# Tag Info

5

Your conclusion appears correct to me. It seems that Eq.(23), modified with your proposed change to the RHS, can be verified by combining Eq.(3), for the upper bound on the $M$ unentangled particles, with Eq.(5), for the upper bound on the $N-M$ entangled particles, using the convexity asserted in Section II. With respect to the $M$ unentangled particles, ...

5

Answer to edited question: It's still not true for qubit systems. Consider these two unit vectors, both of which are entangled: $$|\phi\rangle = \frac{1}{\sqrt{2}} | 00\rangle + \frac{1}{\sqrt{2}} | 11\rangle,\\ |\psi\rangle = \frac{1}{\sqrt{2}} | 01\rangle + \frac{i}{\sqrt{2}} | 10\rangle.$$ Let $\rho = |\psi\rangle\langle \psi |$, which of course is not ...

4

The idea behind this expression is indeed a fairly general one. $\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\calS}{\mathcal{S}}$An entanglement witness $\mathcal W$ is defined as an operator such that $\operatorname{Tr}(\mathcal W\rho)\ge0$ for all separable $\rho\in\mathcal S$, while for some entangled $\rho_{ent}$ we have $\operatorname{Tr}(\mathcal ... 3 This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as. Just as an example of what I mean, let $$W=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & ... 3 Dariusz Chruscinski has provided me with an example of a particular such entanglement witness. It takes the form \left( \begin{array}{cccccccccccccccc} 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 &... 2 A partial explanation is motivated by the proof in Theorem 1 (bottom of page 2). Assuming two locally non-commuting stabilizing operators, using the Cauchy-Schwarz inequality, for pure product states it is shown that$$\langle S_{l}^{(GHZ_N)} + S_{m}^{(GHZ_N)} \rangle \leq 1.$\$ But since we also assume these are stabilizer operators (GHZ eigenstate with ...

1

We have, \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \gamma _ { B } \right| ^ { 2 } \right) / \left( \left| \gamma _ { B } \right| ^ { 2 } - 1 \right) } } { 1 - \left| \gamma _ { B } \right| ^ { 2 } } \right) = \...

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