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- TL;DR Summary
- 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?
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?
