Faster than light communication protocol via GHZ-state

Question

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?..

Answer 1

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.

Answer 2

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

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