Quantum entanglement of many particles As soon as the teacher started talking about the quantum entanglement of more than two quantum particles, I ran here to ask a couple of questions.

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*Can there be a state of entanglement between an arbitrarily large number of particles?


*There are two entangled particles. The third particle interacts with them. Will there be a three-particle state of entanglement, or will the entanglement of the first two particles be broken? If the latter, why? Why is entanglement broken when a particle interacts with a many-particle entangled system?


*Can there be an absolutely separable state, that is, a quantum particle that is absolutely not entangled with anything? Or is the separable state only approximate, but to be precise, everything is blurred with everything?
 A: *

*Yes, definitely. Consider for example the $N$-qubit GHZ state
$$\frac{1}{\sqrt{2}}(|0\rangle^{\otimes N}+|1\rangle^{\otimes N}),$$ which is a generalization of the "standard" three-qubit GHZ state
$$\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle).$$ In this states all particles are entangled with all others. Other classes of states that you can look up are $W$-states, Dicke states and cluster states (which are a subclass of graph states).
One important question to ask is: If I "lose" one of the qubits (which corresponds to taking the partial trace of this qubit's degrees of freedom), is the state you are left with still entangled? For the GHZ state, the answer is NO.
Dicke states, for example, are often cited as having entanglement that is robust against loosing of qubits.


*The entanglement between the first two can be maintained. Assume the first two particles are in the entangled state $$|\psi\rangle=\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle),$$
and the third particle in state $$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle.$$
The combined state $|\psi\rangle\otimes |+\rangle$ is $$\frac{1}{\sqrt{2}^3}(|000\rangle+|010\rangle+|100\rangle+|110\rangle+|001\rangle+|011\rangle+|101\rangle+|111\rangle).$$
If you know apply the controlled phase operation ($CZ$) on the qubit pairs (1,2) and (1,3), the third qubit has interacted with the first two and the whole state is entangled
$$\frac{1}{\sqrt{2}^3}(|000\rangle +|010\rangle +|100\rangle -|110\rangle +|001\rangle -|011\rangle -|101\rangle +|111\rangle).$$
Note: This corresponds to the standard construction of a 3-qubit cluster state.
In general there is no definite answer to that question. The entanglement in the final state is dependent on the initial states and the specific type of interaction that you consider.


*The phrasing of the question is a bit unusual, I think. I mean, if you write down a single-qubit state
$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,$$
you consider an isolated qubit that is definitely not entangled with anything else. In an experiment, this would correspond to a qubit that is as isolated from its environment as possible, which is a good approximation (https://arxiv.org/abs/2008.00251). There will always some residual interaction with the environment, which, in a sense, can be seen as entangling the qubit with the environment. Usually however, this process is termed decoherence and is modeled as a decay of the quantum-coherence properties of the state.
