It is often stated, particularly in popular physics articles and videos about quantum entanglement, that if one measures a particle A that is entangled with some other particle B, then this measurement will immediately affect the state of the entangled partner.

For example, if Alice and Bob share an entangled pair of electrons and Alice measures her spin in the $x$ direction, then Bob's spin will also end up spinning in that direction, and similarly if she measures in the $z$ direction. Moreover, the effect will be instantaneous, regardless of the spatial distance between the two particles, which seems at odds with special relativity.

Can I use a scheme like this to communicate faster than light?

  • 7
    $\begingroup$ I do not think this is a duplicate but please use this meta thread to discuss this if you do. $\endgroup$ Commented Feb 25, 2014 at 19:10
  • $\begingroup$ Entanglement doesn't involve FTL influences, detectable or otherwise physics.stackexchange.com/a/203885/28512 $\endgroup$
    – alanf
    Commented Aug 11, 2022 at 13:48
  • 2
    $\begingroup$ @alanf I know. That's why I posted this self-answered Q&A. $\endgroup$ Commented Aug 11, 2022 at 17:57

1 Answer 1


The problem with this sort of scheme is that Alice has no control over the results of her measurements, since those are random. This means that she can control which basis Bob's spin is projected on, but she cannot control which of the basis states gets chosen. Bob will then see a random mix of results which turns out to contain no trace of what Alice was trying to communicate.

To make this more precise, consider the standard case where they share a Bell triplet state $$ \newcommand{\up}{|\!\uparrow\rangle}\newcommand{\down}{|\!\downarrow\rangle} \newcommand{\plus}{|+\rangle}\newcommand{\minus}{|-\rangle} |\Phi\rangle=\up\up+\down\down $$ (ignoring normalization) at the start of the protocol, which they use as a resource state. Alice can choose to measure along the $z$ direction, in the basis $\{\up,\down\}$, or along the $x$ direction, in the basis $\{\plus=\tfrac1{\sqrt{2}}(\up+\down),\minus=\tfrac1{\sqrt{2}}(\up-\down)\}$. Because of the nice properties of the triplet state, whatever state Alice's qubit is projected on (in these two bases) will be identically replicated in Bob's qubit. Both states of either basis come up with equal probabilities.

Alice's only choice in this scheme is what basis she measures in, and she can transmit one bit of information if she can engineer a situation where Bob can determine that basis. Assume, if you want to, that she can repeat this protocol $n$ times, with $n$ possibly greater than one, to help ensure the information gets there.

Suppose, then, that Alice chose to measure in the $z$ direction. How can Bob determine this fact? To put this more explicitly, how can he determine that Alice didn't measure in the $x$ direction? His problem, then, is to determine whether his ensemble of $n$ qubits is in a random mix of $\up$s and $\down$s, or in a random mix of $\plus$s and $\minus$s.

Unfortunately, this is impossible to do. If he measures in the $z$ direction, he will get fifty/fifty results if Alice is sending $\up$s and $\down$s, but he would also get fifty/fifty chances from each $\plus$ or $\minus$, and therefore from the whole set, if Alice had measured in the $x$ direction. Regardless of what basis Alice chose or what basis he himself measures in, both situations look exactly the same to Bob.

  • $\begingroup$ Thanks, this is exactly what I was looking for. Could you (or someone) please point me what I should read to understant why it's not possible to create an entangled pair with known property, e.g. |↑⟩|↑⟩ ? (...then Alice could decide whether or not to measure in different direction, while Bob would always measure in the basis - if there were 1000 such pairs and Bob measured ~all of them as |↑⟩, he would know that Alice chose not to measure, if he got a random mix, he would know Alice measured, thus he would have received 1 bit of information ... which is not possible, but WHY?) $\endgroup$
    – Aprillion
    Commented Apr 5 at 17:59

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.