- #1
jcap
- 166
- 12
Can the Born rule be understood as time running both forwards and backwards simultaneously?
The probability ##P_{i \rightarrow f}## that an initial quantum state ##\psi_i## is measured to be in final quantum state ##\psi_f##, after evolving according to the unitary time-evolution operator ##U_{i \rightarrow f}##, is given by an application of the Born rule:
\begin{eqnarray*}
P_{i \rightarrow f} &=& |\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle|^2 \\
&=& \langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle^*\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle \\
&=& \langle\psi_i|U^\dagger_{i \rightarrow f}|\psi_f\rangle\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle \\
&=& \langle\psi_i|U_{f \rightarrow i}|\psi_f\rangle\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle\tag{1}
\end{eqnarray*}
where ##U^\dagger_{i \rightarrow f}=U_{f \rightarrow i}##.
Thus the probability ##P_{i \rightarrow f}## can be understood as the product of the amplitudes for the system to evolve both forwards from the initial state to the final state and backwards from the final state to the initial state.
Addition
More accurately after applying the forwards and backwards time operators in equation line ##(1)## we have:
$$P_{i \rightarrow f} =\langle\psi_i(t_i)|\psi_f(t_i)\rangle\langle\psi_f(t_f)|\psi_i(t_f)\rangle$$
Thus the probability ##P_{i \rightarrow f}## can be understood as the product of the overlap of the forwards-evolved initial state and the final state at time ##t_f## and the overlap of the backwards-evolved final state and the initial state at time ##t_i##.
The probability ##P_{i \rightarrow f}## that an initial quantum state ##\psi_i## is measured to be in final quantum state ##\psi_f##, after evolving according to the unitary time-evolution operator ##U_{i \rightarrow f}##, is given by an application of the Born rule:
\begin{eqnarray*}
P_{i \rightarrow f} &=& |\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle|^2 \\
&=& \langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle^*\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle \\
&=& \langle\psi_i|U^\dagger_{i \rightarrow f}|\psi_f\rangle\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle \\
&=& \langle\psi_i|U_{f \rightarrow i}|\psi_f\rangle\langle\psi_f|U_{i \rightarrow f}|\psi_i\rangle\tag{1}
\end{eqnarray*}
where ##U^\dagger_{i \rightarrow f}=U_{f \rightarrow i}##.
Thus the probability ##P_{i \rightarrow f}## can be understood as the product of the amplitudes for the system to evolve both forwards from the initial state to the final state and backwards from the final state to the initial state.
Addition
More accurately after applying the forwards and backwards time operators in equation line ##(1)## we have:
$$P_{i \rightarrow f} =\langle\psi_i(t_i)|\psi_f(t_i)\rangle\langle\psi_f(t_f)|\psi_i(t_f)\rangle$$
Thus the probability ##P_{i \rightarrow f}## can be understood as the product of the overlap of the forwards-evolved initial state and the final state at time ##t_f## and the overlap of the backwards-evolved final state and the initial state at time ##t_i##.
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