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Suppose you are given either $\rho_1$ or $\rho_2$, and you also know that the probabilities you got one or the other are $p_1$ and $p_2$, respectively. If you have no prior knowledge of the frequencies with which I'm going to give you one state or the other, you just use $p_1=p_2=1/2$. You are asking what's the measurement that optimally distinguishes ...


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The optimal probability of guessing correctly is $$ \frac12 + \frac12 \Big\|\frac23 \rho_0 - \frac13 \rho_1 \Big\|_1 $$ where $\| X \|_1 = \mathrm{Tr}[\sqrt{X^* X}]$ is the Schatten 1-norm. This success probability is achieved by the POVM with operator $$ E_0 = \Pi_{[\tfrac23 \rho_0 - \tfrac13 \rho_1]_+} \qquad E_1=I-E_0 $$ where $[X]_+$ denotes the positive ...


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Yes. Expression (X) is in principle more general, but it boils down to doing classical post-processing of the outcomes, and this can always be included in the POVM. So as long as you do optimize over the POVM, expression (Y) is completely general. There are four possibilities: $\lambda_0 p_0(0)>\lambda_1 p_1(0)$ and $\lambda_1 p_1(1)>\lambda_0 p_0(1)$....


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No, there is no way to do this with probability better than 1/2. Basically, you are saying that, ideally, you'd like a protocol where $$ |0\rangle\mapsto|0\rangle,\quad |1\rangle\mapsto|0\rangle,\quad \frac{|0\rangle+e^{i\theta}|1\rangle}{\sqrt2}\mapsto|1\rangle\ \forall\theta. $$ We can get an upper bound on the achievable fidelity using the operator $$ R=\...


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I'll write $\rho_1 = |\psi_1\rangle \langle \psi_1|$ and $\rho_2 = |\psi_2\rangle \langle \psi_2|$. We want the discrimination to be unambiguous so we want, $$ \mathrm{tr}[\rho_1 \Pi_2] = 0 = \mathrm{tr}[\rho_2 \Pi_1]. $$ That is, when we get outcome $i\in \{1,2\}$ we know that we received $\rho_i$ as the other state has a zero probability of obtaining that ...


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Very roughly speaking, yes, "entangled measures" (that is, global measures on multiple copies) make it easier to distinguish states. The intuitive reason is that, if $\langle\rho,\sigma\rangle\equiv\operatorname{Tr}(\rho\sigma)<1$, then $\langle\rho^{\otimes n},\sigma^{\otimes n}\rangle=\langle\rho,\sigma\rangle^n$, which decreases with ...


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It is not possible for a measurement to deterministically give one outcome or the other depending on whether two states are equal or orthogonal. Such a measurement would be some two-outcome POVM $\mu$ such that $$\langle\mu(\text{yes}),\rho\otimes\rho\rangle=1, \qquad \langle\mu(\text{no}),\rho\otimes\sigma\rangle=1,$$ for all states $\rho$ and $\sigma$ with ...


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Disclaimer: Just few ideas, this is not full answer. XOR function is implemented by CNOT gate since: $|00\rangle \rightarrow |00\rangle$ $|01\rangle \rightarrow |01\rangle$ $|10\rangle \rightarrow |11\rangle$ $|11\rangle \rightarrow |10\rangle$ Assuming the result is on second qubit, you can see it is the XOR of $q_1$ and $q_2$, the first qubit of the ...


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$\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\PP}{\mathbb{P}}$Given an arbitrary state $\ket a$, let us write the corresponding density matrix/projector as $\PP_a$. Any such density matrix can be decomposed using a basis of Hermitian traceless operators (think the Bloch sphere representation for qubits) as $\PP_a\equiv 1/2(I + \sum_i a_i\sigma_i)$, ...


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