6

Such an example is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).


4

I assume the question refers to how LOCC is used to distinguish the two states $$ (|00\rangle+|11\rangle)/\sqrt{2}\qquad(|01\rangle+|10\rangle)/\sqrt{2} $$ when Alice and Bob each hold one qubit, and are separated by a great distance. There are many different measurement operators that could achieve the same task. If Alice just held both qubits herself, she ...


3

Let $\rho_{d}, \sigma_{d}$ be the (simultaneously diagonal) density matrices whose eigenvalues are $\{ p_{j} \}, \{ q_{j} \}$, respectively (represented as probability vectors below). Then, if $\vec{p} \succ \vec{q}$, the following sequence of arguments can be observed: There exists a bistochastic matrix $M$ such that $M \vec{p} = \vec{q}$ (basic result of ...


3

The form (A) above is known as a separable superoperator. The effect of any LOCC protocol can be described as a separable superoperator, or as a separable POVM with POVM elements $N_i = A_i\otimes B_i$. This can be seen as follows (adapted from this answer - I will focus on the POVM case, the superoperator is obtained by ignoring the final outcome of the ...


2

If you look at these states in the $X$ basis, they are $$ |++\rangle+|--\rangle,\qquad |+-\rangle+|-+\rangle. $$ Thus, by both measuring in the $X$ basis and computing the parity of the answers, you can tell $C$'s bit value.


2

Consider these two states $$ \sigma_0 = \frac{1}{2}(|11\rangle\langle 11| + |++\rangle\langle ++|) $$ $$ \sigma_1 = \frac{1}{2}(|0-\rangle\langle 0-| + |-0\rangle\langle -0|) $$ I believe they are indistinguishable (with certainty), though to be sure it's better to find the exact proof. Also check this paper https://arxiv.org/abs/quant-ph/9804053. It's ...


1

We can take, for example, $ M = |0 \rangle \langle 0| \otimes I $, right? But then: $$ \Big|\Big| M(\rho - \sigma) \Big|\Big|_1 = \Big|\Big| |0 \rangle \langle 0| \otimes (p_0 \rho_0 - q_0 \sigma_0) \Big|\Big|_1 = \Big|\Big| p_0 \rho_0 - q_0 \sigma_0 \Big|\Big|_1 = 0 \implies p_0 \rho_0 = q_0 \sigma_0 $$ Similary, taking $ M = |x \rangle \langle x| \...


1

We can distinguish these states using SingleQubit Unitaries and Measurements. Lets the 2 states be $|\psi^{00}\rangle = \frac{1}{\sqrt2}(|00\rangle + |11\rangle)$ and $|\psi^{01}\rangle = \frac{1}{\sqrt2}(|00\rangle - |11\rangle)$. Let Alice be in possession of the 1st qubit and Bob be in possession of the 2nd qubit. Bob applies a $H$ gate on his qubit. This ...


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