I was discussing quantum entanglement with a friend and explaining that faster than the speed of light data transfer isn't a reality with our current understanding of physics. He brought up quantum entanglement and I explained that while they are entangled over a distance and in a sense are "exchanging data" (eh) faster than the speed of light, you can't really exchange custom data faster than light because trying to impose a state change on the particles would break the entanglement.

He later sent me this journal entry in PRX Quantum which boasts faster than light data transfer with 90% fidelity via quantum entanglement. Is my understanding of how entanglement works way off? Is this journal being interpreted out of context? Or is there something else entirely that's gone awry?

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    $\begingroup$ Where is that paper do you think they claim to transmit information faster than light? $\endgroup$
    – A. P.
    Dec 23 '20 at 2:38
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    $\begingroup$ the answer to your actual question is indeed "this journal being interpreted out of context by your friend" $\endgroup$
    – Fattie
    Dec 23 '20 at 13:01
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    $\begingroup$ synchronize 2 clocks. Give one to a friend. You can both read the same value on your respective clocks, but you are not transferring any information. $\endgroup$
    – njzk2
    Dec 23 '20 at 21:53

The paper is about quantum teleportation, which is a process enabled by entanglement. However, it does not provide faster-than-light communication. The complete process of teleportation always needs a conventional communication link without which the process cannot be completed. In the theoretical description of entanglement, the outcome of the joint measurement needs to be communicated to the person receiving the final state to know which unitary operation to perform on that state to produce the original state. In practical implementations of teleportation, such a conventional communication link is used to register coincidences in detections.

So, one can emphatically state that faster-than-light communication is not possible, not even with entangled states. There is a theorem, called the no-signaling theorem, that shows that entangled states do not allow faster-than-light communication.


When you prepare an entangled state you have just put some information into it. For example, in Bell state $$ \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) $$ you know that both qubits have to be in same state, either 0 or 1, after measurement.

Imagine that you prepared the Bell state, one qubit remains with you and second one is sent to your friend. When you measure your qubit and get say 0, your friend gets always 0 as the Bell state collapsed to 00 (instantenously at both sides).

However, this does not mean faster than light communication. Firstly, you had to deliver one qubit to your friend. As qubit is realized by some particle, it can move maximally at speed of light (for example photon). Secondly, when you measure your qubit, you do not transmit any information as you cannot pre-select results of measurement. You get either 0 or 1 with 50 % probability and the same is true for yours friend side.

EDIT: Based on comment by KRyan, it is worth noting that you can get 0 (or 1) either as a result of your own measurement or as a result of measurement done by your friend. There is no way how to distinguish these two envent. This is another explanation why no information is transmitted in measuring on entagled qubits (particles).

  • $\begingroup$ Am I right in understanding that the friend also cannot tell, when they measure the qubit and get 0, whether they got 0 because you had already measured it and got 0, or if they have instead measured first and collapsed things to 0 themselves (and so know only that you either did, or will, get 0, but not whether or not you have performed the measurement yet)? This might be worth adding, if so, because if not it seems like it might be plausible to communicate information through that—if nothing else, the information as to whether or not the other qubit has been measured. $\endgroup$
    – KRyan
    Dec 23 '20 at 21:06
  • $\begingroup$ @KRyan: Yes, you are absolutely right. There is no way how to distinguish if 0 is result of your measurement or your friend measurement. $\endgroup$ Dec 24 '20 at 7:25
  • $\begingroup$ @KRyan: I edited the answer. Please feel free to correct it seems unsufficient to you. $\endgroup$ Dec 24 '20 at 7:29

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