# Tag Info

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### Understanding Controlled Operation

The best way to see this is through an example. Consider the circuit below: qubit that is in an eigenstate of the unitary Note that $|-\rangle$ is an eigenstate ...
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### Entanglement test in Pauli Representation

We want to prove that is the sum is 1 or larger, the state is entangled. I'm just going to do it for larger than 1. To do this, observe that under local operations and classical communication, we can ...
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### What is the difference between "maximally entangled" and "entangled" states?

A maximally entangled state is a state that maximises some entanglement measure. In the case of bipartite states, this generally means a state that maximises the entanglement entropy, that is, the von ...

### Depth circuit optimization for 6-qubits GHZ state

If you can do adaptive measurements inside your circuit (i.e. measure a qubit and apply a correction to other qubits depending on the outcome of the measurement like in Measurement Based Quantum ...

### Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere

I'm late to the party, but here's my take: Pure qubit states As you said, the space of pure states of a single qubit can be described as a complex projective line $\mathbb{C}P^1$, which is ...
I think you might just be misunderstanding the Hadamard gate slightly. In the computational basis, $H= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}$ maps $|0\rangle \... 1 vote ### Do unitary matrices acting on entangled states always give a quantum state? Yes, you can absolutely simplify down. The problem you're having isn't one of simplification, rather one of counting all the terms in the normalisation, although you appear to have picked up a stray$\...
$\alpha\gamma|00\rangle + \alpha\lambda|01\rangle +\beta\gamma|10\rangle+\beta\lambda|11\rangle = (\alpha|0\rangle+\beta|1\rangle)\otimes(\gamma|0\rangle+\lambda|1\rangle)$ is a separable state, so it'...