# 3ntangled

Can three or more particles be quantum entangled? And if they can, what will the two particles opposite state, like opposite spin, translate to for three (or more) entangled particles?

Certainly! Any n-particle state that cannot be written as a pure tensor product of n one-particle states is “entangled” to some extent, though in many cases the entanglement may be “partial” in the sense of only involving some of the particles, and the question probably wants a state which is “fully” entangled in the sense that there is no way of writing the state as a pure tensor product of states for any partition of the n particles into subsets. [Note: I am using “partial” and “full” here to refer to the breadth or scope of the entanglement rather than to some measure of its pairwise strength or depth.]

A system of three spin 1/2 particles is in a completely unentangled state if and only if the state vector can be written as a pure tensor of the form $|\Psi\rangle =|\Psi_1\rangle \otimes |\Psi_2\rangle \otimes|\Psi_3\rangle$ where each $|\Psi_i\rangle$is a pure state of the one particle system (which may or may not be an eigenstate of some spin direction). The reason for this definition is because if a state can be written this way, then the effective state of the remaining particles after measurement of any one of them can be shown to always be independent of the measurement value obtained on the observed particle.

Note: A state that is given as a sum of several pure tensors may nonetheless be rewritable as a pure tensor and so be unentangled. For example, if $|\uparrow\rangle$ and $|\downarrow\rangle$ are the eigenstates for spin up and down in, say, the z-direction, then \begin{align}|\Psi\rangle &=\frac{1}{\sqrt{2}}(|\uparrow, \downarrow ,\uparrow \rangle +|\uparrow ,\downarrow ,\downarrow \rangle)\\ &=\frac{1}{\sqrt{2}}(|\uparrow\rangle \otimes |\downarrow\rangle \otimes |\uparrow \rangle +|\uparrow \rangle \otimes |\downarrow \rangle \otimes |\downarrow \rangle) \\ &=|\uparrow\rangle \otimes |\downarrow\rangle \otimes(\frac{|\uparrow\rangle +|\downarrow\rangle}{\sqrt{2}})\end{align}

And this is unentangled because it can be written as a pure tensor and we can see that measurement of spin 3 can “collapse” the $|\Psi_3\rangle$ factor to either $|\uparrow\rangle$ or $|\downarrow\rangle$ without affecting either of the others.

But $|\Psi\rangle =\frac{1}{\sqrt{2}}(|\uparrow\rangle \otimes |\uparrow\rangle +|\downarrow\rangle \otimes |\downarrow\rangle)\otimes(\frac{|\uparrow\rangle +|\downarrow\rangle}{\sqrt{2}})$ is partially entangled because of the connection between the first and second spins. It is not fully entangled though, since we can measure the third spin without affecting the other two.

The state $|\Psi\rangle=\frac{1}{\sqrt{2}}(|\uparrow,\uparrow,\uparrow\rangle +|\downarrow,\downarrow,\downarrow\rangle )$ is fully entangled because finding any of its spins forces both of the others to be the same. This isn’t the only way to achieve full entanglement though, and it is not necessary to force the other spins to all be the same as the one measured (though of course, as mentioned in the question, more than two can’t all be “opposite” to one another).

For example $|\Psi\rangle=1/2(|\uparrow,\uparrow,\uparrow\rangle +|\downarrow,\downarrow,\uparrow\rangle +|\uparrow,\downarrow,\downarrow\rangle +|\downarrow,\uparrow,\downarrow\rangle)$ is a fully entangled state because “collapsing” onto the part with any one of the $|\Psi_i\rangle=|\uparrow\rangle$ gives only cases with [the other two parallel, whereas $|\Psi_3\rangle=|\downarrow\rangle$ corresponds to them being anti-parallel.

P.S. There are various ways of defining the “amount” or “strength” of entanglement between two particles; and it turns out for some of these measures that in a large class of systems of many particles, the strength of entanglement between any two particles is never more than it could be if they were only entangled with one another. One thing that may have given rise to this question is the unfortunate use of “entanglement monogamy” to refer to this phenomenon. (But as Percy Bridgeman points out in answer to another question, the fact that being fully polygamous reduces the amount of time available for marital relations with any one partner does not actually in any way enforce monogamy.)