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Consider the following two states:

  1. $\left|\psi\right\rangle = \frac{1}{\sqrt{2}} \left|\psi_1\right\rangle\otimes\left|\chi_1\right\rangle+\frac{1}{\sqrt{2}}\left|\psi_2\right\rangle\otimes\left|\chi_2\right\rangle$
  2. $\left|\psi\right\rangle = $ a mixed state which has 1/2 probability of being $\left|\psi_1\right\rangle\otimes\left|\chi_1\right\rangle$and 1/2 probability of being $\left|\psi_2\right\rangle\otimes\left|\chi_2\right\rangle$

The setting is such that $\psi_1$ and $\psi_2$ are eigenstates of a certain observable $\Psi$ with eigenvalues 1 and 2, while $\chi_1$ and $\chi_2$ are the states of a measurement apparatus. So $\left|\psi_1\right\rangle\otimes\left|\chi_1\right\rangle$ means that the object+apparatus state is such that the apparatus measures the object to be in state 1, and similarly for $\left|\psi_2\right\rangle\otimes\left|\chi_2\right\rangle$.

There is no doubt that the mixed state option is qualitatively different from the pure state option. But Eugene Wigner claims in his essay "Remarks on the Mind-Body Questions" that when the apparatus is complex (meaning that $\chi_1$ and $\chi_2$ are states with many more degrees of freedom than $\psi_1$ and $\psi_2$), it is difficult to make the distinguishment.

So my question is: how and why is it difficult to distinguish mixed state from pure state when $\chi_1$ and $\chi_2$ are complex? Maybe a complexity theory viewpoint would be of help here? I don't know.

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  • $\begingroup$ Questions of the kind "What does author X mean" are somewhat tricky to answer. But I would guess what is meant is that if one cannot measure in a basis which is a linear combination of $|\chi_1\rangle$ and $|\chi_2\rangle$ -- e..g, since the two states involve many particles, and large superpositions are hard to create/measure (but the same could be the case, e.g., due to some superselection rule) -- such a superposition is indistinguishable from a mixture. $\endgroup$ – Norbert Schuch Aug 23 '16 at 14:54
  • $\begingroup$ I agree with @NorbertSchuch, but perhaps the point is that in order to calculate the expectation value of $\psi$ one would have to trace over an enormous number of states of $\chi$, the result looking very much like a mixed state. $\endgroup$ – garyp Aug 23 '16 at 15:24
  • $\begingroup$ @garyp If you want to compute a general expectation value w.r.t. $\psi$, you better not trace over a large number of particles! $\endgroup$ – Norbert Schuch Aug 23 '16 at 15:26
  • $\begingroup$ @NorbertSchuch One traces over unobserved variables, no? Perhaps I've misinterpreted the question. $\endgroup$ – garyp Aug 23 '16 at 15:32
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    $\begingroup$ The statement may also have to do with the fact that for $\psi_1$, $\psi_2$ orthogonal and $\chi_1$, $\chi_2$ with many degrees of freedom, a statistical superposition of $\psi_1\otimes \chi_1$, $\psi_2\otimes \chi_2$ has negligible entropy compared to any thermal (or simply mixed) states of the apparatus. In this sense the statistical superposition is "very close" to the pure state superposition of zero entropy. $\endgroup$ – udrv Aug 24 '16 at 5:37

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