A Born's probability in decoherence

[Sr1]

TL;DR

How does Born's probability enter the decoherence theory?

Suppose we have a quantum system $S$, an apparatus $A$, and an observer, say Alice, $O$. WLOG let $S$ be a spin-$1/2$particle in a state $0.6|\uparrow\rangle+0.8|\downarrow\rangle$. The apparatus measure it in the $\sigma_z$ basis. Then the observer sees the result.

According to the decoherence theory, what happens in this measurement process is:

$$

\begin{align}

&\quad \psi_{before}=(0.6|\uparrow\rangle+0.8|\downarrow\rangle)\otimes|A_{init}\rangle\otimes|O_{init}\rangle\\

&\overset{\text{S,A interaction}}{\to}

\big(0.6|\uparrow\rangle(|A_{pointer \uparrow}\rangle+\delta |A_{pointer \downarrow}\rangle)+0.8|\downarrow\rangle(|A_{pointer \downarrow}\rangle+\delta |A_{pointer \uparrow}\rangle)\big)\otimes|O_{init}\rangle \\

&\overset{\text{A,O interaction}}{\to}

\psi_{after}=0.6|\uparrow\rangle|A_{pointer \uparrow}\rangle|\text{Alice thinks she observes}\uparrow\rangle

\\ &\quad\qquad\qquad\qquad\quad

+0.8|\downarrow\rangle|A_{pointer \downarrow}\rangle|\text{Alice thinks she observes}\downarrow\rangle

\\

&\qquad\qquad

+\delta|\uparrow\rangle|A_{\uparrow}\rangle|O_\downarrow\rangle

+\delta|\uparrow\rangle|A_{\downarrow}\rangle|O_\downarrow\rangle

+\delta|\uparrow\rangle|A_{\downarrow}\rangle|O_\uparrow\rangle

+\cdots

+\delta|\uparrow\rangle|A_{\uparrow}\rangle|O_{neither \uparrow nor \downarrow }\rangle

+\cdots

\end{align}
$$
I have included error terms since decoherence takes nonzero time, and $\delta$ may differ at different places. If Alice tells Bob the result of the experiment, then Bob will be entangled with the state written above.

My first question: where does Born's rule play its role in claiming that $Pr(O_\uparrow)=0.36$ and $Pr(O_\downarrow)=0.64$? In the first step S,A interaction, or in the second step A,O interaction? Suppose afterward Alice communicates with Bob, should we apply Born's rule in the third step Alice,Bob interaction?

My second question: Born's rule claims that if $\psi_{after}=0.6|\uparrow\rangle|A_{\uparrow}\rangle|O_\uparrow\rangle +0.8|\downarrow\rangle|A_{\downarrow}\rangle|O_\downarrow\rangle$, then $Pr(O_\uparrow)=0.36$ and $Pr(O_\downarrow)=0.64$ hold. However, in reality, there are error terms of forms $|\uparrow\rangle|A_{\uparrow}\rangle|O_\downarrow\rangle$, or even $|\uparrow\rangle|A_{\uparrow}\rangle|\text{Alice is neither }O_{\uparrow} \text{ nor } O_{\downarrow}\rangle$. How should Born's rule deal with those error terms?

 

[Demystifier]

In practice, there is no much difference between applying the Born rule at the apparatus level and at the (conscious) observer level. So you can choose either. If the error terms are sufficiently small, you don't need to worry about them too. Of course, there are always measurement errors, even in classical physics, but if you are not able to quantitatively compute them, you just accept that there are various uncontrollable measurement errors when you compare theory with experiments, and learn to live with them.

 

[Sr1]

I'm actually asking about theory, not practice

 

[Demystifier]

I'm actually asking about theory, not practice

Well, the Born rule is a theory telling how to compute the probability in practice.
More seriously, the quantum theory in its standard minimal form does not tell precisely whether the Born rule should be applied at the apparatus level or the observer level.