# quantum entanglement of 3 or more quanta [closed]

I have 2 questions i would really like to find an answer to: if we have a quantum pair, we can determine whether 2 quanta are entangled or not assuming that we have access to the information of both of them. so if we have a quantum "trio" instead of a quantum pair, does the information about 2 out of 3 quanta suffice to determine whether those two quanta are entangled? if not, why not. the only entanglement i am interested in is the one of the 2 quanta that are tested. the 3rd one is not important for question one. the second question is: if one out of those 3 entangled particles wave functions collapses, this should break the entanglement of all 3 particles and collapse their wave-functions, right?

this is a rephrased version of my original question, in hopes that there would be less confusion.

original question was:

It is my understanding that you can entangle multiple quanta, so that you don't get "quantum pairs" but for example "quantum-trios" etc. I base this assumption on the following study: https://www.nature.com/articles/ncomms13251 Further I believe to know that once the wave function of an entangled particle is collapsed, the entanglement is "broken". I know that there are several resctrictions for faster than light communication with a quantum pair/quantum pairs. I do base following thought experiment on the assumption that one does need only access to the information about 2 out of three quanta to determine whether an entanglement is existent or not.

If we assume a entanglement of atleast 3 quanta(trio) and person (A) has 1 quantum, person (B) has two quanta and person (A) collapses the wave-function, (B) should be able to determine that his 2 quanta are no longer entangled, and faster than light communication would be possible. I don't think it is that simple, but I lack the knowledge. What am I missing?

## closed as unclear what you're asking by WillO, Jon Custer, Daniel Griscom, Void, knzhouNov 8 '17 at 10:08

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• What does "having access to the information about them" mean? Does it mean full knowledge of their state? If so, then the answer is that yes, given a state it is quite easy to determine whether that state is entangled. Otherwise, your question is incomprehensible until you reveal what "the information" means. – WillO Oct 29 '17 at 12:27
• As for your second question, what does "collapse their wave functions" mean here? Obviously the observation of #3 affects the state of #1 and #2. What properties would the new state have to have before you considered it "collapsed"? – WillO Oct 29 '17 at 12:34
• lets say we have an array of entangled photon-trios for each bit of information we want to send(per bit an array of lets say 100 trios). now we can use stochastic to determine entanglement. we collapse the wave-function by measuring the spin. if the entanglement is: 1_2, 2_3 and 1_3 for each trio (numbers represent photons), and person (A) has photons 1 and (B) has photons 2+3. (B) can now measure the spin of his 2's and 3's, and if his statistics tell him that they have the expected spins he knows the entanglement was intact. or not? – user373657 Oct 29 '17 at 12:59
• Again: the state of the 1's is not important for (B) according to my thought experiment. but when the wave-function of 1's is collapsed, the entanglement 1_2 and 1_3 is broken. – user373657 Oct 29 '17 at 13:01
• in this case, 2's and 3's wave-function collapses, and entanglement 2_3 is broken too. this can be detected by testing 2's and 3's for entanglement 2_3. – user373657 Oct 29 '17 at 13:02

## 1 Answer

If we have a quantum "trio" instead of a quantum pair, does the information about 2 out of 3 quanta suffice to determine whether those two quanta are entangled?

No, it is not sufficient.

A simple counter-example is given by GHZ states, $$|\Psi\rangle = \frac{|000\rangle + |111\rangle}{\sqrt{2}},$$ which have maximal tripartite entanglement but for which the tracing out of any of the three systems leaves behind a maximally mixed bipartite state $$\rho_{12} = \mathrm{Tr}_3(|\Psi\rangle\langle\Psi|) = \frac{|00\rangle \langle00|+|11\rangle\langle11|}{2} \tag{*}$$ which has exactly zero bipartite entanglement. That means that there is provably no measurement that's confined to particles 2 and 3 that's going to be able to distinguish whether the tripartite state is a pure GHZ with maximal tripartite entanglement or the maximally mixed state $$\rho_\mathrm{maximally\ mixed} = \frac{|000\rangle \langle000|+|111\rangle\langle111|}{2}$$ with zero tripartite entanglement.

As for your second question,

if one out of those 3 entangled particles wave functions collapses

that's an undefined statement and as posed it is essentially meaningless. Wavefunction collapse is not spontaneous, and it is always relative to a given basis (i.e. relative to a given projective measurement). In this situation, a projective measurement on the third particle can, depending on the choice of measurement basis, either kill all the entanglement, or instate maximal bipartite entanglement (given classical communication from party 3 to parties 1 and 2).

• If you measure along the computational basis $\{|0\rangle, |1\rangle\}$ of the third qubit, then that reduces the 1-2 bipartite states to either $|00\rangle$ or $|11\rangle$ (that's if you communicate the measurement results from 3 to 1 and 2; otherwise, you just get the maximally mixed state $(*)$), neither of which are entangled.

• Alternatively, if you measure along the $\sigma_x$ basis $\{|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle),|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)\}$, then a result along $|+\rangle$ will project the 1-2 bipartite state onto the Bell state $$|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$$ and a result along $|-\rangle$ will project the it onto $$|\Phi^-\rangle = \frac{|00\rangle - |11\rangle}{\sqrt{2}},$$ both of which have maximal bipartite entanglement.

However, it is important to note that for this bipartite entanglement to be useful, you need to know which of the two states you have (i.e. party 3 needs to communicate its measurement outcome to the other two parties along a classical (and therefore slower-than-light) communication channel). If you don't, then you're reduced to the incoherent addition of their density matrices, $$\rho = \frac{|\Phi^+\rangle \langle\Phi^+|+|\Phi^-\rangle \langle\Phi^-|}{2} = \frac{|00\rangle \langle00|+|11\rangle\langle11|}{2},$$ i.e. the maximally-mixed bipartite state with zero bipartite entanglement.

And, finally, touching on one of your remarks,

... and faster than light communication would be possible...

please burn this into the inside of your skull:

There's a reason why the No-Communication Theorem is called a theorem.

• thanks a lot for your detailed answer, i appreciate it very much. "And, finally, touching on one of your remarks, ... and faster than light communication would be possible..." i think essential here is to read that sentence in context, and especially pay attention to the last sentence: "... and faster than light communication would be possible. I don't think it is that simple, but I lack the knowledge. What am I missing?" I knew right from the start that it is not possible, and implicated so three times: 1.i don't think it is that simple 2. i lack the knowledge 3. what am i missing – user373657 Oct 30 '17 at 19:31
• 4. what am i missing just want to point that out. apart from your last remark, your contribution was very useful though. would've been nice if i got an answer like yours right from the start, would've made things much easier and less frustrating. – user373657 Oct 30 '17 at 19:33
• There's nothing wrong with lacking the knowledge, that's why you asked and that's why I answered. And the very first bit of relevant knowledge is the No-Communication Theorem, whose importance is hard to overstate. – Emilio Pisanty Oct 30 '17 at 20:06
• Also, I appreciate that some of the responses you got were suboptimal, but to be frank the question's phrasing was hardly helpful and you got pretty abrasive pretty fast; I was very close to skipping this one based on the assessed risk that it would lead to a belligerent and abrasive comment thread. I mean, in case you find that feedback useful for future threads. – Emilio Pisanty Oct 30 '17 at 20:25