# Entangle more than two atoms?

Alright so we know that we can entangle any two atoms, however, what exactly is stopping us from entangling more? In theory, shouldn't we be able to entangle a third to the already entangled pair?

And if that's possible, then couldn't we theoretically be able to entangle a quintillion, or some other relatively large quantity of atoms?

In one quite real way you have already answered your own question, since atoms are themselves collections of smaller particles. Thus two entangled hydrogen atoms actually involves four particles (2 protons + 2 electrons), or 8 if you further break down the protons into 6 quarks.

That brings up a point, though: Because quarks bind at energies that are many orders of magnitude more intense than how electrons bind to proton, the quite small and subtle effect of, say, electron orbital spin entanglement has no discernable impact on them.

Similarly, entanglement in general gets harder to detect as you spread it out over larger and more massive systems of particles. For spin entanglement (there are other forms) the best violation of Bell's Inequality — that is, the case where the experimental results most conspicuously violate classical concepts of how causality should work — is with two light-weight spin 1/2 fermions (that would be electrons and/or positrons) measured at 45° from the plane of polarization. After that, it gets increasingly difficult to distinguish entanglement outcomes from ordinary classical or "memory based" causality.

• So what you're suggesting is that while it's theoretically possible, it would be much much harder to create different forms of entanglement, depending on the type? – Oleg Silkin Jan 5 '16 at 5:50
• entangled states, not entangled objects ... – user46925 Jan 5 '16 at 7:05
• @igael, correct, it would have been more precise if I had said "entangled states of composite, multi-particle objects". – Terry Bollinger Jan 5 '16 at 19:12
• @OlegSilkin yes, experimentally meaningful entanglement — the kind that is both detectable with real equipment and not explainable using speed-of-light constrained casualty — is a lot like a very tiny, rather precious resource that once created cannot be replicated, only divided up or spread around. – Terry Bollinger Jan 5 '16 at 19:21
• @TerryBollinger So then if it's theoretically possible to entangle more than two particles, could this mean that there are webs of entangled particles all throughout the cosmos, as well as our own planet that we are completely unaware of? And if all of the above is true, then could this possibly give explanation to our understanding of Gravity as a whole? – Oleg Silkin Jan 5 '16 at 22:23

Two things could be entangled not at all, e.g. $$|+\rangle_1\otimes|+\rangle_2.$$

Two things could be entangled a little bit, e.g. $$\frac{24|+\rangle_1\otimes|+\rangle_2} {25}+\frac{7|-\rangle_1\otimes|-\rangle_2}{25}.$$

Two things could be maximally entangled, e.g. $$\frac{|+\rangle_1\otimes|+\rangle_2} {\sqrt 2}+\frac{|-\rangle_1\otimes|-\rangle_2}{\sqrt 2}.$$

But what happens when you try to entangle three things together like: $$\frac{|+\rangle_1\otimes|+\rangle_2\otimes|+\rangle_3} {\sqrt 2}+\frac{|-\rangle_1\otimes|-\rangle_2\otimes|-\rangle_3}{\sqrt 2}?$$

Well, when you measure the first particle in the $|+\rangle,$ $|-\rangle$ basis it either becomes $|+\rangle_2\otimes|+\rangle_3$ or it becomes $|-\rangle_2\otimes|-\rangle_3$ and in either case particles two and three are no longer entangled.

There just isn't a way to maximally entangle three particles. This is often called monogamy of entanglement. Any entanglement of three particles fails to have every part of all three be equally entangled with the other two. In the example we tried, a measurement of one destroyed the entanglement between the other two remaining ones. The first one basically was only as entangled with those unentangled pairs $|+\rangle_2\otimes|+\rangle_3$ or $|-\rangle_2\otimes|-\rangle_3$ as it normally was with the single particle states $|+\rangle_2$ or $|-\rangle_2$.

You could imagine that particle one only has two degrees of freedom, so it just doesn't have enough freedom to match up with all the ways the other two could be. If the one particle has two degrees of freedom, $|+\rangle_1$ or $|-\rangle_1$ then it can't match up with all four degrees of freedom of $|+\rangle_2\otimes|+\rangle_3,$ or $|+\rangle_2\otimes|-\rangle_3,$ or $|-\rangle_2\otimes|+\rangle_3,$ or $|-\rangle_2\otimes|-\rangle_3.$