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In general, "operating on a state with an observable" does not have direct physical meaning (i.e. you cannot think of it as evolving the state doing something to it). What does have physical meaning, is applying a unitary operation to a state. Every unitary operator corresponds to a physical operation that you can (in principle) implement, transforming (...

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You can calculate the probability of a given answer $\pm 1$ to each measurement by evaluating $$\langle B|\frac{I+(-1)^{\eta_1}\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+(-1)^{\eta_2}\vec{n}_2\cdot\vec{\sigma}}{2}|B\rangle$$ Thus, the probability of equal measurement outcomes is $$\langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+\vec{... 1 Intuitively, you can rotate \vec{n}_1 to Z. As Z axis has two antipodal points |0\rangle and |1\rangle, let \vec{n}_1 have two antipodal points |b_0\rangle and |b_1\rangle. Now the Bell state can be rewritten as \frac{1}{\sqrt{2}}(|b_0b_0\rangle+|b_1b_1\rangle). Now in this new basis, the calculation shall be much easier. To be precise, ... 2 Your understanding is correct. In the theory of photon polarization, the parametrization of the Bloch sphere (or its surfave) has traditionally another name. On the wikipedia page for the Jones calculus (the parametrization of the Bloch sphere surface), you'll find a table for the correspondence between kets and polarizations. To summarize, eigenstates ... 1 I see the heart of your question. I'd like to clarify a bit, before answering your question though. Matrices (aka operators) do not measure quantum states--they operate on them. Specifically, they project the state into the matrix's eigenvectors. We can then measure that projected state in a particular basis that may be the same or different than what the ... 4 Write down the two reduced density matrices of the single qubits that you have access to. Apply the Helstrom measurement (there are several descriptions of this on the site already). The problem is that, in this case, the two reduced density matrices are the same. That means that you cannot tell them apart. More explicitly,$$ |\varphi_2\rangle=(I\otimes X)|...

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If $|z\rangle$ are orthogonal to each other, then $$\log(\sum_z |z\rangle\langle z| \cdot b_z) = \sum_z |z\rangle\langle z| \cdot \log(b_z)$$ So $$\mathrm{trace}(\sum_z |z\rangle\langle z| \cdot a_z \cdot \log(\sum_{z^\prime} |z^\prime\rangle\langle z^\prime| \cdot b_{z^\prime}))$$ $$=\mathrm{trace}(\sum_z \sum_{z^\prime} |z\rangle\langle z| \cdot |z^... 3 If you wish to distinguish two states |\psi\rangle and |\phi\rangle, you can only guarantee to do this if \langle\psi|\phi\rangle=0. You do this by measuring in a basis defined by the two states (alternatively, you apply a unitary U such that$$ U|\psi\rangle=|0\rangle,\qquad U|\phi\rangle=|1\rangle,  and then measure in the standard $Z$ basis. ...

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