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Are qubits just analog, continuous classical bits?

Are qubits just analog, continuous classical bits? No. The continuous (vs. discrete) nature of qubit states is not the secret sauce that makes them so powerful. The secret sauce that is missing from ...
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2 votes

Do there already exists solutions to quantum gravity which are mathematically consistent and consistent with PRESENT observations?

Your requirement can be trivially satisfied by taking quantum field theory + general relativity and adding a small modification which cannot be measured at current times. For instance adding a very ...
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2 votes
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Coupled and uncoupled qubits - Hilbert space representation

This isn't my realm of expertise, so my terminology & notation may be unorthodox and possibly off-base. But since this question hasn't garnered any attention so far, I'll give it a go. An "...
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2 votes

A variation of a Bekenstein's thought experiment

Yes, information loss is both about classical and qubits. Consider two glasses, one hot and one cold but with total mass tuned to be exactly the same. By the no hair theorem both generate ...
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2 votes
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Entanglement across a bi-partition of a many particle system

Short answer: yes it is correct to think about qudits. In general, if you are looking at the amount of entanglement across a bipartition, you are ignoring the structure of whatever is within one side ...
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1 vote

Complex Coupling Strength in Light-Matter Interaction Hamiltonian

You can redefine the ladder operators to absorb the phase. The mapping $$a \rightarrow e^{i\phi} a$$ is canonical, since it preserves the commutation relations. So there is no mistake in your logic, ...
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1 vote

What is the difference between "cluster states" and "graph states"?

I don't think there is really a difference, though in general, for "cluster state" people would think of a regular lattice (not only 2D - e.g. 3D cluster states are used for fault-tolerant ...
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1 vote

What is the measure of decoherence?

In time domain decoherence is usually characterized by the rate of decay of coherences, i.e., the non-diagonal density matrix elements. This, of course, is dependent on the mechanism that causes ...
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1 vote

Coherence and superposition principle - equivalent properties?

Imagine a generic pure state$^\dagger$, which is a superposition of some basis states \begin{equation} |\Psi\rangle = \sum_{k=1}^N c_k | k \rangle \end{equation} To be concrete, we can take the basis ...
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1 vote
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Under what conditions is a POVM a von Neumann measurement?

You only need the first two requirements for a POVM. You need Hermitian operators $\{E_i\}$ such that $E_i\ge0$ and $\sum_i E_i=I$. I'm not sure what you mean with them being "probability ...
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1 vote
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Negativity of the real part of eigenvalues of Lindblad operators

It is enough to prove that $Re( \mathcal{L}) \leq 0$. Since the Hamiltonian part is skew it does not matter for the proof of dissipativity. We can look at the single Lindblad operators individually. \...
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