- #106
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A. Neumaier said:In (1) [equation label added by me] you assume without justification that the ##\Psi_k(\vec{x},\vec{y})## with different ##k## have approximately disjoint support. This is unwarranted without a convincing analysis.
You could explain in more detail why (1) follows from your argument for the second equality.Demystifier said:I totally disagree, but if you think so I don't know what argument to offer without repeating myself.
The only way this can be concluded seems to me by assume that the ##\Psi_k(\vec{x},\vec{y})## with different ##k## have approximately disjoint support. But you didn't give an argument for the latter, you just said that we must assume it:
You actually assume that ##A_{kq}(\vec{x})## and ##A_{k'q'}(\vec{x})## have essentially disjoint support whenever ##k\ne k'##. How does this follow from the definition of the ##A_{kq}## in (3') above?Demystifier said:Instead of my Eq. (3), more generally we have
$$|k\rangle|A_0\rangle \rightarrow \sum_q a_q |q\rangle |A_{kq}\rangle~~~~ (3')$$
[...]
In the multi-position representation we have
$$\Psi(\vec{x},\vec{y})=\sum_k c_k\Psi_k(\vec{x},\vec{y})$$
where
$$\Psi_k(\vec{x},\vec{y})=\sum_q a_q \psi_q(\vec{y}) A_{kq}(\vec{x})$$
Using the Born rule in the multi-position space we have
$$\rho(\vec{x},\vec{y}) =|\Psi(\vec{x},\vec{y})|^2
\simeq \sum_k|c_k|^2 |\Psi_k(\vec{x},\vec{y})|^2$$
In the second equality we have assumed that ##A_{kq}(\vec{x})## are macro distinct for different ##k##, which we must assume if we want to have a system that can be interpreted as a measurement of ##k##.