7 votes
Accepted

Truthfulness of Entanglement Fidelity

Initial answer The entanglement fidelity isn't intended to measure a change in entanglement. Rather, it measures how coherently a channel preserves a given state, meaning how well it preserves the ...
John Watrous's user avatar
  • 5,703
4 votes

Is it possible to derive a Schmidt decomposition for a mixed state?

Usually, the Schmidt rank of a mixed state $\rho$ is defined as the minimum of the maximal Schmidt rank of an element in any decomposition of $\rho$ into a mixture of pure states (there are many ...
Danylo Y's user avatar
  • 6,989
3 votes

Why are all Rényi entropies equal for Clifford dynamics?

In stabilizer states (which are the states produced by clifford dynamics) the entanglement spectrum is flat, i.e., all the non-zero Schmidt coefficients are equal. Starting from there, it is ...
Norbert Schuch's user avatar
3 votes

Why are all Rényi entropies equal for Clifford dynamics?

Let me report here a brief demonstration that I used to convince myself. The R'enyi entropies are \begin{equation*} S_\mathrm{A}^\alpha = \frac{1}{1-\alpha} \log_2 \text{Tr} (\rho_\mathrm{A}^{\...
gcemin's user avatar
  • 31
2 votes
Accepted

What happens to $|y\rangle \sum_{x}|x\rangle|f(x) + g(y)\rangle$ when we throw away the first register?

After throwing the first register away the state of the second and third registers becomes $$|\psi_{23}\rangle=\sum_x|x\rangle|f(x)+s\rangle\tag1$$ where $s$ is a fixed value. More precisely, $s:=g(y)$...
Adam Zalcman's user avatar
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2 votes

What happens to $|y\rangle \sum_{x}|x\rangle|f(x) + g(y)\rangle$ when we throw away the first register?

If $y$ is a bitstring, meaning $|y\rangle$ is a basis state, yes, you can just throw it away since it is not entangled with the rest of the state. However, if you had something like $\sum_y|y\rangle\...
Tristan Nemoz's user avatar
2 votes

Is there a measure for entanglement of $n$-qubit system?

There are many measures of entanglement, such as quantum discord partial positive transpose positvity of coherent information concurrence, and more.
FDGod's user avatar
  • 1,843
2 votes
Accepted

How can a third party learn the coefficient of a shared $2n$-qubits state using a classical message from each one?

Get Alice and Bob to measure all their qubits in the $X$ basis, and send the $\pm 1$ results to Charlie. The combined answer is a bit string $w$. Charlie discards all the results associated with ...
DaftWullie's user avatar
  • 57.1k
2 votes

What are toy examples of single-copy entanglement conversion?

I always rather enjoyed the examples in this paper: https://arxiv.org/abs/quant-ph/9905071 They are more stated in terms of deterministic conversion, but you could just as well calculate probabilities ...
DaftWullie's user avatar
  • 57.1k
1 vote

cirq entanglement qubits

The problem is that you negate qubit q1 before the measurement. As both qubits are opposite each other under original settings (see the first circuit results), negation gate you added on q1 leads ...
Martin Vesely's user avatar
1 vote
Accepted

Two-photon N00N state through Mach-Zehnder interferometer

There are some corrections to the calculations, but it is more important to focus on the motivation: why are you sending a NOON state through a Mach-Zehnder interferometer? The idea of the NOON state ...
Quantum Mechanic's user avatar
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

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