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First, is entanglement of three particles in $W$-like state deliberately possible (and not by chance)? Second, is the following statement correct?

In the doubly entangled $W$ state, represented as

$$ |\psi\rangle = \frac{1}{\sqrt{2}}(|100\rangle - |010\rangle), $$

when you measure the spin of one particle (let's say the first particle) and find it in the state $|1\rangle$ (spin-up), the states of the other two particles will be $|0\rangle$ and $|1\rangle$ (opposite spins). Similarly, if you measure the first particle and find it in the state $|0\rangle$ (spin-down), the states of the other two particles will be $|1\rangle$ and $|0\rangle$ (opposite spins).

In short, can you have deliberate entanglement of three particles so that if you measure one of them, the remaining two are GUARANTEED to have opposite spins one with another?

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    $\begingroup$ when you measure the spin of … the first particle … and find it in the state $|1\rangle$ …, the states of the other two particles will be $|0\rangle$ and $|1\rangle$ How do you conclude that from your $|\psi\rangle$? Is there a typo? $\endgroup$
    – Ghoster
    Commented Mar 15 at 21:58

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Are talking about the "usual" 3 particle W states? Your question seems to be about this, not doubly entangled W states (which are extremely complicated) per the title. Assuming I have not misunderstood you:

  1. The order of measurement of the 3 particles is not relevant.
  2. After the first is measured, the remaining 2 will retain a degree of entanglement. (That is a major difference as compared to the 3 particle GHZ state.)
  3. I don't believe there is a W state quite like the formula you presented. Usually there are 3 terms (as @Ghoster says); and it is the square root of three; and there is a plus sign; like this:

|W> = 1√3(|001> + |010> + |100>)

If so: You cannot say that the remaining spins/polarizations will or will not be the same if you measure them on the same basis as the first. Even though they are entangled, that entanglement will be one of two possible types. You may find this reference useful, which compares W states and GHZ states:

Bell's theorem with and without inequalities for the three-qubit Greenberger-Horne-Zeilinger and W states

W states are usually created probabilistically. However, I saw this piece proposing deterministic creation of such states. I don't know if it has been tested.

https://arxiv.org/pdf/1602.04166.pdf

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