Let the apparatus begin in a ready state $|A_0\rangle$. An ideal unitary measurement interaction is defined on the relevant subspace by
$$
U(|e_i\rangle|A_0\rangle)=|e_i\rangle|i\rangle.
$$
Thus an initial $|\phi\rangle=\sum_i a_i|e_i\rangle$ evolves to the entangled state
$$
|\Psi\rangle=\sum_i a_i|e_i\rangle|i\rangle,
$$
and orthogonality of the pointer states gives
$$
\boxed{\rho_S=\operatorname{Tr}_A|\Psi\rangle\langle\Psi|
=\sum_i|a_i|^2|e_i\rangle\langle e_i|}.
$$

Before the interaction, the <Born rule> gives
$$
\boxed{\Pr(P_{ij}=1)
=|\langle\psi_{ij}|\phi\rangle|^2
=\frac12|a_i+a_j|^2}.
$$
Afterward,
$$
\boxed{\Pr(P_{ij}=1)
=\operatorname{Tr}(P_{ij}\rho_S)
=\frac12(|a_i|^2+|a_j|^2)}.
$$
The missing cross term is the lost interference between the $i$ and $j$ branches. Entanglement with orthogonal pointer records therefore explains <quantum decoherence> and the appearance of a classical mixture to the subsystem. It does not solve the <quantum measurement problem>: unitary evolution alone does not explain why one definite pointer value is observed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 325 / 1 / b / Solution

{wiki=Measurement_problem}

The quantum measurement problem asks how a unitary superposition of system and apparatus outcomes gives rise to one definite observed outcome. <Quantum decoherence> explains the suppression of interference between outcome branches, but by itself does not select one branch.

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