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3

The whole point of an EPR pair is that you cannot write (without losing some information) "This is what Alice has" and "This is what Bob has". Partial descriptions can be given using reduced density matrices. However, if you want to identify which bits Alice and Bob each have, we often use a notation like $$(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)/\... 0 One possible answer I can think of is that two particles that "collide" don't absolutely create an unavoidable entanglement. That was what I was alluding to in the previous question I posted. If entanglement is natural and unavoidable in two colliding particles, then my idea posted here gains validity (I think). But perhaps when two particles collide ... 1 You're asking how to prove \begin{equation*} \max_{|φ\rangle_A,|ψ\rangle_B} |\langleφ|_A ⊗ \langleψ|_B |ϕ\rangle_{AB}|^2 < 1 \end{equation*} as opposed to actually answering the question? To prove this, consider the Schmidt decomposition of |ϕ\rangle_{AB}=\sum_i\lambda_i|\phi^i_A\rangle|\phi^i_B\rangle, and let$$ |\gamma_A\rangle=\sum_i\lambda_i|i\...

3

I completely agree with kludg's answer. To follow it up: entanglement is natural and unavoidable the entanglement bond is flimsy and easy to break These two things are intrinsically linked. Entanglement has a property called monogamy. If particle $A$ is entangled to both particles $B$ and $C$, the more it is entangled with $B$, the less it is ...

4

There is no contradiction. Entanglement indeed is a natural property; it is difficult to keep pure states for a long time, because they tend to interact and get entangled with environment; but it also difficult to obtain a particular entangled state, such as Bell state. Uncontrolled entanglement is natural, controlled entanglement requires high technology, ...

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I'd ignore that hint and instead try to prove the contrapositives: show that a pure state is a product state iff it has one (nonzero) Schmidt coefficient.

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Your state is a generalization of so-called W-state. Here is a implementation of the state for three qubits. You can also use method described in paper Transformation of quantum states using uniformly controlled rotations for preparing W-state with any qubits you want. When you use this method, set probability of states $|10\dots0 \rangle$, $|01\dots 0\... 4 The coefficient$\frac{1}{\sqrt{2^3}}$is the normalization factor: if you have a 3-qubit state that is an equal superposition of 8 basis states, its norm still has to be 1; thus the squared amplitude of each basis state has to be$\frac{1}{8}$, and the amplitude will be square root of that, i.e. exactly$\frac{1}{\sqrt{2^3}}$. You can also obtain that ... 3 This depends on the context in which you're using the operators. You're talking about multiplying them, so I guess you're thinking of, for example, unitaries (and other circuit model elements). In this case, not only are terms$A\otimes I$local operators, but$A\otimes B$is also a local operator. For example, on a quantum circuit, two Hadamard gates ... 1 There are two inequivalent definitions of "local operator" used in quantum information theory. The first definition is used in the context of communication over a classical channel (e.g. LOCC). In this context, you have a fixed partition of the complete Hilbert space into a tensor product of$k$different subsystems, and the subsystems are assumed to be so ... 2 A qubit is a two-level quantum system and these two-levels in some hardware can be implemented by spins of the electron, but it's not the only option (can be just two "stabile" quantum states not essentially associated with spins). In this sense it is not only about the spin (directional) correlation between qubits, it's about state correlation between them. ... 2$\mathcal{H}_A \neq \mathcal{H}_B$, they are two distinct physical systems (even if they have the same dimension). That formula for a reduced density matrix is a shortcut for $$\rho_A = \sum_j \big(I_A \otimes \langle j |_B\big) \cdot | \Psi\rangle \langle\Psi| \cdot \big(I_A \otimes |j\rangle_B \big)$$ You can check the dimensions. If$d_A = \text{dim}...

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