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Lets assume I have a particle A which I want to measure it's spin. Based on what I have read about entanglement and measurement we can entangle a particle B with A so that B somehow registers(measures) particle A's spin. Can you describe the detail of the process in terms of bra-ket notation and also how we can extract such information from the final entangled state?

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Easy. You start off with your particle in the spin state $$ |\psi⟩=\alpha\left|\uparrow\right> + \beta\left|\downarrow\right>, $$ and you couple it to your pointer, which at the moment has a blank reading: $$ |\psi⟩\otimes|\mathrm{blank}⟩=\alpha\left|\uparrow\right>\otimes|\mathrm{blank}⟩ + \beta\left|\downarrow\right>\otimes|\mathrm{blank}⟩. $$ You then apply an entangling unitary $\hat U$ that takes the spin state and copies it onto the pointer, \begin{align} \hat U\left|\uparrow\right>\otimes|\mathrm{blank}⟩ & = \left|\uparrow\right>\otimes|\mathrm{up}⟩ \\ \hat U\left|\downarrow\right>\otimes|\mathrm{blank}⟩& = \left|\downarrow\right>\otimes|\mathrm{down}⟩, \end{align} so the state goes to $$ \hat U|\psi⟩\otimes|\mathrm{blank}⟩=\alpha\left|\uparrow\right>\otimes|\mathrm{up}⟩ + \beta\left|\downarrow\right>\otimes|\mathrm{down}⟩, $$ and finally you do a projective measurement on the pointer, which can project it onto the $|\mathrm{up}⟩$ pointer state, giving $$ \frac{1}{\sqrt{p_\mathrm{up}}}(\mathbb I \otimes |\mathrm{up}⟩⟨\mathrm{up}|)\hat U|\psi⟩\otimes|\mathrm{blank}⟩= \left|\uparrow\right>\otimes|\mathrm{up}⟩ $$ with probability $p_\mathrm{up}=|\alpha|^2$, or onto the $|\mathrm{down}⟩$ state, giving $$ \frac{1}{\sqrt{p_\mathrm{down}}}(\mathbb I \otimes |\mathrm{down}⟩⟨\mathrm{down}|)\hat U|\psi⟩\otimes|\mathrm{blank}⟩=\left|\downarrow\right>\otimes|\mathrm{down}⟩ $$ with probability $p_\mathrm{down}=|\beta|^2$.

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  • $\begingroup$ Thank you very much for the answer. What would the blank state $|blank\rangle$ be if our pointer is another particle with two possible spins, namely $|\uparrow\rangle$ and $|\downarrow\rangle$? $\endgroup$ – user650585 Sep 1 '17 at 20:44
  • $\begingroup$ @user650585 It can be anything; the formalism in this answer is completely general. $\endgroup$ – Emilio Pisanty Sep 1 '17 at 20:56

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