7
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How to convert a partially entangled state into maximally entangled using SLOCC
This case is pretty straightforward, if you're already familiar with generalised measurements. Assume $\alpha>\beta$ are real numbers. We can define
$$
M_1=\left(\begin{array}{cc} \sqrt{\frac{\beta}...
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4
votes
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What is the relation between maximally entangled and maximally mixed?
What is a maximally mixed state?
Intuitively, think of a maximally mixed state as a quantum state where all possible measurement outcomes are equally likely to occur at random upon measurement. If ...
- 1,311
4
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Is possible to write a separable state as a finite or countable infinite sum of product states?
Yes, you can always write an integral of separable states as a (finite) convex combination of product states, owing to the fact that the set of separable states is compact.
A quick way to see this is ...
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3
votes
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What is the difference between quantum teleportation and quantum entanglement with dense coding?
Both teleportation and (super)dense coding use entanglement as a resource. I like to think that one is dual to the other, as the quantum circuits are, basically, inverses of each other.
For example, ...
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1
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Can you apply a CNOT gate on qubits in MBQC that aren´t next to each other?
Maybe you should clarify exactly what you mean by MBQC, because to me this seems like a trivial "of course you can".
For example, a CNOT can be implemented up to Pauli tracking by this ...
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How to show that the GHZ state is absolutely maximally entangled?
I assume that you talk about the standard case of a $3$-qubit GHZ state (the generalized versions of $n$-qubit GHZ states are not absolutely maximally entangled):
$$ |GHZ\rangle = \frac{1}{\sqrt{2}} \...
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How to convert a partially entangled state into maximally entangled using SLOCC
Here's a simple circuit construction. Not sure if it's optimal.
By using a CNOT gate onto a partially-rotated ancilla qubit, you get partial information about $|00\rangle$ vs $|11\rangle$. You can ...
- 33.2k
1
vote
Accepted
Creating entangled states via Rydberg blockade
The key insight here is to realize $|gg\rangle$ is only coupled with $|rg\rangle+|gr\rangle$ (bright state) under laser driving. The state with minus phase, $|rg\rangle-|gr\rangle$, is called dark ...
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Creating entangled states via Rydberg blockade
Good question! When we apply our laser pulse we get a $|\psi\rangle = |rg\rangle + |gr\rangle$. Apply another pulse properly can induce stimulated emission which causes a Rydberg state to decay back ...
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