Past exam of the quantum information course of the University of Cambridge / 2015 / Paper 1 / 1 / b

Put $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The detector rotation is real and orthogonal, so its inverse is the same transformation. Substituting it givesThus the requested orthonormal system states areThe pure density operator has not changed: this is only a different Schmidt decomposition of a maximally entangled state. But choosing a nonselective projective measurement in this new detector basis gives a different operation and generally a different state:For an explicit matrix in the original ordered basis, write $q=c^2{s}^2$ and $h=cs(c^2-s^2)/2$. ThenAt $\theta =\pi /2$ the entries $(1,4)$ and $(2,3)$ are $1/4$, whereas they vanish in ${\rho }_{up/down}$. Nevertheless, the system and detector individually each have reduced density matrix $I/2$ for either mixture.

This is the preferred-basis problem: the correlated state alone does not specify which detector observable becomes a stable classical record. Arbitrarily declaring collapse in a Schmidt basis changes the physical prediction. The actual detector interaction and environment must supply the basis selection, rather than a mere choice of notation.