Suppose two entangled particles are far apart. One is with Alice and the other is with Bob. The relative velocity between Alice and Bob is zero (and spacetime is flat), so that we can define a notion of simultaneity that is agreed upon by both observers.

Before measurement, the joint is wavefunction of the particles is $\frac{1}{\sqrt{3}}|up, down\rangle +\frac{\sqrt{2}}{\sqrt{3}} |down,up\rangle$. The second spin label is of Bob's particle

Suppose, at time $t_0$, Bob measures his particle and observes $|down\rangle$.

Can we say that, after time $t_0$, Alice's description of the joint system should become : $|up, down \rangle$?

Or should it become:

$$\frac{1}{\sqrt{3}}|up, down, \text{Bob measured up}\rangle +\frac{\sqrt{2}}{\sqrt{3}} |down,up, \text{Bob measured down}\rangle$$

If the first option is correct, how does it not violate locality? I am thinking that the first option involves the information, that Bob has made a measurement, to instantaneously travel to Alice's end.


1 Answer 1


If Bob observes "down", due to the wavefunction collapse the full system will be described by : $$ \left|down,up\right\rangle $$

There is indeed some sort of "spooky action at a distance", but this action cannot be used to transfer information.

When Bob observes "down", he will instantaneously know the state of the particle in Alice's possession, and yes, that state will instantaneously change for Alice. However, this cannot be used to transfer information. This is mainly because Bob cannot choose what he observes. He will observe $1/3$ of the time "down", and $2/3$ of the times "up".

Using entanglement, you can create correlations which are not possible to make without using the entangled pair (this is at the core of Bell's inequalities), but this does not violate "locality" in the sense that no information or signal can be transmitted faster than light. You are right that there is "something" non-local happening at the moment of the wavefunction collapse.


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