8 votes
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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}...
DaftWullie's user avatar
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7 votes
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Is LOCC equivalence the same as LU equivalence?

LOCC is a setting which asks about, for example, transforming $|\psi\rangle$ to $|\phi\rangle$, and what is the maximum probability with which that is achieved. If you restrict to demanding perfect ...
DaftWullie's user avatar
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7 votes
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Can EPR pairs be established with only classical communication?

No, it is not possible to entangle two qubits using only classical communication, more generally, LOCC can not create entanglement. One can create a shared Bell pair, for example, if Alice makes two ...
Nikita Nemkov's user avatar
7 votes

What is an example of a separable measurement that is not LOCC?

Such an example is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).
Norbert Schuch's user avatar
6 votes
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How to transfer non maximally entangled state to maximally entangled?

The first question that we have to deal with is what is meant by "maximally entangled" in this context. There's no single straightforward notion. In particular, for three qubits, there are two ...
DaftWullie's user avatar
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5 votes
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Can LOCC operations take product states to non-product states?

Sure, the classical communication allows you to generate the necessary correlations between two sites. For instance, suppose that Alice flips a coin and communicates the outcome to Bob. When the ...
Rammus's user avatar
  • 5,503
5 votes

Can EPR pairs be established with only classical communication?

There are only two ways to create EPR pairs distributed between spacelike separated Alice and Bob: use a quantum channel or distill EPR pairs from pre-existing entangled states. In other words, local ...
Adam Zalcman's user avatar
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4 votes
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Prove that there are infinitely many two-qubit entanglement classes under LU

TL;DR: This is an application of Schmidt decomposition followed by local basis change on each qubit. By Schmidt decomposition any two-qubit state $|\psi\rangle$ may be written in the form $$ |\psi\...
Adam Zalcman's user avatar
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4 votes
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Prove that $A\preceq B$ implies $A=\Psi(B)$ for some channel $\Psi$

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{...
keisuke.akira's user avatar
4 votes
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How is the measurement described in the LOCC wikipedia a measurement on the product state $\mathbb C^2\otimes\mathbb C^n$?

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 ...
DaftWullie's user avatar
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4 votes
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How to prove that there are only 2 classes in 3 qubits entangled state?

The proof can be found at https://arxiv.org/pdf/quant-ph/0005115.pdf, the original paper by Dür, Vidal and Cirac. It shows that, if you consider a pure 3-qubit entangled state (that is, a 3-qubit ...
GotCarter's user avatar
  • 401
3 votes

Is there a non-deterministic protocol for entanglement generation between distant parties?

No. By definition, the set of states that you can produce under LOCC are the separable states. This includes all possible measurements, post-selection on certain outcomes etc. The whole point of ...
DaftWullie's user avatar
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3 votes
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Why can any LOCC operation be written as $\sum_k (A_k\otimes B_k)\rho(A_k^\dagger\otimes B_k^\dagger)$?

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 ...
Norbert Schuch's user avatar
2 votes
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Are Bell states distinguishable through LOCC?

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 ...
DaftWullie's user avatar
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2 votes
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Are separable, orthogonal states LOCC distinguishable?

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 ...
Danylo Y's user avatar
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2 votes
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Is there a non-deterministic protocol for entanglement generation between distant parties?

No, any such protocol would violate the holevo bound (1 bit of communication per 1 sent qubit, including qubits sent during preparation). You could just keep repeating the process until it gave you ...
Craig Gidney's user avatar
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2 votes
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Show that any two product states of the same dimension are LU-equivalent

For part (a), you can certainly manually construct a unitary that for which $U|\phi\rangle = |\psi\rangle$ (though there may be a more elegant proof of this fact). For example, suppose $|\phi\rangle = ...
forky40's user avatar
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2 votes
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An intuitive definition of "One-way LOCC distance"

$\Lambda_{B\rightarrow M}$ is a channel, so its action on $X$ is that of a superoperator (rather than a unitary matrix). The tensor product indicates that the composite channel acting on $X$ consists ...
Andrew Guo's user avatar
2 votes

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 ...
Craig Gidney's user avatar
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1 vote

What operations are allowed in LOCC?

I think you are basically right. If Alice on Earth has in her possession a number of qubits $|\psi\rangle$, and Bob on Mars has a number of qubits $|\phi\rangle$ in his possession, then even if $|\...
Mark Spinelli's user avatar
1 vote

What's the best entangling circuit to measure the Peres-Wootters double-trine state?

Following on from my previous answer about unambiguous discrimination, there is also an optimal solution for the minimum error probability, which I found here. The trick is to introduce a state $$|\...
DaftWullie's user avatar
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1 vote
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What's the best entangling circuit to measure the Peres-Wootters double-trine state?

There's no simple answer to what is the "best". You need to define what you mean a bit more carefully. There are two common scenarios under which one optimises: minimum error: we might get ...
DaftWullie's user avatar
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1 vote
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perform a SWAP measurement using local operations and classical feedback

Measure SWAP from 7 CNOTs, linear connectivity, ancilla outside: (Note: improved this to not need any feedback anymore, and to just generally be cleaner.)
Craig Gidney's user avatar
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1 vote

Proof that Bell state can be transformed into any 2-qubit state $\vert\psi\rangle$ under LU

What you want is the theory of the Schmidt decomposition. Let's say you have a two-qubit state $|\psi\rangle$ which is written $$ |\psi\rangle=\alpha_{00}|00\rangle+\alpha_{01}|01\rangle+\alpha_{10}|...
DaftWullie's user avatar
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1 vote

How Quickly Can We Entangle a Pair of Unentangled Qubits Without Using Pre-existing Entanglement?

You can parallelise your proposal. Imagine you have $k+1$ points along a line where points 0 and $k$ are locations $a$ and $b$. Let $k$ be even for the sake of argument. At each even-numbered point $i$...
DaftWullie's user avatar
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1 vote
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Filtering operation is trace decreasing?

Yes. Generally you should think of filtering as being a measurement. You're only describing the effect of one measurement outcome. There's generally a second one such that the net effect is trace ...
DaftWullie's user avatar
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1 vote
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Universal resource for measurement based quantum computation

That only works if you consider the bipartition of the bit that's going to be $|\phi\rangle$, versus the other $m-n$ qubits. Thus, to make that work, you have to be able to do untaries over the ...
DaftWullie's user avatar
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1 vote

Are mixtures of pairs of Bell states perfectly distinguishable by local operations?

The states $\sigma_0$ and $\sigma_1$ can be seen to have the form $$\sigma_0 = N[\mathbb P(e_1+e_4) + \mathbb P(e_2+e_3)], \\ \sigma_1 = N[\mathbb P(e_1-e_4) + \mathbb P(e_2-e_3)],$$ where $N$ are ...
glS's user avatar
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1 vote

Are perfectly LOCC-indistinguishable states necessarily identical?

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) ...
tsgeorgios's user avatar
  • 1,406
1 vote

Are Bell states distinguishable through LOCC?

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}{\...
vasjain's user avatar
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