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6 votes
Accepted

Understanding the $M$ upper bound in the paper: "Multipartite entanglement and high-precision metrology"

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, ...
Jonathan Trousdale's user avatar
5 votes
Accepted

Are entanglement witnesses of this form optimal?

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}} | ...
John Watrous's user avatar
  • 6,087
5 votes
Accepted

How is the expression for the optimal entanglement witness derived?

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 ...
glS's user avatar
  • 25.2k
4 votes
Accepted

Is purity an entanglement witness?

It is not an entanglement witness for two reasons: Entanglement witnesses are linear maps in $\rho$. $\mathrm{tr}\,\rho^2$ is not a linear map. Entanglement witnesses, by definition, detect ...
Norbert Schuch's user avatar
4 votes
Accepted

How does a map being "only" positive reflect on its Choi representation?

If $\Phi$ is positive but not completely positive, then it gives an operator that has positive trace with separable quantum states, that is, an entanglement witness. To see that, let $|\Omega\rangle :=...
Mateus Araújo's user avatar
3 votes
Accepted

How are witness operators physically implemented?

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 ...
DaftWullie's user avatar
  • 58.7k
3 votes

Explicit 16⨯16 matrix representations of two-qudit entanglement witnesses

Dariusz Chruscinski has provided me with an example of a particular such entanglement witness. It takes the form \begin{equation} \left( \begin{array}{cccccccccccccccc} 1 & 0 & 0 & 0 &...
Paul B. Slater's user avatar
3 votes

Is purity an entanglement witness?

It is not an entanglement witness per se. A subsystem having purity = 1 does mean it is separable, i.e., it is a product state with the rest of the system. However, a subsystem having purity $\neq$ 1 ...
FDGod's user avatar
  • 2,401
3 votes

Can we characterise the general structure of two-qubit witness operators?

You can try to use the Størmer-Woronowicz theorem for that (it's used to prove the sufficiency of the Peres–Horodecki criterion in $2 \times 2$ and $2 \times 3$ cases). The theorem states that any ...
Danylo Y's user avatar
  • 7,332
2 votes

Entanglement Witnesses close to GHZ states

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 ...
John Doe's user avatar
  • 911
1 vote

How to prove the following bosonic entanglement expression?

We have, \begin{equation} \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \...
EnthusiastiC's user avatar
1 vote
Accepted

What are examples of weakly optimal witnesses?

Consider the optimal witness $$W = I-2|\psi^-\rangle \langle \psi^-|,$$ and now add some amount of $|01\rangle\langle01|$ to it, i.e., let $$W' = W + \alpha |01\rangle\langle01|,$$ for some $\alpha \...
Mateus Araújo's user avatar
1 vote

Prove that an entanglement witness is optimal iff it's zero on a spanning set of product states

Definition of optimal witness — Start recalling that an entanglement witness $W^{\rm opt}$ is said to be optimal if there is no other witness $W'$ with detects a strict superset of entangled states — ...
glS's user avatar
  • 25.2k
1 vote

Prove that an entanglement witness satisfies $\operatorname{tr}(W)>0$ and $\operatorname{tr}(W)^2\ge \operatorname{tr}(W^2)$

(a) holds since $\mathrm{tr}(W)$ is the value the witness takes on the maximally mixed state. Since there is an open ball of separable states around the maximally mixed state, this implies that we can ...
Norbert Schuch's user avatar
1 vote

Prove that an entanglement witness satisfies $\operatorname{tr}(W)>0$ and $\operatorname{tr}(W)^2\ge \operatorname{tr}(W^2)$

For (a), you've started well - taking a separable orthonormal basis and showing that the sum over all those expectations is non-negative. And, yes, you're right to worry that all those diagonal ...
DaftWullie's user avatar
  • 58.7k

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