# Thought experiment on graviton breaking the speed of light [closed]

The effect of gravity travels at the speed of light. Suppose we can entangle a pair of gravitons (which are only theoretical, but who knows for sure?) and separate them over a vast distance. Hold on - observing their quantum states does not violate relativity, does it? If an entangled particle should decay into a graviton, does it also affect the other?

• Why would two entangled gravitons be different to two entangled photons? Since the latter doesn't involve superluminal communication, why should the former? – John Rennie Mar 17 '15 at 12:26
• Could you explain the process involved in the decay of a gravition into a graviton. (Assuming they exist, as you say) or just provide a link to a source would do Regards and thanks – user74893 Mar 17 '15 at 12:29
• I don't see what this would have to do with superluminal speed, if the other particles don't do the same. – Jimmy360 Mar 17 '15 at 12:53
• @Jimmy360 my thought experiment goes like this: 2 gravitons gets entangled and separate them many many light years apart, then measure the quantum states of 1 of the pair and immediately we would already know the quantum states of the other pair with very high probability (higher than guessing) conclusion: spooky action at a distance as famously quote by einstein – user6760 Mar 17 '15 at 14:03
• @irishphysics I mentioned "an entangled particle" instead of a graviton decaying into itself. I'm no expert and I don't know if anything is capable of decaying into a graviton and vice versa however I'm sure higgs boson can decay into other particles. – user6760 Mar 17 '15 at 14:10

Entanglement isn't about interaction or information transfer betweeen entangled particles.

Consider spin-entaglement of two spin-$\frac{1}{2}$ particles: Let them be in singulet-state relative to an arbitrary axis (say z-axis):

$$|\Psi \rangle = \frac{1}{\sqrt{2}} (\ |\uparrow_z, \downarrow_z \rangle - |\downarrow_z,\uparrow_z\rangle \ )$$

The propability $P$ to measure both particles in state $|i,j \rangle$ with $i,j \in \{ \uparrow, \downarrow \}$ where the axis of both measurments enclose the angle $\theta$ is given by: $$P_{i,j} = \| \langle i,j | \Psi \rangle \|^2 = \frac{1}{4} (1 - i \cdot j \cdot \cos \theta )$$ if we take $i,j$ to be 1 and -1 for $\uparrow$ and $\downarrow$, respectively.

The reduced propability $p_i$ of measuring only one particle (e.g. if we don't care about the other) is given by: $$p_i = \sum_{j \in \{1,-1\}} P_{i,j} = \frac{1}{2}$$

The conditional propability of measuring the other particle (after we already know about the measurment of the first particle) is given by: $$\tilde{p}_{j|i} = \frac{P_{i,j}}{p_i} = \frac{1}{2} (1 - i \cdot j \cdot \cos \theta )$$

This does involve the angle $\theta$ and usually one starts here to argue about non-locality and instantanious actions changing the outcome of experiment when we change the angle $\theta$ at the first measurment apparatus.

This is however not true. If we are talking about conditional propabilities we have already performed a measurment and set the measurment axis of the first measurment. Changing this axis afterwards will not affect the propability as the angle $\theta$ is relative to the measured axis. Changing the axis of the second measurment only changes the propability predicting the outcome of the later measurment for the first observer because he has that extra knowledge.

The propability for the second observer stays the same, as this is the reduced propability (he doesn't know about the first measurment): $$p_j = \sum_{i \in \{1,-1\}} P_{i,j} = \frac{1}{2}$$

In short: Without the extra knowledge of the first measurment, entanglement is not important for the second observer. To gain that extra knowledge there must be an additional information transfer to the second observer and this is restricted by means of relativity-causality ($v\le c$ etc.). So entanglement neither breaks causality nor can it transfer any information.