I had a chat with a guy recently who claimed that multipartite entanglement might actually be used for superluminal communication, and his argument is very simple.

Suppose that 2 observers (Alice and Bob, $AB$) are close to each other (say, in the same lab) while the 3rd observer (Clark, $C$) is far away, and if all their devices are oriented along the same direction (say, Z) - it is possible for $AB$ to tell, whether $C$ has changed the way he's measuring his particles, or not.

For instance, given the GHZ-state: $$|GHZ\rangle={\frac{1}{{\sqrt{2}}}}(|11\rangle_{AB}|1\rangle_{C} + |00\rangle_{AB}|0\rangle_{C})$$

There are 2 possible results for a Z direction: $|111\rangle + |000\rangle$; and 8 more results for another directions:

  • $|111\rangle + |100\rangle + |001\rangle + |010\rangle$;
  • $|000\rangle + |110\rangle + |011\rangle + |101\rangle$.

Suppose that there's a constant stream of entangled photons (a laser) between $A$, $B$ and $C$ - so one could imagine each observer reading millions results (such as listed above) per second (providing, that all the noises and errors are reduced to a practical minimum); I should also point out, that if this laser is originated from $AB$ 's location - then it's also delayed at their ends, so that photons from each entangled triplet would arrive to all parties simultaniously (in their mutual rest frame of reference at least); hence, I'm expecting a "quantum" mistake to be revealed in discussions - not a "relativistic" one!

...Now, the "GHZ-FTL" idea from my opponent suggessts, that if $AB$ recieved few millions of $|11\rangle_{AB}$ and $|00\rangle_{AB}$ ("the same") results in a row - they could claim, that $C$ has his device in the same Z direction as theirs.

Therefore, if in another instant $AB$ recieved $|10\rangle_{AB}$ and $|01\rangle_{AB}$ results as often as "the same" ones - they could claim, that $C$ has now his device in "the different" (say, orthogonal - for a maximal diversity) direction.

Thus, switching between "the same" and "the different" respective positions of his device - (it seems, that) $C$ is able to encode classical information into, say, a Morse code, or something even better.

Can such scenario be expected from multipartite entangled states?..

  • $\begingroup$ Doesn't this argument just say that when A and B receive the other results often that means they now know that C reoriented his device when those photons were sent, and thus they know something about C's state exactly 1 light-transmission period ago? $\endgroup$
    – Cort Ammon
    Commented Feb 12, 2021 at 23:55
  • $\begingroup$ @CortAmmon yeah, but I pointed out that photons are delayed at A and B ends in order to hit AB and C simultaneously (in their mutual rest frame). $\endgroup$ Commented Feb 13, 2021 at 0:03
  • $\begingroup$ @VictorNovak I don't see how adding a delay gets around the fact that A&B have to wait at least a light-travel-time to know what C is doing. I also think that calling other users "arrogant pricks" is highly inappropriate for this site. $\endgroup$
    – d_b
    Commented Feb 13, 2021 at 0:36
  • $\begingroup$ @d_b the whole concept was complete nonsense for me from the start: it was only my inability to read\write the math to derive that. Now that I've read GHZ-state article more carefully\patiently - I can (almost) tell the difference) as for the curses... well, if anyone seems appropriate to throw a stone at me without any reasoning\feedback whatsoever - I don't mind if our path won't cross at later times. $\endgroup$ Commented Feb 13, 2021 at 1:07
  • $\begingroup$ I'm not sure where the particular issue is here, other than that I get the really strong smell of the Delayed Choice Quantum Eraser in this. Particularly, we find that DCQE shows this curious consistency between the readings, but you can't actually determine the state of the other devices until notes are exchanged (at the speed of light or slower). My gut instinct is that, were we to fill in the missing details in your opponent's design, he would fill them in to be basically the same as the quantum eraser. $\endgroup$
    – Cort Ammon
    Commented Feb 13, 2021 at 19:43

2 Answers 2


So this is actually a great question and indeed Alice and Bob cannot observe these 01 and 10 states, no matter how Carol orients her detector! Quantum entanglement is only observed in correlations, and defeats the FTL requirements mostly by “you have to bring the data back into a shared place to see the correlations.”

In this case, if we are talking standard conventions for qubits, the +$x$-direction and -$x$-directions are $$ \begin{align} |{+}\rangle&= \sqrt{\frac12}\big(|0\rangle + |1\rangle\big)\text{ and}\\ |{-}\rangle&= \sqrt{\frac12}\big(|0\rangle - |1\rangle\big)\\ \end{align} $$respectively, then the inverse transform is $$ \begin{align} |0\rangle&= \sqrt{\frac12}\big(|{+}\rangle + |{-}\rangle\big),\\ |1\rangle&= \sqrt{\frac12}\big(|{+}\rangle - |{-}\rangle\big).\\ \end{align} $$ This means that the relevant GHZ state can be expressed as, $$ \sqrt{\frac12}\big(|000\rangle + |111\rangle\big) = \sqrt{\frac14}\big(|00{+}\rangle + |00{-}\rangle + |11{+}\rangle - |11{-}\rangle\big).$$ So, if the measurement happens along the $z$-axis and we are not allowed to see it, we get a 50/50 statistical mixture of $|00\rangle$ and $|11\rangle$, whereas if it happens on the $x$-axis, the mixture is instead of $\sqrt{\frac12}\big(|00\rangle\pm|11\rangle\big)$. It is therefore obvious that the 01 and 10 states cannot occur here. They could not have occurred no matter how I changed the axis that I measure that last qubit on.

