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How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

So I started with rewriting state $|x+\rangle$ in z-representation: $$ |\psi \rangle = \frac{1}{\sqrt{2}} |+\rangle|+\rangle + a|+\rangle \Bigl( \frac{1}{\sqrt{2}} \bigl( |+\rangle + |-\rangle \bigr) \...
hft's user avatar
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How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

The best way to quantify the entanglement for a 2-qubit state is something called the "concurrence"; it goes to zero for a separable state and 1 for a maximally entangled state (like a Bell ...
Ken Wharton's user avatar
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What happens to entangled particles at a Black Hole?

Anything you could learn from an observation on one particle of a maximally entangled particle pair, you could still learn even after the other enters a black hole. To understand why this is true: ...
DrChinese's user avatar
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-1 votes

Can you split maximally entangled qudits into maximally entangled lower-dimensional qudits?

It is possible, under certain conditions: If the goal is to preserve the original state, so that it may be recovered from the split system, then the total dimensionality of the split system must be ...
user34722's user avatar
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2 votes

How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

A pure bipartite state $$\left|\psi\right> = \sum_{i = 1}^{d_A} \sum_{j = 1}^{d_B} c_{ij} \left|i\right>\!\left|j\right> \in H_A \otimes H_B$$ is separable if and only if the matrix $(c_{ij})$...
Vladimir Lysikov's user avatar
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How can resolution of particle entanglement be detected, and why can't it be faster than light communication?

You asked the questions, and then made a few comments too. So I will try to piece together a response. There is no way to look at an individual particle and detect whether it is or was entangled ...
DrChinese's user avatar
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How to measure the entanglement of three entangled qubits?

The three-tangle https://arxiv.org/abs/quant-ph/9907047 is such a measure. It is in a sense a generalization of the concurrence and has nice properties.
Karl Pilkington's user avatar
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Understanding operator product of three mixed states with three projector operators

Each party holds two quantum systems, which are parts of two bipartite systems. For example, the Hilbert space $H_A$ corresponding to the party $A$ is the tensor product $H_A = H_{A1} \otimes H_{A2}$ ...
Vladimir Lysikov's user avatar
1 vote

Path Integrals for entangled states

Yes, I've worked out exactly how one can analyze entanglement using the path integral in a pair of published papers; first, second. Note that here we're using the discrete sort of path integral more ...
Ken Wharton's user avatar
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1 vote

What happens when two entangled particles are not measured synchroneously?

Suppose we have a pair of entangled photons, one photon sent to Alice and the other to Bob. A and B are at rest in the same reference frame. When A receives and measures the polarization of her photon,...
alanf's user avatar
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What happens when two entangled particles are not measured synchroneously?

I think remarkably in order to uphold Heisenberg Uncertainty and Quantum Mechanics Together, It must be that given particle A and Particle B are entangled: When one measures a property of particle A, ...
ESponge2000's user avatar
-1 votes

Is there a class of Bell inequalities that are violated by certain mixtures of known orthogonal entangled states?

This question was bumped by the Community bot, so I am supplying an answer. Better late than never? This is a clever question and has some good points mixed in. To discuss the main question: Note: ...
DrChinese's user avatar
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1 vote

Proving that a positive partial transpose (PPT) state is non-distillable

(1) is not true -- you have to transpose Q as well. (2) That should be obvious, since (a) $X\ge0\ \Rightarrow X^{\otimes n}\ge 0$, and (b) $Y\ge0\ \Rightarrow\ AYA^\dagger\ge0$.
Norbert Schuch's user avatar

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