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Given a quantum system $|\psi\rangle=\alpha_0|\psi_0\rangle\otimes |0\rangle+\alpha_1|\psi_1\rangle\otimes |1\rangle$, such that each subsystem $|\psi_i\rangle$ is entangled with a qubit is state $|i\rangle$.

Is it correct that applying the unambiguous quantum state discrimination operator (USD) $P=I\otimes|0\rangle\langle0|$ on $|\psi\rangle$, where $I$ is the identity operator gives,

$P|\psi\rangle=|\psi_0\rangle\otimes |0\rangle$

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We have $$P\left|\psi\right\rangle =I\otimes\left|0\right\rangle\left\langle0\right|\left(\alpha_0|\psi_0\rangle\otimes |0\rangle+\alpha_1|\psi_1\rangle\otimes |1\rangle\right)$$ $$=\alpha_0|\psi_0\rangle\otimes\left|0\right\rangle\left\langle0\right|0\rangle +\alpha_1|\psi_1\rangle\otimes\left|0\right\rangle\left\langle0\right|1\rangle =\alpha_0|\psi_0\rangle\otimes\left|0\right\rangle$$ if the states $\left|0\right\rangle$ and $\left|1\right\rangle$ are orthogonal (and then $\left\langle0\right|1\rangle=0$).

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  • $\begingroup$ The result cannot be $\alpha_0|\psi_0\rangle\otimes |0\rangle$ because $|\alpha_0|^2+|\alpha_1|^2=1$ unless $\alpha_0=1$ before applying $P$ and in this case, $P$ didn't do any change of $|\psi\rangle$. $\endgroup$ – Ahmed Younes May 5 '15 at 13:21
  • $\begingroup$ Yet the answer is mathematically correct. Since $P$ is a linear operator, since the tensor product is bilinear, the coefficient $\alpha_0$ must appear at the end of the calculus. $\endgroup$ – Nicolas May 5 '15 at 13:23
  • $\begingroup$ The operator is a projective measurement, so we should normalize. $\endgroup$ – Ahmed Younes May 5 '15 at 13:24
  • $\begingroup$ Nothing tells us to normalize mathematically, and I do not see a physical reason to do this. $|\alpha_0|^2$ will be the probability to have the above state, and it can be less than $1$. $\endgroup$ – Nicolas May 5 '15 at 13:29
  • $\begingroup$ For example, $\left|0\right\rangle$ and $\left|1\right\rangle$ can be two states of polarization of a photon, and $|\alpha_0|^2$ gives you the probability that your photon will be polarized with respect to $\left|0\right\rangle$, and it can be $1/2$ if, for example, the both directions of polarization are $\pi/4$ compared to the original direction. $\endgroup$ – Nicolas May 5 '15 at 13:36

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