensions, the length of a sum of $n$ mutually-orthogonal unit vectors is $\sqrt{n}$, so this would match our original expression. Then we can simply count branches to find the probability. There are $a+b$ independent branches in total, of which $a$ give an $\left|x\right>$ and $b$ give a $\left|y\right>$.

The trick is to make those apparently identical outcome states orthogonal. That is done by combining each with a different environment-observer basis state.

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