# The choice of measurement basis on one half of an entangled state affects the other half. Can this be used to communicate faster than light?

It is often stated, particularly in popular physics articles and videos, 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?

• I do not think this is a duplicate but please use this meta thread to discuss this if you do. – Emilio Pisanty Feb 25 '14 at 19:10

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.