= Solution
At $A=\pi/4$, define
$$
\boxed{|E_\leftarrow\rangle=\frac{|E_\uparrow\rangle-i|E_\downarrow\rangle}{\sqrt2},\qquad
|E_\rightarrow\rangle=\frac{|E_\uparrow\rangle+i|E_\downarrow\rangle}{\sqrt2}.}
$$
They are normalized and orthogonal. The full state is the pure superposition $(|\uparrow d_\uparrow E_\leftarrow\rangle+|\downarrow d_\downarrow E_\rightarrow\rangle)/\sqrt2$. A nonselective <von Neumann process 1> selecting the original detector records gives
$$
\boxed{\rho^{(1)}_{SDE}=\frac12\left(
|\uparrow d_\uparrow E_\leftarrow\rangle\langle\uparrow d_\uparrow E_\leftarrow|
+|\downarrow d_\downarrow E_\rightarrow\rangle\langle\downarrow d_\downarrow E_\rightarrow|\right).}
$$
Without imposing any projection on the full state, tracing out the orthogonal environment states already gives $\rho_{SD}=\rho_{\rm up/down}$ from part (a). These two global descriptions share that reduced state but differ in their global coherence. <Quantum decoherence> does not make a pure global state literally equal to the corresponding statistical mixture.
Back to article page