It is still not immediately obvious that these are perfectly indistinguishable for Alice and Bob, and requires a slight elaboration on traditional wave functions called state matrices: these can handle these 50/50 mixtures just fine. But according to the state matrices, these two mixtures are indeed the same for any experiment that is local to just Alice and Bob, $$ \frac12\begin{bmatrix}1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix} + \frac12\begin{bmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&1\end{bmatrix}= \frac12\begin{bmatrix}1/2&0&0&1/2\\ 0&0&0&0\\ 0&0&0&0\\ 1/2&0&0&1/2\end{bmatrix}+ \frac12\begin{bmatrix}1/2&0&0&-1/2\\ 0&0&0&0\\ 0&0&0&0\\ -1/2&0&0&1/2\end{bmatrix}.$$ Since these are the same “reduced” state matrix there is no experiment that they can do “locally” to discover this change in measurement.

  • $\begingroup$ (after also processing your starting post comments) ...okay, but we can express GHZ instead as $$\sqrt{\frac12}\big(|0\rangle(|+\rangle+|-\rangle)(|+\rangle+|-\rangle)+|1\rangle(|+\rangle-|-\rangle)(|+\rangle-|-\rangle)\big)\\=\sqrt{\frac18}\big(|0++\rangle+|0+-\rangle+|0-+\rangle+|0--\rangle+|1++\rangle-|1+-\rangle-|1-+\rangle+|1--\rangle\big)$$ and it's physics seems like that 1 device is in Z axis, and another 2 are in X; and here we see the whole spectrum of 8 possible outcomes. My question then is... how ZZX case (from your answer) can be physically different from ZXX - ?)) $\endgroup$ Commented Feb 15, 2021 at 14:06
  • $\begingroup$ reffering to my comment above: it's obvious that the answer is that in your ZZX we have 1 "expressed" quibit (wrote in the different basis), while in mine ZXX "expressed" quibits count is 2; and the key of that "expression" is that the left part of these quibits is "expressed" with "+" (like a+b), while the right part is with "-" (like a-b). The real (for me) question then is what's the physical difference between the signs (in terms of devices alignments) - ? $\endgroup$ Commented Feb 15, 2021 at 14:50
  • $\begingroup$ I just realized (probably right), that 4 outcomes in your "ZZX" case (as I named it) are not about "what ZZX devices would give us", but instead about "what ZZX and ZZZ devices would give us", for the X quibit is encoding information from 2 different physical situations - wheter it's respective 3rd device is in the same axis as the other 2, or not. And by expressing GHZ in XXX basis - we simply mean, that this "the same, or not" question is "expressed" onto all 3 devices, instead of 1 or none: that's why it gives us $2^3 = 8$ logical (not physical) answers. $\endgroup$ Commented Feb 15, 2021 at 16:10
  • $\begingroup$ The physics of XXX case is that "all 3 devices could give us 3 same outcomes" - which 111 state represents, or "1 of them will be different from another 2" - which the other 3 states means. The point is that it doesn't mean that 000 state is impossible in XXX case: it's just that the physics of 000 is indistinguishable from 111 - and the same goes for "the other 3" states, by just swapping up their sign (0 to 1 and otherwise). Whoa! XD $\endgroup$ Commented Feb 15, 2021 at 16:15
  • 1
    $\begingroup$ I will say that the minus signs here are kind of the fundamental point of quantum mechanics: a probability theory with destructive “wavy” interference. E.g. in the “XXX” configuration, where you only get 4 possibilities, all of which have an even number of bits set; four of the eight “ZXX” possibilities destructively interfere. But yeah, qubits are kinda trippy, as the Bloch sphere shows: the idea that you have more spin $\hbar\sqrt{3/4}$ than can be measured along any axis $\hbar/2$ so that in the $+z$ state you get non-zero measurement along the $x$-axis, very counterintuitive to me. $\endgroup$
    – CR Drost
    Commented Feb 15, 2021 at 19:50

I am not an expert in quantum mechanics but consider this:

Isn't quantum entanglement of 2 particles destroyed when they are measured?

  • $\begingroup$ Yeah, I guess you're right; Wikipedia says: On the other hand, if we were to measure one of the subsystems in such a way that the measurement distinguishes between the states 0 and 1, we will leave behind either {\displaystyle |00\rangle }|00\rangle or {\displaystyle |11\rangle }|11\rangle , which are unentangled pure states. This is unlike the W state, which leaves bipartite entanglements even when we measure one of its subsystems. $\endgroup$ Commented Feb 13, 2021 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.