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For your first questions: Observables correspond to hermitian (or self-adjoint) operators. As such, the eigenvalues are real, and these values are the possible outcomes. Also because of the hermitian/self-adjointness of the operator, when you have eigenvectors with different eigenvalues, the eigenvectors are orthogonal. So you know the values, but what ...

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Hints: What happens to the density operator and its eigenvalues under a change of the orthonormal basisvectors $|0\rangle$, $|1\rangle$, $|2\rangle$ by phase factors? More generally, what happens to the density operator and its eigenvalues under a unitary transformation $\rho\longrightarrow U^{\dagger}\rho U$?

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It is confusing because they say superposition when they mean entanglement. The yellow beam containing the information about the cutout got it from an entangled red beam which made contact with the cutout. In fact, National Geographic has a much better article: ...

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The things are simpler than you say. Though, let me work with the polarization singlet of photons, $$|S\rangle = \frac {|+\rangle |+\rangle + |-\rangle |-\rangle}{\sqrt {2}}.$$ Let Bob prepare 4 sets of polarized photons: (1) $|+\rangle$, (2) $|-\rangle$, (3) $|u\rangle = \frac {|+\rangle + |-\rangle}{\sqrt{2}}$, (4) $|v\rangle = \frac {|+\rangle - ... 1 The second density matrix is actually a rank-1 projection (if normalised) hence a dyadic product and therefore a pure state. It is enough then to measure against a state which is perpendicular to this vector (i.e.$(1/\sqrt 2,1/\sqrt 2)$) to say whether the qubit is not coming from the second source. 0$a$and$a^\dagger$are operators.$\alpha$is a pure imaginary number which we then call$i\alpha$with$\alpha$now real, this is why from eq1 to eq2 we use$\alpha→i\alpha$and$−\alpha^*→i\alpha$. Now from eq2 to eq3 we simply substitute$a+a^\dagger$for$x$operator. I don't understand why you want to write$\alpha=\ldots$, its just some real constant. ... 0 Mathematically, Norbert Schuch pointed out "purification of quantum states": Given an ensemble on a Hilbert space$\mathcal{H}$, you can always write it as a pure state on a bigger space$\mathcal{H}\otimes \mathcal{H}^{\prime}$such that the restriction of the pure state to$\mathcal{H}\$ results in the ensemble description. There is maybe a second and ...

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