Write $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$. Before any interaction the system is in the <pure state> $|\psi_S\rangle=\alpha|0\rangle+\beta|1\rangle$, so
$$
\rho_S^{(0)}=
\begin{pmatrix}
|\alpha|^2&\alpha\beta^*\\
\alpha^*\beta&|\beta|^2
\end{pmatrix}.
$$
One interaction leaves the environmental state $|+\rangle$ unchanged on the system branch $|0\rangle$, while on the branch $|1\rangle$ it produces
$$
|e\rangle=\frac{(\cos\epsilon-\sin\epsilon)|0\rangle+(\cos\epsilon+\sin\epsilon)|1\rangle}{\sqrt2},
\qquad \langle e|+\rangle=\cos\epsilon.
$$
After interactions with $n$ distinct environment qubits, the joint state is
$$
|\Psi_n\rangle
=\alpha|0\rangle|+\rangle^{\otimes n}
+\beta|1\rangle|e\rangle^{\otimes n},
$$
and the <decoherence factor> is the product $(\cos\epsilon)^n$. Taking the <partial trace> gives
$$
\boxed{\rho_S^{(n)}=
\begin{pmatrix}
|\alpha|^2&\alpha\beta^*(\cos\epsilon)^n\\
\alpha^*\beta(\cos\epsilon)^n&|\beta|^2
\end{pmatrix}}.
$$
For small $\epsilon$, $(\cos\epsilon)^n\simeq e^{-n\epsilon^2/2}$: the populations remain fixed while the phase coherence decays. This is <environmental decoherence>, caused by entanglement with unobserved records rather than by a nonunitary evolution of the complete state.

A measurement in the computational basis cannot reveal the decay because its probabilities remain $|\alpha|^2$ and $|\beta|^2$. An interference measurement can. For example, measuring in the <Hadamard basis> gives
$$
P(\mathord\pm)=\frac12\left[1\pm2\operatorname{Re}(\alpha\beta^*)(\cos\epsilon)^n\right].
$$
Changing the measurement phase similarly accesses the imaginary part, so the shrinking interference visibility directly displays decoherence.

If the same environment qubit is reused, the conditional rotation accumulates coherently. After $n$ interactions the overlap is $\cos(n\epsilon)$ rather than $(\cos\epsilon)^n$, and the off-diagonal entries oscillate. Coherence vanishes at some times but returns periodically: this finite environment exhibits <quantum recoherence> rather than effectively irreversible decay.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 325 / 1 / b / Solution

If a joint pure state is $\alpha|0\rangle|E_0\rangle+\beta|1\rangle|E_1\rangle$, the off-diagonal entry of the system's <reduced density matrix> is multiplied by the environmental overlap $\langle E_1|E_0\rangle$. Independent repeated interactions multiply these overlaps.

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 Incoming links: Decoherence factor